ED_explained

Semiconductor Device Fundamentals: Explained

A chapter-by-chapter guided tour of Pierret, Semiconductor Device Fundamentals, rebuilt from OCR notes with full step-by-step solutions.

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Chapter 1: Semiconductors: A General Introduction

Source note. This chapter follows Pierret, Semiconductor Device Fundamentals, Chapter 1 (book pp. 3-22). Where the OCR scan lost a figure or an overbar, I reconstruct it from the physics and say so explicitly.

The big question of this chapter. Almost every phone, laptop, car ignition and LED on Earth is built from a handful of strange solids that are neither good conductors nor good insulators. What are these solids made of, how clean must they be, how are their atoms arranged, how do we give names to the planes and directions inside them, and how on Earth do we grow a perfect crystal the size of a baseball bat? That is the whole chapter. Think of it as the “meet the cast and see the stage” chapter: later chapters will let the actors (electrons and holes) perform.

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1.1 General Material Properties

The overwhelming majority of solid-state devices on the market today are made from semiconductors. So before any device physics, let us ask: what do semiconducting materials look like in terms of (1) composition, (2) purity and (3) structure?

1.1.1 Composition

Table 1.1 lists the atomic compositions of semiconductors you are likely to meet in the device literature. There are three big families.

Classification	Symbol	Name
(1) Elemental	Si	Silicon
	Ge	Germanium
(2) Compounds
(a) IV-IV	SiC	Silicon carbide
(b) III-V	AlP	Aluminum phosphide
	AlAs	Aluminum arsenide
	AlSb	Aluminum antimonide
	GaN	Gallium nitride
	GaP	Gallium phosphide
	GaAs	Gallium arsenide
	GaSb	Gallium antimonide
	InP	Indium phosphide
	InAs	Indium arsenide
	InSb	Indium antimonide
(c) II-VI	ZnO	Zinc oxide
	ZnS	Zinc sulfide
	ZnSe	Zinc selenide
	ZnTe	Zinc telluride
	CdS	Cadmium sulfide
	CdSe	Cadmium selenide
	CdTe	Cadmium telluride
	HgS	Mercury sulfide
(d) IV-VI	PbS	Lead sulfide
	PbSe	Lead selenide
	PbTe	Lead telluride
(3) Alloys
(a) Binary	${\mathrm{Si}}_{1-x}{\mathrm{Ge}}_x$
(b) Ternary	${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$ (or ${\mathrm{Ga}}_{1-x}{\mathrm{Al}}_x\mathrm{As}$)
	${\mathrm{Al}}_x{\mathrm{In}}_{1-x}\mathrm{As}$
	${\mathrm{Cd}}_{1-x}{\mathrm{Mn}}_x\mathrm{Te}$
	${\mathrm{GaAs}}_{1-x}{\mathrm{P}}_x$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}\mathrm{As}$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}\mathrm{P}$
	${\mathrm{Hg}}_{1-x}{\mathrm{Cd}}_x\mathrm{Te}$
(c) Quaternary	${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}{\mathrm{As}}_y{\mathrm{Sb}}_{1-y}$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}{\mathrm{As}}_y{\mathrm{P}}_{1-y}$

(The scan of the last rows is garbled; the entries above are the standard ones the book lists.)

The alloy subscript. In an alloy formula the $x$ (or $y$) is a fraction between 0 and 1. For instance ${\mathrm{Al}}_{0.3}{\mathrm{Ga}}_{0.7}\mathrm{As}$ means 3 Al atoms and 7 Ga atoms for every 10 As atoms. The Al and Ga together share one group of lattice sites; As occupies the other.

Who are the stars?

• Silicon (Si) is, by a wide margin, king. Thanks largely to the maturity of its fabrication technology, it totally dominates the market: most discrete devices and integrated circuits (ICs), including the CPU in your computer and the ignition module in a car, are silicon.
• Gallium arsenide (GaAs) has superior electron transport (electrons move faster) and special optical properties, so it appears in laser diodes and high-speed ICs.
• The rest live in “niche” roles that are almost always high-speed, high-temperature, or optoelectronic (light-related).

Because of Si’s dominance, the text focuses on Si, gives GaAs comparable attention where feasible, and brings in others as needed.

Why is the list so short? Think of the periodic table as a huge menu. You could combine elements in countless ways, yet only a tiny group give semiconductors. Look at the abbreviated periodic chart (Table 1.2):

II	III	IV	V	VI
4 Be	5 B	6 C	7 N	8 O
12 Mg	13 Al	14 Si	15 P	16 S
30 Zn	31 Ga	32 Ge	33 As	34 Se
48 Cd	49 In	50 Sn	51 Sb	52 Te
80 Hg	81 Tl	82 Pb	83 Bi	84 Po

(Table 1.2: abbreviated periodic chart; the column numbers are the Roman numerals, the numbers are atomic numbers.)

Except for the IV-VI compounds, every semiconductor in Table 1.1 is either built from Column IV elements, or from a pair of elements sitting equal distances on either side of Column IV:

• Column III (Ga) + Column V (As) $\to$ III-V compound GaAs.
• Column II (Zn) + Column VI (Se) $\to$ II-VI compound ZnSe.
• A fractional mix of Column III elements (Al and Ga) + Column V (As) $\to$ alloy ${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$.

The hidden reason (an intuition you should keep). Average the valence electrons: Ga has 3, As has 5; $(3+5)/2=4$. Zn has 2, Se has 6; $(2+6)/2=4$. Whatever the recipe, the average is four valence electrons per atom. That magic number controls the chemical bonding, as we will see when we find that each atom has four neighbors, and it is why the semiconductor club has such exclusive membership. Think of it like a dance where every dancer must hold exactly four hands: you can swap partners of different “hand-counts” only as long as the average stays at four.

1.1.2 Purity

As Chapter 2 will explain, extremely minute traces of impurity atoms called dopants can drastically change a semiconductor’s electrical properties. So the composition must be controlled with extraordinary care, and modern semiconductors are among the purest solids ever made.

In silicon the unintentional dopant content is routinely below

$$\frac{1\text{ impurity atom}}{10^9\text{Si atoms}}$$

A way to feel this number. Plant a forest of maple trees from coast to coast and border to border across the entire United States (Alaska included), one tree every 50 ft. An impurity level of one part in $10^9$ would correspond to finding only about 25 crabapple trees in that whole continental maple forest. That is the purity of the raw material.

Do not be confused: this refers to unintentional, undesired impurities. Dopant atoms are purposely added at levels ranging from about one part in $10^8$ up to one impurity atom per $10^3$ semiconductor atoms to control the electrical behavior. So the recipe is: make it almost perfectly pure, then sprinkle in exactly the amount of “spice” you want.

1.1.3 Structure

The spatial arrangement of atoms strongly influences material properties. Figure 1.1 (three sketches of atoms) classifies solids into three broad types.

Figure 1.1 (described). (a) Amorphous: dots scattered with no repeating pattern. (b) Polycrystalline: several patches, each of which is an ordered array, but each patch is tilted relative to its neighbors. (c) Crystalline: one perfectly regular array filling the whole picture.

Type	Description	Analogy
Amorphous	No recognizable long-range order; the arrangement in one section looks different from any other section.	A pile of marbles dumped into a box.
Polycrystalline	Composed of crystalline subsections (grains) that are disjointed or misoriented relative to each other.	A patchwork quilt of tiled floors laid by workers who never agreed on orientation.
Crystalline	Atoms form an orderly 3-D array; you can reproduce the arrangement in any section from any other section.	A perfectly stacked orchard, trees in flawless rows in every direction.

Examples from real devices: an amorphous-Si thin-film transistor is the switch inside liquid crystal displays (LCDs); polycrystalline-Si gates are used in MOSFETs. But in the vast majority of devices, the active region sits in a crystalline semiconductor. Hence the next section.

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1.2 Crystal Structure

Since working semiconductors are typically crystalline, we want a more detailed picture of the atomic arrangement in the principal semiconductors. The plan of this section: (1) how to describe atom positions in crystals (unit cells); (2) practice with simple 3-D lattices; (3) the semiconductor lattices themselves; (4) Miller indices, a shorthand for planes and directions.

1.2.1 The Unit Cell Concept

Definition. A unit cell is a small portion of a crystal that could be used to reproduce the crystal.

Figure 1.2 (described). (a) A sample 2-D lattice of atoms. (b) A unit cell: a square of side $a$ with a quarter-atom at each corner. (c) The original lattice rebuilt by copying the cell and stacking copies edge to edge. (d) An alternative, shifted choice of unit cell that works equally well.

Think of wallpaper: you never need the whole wall’s pattern, only one repeating tile. Copy the tile and tessellate it, and the wall appears. That tile is the unit cell.

Two common confusions:

1. Unit cells are not unique. The cell of Fig. 1.2(d) specifies the lattice just as well as that of Fig. 1.2(b). Any tile that repeats to fill space works.
2. A unit cell need not be primitive (the smallest possible cell). It is usually better to use a slightly larger cell with orthogonal (right-angle) sides rather than a primitive cell with slanted sides, especially in 3-D, where non-cubic cells are hard to describe and visualize.

1.2.2 Simple 3-D Unit Cells

Semiconductor crystals are three-dimensional, so we use 3-D cells.

Figure 1.3 (described). Four cubes with side length (the lattice constant) $a$: (a) simple cubic: atoms at the 8 corners; (b) the “pedantically correct” simple cubic showing only the $1/8$ of each corner atom inside the cell; (c) body-centered cubic (bcc): corners plus one atom at the cube center; (d) face-centered cubic (fcc): corners plus one atom at the center of each of the 6 faces.

Simple cubic. An equal-sided box with an atom at each corner. Only $1/8$ of each corner atom lies inside the cell (the other seven octants belong to neighboring cells). Stack copies like nursery blocks to get the simple cubic lattice: planes of atoms separated by the lattice constant $a$, and viewed from above, atoms in each plane sit directly over atoms in the plane below.

bcc adds one atom at the cube center. fcc adds one atom at the center of each face, of which only half lies inside the cell.

Counting atoms per cell (verify these yourself!):

$$\text{simple cubic: }8\times \frac{1}{8}=1$$ $$\text{bcc: }8\times \frac{1}{8}+1=2$$ $$\text{fcc: }8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$$

Intuition for the counting: a corner atom is shared by 8 cubes (so contributes $1/8$), a face atom is shared by 2 cubes (so $1/2$), and a body-center atom belongs to one cube (so $1$).

Exercise 1.1 (computer-based, visualization)

Task. The book’s software contains a directory MacMolecule with ASCII input files for a Macintosh program (also named MacMolecule, from the University of Arizona) that draws “ball-and-stick” 3-D colour renderings of molecules, unit cells and lattices. The files visualize the simple cubic, bcc and fcc cells and the diamond and zincblende cells.

What to do and why. The first display is a $z$-direction view, so it looks nearly two-dimensional. Rotate the model by grabbing and dragging its edges to see the depth. Advanced users may edit or write input files. There is no equation or MATLAB code here; the exercise trains 3-D intuition, which is exactly the skill you need for the diamond lattice in the next subsection. (If you do not have a Mac, any free crystal viewer or a short script that plots atom coordinates can serve the same purpose; see the MATLAB code in Exercise 1.3 for a starting point on coordinates.)

1.2.3 Semiconductor Lattices

Now we position atoms in the real semiconductors.

Diamond lattice (Si and Ge). Figure 1.4(a) shows the diamond lattice unit cell (named after diamond, carbon, which is also a Column IV element).

Figure 1.4 (described). (a) A cube with atoms at all corners and face centers (like fcc) plus four additional atoms inside the cube. (b) The zincblende cell drawn with GaAs: same geometry, but two kinds of atoms (Ga and As). (c) An enlarged top corner of (a) showing one atom at the cube corner bonded to four nearest neighbors, which sit at the vertices of a tetrahedron; the corner-to-neighbor distance is one quarter of the body diagonal; marks $a/2$ indicate half the cube side. Cube side lengths: $a=5.43\text{A˚}$ for Si and $a=5.65\text{A˚}$ for GaAs at $T=300\text{K}$.

Where are the four interior atoms? In coordinates $(x,y,z)$ in units of $a$, with the origin at the top-front-left corner and measuring along the body diagonal:

• One interior atom lies along the body diagonal, exactly $1/4$ of the way down from the top-front-left corner.
• The other three are displaced by the same quarter-body-diagonal vector from the front, top and left face-center atoms, respectively.

Cleaner way to see it: the diamond lattice is just two interpenetrating fcc lattices. The corner and face atoms belong to one fcc lattice; the four interior atoms belong to a second fcc lattice, displaced by one-quarter of a body diagonal along the body-diagonal direction relative to the first.

Atoms per diamond cell: fcc contributes 4, the interior gives 4 more:

$$N_{\text{diamond}}=4+4=8$$

Zincblende lattice (most III-V, including GaAs). Essentially identical to diamond, except the lattice sites are shared between two different atoms. Ga occupies one of the interpenetrating fcc sublattices; As occupies the other.

Why bother? A useful application: atomic density. With the unit cell we can do geometry. At room temperature the Si lattice constant is $a=5.43\text{A˚}$ (where $1\text{A˚}=10^{-8}\text{cm}$). There are 8 atoms per cell and the cell volume is $a^3$, so

$$n_{\text{Si}}=\frac{8}{a^3}$$ $$=\frac{8}{(5.43\times 10^{-8}\text{cm}{)}^3}$$ $$\approx 5\times 10^{22}{\text{atoms/cm}}^3$$

(almost exactly; the precise value is $4.9968\times 10^{22}$ as the MATLAB run below shows). Similar calculations give atomic radii, distances between atomic planes, and so on.

The key physical fact. For our purposes the most important message is (Fig. 1.4(c)) that atoms in the diamond and zincblende lattices have four nearest neighbors. The chemical bonding (“crystalline glue”) in the major semiconductors is therefore dominated by the attraction between each atom and its four closest neighbors, matching the four valence electrons per atom from Section 1.1.1. Each atom shares one electron with each of four neighbors: four bonds, four electrons, a stable arrangement. Store this away; Chapter 2 builds on it.

Exercise 1.2

Problem. If the lattice constant of Si is $a=5.43\times 10^{-8}$ cm, what is the distance $d$ from the center of one Si atom to the center of its nearest neighbor?

Solution.

The atom at the top-front corner and the interior atom lying one quarter of the way down the body diagonal are nearest neighbors (Fig. 1.4 caption).

The body diagonal of a cube of side $a$ is

$$L_{\text{diag}}=\sqrt{a^2+a^2+a^2}=\sqrt{3}a\text{(Pythagoras in 3-D)}$$

The nearest neighbor sits a quarter of the way along it:

$$d=\frac{1}{4}{L}_{\text{diag}}=\frac{\sqrt{3}}{4}a$$ $$=\frac{\sqrt{3}}{4}(5.43\times 10^{-8}\text{cm})$$ $$=0.4330\times 5.43\times 10^{-8}\text{cm}$$ $$\boxed{d\approx 2.35\times 10^{-8}\text{cm}=2.35\text{A˚}}$$

Exercise 1.3 (computer-based, MATLAB)

Task. Write a MATLAB program that computes the number of atoms per cm${}^3$ in a cubic crystal. Use the input function to enter the number of atoms per unit cell and the lattice constant $a$ for a specific crystal. List the program and record the output for silicon.

Logic. The atom density is (atoms per cell) divided by (cell volume):

$$n=\frac{N}{a^3}$$

If $a$ is entered in angstroms, convert to cm: $a_{\text{cm}}=a_{\text{A˚}}\times 10^{-8}$, so $a_{\text{cm}}^3=a_{\text{A˚}}^3\times 10^{-24}$, and therefore

$$n=\frac{N(10^{24})}{a_{\text{A˚}}^3}{\text{cm}}^{-3}$$

That is exactly the line in the program (the factor $10^{24}$ does the unit conversion so the user can type the convenient number 5.43).

% Exercise 1.3
% Computation of the number of atoms/cm^3 in a cubic lattice
N = input('input number of atoms/unit cell, N = ');
a = input('lattice constant in angstrom, a = ');
atmden = N*(1.0e24)/(a^3)   % number of atoms/cm^3 (no semicolon, so MATLAB prints it)

Program output for Si:

input number of atoms/unit cell, N = 8
lattice constant in angstrom, a = 5.43
atmden =
   4.9968e+22

Check by hand: $5.43^3=160.10$, so $8\times 10^{24}/160.10=4.9968\times 10^{22}{\text{cm}}^{-3}$. Matches.

1.2.4 Miller Indices

Device-grade silicon comes as thin, round wafers (Fig. 1.5): 150 mm (6”) and 200 mm (8”) diameter wafers about 0.625 mm and 0.725 mm thick. The face is polished and etched to a mirror finish. Crucially:

• The wafer surface is carefully pre-oriented along a specific crystallographic plane, e.g. a $(100)$ plane.
• A flat or notch ground along the periphery marks a reference direction in the surface plane, e.g. a $[011]$ direction (Fig. 1.5 labels the $(100)$ surface and the $[011]$ directions for the notch and the flat).

Why care? Surface orientation affects processing steps and the characteristics of devices. The flat is used, for example, to orient device arrays on the wafer so that, when the wafer is sawed up (separated into individual dice), you get high yield without shattering chips along the “wrong” crystal direction. Miller indices are the standard language for planes and directions.

Indexing procedure for planes (four steps). I carry out a sample with the plane of Fig. 1.6(a).

Figure 1.6 (described). (a) A plane cutting the $x$-axis at $1a$, the $y$-axis at $2a$ and the $z$-axis at $3a$. (b) The $(001)$ plane, parallel to the $xy$ plane, cutting only the $z$-axis at $a$. (c) The $(2\bar{2}1)$-type plane of the text; see the caution below.

Step	Procedure	Sample (Fig. 1.6(a))
1	Set axes along the unit-cell edges. Note where the plane intercepts each axis; divide each intercept by the cell length along that axis; record the normalized intercepts in order $x,y,z$.	$1,2,3$
2	Invert: form $[1/\text{intercept}]$s.	$1,\frac{1}{2},\frac{1}{3}$
3	Multiply by the smallest number that turns the set into the smallest possible whole numbers.	$6,3,2$
4	Enclose in parentheses.	$(632)$

Step 3 detail: the least common multiple of the denominators $1,2,3$ is $6$, so $6\times (1,\frac{1}{2},\frac{1}{3})=(6,3,2)$.

Special facts.

1. A plane parallel to an axis intercepts it at infinity, so the reciprocal is $0$. Example: Fig. 1.6(b) intercepts the axes at $\infty ,\infty ,1$, inverse $0,0,1$, so it is a $(001)$ plane.
2. If the plane crosses the negative side of an axis, a minus sign (written as an overbar) goes over that index. The Fig. 1.6(c) plane, with one negative intercept, is the type of plane written $(2\bar{2}1)$ in the book (the overbar was lost in the scan; the idea is that the plane has a negative intercept on one axis).
3. Equivalent planes. In the diamond lattice the six planes that coincide with the cube faces contain identical atomic arrangements, so crystal symmetry makes $(100),(010),(001),(\bar{1}00),(0\bar{1}0),(00\bar{1})$ indistinguishable. A group of equivalent planes is written with braces: $\{100\}$.
4. Planes through the origin cannot be indexed (their intercepts would be $0$ and the reciprocals infinite). Move the origin to a lattice point not on the plane; this is legitimate because parallel planes are equivalent.

Figure 1.7 (described). (a) Unit cubes showing the $(100)$ plane (the face perpendicular to $x$), the $(110)$ plane (a diagonal slice parallel to $z$, hitting $x$ and $y$ at $a$), and the $(111)$ plane (the triangle through $(a,0,0),(0,a,0),(0,0,a)$). (b) Direction vectors $[100]$ (along $x$), $[110]$ (face diagonal in the $xy$ plane), $[111]$ (body diagonal).

Directions. Treat a direction as a vector:

1. Draw a vector of arbitrary length along the direction.
2. Find its projections (components) on the axes.
3. Multiply to get the smallest whole numbers (length changes, direction does not).
4. Enclose in square brackets $[]$.

Table 1.3 Miller convention summary.

Convention	Interpretation
$(hkl)$	Crystal plane
$\{hkl\}$	Equivalent planes
$[hkl]$	Crystal direction
$⟨hkl⟩$	Equivalent directions

Practical tips.

• The inverse job (given indices, picture the plane) is easy for the low-index set, e.g. $(111)$, $(110)$, $[001]$. Memorize those.
• For cubic crystals, the direction $[hkl]$ is normal to the plane $(hkl)$. E.g. $[110]\perp (110)$.

Exercise 1.4

Problem. For a cubic crystal lattice: (a) determine the Miller indices of the plane and direction vector shown in Fig. E1.4(a); (b) sketch the plane and direction vector characterized by $(011)$ and $[011]$.

Solution (a). From the figure, the plane intercepts the $x$, $y$, $z$ axes at $-1,1,2$ (in units of $a$). The direction vector has projections $2a,a,0$.

Plane:

$$\text{intercepts: }(-1,1,2)$$ $$[1/\text{intercept}]\text{s: }\left(-1,1,\frac{1}{2}\right)$$ $$\times 2\text{ (lowest whole numbers): }(-2,2,1)$$ $$\boxed{(\bar{2}21)}$$

Direction:

$$\text{projections: }(2,1,0)\text{(already whole, no common factor)}$$ $$\boxed{[210]}$$

Solution (b). For the $(011)$ plane:

$$[1/\text{intercept}]\text{s}=(0,1,1)\Rightarrow \text{intercepts}=(\infty ,1,1)$$

The plane meets the $y$-axis at $a$, the $z$-axis at $a$, and is parallel to the $x$-axis. In a cubic crystal, $[011]$ is normal to the $(011)$ plane, so the direction vector is drawn from the origin to the point $(0,a,a)$ (perpendicular to the plane, which itself is the slanted rectangle containing the $x$ direction and the line from $(0,a,0)$ to $(0,0,a)$). These are the sketches of Fig. E1.4(b).

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1.3 Crystal Growth

1.3.1 Obtaining Ultrapure Si

Where does device silicon come from? Not sandstone deposits? No. A by-product of South African diamond mines? No. Scooped from the ocean floor by special submarines? No. Although Si is the second most abundant element in the Earth’s crust and a component of many compounds (chiefly silica, impure $\mathrm{Si}{\mathrm{O}}_2$, and the silicates, Si + O + another element), silicon never occurs alone in nature. Single-crystal Si is a man-made material.

So step one is to separate Si from its compounds and purify it (Fig. 1.8).

Figure 1.8 (flow chart described).

$$\text{Silica}\underset{\text{in presence of C}}{\overset{\text{reduced}}{\to }}\text{very impure Si}\stackrel{\text{chlorinated}}{\to }{\text{SiCl}}_4\text{(liquid)}\stackrel{\text{distilled, etc.}}{\to }{\text{ultrapure SiCl}}_4\underset{\text{atmosphere}}{\overset{{\text{reduced in H}}_2}{\to }}\text{ultrapure polycrystalline Si}$$

(In the scan the intermediate labels read “SiC”; the text clarifies that the chlorinated products are $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$.)

1. Reduction. Silica is heated with carbon in an electric furnace. Carbon pulls the oxygen away from $\mathrm{Si}{\mathrm{O}}_2$ (reduces it), leaving impure (low-grade, “ferrosilicon”) Si.
2. Chlorination. The impure Si is chlorinated to give $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$, which are liquids at room temperature.
3. Distillation. Liquids are much easier to purify than solids (standard methods like repeated distillation exist). The result is ultrapure $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$. Turning the solid into a liquid is the clever maneuver: it is like washing a dusty rug by unraveling it into thread instead of beating it.
4. Chemical reduction. The ultrapure halide is reduced to elemental Si, e.g. by heating in hydrogen:

$$\mathrm{SiC}{\mathrm{l}}_4+2{\mathrm{H}}_2⟶4HCl+\mathrm{Si}$$

1.3.2 Single-Crystal Formation

The silicon so obtained is ultrapure but polycrystalline. To make the large single crystals used in devices, we use the Czochralski method.

Figure 1.9 (described). (a) Photo of a computer-controlled Czochralski crystal puller. (b) Schematic: a closed chamber filled with argon (Ar) with vacuum pumps and a gas valve; inside, a graphite crucible holding a silica (quartz) crucible with the Si melt, surrounded by a graphite heater and a thermal shield; a crucible shaft below rotates the crucible; above, a seed shaft carries a seed holder with a seed crystal that dips into the melt; a viewing port with an optical system lets operators monitor; exhaust gas is Ar + SiO + CO. (c) Photo of the resulting silicon ingot.

Steps.

1. Place ultrapure polycrystalline Si in a quartz crucible and heat it in an inert atmosphere (argon) to form a melt.
2. Take a small seed crystal of single-crystal Si whose bottom face normal is carefully aligned along a chosen direction (typically a $⟨111⟩$ or $⟨100⟩$ direction), clamp it to a metal rod, and dip it into the melt.
3. Once thermal equilibrium is reached, lower the temperature of the melt near the seed. Silicon freezes onto the seed, and the added material is a structurally perfect extension of the seed’s lattice.
4. Slowly rotate and withdraw the seed. More and more Si freezes onto the bottom of the growing crystal.

The result is a large, cylindrical single crystal, an ingot, routinely up to 200 mm (8 in) in diameter and 1 to 2 m long. Wafers (Fig. 1.5) are made by slicing the ingot into thin discs with a diamond-edged saw.

Intuition: it is like growing rock candy on a string. The string (seed) tells the new sugar (silicon) exactly how to line up, so the grown crystal continues the same orientation forever. The slow rotation averages out temperature irregularities so the ingot grows round and uniform.

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1.4 Summary

• Semiconductor compositions: elemental (Si, Ge), compounds (III-V, II-VI, etc.) and alloys. On average there are four valence electrons per atom.
• Device-quality semiconductors are typically single crystals of very high compositional purity.
• Si crystallizes in the diamond lattice; GaAs in the zincblende lattice; in both cases each atom has four nearest neighbors.
• Miller indices are the accepted means of identifying planes and directions within a crystal.
• Large single crystals of device-grade Si are commonly produced by the Czochralski method.

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Problems (Chapter 1)

Problem information table (as in the book).

Problem	Complete after	Difficulty	Points (weighting)	Short description
1.1	1.4	-	10 (a-h: 1 each, i: 2)	Short answer review
1.2	1.2.3	1	8	Unit cell for alloy
1.3	-	-	10 (5 each part)	Ge atomic density
1.4	-	-	8 (4 each part)	Nearest-neighbor distances
1.5	1.2.4	2-3	15 (a-2, b-2, c-3, d-3, e-5)	Atoms/cm${}^2$ on planes
1.6	-	1	8 (2 each subpart)	Determine Miller indices
1.7	-	1	16 (2 each part)	Given (hkl), sketch plane
1.8	-	1	16 (2 each part)	Given [hkl], sketch direction
1.9	-	1	8 (4 each part)	Perpendicular directions
1.10	-	2	10	Direction on wafer surface
1.11	-	2	10 (a-3, b-2, c-3, d-2)	Combination question
1.12	-	2	12 (2 each part)	Equivalent planes/directions
1.13	1.2.3	3	20 (5 each part)	% of volume occupied

Problem 1.1 Quick Quiz

Problem. Answer concisely. (a) Name (i) one elemental semiconductor and (ii) one III-V compound semiconductor. (b) Difference between crystalline and polycrystalline? (c) Word definition of unit cell. (d) How many atoms in a simple cubic, bcc, fcc and diamond unit cell? (e) $1\text{A˚}=$ ? cm. (f) In terms of $a$, distance between nearest neighbors in a simple cubic lattice? (g) How many nearest neighbors in the diamond and zincblende lattices? (h) What is being indicated by the bracket sets $()$, $[]$, $\{\}$, $⟨⟩$ in Miller indexing? (i) Describe the Czochralski method.

Solution.

(a) (i) Silicon (Si) or germanium (Ge). (ii) GaAs (also InP, GaP, etc.).

(b) A crystalline solid has atoms in one orderly 3-D array throughout; a polycrystalline solid is made of many crystalline grains that are misoriented/disjointed relative to each other, so order exists only within each grain.

(c) A unit cell is a small portion of a crystal that can be used to reproduce the entire crystal by stacking copies of it.

(d) Counting corner ($1/8$), face ($1/2$), interior (1):

$$\text{simple cubic: }8\left(\frac{1}{8}\right)=1,\text{bcc: }8\left(\frac{1}{8}\right)+1=2$$ $$\text{fcc: }8\left(\frac{1}{8}\right)+6\left(\frac{1}{2}\right)=4,\text{diamond: }4(\text{fcc})+4(\text{interior})=8$$

(e) $1\text{A˚}=10^{-8}\text{cm}$.

(f) In a simple cubic lattice the nearest neighbors are one cell edge apart: $d=a$.

(g) Four nearest neighbors in both diamond and zincblende lattices.

(h) $(hkl)$: a specific crystal plane; $[hkl]$: a specific crystal direction; $\{hkl\}$: a set of equivalent planes; $⟨hkl⟩$: a set of equivalent directions.

(i) Ultrapure polycrystalline Si is melted in a quartz crucible in an inert atmosphere. A small oriented single-crystal seed on a rod is dipped in the melt. The melt near the seed is cooled slightly so Si freezes onto the seed as a perfect extension of its lattice. The seed is slowly rotated and withdrawn, pulling out a large cylindrical single-crystal ingot (up to about 200 mm diameter, 1-2 m long) that is later sawed into wafers.

Problem 1.2 Unit cell of ${\mathrm{Al}}_{0.5}{\mathrm{Ga}}_{0.5}\mathrm{As}$

Problem. The GaAs unit cell is pictured in Fig. 1.4(b). Describe (or sketch) the unit cell for ${\mathrm{Al}}_{0.5}{\mathrm{Ga}}_{0.5}\mathrm{As}$.

Solution.

The zincblende cell contains 8 atoms: 4 on one fcc sublattice and 4 on the other. In GaAs, the 4 Ga atoms sit on the sublattice (cube corners and face centers, if we choose that one) and the 4 As atoms sit on the other (the four interior positions).

In ${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$ the Group III sublattice is shared between Al and Ga, and the As sublattice stays purely As. With $x=0.5$:

$$N_{\text{group III sites}}=4,N_{\mathrm{Al}}=x\cdot 4=0.5\times 4=2,N_{\mathrm{Ga}}=(1-x)\cdot 4=2,N_{\mathrm{As}}=4$$

So the cell is the same zincblende cube (side about 5.65 Å, nearly the same as GaAs), with 4 As atoms on one fcc sublattice and, on the other fcc sublattice, 2 Al and 2 Ga atoms that are randomly distributed over those 4 sites. (For a sketch: replace two of the four Ga atoms in Fig. 1.4(b), for instance two of the face-center or corner Ga atoms, with Al.) On average, of course, every Group III site is “half Al, half Ga”.

Problem 1.3 Germanium atomic density

Problem. (a) The lattice constant of Ge at room temperature is $a=5.65\times 10^{-8}$ cm. Determine the number of Ge atoms per cm${}^3$. (b) Run the Exercise 1.3 program to verify your result from (a).

Solution (a). Ge has the diamond structure: 8 atoms per cell.

$$n=\frac{8}{a^3}$$ $$a^3=(5.65\times 10^{-8}{)}^3=180.36\times 10^{-24}{\text{cm}}^3=1.8036\times 10^{-22}{\text{cm}}^3$$ $$n=\frac{8}{1.8036\times 10^{-22}{\text{cm}}^3}$$ $$\boxed{n\approx 4.44\times 10^{22}{\text{atoms/cm}}^3}$$

Solution (b). Run the program with N = 8, a = 5.65:

input number of atoms/unit cell, N = 8
lattice constant in angstrom, a = 5.65
atmden =
   4.4356e+22

since $8\times 10^{24}/5.65^3=8\times 10^{24}/180.36=4.4356\times 10^{22}$. This agrees with part (a).

Problem 1.4 Nearest-neighbor distances

Problem. In terms of the lattice constant $a$, what is the distance between nearest-neighbor atoms in (a) a bcc lattice? (b) an fcc lattice?

Solution (a), bcc. The center atom touches the 8 corner atoms. The distance from the cube center to a corner is half a body diagonal:

$$d_{\text{bcc}}=\frac{1}{2}\sqrt{a^2+a^2+a^2}=\frac{\sqrt{3}}{2}a\approx 0.866a$$

(Check that this beats the corner-to-corner distance $a$: $0.866a<a$, so yes it is the nearest.)

Solution (b), fcc. A corner atom and the atom at the center of an adjacent face are separated by half a face diagonal:

$$d_{\text{fcc}}=\frac{1}{2}\sqrt{a^2+a^2}=\frac{\sqrt{2}}{2}a=\frac{a}{\sqrt{2}}\approx 0.707a$$

(Again less than $a$.)

Problem 1.5 Atoms per cm${}^2$ on wafer surfaces

Problem. The surface of a Si wafer is a $(100)$ plane. (a) Sketch the placement of Si atoms on the surface. (b) Determine the number of atoms per cm${}^2$ at the surface. (c)/(d) Repeat (a) and (b) for a $(110)$ surface. (e) Write a MATLAB program that computes the surface density of atoms on $(100)$ planes of cubic crystals, with the number of atoms centered on the $(100)$ face of the unit cell and the lattice constant in angstrom as inputs; confirm the result of (b).

Solution (a). The $(100)$ plane is one face of the cubic cell, a square of side $a$. In the diamond lattice, the atoms with centers on that face are the four corner atoms and the one atom at the face center (a square pattern with an additional atom in the middle; no interior atoms lie in the face). The interior atoms are at $a/4$ and $3a/4$ depths, not on the face.

Solution (b). Atoms per face (corner atoms are shared among 4 squares in the plane, so each counts $1/4$):

$$N_{\text{face}}=4\left(\frac{1}{4}\right)+1=2$$

Area of the face:

$$A=a^2=(5.43\times 10^{-8}\text{cm}{)}^2=2.949\times 10^{-15}{\text{cm}}^2$$ $${\sigma }_{(100)}=\frac{2}{a^2}=\frac{2}{2.949\times 10^{-15}{\text{cm}}^2}$$ $$\boxed{{\sigma }_{(100)}\approx 6.78\times 10^{14}{\text{atoms/cm}}^2}$$

Solution (c). The $(110)$ plane is a rectangle cutting the cube diagonally. With $x+y=a$ it contains the four points $(a,0,0),(0,a,0),(a,0,a),(0,a,a)$, so its sides are $a$ (along $z$) and $\sqrt{2}a$ (along the face diagonal). Atoms with centers on this plane:

• the 4 corner points of the rectangle, which are fcc lattice sites: each shared by 4 rectangles in the plane, so $4\times \frac{1}{4}=1$;
• two face-center atoms located at the midpoints of the two long edges, e.g. $(a/2,a/2,0)$ and $(a/2,a/2,a)$, each shared by 2 rectangles: $2\times \frac{1}{2}=1$;
• two of the diamond interior atoms, $(3a/4,a/4,3a/4)$ and $(a/4,3a/4,3a/4)$, which both satisfy $x+y=a$ and lie fully inside the rectangle: $2$.

(For the sketch, draw a rectangle $a$ tall by $\sqrt{2}a$ wide with atoms at the four corners, the midpoints of the top and bottom edges, and two atoms inside, positioned at one-quarter and three-quarters of the width, at three-quarters of the height.)

Solution (d).

$$N_{(110)}=1+1+2=4,A=\sqrt{2}{a}^2$$ $${\sigma }_{(110)}=\frac{4}{\sqrt{2}{a}^2}=\frac{2\sqrt{2}}{a^2}=\frac{2.828}{2.949\times 10^{-15}{\text{cm}}^2}$$ $$\boxed{{\sigma }_{(110)}\approx 9.59\times 10^{14}{\text{atoms/cm}}^2}$$

Solution (e), MATLAB. The surface density is (atoms on the face)/(face area). With $a$ in angstrom, $a^2$ in cm${}^2$ equals $a_{\text{A˚}}^2\times 10^{-16}$, so $\sigma =N_{\text{face}}\times 10^{16}/a_{\text{A˚}}^2$.

% Problem 1.5(e)
% Surface density of atoms on (100) planes of cubic crystals
N = input('number of atoms with centers on the (100) face, N = ');
a = input('lattice constant in angstrom, a = ');
sdens = N*(1.0e16)/(a^2)   % atoms/cm^2

Run for Si:

number of atoms with centers on the (100) face, N = 2
lattice constant in angstrom, a = 5.43
sdens =
   6.7831e+14

This confirms the part (b) result ($6.78\times 10^{14}{\text{cm}}^{-2}$).

Problem 1.6 Determining Miller indices

Problem. Record all steps. (a) As shown in Fig. P1.6(a), a crystalline plane has intercepts of $1a$, $3a$ and $1a$ on the $x$, $y$ and $z$ axes ($a$ = cubic cell side). (i) Miller index notation for the plane? (ii) Miller indices of the direction normal to the plane? (b) Assuming a cubic structure, determine the Miller indices of (i) the plane and (ii) the vector pictured in Fig. P1.6(b).

Assumption note: the figure labels in the scan are garbled. I read Fig. P1.6(a) as intercepts $(1,3,1)\times a$, and Fig. P1.6(b) as a plane with intercepts $x=a$, $y=a$, $z=-2a$ together with a vector whose projections are $(a,a,-a/2)$. If your figure differs, apply exactly the same steps to the values you read.

Solution (a)(i).

$$\text{normalized intercepts: }(1,3,1)$$ $$[1/\text{intercept}]\text{s: }\left(1,\frac{1}{3},1\right)$$ $$\times 3:(3,1,3)$$ $$\boxed{(313)}$$

Solution (a)(ii). In a cubic crystal the normal to $(hkl)$ is $[hkl]$:

$$\boxed{[313]}$$

Solution (b)(i).

$$\text{intercepts: }(1,1,-2)$$ $$[1/\text{intercept}]\text{s: }\left(1,1,-\frac{1}{2}\right)$$ $$\times 2:(2,2,-1)$$ $$\boxed{(22\bar{1})}$$

Solution (b)(ii).

$$\text{projections: }\left(1,1,-\frac{1}{2}\right)\times 2=(2,2,-1)$$ $$\boxed{[22\bar{1}]}$$

(consistent: for a cubic crystal this direction is normal to the plane of part (i)).

Problem 1.7 Sketching planes

Problem. Assuming a cubic crystal system, sketch the following planes: (a) $(001)$, (b) $(111)$, (c) $(123)$, (d) $(\bar{1}10)$, (e) $(010)$, (f) $(\bar{1}\bar{1}\bar{1})$-type plane as printed, (g) $(2\bar{2}1)$-type plane as printed, (h) $(0\bar{1}0)$-type plane as printed.

Assumption note: overbars were lost in the OCR for parts (f)-(h). I take (f) $=(\bar{1}11)$, (g) $=(\bar{2}21)$, (h) $=(0\bar{1}0)$, and give the general recipe so any other sign choice follows immediately.

Method for every plane. Invert the indices to get intercepts, $x=a/h,y=a/k,z=a/l$ (an index $0$ means parallel to that axis), then mark the three intercepts on the axes and join them. Negative indices put the intercept on the negative axis.

Part	$(hkl)$	Intercepts $(x,y,z)$ in units of $a$	Description of the sketch
(a)	$(001)$	$(\infty ,\infty ,1)$	The top face of the unit cube: horizontal square at $z=a$, parallel to the $xy$ plane.
(b)	$(111)$	$(1,1,1)$	The equilateral triangle joining $(a,0,0)$, $(0,a,0)$ and $(0,0,a)$.
(c)	$(123)$	$(1,\frac{1}{2},\frac{1}{3})$	Triangle joining $(a,0,0)$, $(0,a/2,0)$, $(0,0,a/3)$; a steep plane.
(d)	$(\bar{1}10)$	$(-1,1,\infty )$	A vertical plane parallel to $z$, cutting the $-x$ axis at $-a$ and the $+y$ axis at $+a$.
(e)	$(010)$	$(\infty ,1,\infty )$	The face at $y=a$, parallel to the $xz$ plane.
(f)	$(\bar{1}11)$	$(-1,1,1)$	Triangle joining $(-a,0,0)$, $(0,a,0)$, $(0,0,a)$.
(g)	$(\bar{2}21)$	$(-\frac{1}{2},\frac{1}{2},1)$	Triangle joining $(-a/2,0,0)$, $(0,a/2,0)$, $(0,0,a)$.
(h)	$(0\bar{1}0)$	$(\infty ,-1,\infty )$	The face at $y=-a$, parallel to $xz$ (the opposite face from (e)).

If a plane would pass through the origin (all indices not zero cannot, but a sketch from shifted indices might), translate the origin to a neighboring lattice point first, as noted in special fact 4.

Problem 1.8 Sketching directions

Problem. Assuming a cubic system, use an appropriately directed arrow to identify each direction.

Assumption note: the printed list is scrambled by the OCR and its overbars are lost, so I treat the eight directions as (a) $[\bar{1}\bar{1}\bar{1}]$, (b) $[\bar{1}00]$, (c) $[0\bar{1}1]$, (d) $[00\bar{1}]$, (e) $[\bar{1}10]$, (f) $[\bar{2}\bar{1}0]$, (g) $[\bar{1}21]$, (h) $[1\bar{1}0]$. The method works for any triple.

Method. The direction $[hkl]$ is the arrow from the origin to the point $(h,k,l)$ (in any common unit). Draw the arrow from the origin to that point.

Part	Direction	Arrow from origin to	Description
(a)	$[\bar{1}\bar{1}\bar{1}]$	$(-1,-1,-1)$	Body diagonal, pointing toward the rear-bottom-left corner.
(b)	$[\bar{1}00]$	$(-1,0,0)$	Along the negative $x$-axis.
(c)	$[0\bar{1}1]$	$(0,-1,1)$	Face diagonal of the $yz$ face, toward $-y$ and $+z$.
(d)	$[00\bar{1}]$	$(0,0,-1)$	Straight down the $-z$ axis.
(e)	$[\bar{1}10]$	$(-1,1,0)$	Face diagonal in the $xy$ plane toward $-x,+y$.
(f)	$[\bar{2}\bar{1}0]$	$(-2,-1,0)$	In the $xy$ plane, twice as far along $-x$ as along $-y$.
(g)	$[\bar{1}21]$	$(-1,2,1)$	Component $-1$ in $x$, $+2$ in $y$, $+1$ in $z$.
(h)	$[1\bar{1}0]$	$(1,-1,0)$	Face diagonal in the $xy$ plane toward $+x,-y$.

Problem 1.9 Perpendicular directions

Problem. Identify two crystalline directions in a cubic crystal that are perpendicular to (a) the $[100]$ direction, (b) the $[111]$ direction.

The note in the book: for directions $[h_1{k}_1{l}_1]$ and $[h_2{k}_2{l}_2]$ in a cubic crystal,

$$\cos \theta =\frac{h_1{h}_2+k_1{k}_2+l_1{l}_2}{\sqrt{(h_1^2+k_1^2+l_1^2)(h_2^2+k_2^2+l_2^2)}}$$

so perpendicularity ($\cos \theta =0$) requires

$$h_1{h}_2+k_1{k}_2+l_1{l}_2=0$$

Solution (a). With $[h_1{k}_1{l}_1]=[100]$:

$$1\cdot h_2+0\cdot k_2+0\cdot l_2=0\Rightarrow h_2=0$$

Any $[0kl]$ works. Two choices:

$$\boxed{[010]\text{and}[001]}$$

(also $[011]$, $[01\bar{1}]$, and so on).

Solution (b). With $[111]$:

$$h_2+k_2+l_2=0$$

Two simple solutions:

$$[1\bar{1}0]:1-1+0=0✓[01\bar{1}]:0+1-1=0✓$$ $$\boxed{[1\bar{1}0]\text{and}[01\bar{1}]}$$

(also $[11\bar{2}]$: $1+1-2=0$.)

Problem 1.10 Grooves on a wafer surface

Problem. As in Fig. 1.5, $[011]$ is the direction in the surface plane normal to the major flat when the wafer surface is $(100)$. To construct a device structure, a parallel set of grooves must be etched in the $(100)$ surface plane along the $[01\bar{1}]$ direction (OCR lost the overbar; I assume $[01\bar{1}]$). Sketch how the grooves are oriented relative to the flat and explain.

Solution.

The surface is $(100)$, so every in-plane direction $[0kl]$ has $h=0$. Check that $[01\bar{1}]$ is in the surface:

$$h=0✓$$

Angle with the flat’s normal direction $[011]$:

$$\cos \theta =\frac{(0)(0)+(1)(1)+(-1)(1)}{\sqrt{(0+1+1)(0+1+1)}}=\frac{0}{2}=0\Rightarrow \theta =90^{\circ }$$

Since $[011]$ is perpendicular to the flat (normal to it), a direction perpendicular to $[011]$ within the surface is parallel to the flat. Sketch: draw the circular wafer with the flat at the bottom edge; draw parallel grooves as straight lines running parallel to the flat (left-right). (If the problem actually asked for the grooves along $[011]$ itself, they would be perpendicular to the flat: top-to-bottom lines.)

Problem 1.11 Combination question (single atom at the cube center)

Problem. A crystalline lattice is characterized by the cubic unit cell of Fig. P1.11, which has a single atom positioned at the center of the cube. (a) What is the name of the lattice generated by the cell? (b) Number of atoms per unit volume (in terms of $a$). (c) If the crystal has a $(110)$ surface plane, number of atoms per unit area whose centers lie on it. (d) A direction vector is drawn through the center of the atom in the unit cell; specify its Miller indices.

Assumption for (d): the vector drawn in the figure is lost; I take it to pass from the cube corner (origin) through the atom at the center, i.e. along the body diagonal. Any other vector is handled by the projection method.

Solution (a). An atom at the center of the cube and nothing else. Translating the cell by $a/2$ along each axis puts the atom at a corner and gives atoms at the 8 corners: this is the simple cubic lattice (the atoms are separated by $a$ in each direction).

Solution (b). One atom per cell of volume $a^3$:

$$\boxed{n=\frac{1}{a^3}}$$

Solution (c). The $(110)$ plane is a rectangle of sides $a$ and $\sqrt{2}a$ whose four corners are lattice points (each atom shared by four rectangles; no other lattice points are on it):

$$N=4\left(\frac{1}{4}\right)=1,A=\sqrt{2}{a}^2$$ $$\boxed{{\sigma }_{(110)}=\frac{1}{\sqrt{2}{a}^2}\approx \frac{0.707}{a^2}}$$

Solution (d). From the corner at the origin to the center at $(a/2,a/2,a/2)$:

$$\text{projections: }\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)\times \frac{2}{a}=(1,1,1)$$ $$\boxed{[111]}$$

Problem 1.12 Equivalent planes and directions

Problem. For a cubic lattice, how many equivalent planes (a)-(c) or equivalent directions (d)-(f) belong to each designation: (a) $\{100\}$, (b) $\{110\}$, (c) $\{111\}$, (d) $⟨100⟩$, (e) $⟨110⟩$, (f) $⟨111⟩$?

Solution. Count all index permutations with all sign choices (the book counts $(100)$ and $(\bar{1}00)$ as different planes, as in its remark about the six cube faces).

(a) $\{100\}$: one index is $\pm 1$ and two are $0$. Choose which index is nonzero (3 ways) times sign (2):

$$3\times 2=\boxed{6}$$

(b) $\{110\}$: two indices $\pm 1$, one $0$: choose which is zero (3) times signs ($2^2=4$):

$$3\times 4=\boxed{12}$$

(c) $\{111\}$: all indices $\pm 1$:

$$2^3=\boxed{8}$$

(d) $⟨100⟩$: same counting as (a): $\boxed{6}$.

(e) $⟨110⟩$: same as (b): $\boxed{12}$.

(f) $⟨111⟩$: same as (c): $\boxed{8}$.

(Equivalent directions: the 6 along the cube edges, 12 along face diagonals, 8 along body diagonals. If one counts only distinct orientations up to reversal, divide each by 2: 3, 6, 4.)

Problem 1.13 Packing fractions

Problem. Treating atoms as rigid spheres with radii equal to one half the distance between nearest neighbors, show that the ratio of volume occupied by atoms to the total available volume is (a) $\pi /6$ ($52\%$) for simple cubic, (b) $\sqrt{3}\pi /8$ ($68\%$) for bcc, (c) $\sqrt{2}\pi /6$ ($74\%$) for fcc, (d) $\sqrt{3}\pi /16$ ($34\%$) for diamond.

General set-up. If the nearest-neighbor distance is $d$, the sphere radius is $r=d/2$, the volume of one sphere is $\frac{4}{3}\pi r^3$, and the fraction is

$$f=\frac{N\cdot \frac{4}{3}\pi r^3}{a^3}$$

with $N$ atoms per cell.

(a) Simple cubic. $N=1$, $d=a$, $r=a/2$.

$$f=\frac{1\cdot \frac{4}{3}\pi {\left(\frac{a}{2}\right)}^3}{a^3}=\frac{\frac{4}{3}\pi \cdot \frac{1}{8}}{1}=\frac{\pi }{6}=0.5236\Rightarrow \boxed{52\%}$$

(b) bcc. $N=2$, $d=\frac{\sqrt{3}}{2}a$ (Problem 1.4), so $r=\frac{\sqrt{3}}{4}a$.

$$r^3=\frac{3\sqrt{3}}{64}{a}^3$$ $$f=\frac{2\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{64}{a}^3}{a^3}=2\cdot \frac{4\cdot 3\sqrt{3}\pi }{3\cdot 64}=2\cdot \frac{\sqrt{3}\pi }{16}=\frac{\sqrt{3}\pi }{8}=0.6802\Rightarrow \boxed{68\%}$$

(c) fcc. $N=4$, $d=\frac{\sqrt{2}}{2}a$, $r=\frac{\sqrt{2}}{4}a$.

$$r^3=\frac{2\sqrt{2}}{64}{a}^3=\frac{\sqrt{2}}{32}{a}^3$$ $$f=\frac{4\cdot \frac{4}{3}\pi \cdot \frac{\sqrt{2}}{32}{a}^3}{a^3}=4\cdot \frac{\sqrt{2}\pi }{24}=\frac{\sqrt{2}\pi }{6}=0.7405\Rightarrow \boxed{74\%}$$

(d) Diamond. $N=8$, $d=\frac{\sqrt{3}}{4}a$ (Exercise 1.2), $r=\frac{\sqrt{3}}{8}a$.

$$r^3=\frac{3\sqrt{3}}{512}{a}^3$$ $$f=\frac{8\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{512}{a}^3}{a^3}=8\cdot \frac{\sqrt{3}\pi }{128}=\frac{\sqrt{3}\pi }{16}=0.3401\Rightarrow \boxed{34\%}$$

Take-away. Diamond is a very “empty” lattice (only 34 percent full) compared to fcc (74 percent, the densest possible packing of equal spheres). The emptiness is the price of strong, directional four-fold bonding: atoms hold onto exactly four neighbors rather than packing as tightly as possible. This sets the stage for Chapter 2, where those four bonds determine how electrons behave.

Chapter 2: Carrier Modeling

The big question of this chapter. A silicon crystal contains about $5\times 10^{22}$ atoms in every cubic centimetre, each with 14 electrons. Yet a piece of silicon conducts electricity only through a tiny, controllable handful of them (roughly $10^{10}$ per cm${}^3$ in pure Si). Who are those few special electrons, what is the strange “anti-particle” called a hole, how many of each are there, and how can we dial that number up or down on purpose? By the end of the chapter you will be able to answer all of that with a single, tidy set of equations.

Carriers are the entities that carry charge from place to place inside a material and hence give rise to electrical currents. In daily life the carrier you meet is the electron in a copper wire. In semiconductors you also meet a second, equally important carrier, the hole.

Equilibrium, the “rest” condition. Everything in this chapter assumes equilibrium: no external voltages, magnetic fields, stresses, light, or other perturbations, and every observable is independent of time. Think of a perfectly still pond. If you understand the still pond, you can predict what happens when you throw a stone in. Equilibrium is the still pond of semiconductor physics. Whenever a formula in this chapter is only valid at equilibrium, remember that it will not hold once we apply a voltage or shine light (later chapters).

The author warns that some results are quoted without proof; the philosophy is that knowing how to interpret and use a result matters more than being able to derive it. (In this study version I have nevertheless filled in every algebraic step that the book leaves out.) Appendix A of the book gives a deeper account of quantization, and Section 2.1 may be replaced by it without loss of continuity.

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2.1 The Quantization Concept

Instead of diving straight into crystalline silicon (14 electrons per atom, $5\times 10^{22}$ atoms/cm${}^3$), we first learn the ground rules on the simplest possible atom, hydrogen.

The puzzle that started it all

In the early 1900s, scientists knew a hydrogen atom was a light negative particle orbiting a much heavier positive nucleus. What they could not explain: when hydrogen gas is heated it glows only at certain discrete wavelengths (sharp colored lines), whereas classical physics predicted a smooth rainbow (continuum).

Analogy. A violin string clamped at both ends can only sustain certain notes (standing waves); you cannot get a pitch “in between” the allowed harmonics. The hydrogen electron, it turns out, is also a wave trapped around a nucleus, so only certain “notes” (energies) are allowed.

Bohr’s solution (1913)

Niels Bohr postulated that the electron can only occupy certain well-defined orbits; equivalently, its angular momentum can only take the values $n\hslash$. That restriction is quantization. Combining it with Coulomb attraction and circular-orbit mechanics gives the binding energy of the electron:

$$E_H=-\frac{m_0{q}^4}{2(4\pi {\epsilon }_0\hslash n{)}^2}=-\frac{13.6}{n^2}\text{eV},n=1,2,3,\dots \text{\hspace{1em}(2.1)}$$

where

• $E_H$ is the electron energy (binding energy; negative means “bound”),
• $m_0$ is the free-electron rest mass,
• $q$ is the magnitude of the electron charge,
• ${\epsilon }_0$ is the permittivity of free space,
• $\hslash =h/2\pi$ ($h$ = Planck’s constant),
• $n$ is the energy quantum number (orbit identifier).

The electron-volt is the energy gained by one electron falling through 1 V: $1\text{eV}=1.6\times 10^{-19}\text{J}.$

Energy ladder (Fig. 2.1). The figure shows the first allowed orbits and their energies:

$n$	$E_H$
1	$-13.6$ eV
2	$-3.4$ eV
3	$-1.51$ eV

(Check: $13.6/4=3.4$, $13.6/9=1.51$.)

Why spectral lines are discrete. When the electron drops from a higher $n$ to a lower $n$, the atom releases the energy difference as a photon. Because only certain energies exist, only certain differences (and therefore only certain wavelengths) exist. For instance $n=2\to 1$ releases $\Delta E=-3.4-(-13.6)=10.2\text{eV}.$

Ionization energy. To free a ground-state ($n=1$) electron entirely you must supply $|E_H(n=1)|=13.6$ eV.

From hydrogen to silicon (Fig. 2.2)

The one big lesson: the energy of an electron in an atomic system is restricted to a limited set of values. A multi-electron atom is more complicated, but its energy-level structure is still easy to describe for silicon.

• Of silicon’s 14 electrons, 10 occupy very deep-lying, tightly bound levels. They stay essentially undisturbed in chemical reactions or normal atom-atom interactions. The nucleus plus these ten electrons is called the core.
• The remaining 4 electrons are weakly bound and take part strongly in chemistry; they are the valence electrons.
• Figure 2.2: the four valence electrons, if unperturbed, occupy four of the eight allowed states of the next-higher shell ($n=3$): two states at the “s” energy and six states at the (slightly higher) “p” energy. Deeper, the $n=1$ shell holds 2 electrons and the $n=2$ shell holds 8, totalling 10 core electrons.
• Germanium (32 electrons) has essentially the same valence structure with a 28-electron core.

(Footnote of the book: not only energy, but many observables of atomic-size particles are quantized; Quantum Mechanics is the field developed to describe them.)

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2.2 Semiconductor Models

This section introduces two visualization aids used constantly in device analysis. It is appropriate in a chapter on carrier modeling because we are modeling the carrier “container,” the crystal.

2.2.1 Bonding Model

An isolated Si atom has 4 valence electrons. In the diamond lattice (Fig. 1.4c) each atom has 4 nearest neighbors and shares one valence electron with each. This equal sharing is covalent bonding.

Fig. 2.3 (the bonding model). A flat grid in which

• each circle is the core of a Si atom,
• each line is a shared valence electron. Eight lines touch each atom because an atom both contributes four shared electrons and accepts four from neighbors (two lines per bond, one from each partner).

The 2-D picture is an idealization to help mental visualization.

Sample uses (Fig. 2.4):

• (a) A missing atom (point defect) is drawn by erasing one circle with its eight line-ends.
• (b) A broken bond is drawn by deleting one line segment (one shared electron) and showing that electron wandering free.

Strictly, the perfect picture of Fig. 2.3 is valid only at $T\approx 0$ K with no defects and no impurities, since bond breaking (at $T>0$) and defects occur naturally.

What the bonding model cannot do: it says nothing about electron energies. For that we need the energy band model.

2.2.2 Energy Band Model

Building the band picture step by step (Fig. 2.5).

1. Start with $N$ isolated Si atoms very far apart. Each has the same discrete levels. Ten electrons per atom are core, so ignore them. Four valence electrons per atom occupy 4 of the 8 states of the outer shell. In total: $2N$ s-states and $6N$ p-states (that is $8N$ states) and $4N$ valence electrons.

2. Now bring the atoms closer together. Their wavefunctions overlap and interatomic forces shift the energy levels. Because the Pauli exclusion principle forbids two electrons occupying the same state, each of the $N$ nearly identical levels must split into slightly different levels. With $N\sim 10^{22}$ the levels are so closely spaced that they form quasi-continuous energy bands.

   Analogy: $N$ identical tuning forks placed on a shared table begin to couple and ring at slightly different frequencies; the single pure note becomes a smear (a band) of tones.

3. At the actual Si lattice spacing, the allowed states divide into two bands separated by an energy gap:

   • upper band: the conduction band,
   • lower band: the valence band,
   • gap in between: the forbidden gap or band gap $E_G$. No allowed electron states exist there.

4. Each band contains $4N$ states. Since there are exactly $4N$ valence electrons and electrons fall to the lowest available energies (single occupancy per state, Pauli), the valence band is (almost completely) filled and the conduction band is (almost completely) empty. At $T\to 0$ K the valence band is completely full and the conduction band completely empty.

5. A further crucial fact: unlike the valence electrons of an isolated atom, band electrons are not tied to any one atom. On average four electrons are shared between an atom and its neighbors, but which electrons they are changes in time; the electrons roam. The allowed states belong to the crystal as a whole. In a perfect crystal one sees the same allowed-state configuration at every point.

Therefore, a plot of allowed electron energy versus position $x$ along a crystal direction gives horizontal bands (Fig. 2.5, right). We name:

• $E_c$: the lowest possible conduction band energy (conduction band edge),
• $E_v$: the highest possible valence band energy (valence band edge),
• $E_G=E_c-E_v$: the band gap energy.

Fig. 2.6, the energy band diagram. In practice we draw only two horizontal lines $E_c$ and $E_v$ with a gap between them. The line showing the top of the conduction band, the line showing the bottom of the valence band, the filling pattern, the energy axis label (vertical) and the position axis label (horizontal) are all understood but not drawn. This is the “shorthand” version used everywhere afterwards.

2.2.3 Carriers

No broken bonds, no carriers (Fig. 2.7a). In the bonding model a perfect lattice has no mobile charges. Equivalent energy-band statement: valence band completely full, conduction band completely empty means no carriers and no current.

Subtle puzzle. In the band model valence electrons do move around the crystal. Why do they not carry current? Answer: momentum is quantized too. For every state with momentum $+\mathbf{p}$ there is a state with momentum $-\mathbf{p}$ of equal magnitude. If a band is completely filled, all momenta cancel in pairs and the net momentum is exactly zero: ${\sum }_{\text{filled band}}\mathbf{p}=0\Rightarrow \text{no net current}.$ Think of a completely full auditorium: everyone is in motion in their heads, but nobody can go anywhere, so nothing is transported.

The electron (Fig. 2.7b).

• Bonding model: a Si-Si bond breaks and the released electron wanders through the lattice. That wandering electron is a carrier.
• Band model: an electron is excited from the valence band to the conduction band; electrons in the conduction band are carriers.

The energy needed to break a bond is the band gap energy $E_G$. “Free bonding-model electron” and “conduction band electron” are two names for the same thing; from now on “electron” with no modifier means a conduction band electron.

The hole (Fig. 2.7c). Breaking a bond leaves a missing bond (a void).

• Bonding model: a neighboring bound electron can hop into the void, which moves the void to where that electron came from. The void migrates.
• Band model: removing an electron from the valence band leaves an empty state in an otherwise filled sea of states. Like a bubble in a liquid, the empty state moves rather freely because of the cooperative motion of valence electrons.

That mobile missing bond or empty valence-band state is the hole, the second kind of carrier.

Intuition. Picture a parking garage with every space full except one, and cars shift into the vacancy one by one. Watching the cars, you track dozens of moving objects. Watching the empty space, you track a single object moving the opposite way. It is much simpler to follow the hole than the sea of electrons. That is why a hole behaves as a positive particle.

A valence-band hole is on an equal footing with a conduction-band electron. Both participate in most devices, and holes are the main carrier in some devices. With familiarity one thinks of the hole as a subatomic particle.

2.2.4 Band Gap and Material Classification

The band picture of Fig. 2.6 applies, with slight modification, to all materials. What differs is the size of the gap (Fig. 2.8).

Class	Band gap	Result at room temperature
Insulator	very wide, e.g. diamond $\approx 5$ eV, SiO${}_2$ $\approx 8$ eV	thermal energy excites almost no electron across the gap, so few carriers, poor conductor
Semiconductor	moderate: GaAs 1.42 eV, Si 1.12 eV, Ge 0.66 eV (at 300 K)	thermal excitation makes a moderate number of carriers, intermediate conductivity
Metal	very narrow or zero (valence and conduction bands overlap)	abundant carriers always, excellent conductor

Analogy. The gap is a wall between the ground floor (valence band) and the upper floor (conduction band). In an insulator the wall is a skyscraper’s height, in a metal there is no wall, and in a semiconductor it is a wall that thermal “jumps” can occasionally clear.

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2.3 Carrier Properties

We have now met two characters, the electron and the hole. Before any equations, let us interview them: what charge do they carry, how heavy are they, how many of them are there, and how can we change that number? This section is a “collage” of general facts, and every later section leans on it.

2.3.1 Charge

Both carriers are charged, and the magnitude of the charge is the same for each:

• electron charge: $-q$
• hole charge: $+q$
• $q=1.60\times 10^{-19}\text{C}$ (three-place accuracy, MKS units).

The sign is always written explicitly (we never hide it inside $q$). Remember this convention, because a forgotten sign is the most common source of wrong answers in the next chapters.

2.3.2 Effective Mass

Charge is a single number. Mass is not so simple, and this is where the physics gets interesting.

Thought experiment 1: the electron in a vacuum. Put an electron of rest mass $m_0$ between two parallel plates with an electric field $\mathcal{E}$ between them. Newton’s second law gives

$$F=-q\mathcal{E}=m_0\frac{dv}{dt}\text{\hspace{1em}(2.2)}$$

where $v$ is the velocity and $t$ is time. The electron simply accelerates, and the acceleration is set by $m_0$ alone.

Thought experiment 2: the electron in a crystal. Now put the same field across a semiconductor crystal (Fig. 2.9b). Does Eq. (2.2) still describe the motion? The answer is no, for two reasons:

1. The electron collides with atoms and is periodically decelerated.
2. Even between collisions, besides our applied field the electron feels the complicated internal electric fields of the crystal, which Eq. (2.2) knows nothing about.

Strictly, only quantum mechanics can describe the motion. Here is the happy surprise: if the crystal is large compared with atomic dimensions, the complicated quantum description collapses into an equation of the same form as Newton’s law, provided we swap $m_0$ for an effective mass:

$$F=-q\mathcal{E}=m_n^{\ast }\frac{dv}{dt}\text{\hspace{1em}(2.3)}$$

where $m_n^{\ast }$ is the electron effective mass; for holes we write the same with $m_p^{\ast }$. All the crystal’s internal forces and quantum effects are “lumped into” the number $m^{\ast }$.

Analogy. Imagine running through a swimming pool versus through open air. Your actual body mass is the same, but you respond to your own push as if you were much heavier. The effective mass is the “apparent mass” of the carrier after the surroundings have had their say. It lets us treat electrons and holes as ordinary, classical-looking particles.

Caveats the book mentions: the acceleration can depend on direction in the crystal (effective mass has several components); different observables (cyclotron resonance, conductivity, density of states) use different groupings of these components; and $m^{\ast }$ drifts slightly with temperature. In this chapter we use only the density of states effective masses (Section 2.4.1), listed in Table 2.1.

Table 2.1: Density of states effective masses at 300 K

Material	$m_n^{\ast }/m_0$	$m_p^{\ast }/m_0$
Si	1.18	0.81
Ge	0.55	0.36
GaAs	0.066	0.52

Notice the striking number: electrons in GaAs are very light ($0.066m_0$), which is one reason GaAs is used for fast devices.

2.3.3 Carrier Numbers in Intrinsic Material

An intrinsic semiconductor is, loosely, an extremely pure sample with a negligible amount of impurity; more precisely, one whose properties are native to the material and not caused by additives. Define

$$n={\text{number of electrons/cm}}^3,p={\text{number of holes/cm}}^3.$$

In an intrinsic semiconductor at equilibrium

$$n=p=n_i\text{\hspace{1em}(2.4)}$$

and, at room temperature,

$$n_i\approx \{\begin{array}{ll}2\times 10^6{\text{cm}}^{-3}& \text{GaAs}\\ 1\times 10^{10}{\text{cm}}^{-3}& \text{Si}\\ 2\times 10^{13}{\text{cm}}^{-3}& \text{Ge}\end{array}$$

Why $n=p$? In a very pure material carriers are created only in pairs. Break a bond and you simultaneously create a free electron and a missing bond (hole). Lift an electron from the valence to the conduction band and you automatically leave a valence-band hole behind.

How big is $n_i$, really? Put numbers on it:

$$\begin{array}{rlrl}{\text{bonds per cm}}^3& =(5\times 10^{22})(4)& & \text{four bonds per Si atom}\\ & =2\times 10^{23}{\text{cm}}^{-3}& & \text{multiply}\\ \frac{n_i}{\text{bonds}}& =\frac{10^{10}}{2\times 10^{23}}=5\times 10^{-14}& & \text{fraction of bonds broken}\end{array}$$

That is fewer than one bond in $10^{13}$ broken. The book’s vivid picture: cover every university chalkboard in the world with the bonding model and you would probably draw just one broken bond.

2.3.4 Manipulation of Carrier Numbers: Doping

Intrinsic carrier numbers are tiny and fixed by nature. Doping is the deliberate addition of controlled amounts of specific impurity atoms to raise either the electron or the hole concentration. Nearly every semiconductor device is made this way.

Table 2.2: Common silicon dopants (most widely used marked with $\star$)

Donors (electron-increasing), Column V	Acceptors (hole-increasing), Column III
P $\star$	B $\star$
As $\star$	Ga
Sb	In
	Al

Donor action (bonding model, Fig. 2.10a). Substitute a Column V atom (5 valence electrons) for a Si atom. Four of its five electrons fit snugly into the four bonds. The fifth does not fit and is only weakly bound; at room temperature it is easily freed and wanders through the lattice as a carrier. Important: this does not create a hole. The leftover donor ion (${\text{P}}^{+}$) is fixed in the lattice, and no Si-Si bond was broken.

Acceptor action (Fig. 2.10b). Substitute a Column III atom (3 valence electrons). It cannot complete all four bonds, so it readily accepts an electron from a neighboring Si-Si bond. That completes its own bonding (making it a fixed negative ion, ${\text{B}}^{-}$) and leaves behind a wandering hole. Again only one carrier type increases.

Memory trick. Column V donates a spare electron (5 > 4: surplus). Column III accepts one (3 < 4: shortage). The ion left behind is immobile; it is the carrier that moves.

Loose end 1: how weakly is “weakly bound”? Breaking a Si-Si bond takes about 1 eV, and very few break at room temperature. So “weakly” should mean a binding energy of about $0.1$ eV or less. Let us estimate it using the hydrogen atom (Fig. 2.11, the pseudo-hydrogen atom).

Treat the donor core (${\text{P}}^{+}$) plus its fifth electron as a hydrogen atom, but with two substitutions because the electron now moves through a sea of Si atoms rather than a vacuum:

• ${\epsilon }_0\to K_s{\epsilon }_0$ (with Si dielectric constant $K_s=11.8$),
• $m_0\to m_n^{\ast }$.

Applying these to Eq. (2.1) with $n=1$:

$$E_B\approx \frac{m_n^{\ast }}{m_0}\frac{1}{K_s^2}{E}_H{|}_{n=1}\approx -0.1\text{eV}\text{\hspace{1em}(2.5)}$$

Let us verify the number using Table 2.1 ($m_n^{\ast }/m_0=1.18$):

$$\begin{array}{rlrl}{K}_s^2& =11.8^2=139.24& & \text{square the dielectric constant}\\ E_B& \approx 1.18\times \frac{1}{139.24}\times (-13.6\text{eV})& & \text{insert }{m}_n^{\ast }/m_0,K_s,E_H\\ & \approx -0.115\text{eV}& & \text{multiply out}\end{array}$$

So $|E_B|\sim 0.1$ eV. Measured values (Table 2.3) agree, and they are about $1/20$ of the Si band gap.

Table 2.3: Dopant-site binding energies in Si

Donor	$|E_B|$	Acceptor	$|E_B|$
Sb	0.039 eV	B	0.045 eV
P	0.045 eV	Al	0.067 eV
As	0.054 eV	Ga	0.072 eV
		In	0.16 eV

Why the binding is so weak: the dielectric constant enters squared in the denominator. The crystal screens the donor’s pull by a large factor, and the small effective mass shrinks the binding further.

Loose end 2: dopants in the energy band picture. An electron freed from a donor with exactly the binding energy becomes a conduction band electron with the lowest possible energy, $E_c$. Hence the bound electron must sit an energy $|E_B|$ below $E_c$:

$$E_D=E_c-|E_B|.$$

We draw donor levels as short dashes (Fig. 2.12) rather than a line, because a bound electron is localized near its donor site and does not wander through the crystal. Likewise acceptors add allowed levels slightly above $E_v$:

$$E_A=E_v+|E_B|.$$

What happens as temperature rises (Fig. 2.13):

• Donors: at $T\to 0$ K all donor levels are filled (no thermal energy to kick electrons up). With increasing $T$ electrons hop into the conduction band. At room temperature donors are essentially all ionized.
• Acceptors: at $T\to 0$ K all acceptor levels are empty. With increasing $T$ valence-band electrons hop onto acceptor sites, leaving holes behind. At room temperature essentially all acceptors hold an extra electron.

Doping non-elemental semiconductors (GaAs). The same ideas hold but there are two lattice sites:

• Column VI atoms (S, Se, Te) replacing Column V As act as donors.
• Column II atoms (Be, Mg, Zn) replacing Column III Ga act as acceptors.
• Column IV atoms (Si, Ge) are special: Si usually replaces Ga (donor, a popular n-type dopant), but under some conditions replaces As (acceptor). An impurity that can act as either is called amphoteric.

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Exercise 2.1: Energy Quiz

(a) $1\text{eV}$ equals how many joules?

$$\begin{array}{rlrl}1\text{eV}& =(1.60\times 10^{-19}\text{C})(1\text{V})& & \text{energy gained by a charge }q\text{ across 1 V}\\ & =1.60\times 10^{-19}\text{J}& & \text{C}\cdot \text{V}=\text{J}\end{array}$$

(b) $kT$ in eV at $300$ K:

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5}\text{eV/K})(300\text{K})& & \text{Boltzmann constant times }T\\ & =2.585\times 10^{-2}\text{eV}\approx 0.0259\text{eV}& & \text{multiply}\end{array}$$

This number, about $26$ meV, is the natural “thermal energy unit” for the rest of the book. Keep it in your head.

(c) Ionization energy of dopants in Si: it equals $|E_B|$ of the dopant site, roughly $0.1$ eV (measured $0.04$ to $0.07$ eV for common dopants).

(d) $E_G(\text{Si})=1.12$ eV at 300 K (Section 2.2.4).

(e) $E_G({\text{SiO}}_2)\approx 8$ eV (Section 2.2.4).

(f) Ionizing hydrogen from $n=1$ costs, from Eq. (2.1),

$$|E_H(n=1)|=\frac{13.6}{1^2}=13.6\text{eV}.$$

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2.3.5 Carrier-Related Terminology

Terminology is a classic stumbling block, so here is the glossary to memorize:

Term	Meaning
Dopants	Specific impurity atoms added in controlled amounts to raise the electron or the hole concentration
Intrinsic semiconductor	Undoped, extremely pure; properties native to the material
Extrinsic semiconductor	Doped; properties controlled by the added impurities
Donor	Impurity that increases the electron concentration (n-type dopant)
Acceptor	Impurity that increases the hole concentration (p-type dopant)
n-type material	Donor-doped; more electrons than holes
p-type material	Acceptor-doped; more holes than electrons
Majority carrier	Most abundant carrier: electrons in n-type, holes in p-type
Minority carrier	Least abundant carrier: holes in n-type, electrons in p-type

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2.4 State and Carrier Distributions

So far our knowledge has been qualitative. Real engineers need numbers: how many electrons are in a doped sample, and at what energies do they live? We build the answer from two ingredients, like counting people in a stadium:

1. How many seats exist at each energy? (density of states)
2. What fraction of those seats is occupied? (Fermi function)

Multiply the two and you have the carrier distribution.

2.4.1 Density of States

Section 2.2 said each band holds $4N$ states, but not how those states are spread in energy. That spread is the density of states. Quantum mechanics gives, for energies not too far from the band edges,

$$g_c(E)=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }(E-E_c)}}{{\pi }^2{\hslash }^3},E\ge E_c\text{\hspace{1em}(2.6a)}$$ $$g_v(E)=\frac{m_p^{\ast }\sqrt{2m_p^{\ast }(E_v-E)}}{{\pi }^2{\hslash }^3},E\le E_v\text{\hspace{1em}(2.6b)}$$

where $g_c$, $g_v$ are the densities of states in the conduction and valence bands.

Football stadium analogy. The field is the band gap. Seats lie at various distances from the field; the number of seats at a given distance is the number of states at a given energy distance from the band edge. There are no seats on the field itself (no states in the gap) and the stands widen as you climb away from it.

Key points to remember:

• $g_c(E)=0$ at $E=E_c$ and grows like $\sqrt{E-E_c}$ upward; $g_v(E)=0$ at $E=E_v$ and grows like $\sqrt{E_v-E}$ downward (Fig. 2.14).
• The two curves differ only because $m_n^{\ast }\ne m_p^{\ast }$. If the masses were equal the two “stands” would be mirror images.
• $g_c(E)dE$ is the number of conduction band states per cm${}^3$ between $E$ and $E+dE$; likewise $g_v(E)dE$ for the valence band. So $g$ has units of states per cm${}^3$ per eV.

2.4.2 The Fermi Function

The density of states says how many seats exist. The Fermi function $f(E)$ says what fraction is filled, that is, the probability that an available state at energy $E$ is occupied by an electron at equilibrium:

$$f(E)=\frac{1}{1+e^{(E-E_F)/kT}}\text{\hspace{1em}(2.7)}$$

with

• $E_F$: the Fermi energy or Fermi level,
• $k=8.617\times 10^{-5}$ eV/K: Boltzmann’s constant,
• $T$: absolute temperature in K.

Case 1: $T\to 0$ K. Look at the exponent $(E-E_F)/kT$:

$$\begin{array}{rl}E<E_F:& \frac{E-E_F}{kT}\to -\infty \Rightarrow f(E)\to \frac{1}{1+e^{-\infty }}=1\\ E>E_F:& \frac{E-E_F}{kT}\to +\infty \Rightarrow f(E)\to \frac{1}{1+e^{+\infty }}=0\end{array}$$

So at absolute zero every state below $E_F$ is full and every state above is empty: a sharp cutoff (Fig. 2.15a). It is a bucket of water at rest: everything below the water line is wet.

Case 2: $T>0$ K. Three observations:

1. At $E=E_F$: $f(E_F)=\frac{1}{1+e^0}=\frac{1}{2}$. The Fermi level is the energy at which a state is exactly a coin-flip.
2. For $E\ge E_F+3kT$: $e^{(E-E_F)/kT}\gg 1$, so

$$f(E)\approx e^{-(E-E_F)/kT}.$$

The filled probability decays exponentially to zero and most states are empty. 3. For $E\le E_F-3kT$: $e^{(E-E_F)/kT}\ll 1$, so

$$f(E)\approx 1-e^{(E-E_F)/kT},\text{i.e. }1-f(E)\approx e^{(E-E_F)/kT}.$$

The empty-state probability $1-f$ decays exponentially to zero as you go further below, and most states are filled.

Raising $T$ merely “rounds the corner” of the step, over a window of a few $kT$ ($\approx 3kT=78$ meV at 300 K), tiny against the $1.12$ eV Si gap (Fig. 2.15b).

Universality. The Fermi function holds only at equilibrium, but it is universal: it applies equally to insulators, semiconductors, and metals. It is just the statistics of electrons.

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Exercise 2.2

Problem. The probability that a state is filled at the conduction band edge $E_c$ is exactly equal to the probability that a state is empty at the valence band edge $E_v$. Where is the Fermi level?

Solution.

The empty-state probability is $1-f(E)$, so the statement reads

$$f(E_c)=1-f(E_v)\text{given condition}$$

Write each side using Eq. (2.7):

$$f(E_c)=\frac{1}{1+e^{(E_c-E_F)/kT}}\text{definition of }f$$ $$1-f(E_v)=1-\frac{1}{1+e^{(E_v-E_F)/kT}}=\frac{e^{(E_v-E_F)/kT}}{1+e^{(E_v-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}\text{common denominator, then divide top and bottom by }{e}^{(E_v-E_F)/kT}$$

Equating the two:

$$\frac{1}{1+e^{(E_c-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}\text{set equal}$$ $$\frac{E_c-E_F}{kT}=\frac{E_F-E_v}{kT}\text{equal fractions with equal numerators have equal exponents}$$ $$E_c-E_F=E_F-E_v\text{multiply by }kT$$ $$E_F=\frac{E_c+E_v}{2}\text{solve for }{E}_F$$

Result: the Fermi level sits at midgap.

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Exercise 2.3: Fermi function versus temperature

Problem. For $T=100,200,300,400$ K compute and plot $f(E)$ versus $\Delta E=E-E_F$ for $-0.2\le \Delta E\le 0.2$ eV, all curves on one set of axes.

Solution. The book uses a MATLAB script. The steps in words and equations:

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5})T& & \text{thermal energy for each }T\\ \Delta E_j& =-5kT+0.1kT(j-1),j=1,\dots ,101& & \text{energy grid from }-5kT\text{ to }+5kT\\ f_j& =\frac{1}{1+e^{\Delta E_j/kT}}& & \text{Fermi function at each grid point}\end{array}$$

% Fermi Function Calculation, f(dE,T)
k = 8.617e-5;                 % Boltzmann constant (eV/K)
for ii = 1:4
    T  = 100*ii;
    kT = k*T;
    dE(ii,1) = -5*kT;
    for jj = 1:101
        f(ii,jj)    = 1/(1+exp(dE(ii,jj)/kT));
        dE(ii,jj+1) = dE(ii,jj) + 0.1*kT;
    end
end
dE = dE(:,1:jj);              % strip the extra dE value
plot(dE', f); grid;           % transpose to form data columns
xlabel('E-E_F (eV)'); ylabel('f(E)');

What the plot shows. Every curve passes through $f=1/2$ at $\Delta E=0$. The $T=100$ K curve is nearly a step; the $T=400$ K curve is the most gradual. Hotter means a fatter transition region, exactly as the $3kT$ rule predicts (note the grid for each curve spans $\pm 5kT$, so the cold curves reach the plot’s flat parts sooner).

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2.4.3 Equilibrium Distribution of Carriers

Now multiply seats by occupancy:

$$\begin{array}{rl}\text{electrons in conduction band:}& g_c(E)f(E)\\ \text{holes in valence band:}& g_v(E)[1-f(E)]\end{array}$$

(A hole is an unfilled state, hence the factor $1-f$.) Fig. 2.16 shows these for three Fermi-level positions. General features:

• All distributions are zero at the band edges (because $g=0$ there), peak very close to $E_c$ or $E_v$, and then decay rapidly. Most carriers live near the band edges.
• $E_F$ in the upper half of the gap (or higher): electrons far outnumber holes, because $f(E)$ is much larger than $1-f(E)$ at corresponding energies.
• $E_F$ near midgap: about equal numbers of both (assuming $g_c$ and $g_v$ are of comparable size, true for Si and Ge).
• $E_F$ in the lower half of the gap: holes dominate.
• The exponential-tail statements hold provided $E_c-3kT\ge E_F\ge E_v+3kT$.

Shorthand pictures: Fig. 2.17 draws many dots near the band edges and fewer farther in. Fig. 2.18, the “at a glance” picture, draws a dashed line near midgap labeled $E_i$ (the intrinsic Fermi level). A solid $E_F$ line above $E_i$ means n-type; below $E_i$ means p-type. In extrinsic diagrams the dashed $E_i$ is kept as a reference that splits the gap into upper and lower halves.

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2.5 Equilibrium Carrier Concentrations

This is the summit of the chapter. We now turn the qualitative picture into working formulas. The book compares the alternative forms of each formula to a toolbox: open-end, box and ratchet wrenches all turn the same bolt, but in a tight spot only one fits. Do not be alarmed by the algebra; every step is shown.

2.5.1 Formulas for $n$ and $p$

Since $g_c(E)dE$ is the number of conduction band states per cm${}^3$ in $[E,E+dE]$ and $f(E)$ is the probability each is filled, $g_c(E)f(E)dE$ is the number of electrons there. Adding up (integrating) over the whole band gives the electron concentration. The same logic with $g_v(1-f)$ gives holes:

$$n={\int }_{E_c}^{E_{top}}{g}_c(E)f(E)dE\text{\hspace{1em}(2.8a)}$$ $$p={\int }_{E_{bottom}}^{E_v}{g}_v(E)[1-f(E)]dE\text{\hspace{1em}(2.8b)}$$

Derivation of the $n$ formula. Insert Eq. (2.6a) and Eq. (2.7):

$$n=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{E_c}^{E_{top}}\frac{\sqrt{E-E_c}}{1+e^{(E-E_F)/kT}}dE\text{\hspace{1em}(2.9)}$$

Make the substitutions

$$\eta =\frac{E-E_c}{kT},{\eta }_c=\frac{E_F-E_c}{kT},E_{top}\to \infty \text{(i.e. upper limit }\eta \to \infty )\text{\hspace{1em}(2.10)}$$

Why may we push the upper limit to infinity? The integrand dies off exponentially within a few $kT$ of $E_c$, so extending the limit adds a negligible amount. Also, $E-E_F=(E-E_c)-(E_F-E_c)=kT(\eta -{\eta }_c)$ and $\sqrt{E-E_c}dE=(kT{)}^{3/2}\sqrt{\eta }d\eta$. Therefore

$$n=\frac{\sqrt{2}(m_n^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\frac{\sqrt{\eta }d\eta }{1+e^{\eta -{\eta }_c}}\text{\hspace{1em}(2.11)}$$

This integral has no closed form with simple functions; it is a tabulated function, the Fermi-Dirac integral of order 1/2:

$$F_{1/2}({\eta }_c)={\int }_0^{\infty }\frac{\sqrt{\eta }d\eta }{1+e^{\eta -{\eta }_c}}\text{\hspace{1em}(2.12)}$$

Define the effective densities of states

$$N_C=2{\left(\frac{2\pi m_n^{\ast }kT}{h^2}\right)}^{3/2},N_V=2{\left(\frac{2\pi m_p^{\ast }kT}{h^2}\right)}^{3/2}\text{\hspace{1em}(2.13)}$$

(equivalently $2(m^{\ast }kT/2\pi {\hslash }^2{)}^{3/2}$). Checking that the constants combine: $\frac{2}{\sqrt{\pi }}{N}_C=\frac{2}{\sqrt{\pi }}\cdot \frac{2(m_n^{\ast }kT{)}^{3/2}}{(2\pi {)}^{3/2}{\hslash }^3}=\frac{\sqrt{2}(m_n^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}$, exactly the prefactor in Eq. (2.11). Hence

$$n=N_C\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_c),{\eta }_c=\frac{E_F-E_c}{kT}\text{\hspace{1em}(2.14a)}$$ $$p=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v),{\eta }_v=\frac{E_v-E_F}{kT}\text{\hspace{1em}(2.14b)}$$

These hold for any Fermi level position. At 300 K the constants are $N_{C,V}=(2.510\times 10^{19}{\text{cm}}^{-3})(m_{n,p}^{\ast }/m_0{)}^{3/2}$.

Simplification. If $E_F\le E_c-3kT$ (so ${\eta }_c\le -3$), the “+1” in the denominator is negligible and

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_c)\approx e^{{\eta }_c}=e^{(E_F-E_c)/kT}\text{\hspace{1em}(2.15a)}$$

and likewise for $E_F\ge E_v+3kT$:

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\approx e^{(E_v-E_F)/kT}\text{\hspace{1em}(2.15b)}$$

So, if $E_v+3kT\le E_F\le E_c-3kT$:

$$\boxed{n=N_C{e}^{(E_F-E_c)/kT}}\text{\hspace{1em}(2.16a)}$$ $$\boxed{p=N_V{e}^{(E_v-E_F)/kT}}\text{\hspace{1em}(2.16b)}$$

This is the same as replacing $f$ and $1-f$ by simple exponentials, valid when $E_F$ is in the gap and at least $3kT$ from either edge.

Degenerate vs nondegenerate (Fig. 2.19).

• Nondegenerate: $E_v+3kT\le E_F\le E_c-3kT$. Eqs. (2.16) work.
• Degenerate: $E_F$ closer than $3kT$ to a band edge, or inside a band. Then one must use the Fermi-Dirac integral, Eqs. (2.14).

2.5.2 Alternative Expressions for $n$ and $p$

An intrinsic semiconductor is surely nondegenerate, so Eqs. (2.16) apply. Setting $n=p=n_i$ and $E_F=E_i$:

$$n_i=N_C{e}^{(E_i-E_c)/kT}\text{\hspace{1em}(2.17a)}$$ $$n_i=N_V{e}^{(E_v-E_i)/kT}\text{\hspace{1em}(2.17b)}$$

Solve for the constants:

$$N_C=n_i{e}^{(E_c-E_i)/kT}\text{\hspace{1em}(2.18a)}$$ $$N_V=n_i{e}^{(E_i-E_v)/kT}\text{\hspace{1em}(2.18b)}$$

Substitute into Eqs. (2.16):

$$\begin{array}{rlrl}n& =n_i{e}^{(E_c-E_i)/kT}{e}^{(E_F-E_c)/kT}=n_i{e}^{(E_F-E_i)/kT}& & \text{(2.19a)}\\ p& =n_i{e}^{(E_i-E_v)/kT}{e}^{(E_v-E_F)/kT}=n_i{e}^{(E_i-E_F)/kT}& & \text{(2.19b)}\end{array}$$

Only one constant ($n_i$) and two energies appear, and the symmetry is easy to remember: swap $E_F\leftrightarrow E_i$ to go from $n$ to $p$. Intuition: the farther $E_F$ sits above $E_i$, the more electrons; the farther below, the more holes.

2.5.3 $n_i$ and the $np$ Product

Multiply Eqs. (2.17a) and (2.17b):

$$n_i^2=N_C{N}_V{e}^{-(E_c-E_v)/kT}=N_C{N}_V{e}^{-E_G/kT}\text{\hspace{1em}(2.20)}$$ $$n_i=\sqrt{N_C{N}_V}{e}^{-E_G/2kT}\text{\hspace{1em}(2.21)}$$

This tells you $n_i$ is controlled by the exponential of the band gap over thermal energy: wide gap means tiny $n_i$, and increasing $T$ rapidly increases $n_i$ (Fig. 2.20 shows $n_i$ versus $T$ for Ge, Si, GaAs).

Now multiply Eqs. (2.19a) and (2.19b):

$$np=n_i^2{e}^{(E_F-E_i)/kT}{e}^{(E_i-E_F)/kT}=n_i^2\text{\hspace{1em}(2.22)}$$

The $np$ product holds for any equilibrium, nondegenerate sample, doped or not. If you know one carrier concentration, you instantly get the other. Think of a seesaw: push electrons up and holes must go down, with the product pinned at $n_i^2$.

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Exercise 2.4: $n_i$ versus temperature in Si

Combine Eq. (2.21) with Eq. (2.13), using $N_{C,V}=2.510\times 10^{19}(m^{\ast }/m_0{)}^{3/2}(T/300{)}^{3/2}$:

$$n_i=2.510\times 10^{19}{\left(\frac{m_n^{\ast }}{m_0}\frac{m_p^{\ast }}{m_0}\right)}^{3/4}{\left(\frac{T}{300}\right)}^{3/2}{e}^{-E_G/2kT}$$

Problem (a). Confirm the Si curve of Fig. 2.20 using

$$E_G(T)=1.17-\frac{4.730\times 10^{-4}{T}^2}{T+636},\frac{m_n^{\ast }}{m_0}=1.028+6.11\times 10^{-4}T-3.09\times 10^{-7}{T}^2,\frac{m_p^{\ast }}{m_0}=0.610+7.83\times 10^{-4}T-4.46\times 10^{-7}{T}^2,$$

with $E_G$ in the exponent replaced by $E_G-E_{ex}$, $E_{ex}=0.0074$ eV.

Solution. The MATLAB script just evaluates the formula at each temperature. Doing it by hand at $T=300$ K shows what the script does:

$$\begin{array}{rlrl}{E}_G(300)& =1.17-\frac{(4.730\times 10^{-4})(300{)}^2}{300+636}=1.17-\frac{42.57}{936}=1.1245\text{eV}& & \text{band gap fit}\\ E_G-E_{ex}& =1.1245-0.0074=1.1171\text{eV}& & \text{exciton correction}\\ \frac{m_n^{\ast }}{m_0}& =1.028+0.1833-0.0278=1.1835& & \text{electron mass ratio}\\ \frac{m_p^{\ast }}{m_0}& =0.610+0.2349-0.0401=0.8048& & \text{hole mass ratio}\\ (\frac{m_n^{\ast }}{m_0}\frac{m_p^{\ast }}{m_0}{)}^{3/4}& =(0.9524{)}^{0.75}=0.9641& & \text{mass factor}\\ kT& =(8.617\times 10^{-5})(300)=0.025851\text{eV}& & \text{thermal energy}\\ \frac{E_G-E_{ex}}{2kT}& =\frac{1.1171}{0.051702}=21.607& & \text{exponent}\\ e^{-21.607}& =4.13\times 10^{-10}& & \text{exponential}\\ n_i& =(2.510\times 10^{19})(0.9641)(1)(4.13\times 10^{-10})\approx 1.0\times 10^{10}{\text{cm}}^{-3}& & \text{multiply}\end{array}$$

This matches the accepted value; that is exactly why $E_{ex}$ was tuned to $0.0074$ eV. The script repeats this for $T=200,225,\dots ,700$ K.

Problem (b). Compare with the Sproul-Green experimental fit (valid $275\le T\le 375$ K):

$$n_i=9.15\times 10^{19}{\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}$$

Solution. At $T=300$ K:

$$\begin{array}{rlrl}\frac{0.5928}{kT}& =\frac{0.5928}{0.025851}=22.932& & \text{exponent}\\ e^{-22.932}& =1.10\times 10^{-10}& & \text{exponential}\\ n_i& =(9.15\times 10^{19})(1)(1.10\times 10^{-10})\approx 1.006\times 10^{10}{\text{cm}}^{-3}& & \text{multiply}\end{array}$$

Script output (both agree within about 2%):

$T$ (K)	(a) $n_i$ (cm${}^{-3}$)	(b) $n_i$ (cm${}^{-3}$)
200	$5.246\times 10^4$
275	$1.059\times 10^9$	$1.051\times 10^9$
300	$1.000\times 10^{10}$	$1.006\times 10^{10}$
325	$6.798\times 10^{10}$	$6.887\times 10^{10}$
350	$3.565\times 10^{11}$	$3.623\times 10^{11}$
375	$1.518\times 10^{12}$	$1.542\times 10^{12}$
400	$5.449\times 10^{12}$
500	$2.716\times 10^{14}$
600	$3.988\times 10^{15}$
700	$2.865\times 10^{16}$

Look how violent the growth is: from 300 K to 700 K, $n_i$ increases about $2.9\times 10^6$ times.

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2.5.4 Charge Neutrality Relationship

So far no formula knows how much dopant we added. Charge neutrality supplies the missing link.

Take a uniformly doped sample at equilibrium and look at any small piece far from the surface. It must be electrically neutral; otherwise there would be an electric field, hence carrier motion, hence current, contradicting equilibrium. The charged objects are electrons ($-q$), holes ($+q$), ionized donors $N_D^{+}$ ($+q$) and ionized acceptors $N_A^{-}$ ($-q$). Net charge per cm${}^3$:

$$qp-qn+q{N}_D^{+}-q{N}_A^{-}=0\text{\hspace{1em}(2.23)}$$ $$p-n+N_D^{+}-N_A^{-}=0\text{\hspace{1em}(2.24)}$$

At room temperature essentially all shallow dopants are ionized, so $N_D^{+}=N_D$ and $N_A^{-}=N_A$ (total donors and acceptors per cm${}^3$), giving the standard form

$$\boxed{p-n+N_D-N_A=0}\text{\hspace{1em}(2.25)}$$

2.5.5 Carrier Concentration Calculations

Assume nondegenerate and totally ionized. Then we have two equations and two unknowns ($n$, $p$), with $n_i$, $N_A$, $N_D$ known.

$$\begin{array}{rlrl}p& =\frac{n_i^2}{n}& & \text{from }np=n_i^2(2.26)\\ \frac{n_i^2}{n}-n+N_D-N_A& =0& & \text{insert in (2.25)}(2.27)\\ n^2-n(N_D-N_A)-n_i^2& =0& & \text{multiply by }n\text{ and rearrange}(2.28)\end{array}$$

This is a quadratic in $n$. By the quadratic formula (keeping the positive root since concentrations cannot be negative):

$$n=\frac{N_D-N_A}{2}+{\left[{\left(\frac{N_D-N_A}{2}\right)}^2+n_i^2\right]}^{1/2}\text{\hspace{1em}(2.29a)}$$

and, by the same steps with $n\leftrightarrow p$, $N_D\leftrightarrow N_A$,

$$p=\frac{N_A-N_D}{2}+{\left[{\left(\frac{N_A-N_D}{2}\right)}^2+n_i^2\right]}^{1/2}\text{\hspace{1em}(2.29b)}$$

Four special cases:

1. Intrinsic ($N_A=N_D=0$): $n=p=n_i$.
2. Doped, one species dominant ($N_D-N_A\approx N_D\gg n_i$, or $N_A-N_D\approx N_A\gg n_i$). This is the common practical case, since Si dopings are rarely below $10^{14}$ cm${}^{-3}$ while $n_i\approx 10^{10}$ cm${}^{-3}$. The square root becomes $\sqrt{(N_D/2{)}^2}=N_D/2$, so

$$n\approx N_D,p\approx \frac{n_i^2}{N_D}(N_D\gg N_A,N_D\gg n_i)\text{\hspace{1em}(2.30)}$$ $$p\approx N_A,n\approx \frac{n_i^2}{N_A}(N_A\gg N_D,N_A\gg n_i)\text{\hspace{1em}(2.31)}$$

Example. Si at 300 K with $N_D=10^{15}$ cm${}^{-3}$: $n\approx 10^{15}$ cm${}^{-3}$ and $p=n_i^2/n=10^{20}/10^{15}=10^5$ cm${}^{-3}$. The minority carriers are ten orders of magnitude fewer than the majority ones. 3. High temperature ($n_i\gg |N_D-N_A|$): the roots reduce to $n=p\approx n_i$. Every semiconductor becomes intrinsic when hot enough. 4. Compensated ($N_A\approx N_D$, both nonzero): donors and acceptors cancel each other, even producing intrinsic-like material when $N_D-N_A=0$. Both $N_A$ and $N_D$ must be kept in the formulas (common in as-grown GaAs).

Use the full Eqs. (2.29) only when $|N_D-N_A|\sim n_i$; otherwise Eqs. (2.30), (2.31) suffice.

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Exercise 2.5

Problem. A Si sample is doped with $10^{14}$ boron atoms/cm${}^3$. Find the carrier concentrations at (a) 300 K and (b) 470 K.

Solution (a), 300 K.

$$\begin{array}{rlrl}& \text{Boron is an acceptor, so }{N}_A=10^{14}{\text{cm}}^{-3}& & \text{Table 2.2}\\ & N_D\text{not mentioned}\Rightarrow N_D\ll N_A& & \text{inferred}\\ & n_i=1.00\times 10^{10}{\text{cm}}^{-3}\ll N_A& & \text{at 300 K, so (2.31) applies}\\ p& \approx N_A=10^{14}{\text{cm}}^{-3}& & \text{(2.31a)}\\ n& =\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{14}}=10^6{\text{cm}}^{-3}& & \text{(2.31b)}\end{array}$$

Solution (b), 470 K. From Fig. 2.20, $n_i\approx 10^{14}$ cm${}^{-3}$ at 470 K. Now $n_i\sim N_A$, so the full Eq. (2.29b) is needed:

$$\begin{array}{rlrl}p& =\frac{N_A}{2}+{\left[{\left(\frac{N_A}{2}\right)}^2+n_i^2\right]}^{1/2}& & N_D=0\\ & =0.5\times 10^{14}+{\left[(0.5\times 10^{14}{)}^2+(10^{14}{)}^2\right]}^{1/2}& & \text{insert numbers}\\ & =0.5\times 10^{14}+{\left[1.25\times 10^{28}\right]}^{1/2}& & \text{add under root}\\ & =0.5\times 10^{14}+1.118\times 10^{14}=1.62\times 10^{14}{\text{cm}}^{-3}& & \text{take root, add}\\ n& =\frac{n_i^2}{p}=\frac{10^{28}}{1.62\times 10^{14}}=6.18\times 10^{13}{\text{cm}}^{-3}& & \text{np product}\end{array}$$

Notice how heating has erased most of the doping advantage: electrons are now more than a third as numerous as holes.

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2.5.6 Determination of $E_F$

Eqs. (2.14), (2.16), (2.19) give a one-to-one link between $E_F$, $n$ and $p$: know any one and you can compute the other two.

(1) Exact position of $E_i$. In an intrinsic sample $n=p$. Using Eq. (2.16) with $E_F=E_i$:

$$\begin{array}{rlrl}{N}_C{e}^{(E_i-E_c)/kT}& =N_V{e}^{(E_v-E_i)/kT}& & n=p(2.33)\\ \frac{E_i-E_c}{kT}+\ln N_C& =\frac{E_v-E_i}{kT}+\ln N_V& & \text{take }\ln \\ 2E_i& =E_c+E_v+kT\ln \frac{N_V}{N_C}& & \text{collect }{E}_i\\ E_i& =\frac{E_c+E_v}{2}+\frac{kT}{2}\ln \frac{N_V}{N_C}& & (2.34)\end{array}$$

Since $N_V/N_C=(m_p^{\ast }/m_n^{\ast }{)}^{3/2}$ by Eq. (2.13):

$$\boxed{E_i=\frac{E_c+E_v}{2}+\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}}\text{\hspace{1em}(2.36)}$$

So $E_i$ is exactly at midgap only if $m_p^{\ast }=m_n^{\ast }$ or $T\to 0$. In Si at 300 K, $m_p^{\ast }<m_n^{\ast }$ so $E_i$ lies slightly below midgap (computed below, about $0.0073$ eV). That is usually small enough to be ignored when sketching.

(2) Doped semiconductors (nondegenerate, fully ionized). Solve Eqs. (2.19) for $E_F-E_i$:

$$\boxed{E_F-E_i=kT\ln \frac{n}{n_i}=-kT\ln \frac{p}{n_i}}\text{\hspace{1em}(2.37)}$$

Insert the appropriate concentrations. With $n\approx N_D$ (n-type) and $p\approx N_A$ (p-type):

$$E_F-E_i=kT\ln \frac{N_D}{n_i}(N_D\gg N_A,N_D\gg n_i)\text{\hspace{1em}(2.38a)}$$ $$E_i-E_F=kT\ln \frac{N_A}{n_i}(N_A\gg N_D,N_A\gg n_i)\text{\hspace{1em}(2.38b)}$$

More donors push $E_F$ up away from $E_i$; more acceptors push it down (Fig. 2.21). There are maximum nondegenerate dopings in Si at room temperature: about $N_D\approx 1.6\times 10^{18}$ cm${}^{-3}$ and $N_A\approx 9.1\times 10^{17}$ cm${}^{-3}$. Beyond these the material is degenerate, which is why “highly doped” ($n^{+}$, $p^{+}$) and “degenerate” are often used as synonyms.

Practical rule. If unsure, assume nondegenerate and compute $E_F$. If the answer falls inside the degenerate zone ($E_F$ within $3kT$ of an edge), redo the calculation with the full Fermi-Dirac machinery.

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Exercise 2.6

Problem. For each condition of Exercise 2.5, locate $E_i$, compute $E_F-E_i$, and draw a dimensioned band diagram. Use $E_G(\text{Si})=1.08$ eV and $m_p^{\ast }/m_n^{\ast }=0.71$ at 470 K.

Solution (a), 300 K ($N_A=10^{14}$ cm${}^{-3}$).

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}=\frac{3}{4}(0.0259)\ln \frac{0.81}{1.18}& & \text{(2.36), Table 2.1}\\ & =0.01943\times (-0.3761)=-0.0073\text{eV}& & \text{evaluate}\end{array}$$

So $E_i$ is $0.0073$ eV below midgap. Next, since $N_A\gg n_i$, use (2.38b):

$$\begin{array}{rlrl}{E}_i-E_F& =kT\ln \frac{N_A}{n_i}=0.0259\ln \frac{10^{14}}{10^{10}}& & \text{(2.38b)}\\ & =0.0259\times 9.210=0.239\text{eV}& & \ln 10^4=9.21\end{array}$$

($E_F$ is $0.239$ eV below $E_i$.) Dimensions for the diagram with $E_G=1.12$ eV: $E_c-E_i=0.56+0.0073=0.567$ eV; $E_i-E_v=0.553$ eV; $E_F-E_v=0.553-0.239=0.314$ eV.

Solution (b), 470 K. Here $kT=0.0405$ eV, $p=1.62\times 10^{14}$, $n_i=10^{14}$ cm${}^{-3}$.

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}(0.0405)\ln (0.71)=0.0304\times (-0.3425)=-0.0104\text{eV}& & \text{(2.36)}\end{array}$$

$E_i$ is $0.0104$ eV below midgap. Because $N_A\sim n_i$, use Eq. (2.37) with the exact $p$:

$$\begin{array}{rlrl}{E}_i-E_F& =kT\ln \frac{p}{n_i}=0.0405\ln \frac{1.62\times 10^{14}}{10^{14}}& & \text{(2.37)}\\ & =0.0405\times 0.4824=0.0195\text{eV}& & \ln 1.62=0.4824\end{array}$$

$E_F$ is only $0.0195$ eV below $E_i$: the hot sample is nearly intrinsic. Dimensions with $E_G=1.08$ eV: $E_c-E_i=0.54+0.0104=0.550$ eV; $E_i-E_v=0.530$ eV; $E_F-E_v=0.530-0.0195=0.511$ eV.

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2.5.7 Carrier Concentration Temperature Dependence

We have scattered facts about temperature already; now let us stitch them into one story. Consider a phosphorus-doped Si sample with $N_D=10^{15}$ cm${}^{-3}$ and follow its electron concentration $n$ as we heat it from nearly 0 K (Fig. 2.22).

Three regimes:

Regime	Roughly	What happens	$n$
Freeze-out	below about 100 K	Too little thermal energy to free the weakly bound fifth electrons; only a fraction of donors are ionized. Band-to-band excitation is utterly negligible.	$n=N_D^{+}$, rising from $0$ at $T=0$ K
Extrinsic (the normal operating range of most devices)	about 150 K to 450 K	Essentially all donors are ionized; thermally generated band-to-band electrons are still far fewer than $N_D$.	$n\approx N_D$, flat
Intrinsic	high $T$	Band-to-band excitation grows exponentially until it swamps the fixed donor supply.	$n\approx n_i$, rising steeply

Analogy. Imagine a reservoir ($N_D$ donors) with a small dam. At low temperature the dam barely leaks (freeze-out). At moderate temperature it spills fully and the water level sits at a plateau (extrinsic). At very high temperature a flood from the whole river system (band-to-band excitation) overwhelms the reservoir (intrinsic).

Two mechanisms feed the electron count: electrons donated by dopants, and electrons excited from the valence band (broken Si-Si bonds). At $T=0$ K both are zero. Warming first “defrosts” donor electrons; later the band-to-band mechanism takes over.

Why materials with wide gaps run hotter. The larger $E_G$, the larger the energy needed to excite across the gap, so the intrinsic regime begins at a higher temperature. Since the start of the intrinsic region is the top of the normal operating range, GaAs devices can operate hotter than Si devices, which can operate hotter than Ge devices. Taking the onset as $n_i=N_D=10^{15}$ cm${}^{-3}$ and reading Fig. 2.20, the maximum operating temperatures are approximately

$$T_{max}\approx 385\text{K (Ge)},540\text{K (Si)},>700\text{K (GaAs)}.$$

This is why GaAs and SiC ($E_G>2$ eV) are pursued for high-temperature electronics.

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2.6 Summary and Concluding Comments

We have described the carriers in a semiconductor at “rest” (equilibrium). What you now own:

• Two visualization models: the bonding model (great for seeing broken bonds, holes, dopants) and the energy band model (a kind of nonverbal sign language that encodes energies).
• A picture of the carriers: classical-like particles with charge $-q$ (electron) and $+q$ (hole) and effective masses $m_n^{\ast }$, $m_p^{\ast }$. In pure material they are equal in number and few; doping selectively raises one type.
• A chain of quantitative tools: densities of states, the Fermi function, the symmetric nondegenerate forms for $n$ and $p$, the $np$ product, charge neutrality, and the simplified room-temperature results.
• Qualitative feel: carriers cluster near band edges, $n_i$ rises steeply with $T$, and majority carriers sit at $N_D$ (or $N_A$) over the useful temperature range.
• A warning: do not “plug and chug.” Every formula has assumptions (equilibrium, nondegenerate, fully ionized dopants) and you should know their limits before using them.
• Reference numbers worth memorizing: $E_G(\text{Si})=1.12$ eV, $n_i(\text{Si})=10^{10}$ cm${}^{-3}$ at 300 K, $kT=0.0259$ eV at 300 K.

Table 2.4: Carrier Modeling Equation Summary

Density of states and Fermi function

$$g_c(E)=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }(E-E_c)}}{{\pi }^2{\hslash }^3}(E\ge E_c),g_v(E)=\frac{m_p^{\ast }\sqrt{2m_p^{\ast }(E_v-E)}}{{\pi }^2{\hslash }^3}(E\le E_v),f(E)=\frac{1}{1+e^{(E-E_F)/kT}}$$

Carrier concentration relationships

$$\begin{array}{rlrlrl}n& =N_C\frac{2}{\sqrt{\pi }}{F}_{1/2}\left(\frac{E_F-E_c}{kT}\right),& p& =N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}\left(\frac{E_v-E_F}{kT}\right)& & \text{(any }{E}_F)\\ N_C& =2{\left(\frac{2\pi m_n^{\ast }kT}{h^2}\right)}^{3/2},& N_V& =2{\left(\frac{2\pi m_p^{\ast }kT}{h^2}\right)}^{3/2}& & \\ n& =N_C{e}^{(E_F-E_c)/kT},& p& =N_V{e}^{(E_v-E_F)/kT}& & \text{(nondegenerate)}\\ n& =n_i{e}^{(E_F-E_i)/kT},& p& =n_i{e}^{(E_i-E_F)/kT}& & \text{(nondegenerate)}\end{array}$$

$n_i$, $np$ product, and charge neutrality

$$n_i=\sqrt{N_C{N}_V}{e}^{-E_G/2kT},np=n_i^2,p-n+N_D-N_A=0$$

$n$, $p$, and Fermi level computational relationships

$$\begin{array}{rlrl}n& =\frac{N_D-N_A}{2}+{\left[{\left(\frac{N_D-N_A}{2}\right)}^2+n_i^2\right]}^{1/2},& p& =\frac{N_A-N_D}{2}+{\left[{\left(\frac{N_A-N_D}{2}\right)}^2+n_i^2\right]}^{1/2}\\ E_i& =\frac{E_c+E_v}{2}+\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }},& E_F-E_i& =kT\ln \frac{n}{n_i}=-kT\ln \frac{p}{n_i}\\ n& \approx N_D,p\approx \frac{n_i^2}{N_D}(N_D\gg N_A,N_D\gg n_i),& E_F-E_i& =kT\ln \frac{N_D}{n_i}\\ p& \approx N_A,n\approx \frac{n_i^2}{N_A}(N_A\gg N_D,N_A\gg n_i),& E_i-E_F& =kT\ln \frac{N_A}{n_i}\end{array}$$

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Problems: Chapter 2

Here are all 23 end-of-chapter problems, each with a full solution. Constants used throughout: $kT=0.025851$ eV at 300 K ($0.0259$ in rounded form), $n_i(\text{Si},300\text{ K})=10^{10}$ cm${}^{-3}$, $q=1.602\times 10^{-19}$ C.

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Problem 2.1: $E_G$ versus $T$

Band gap model:

$$E_G(T)=E_G(0)-\frac{\alpha T^2}{T+\beta }$$

with, for Si, $E_G(0)=1.170$ eV, $\alpha =4.730\times 10^{-4}$ eV/K, $\beta =636$ K.

(a) Plot $E_G$ versus $T$ for $0\le T\le 600$ K and note $E_G(300)$.

$$\begin{array}{rlrl}{E}_G(0)& =1.170& & \alpha T^2=0\\ E_G(100)& =1.170-\frac{(4.73\times 10^{-4})(10^4)}{736}=1.170-0.0064=1.1636\text{eV}& & \\ E_G(200)& =1.170-\frac{(4.73\times 10^{-4})(4\times 10^4)}{836}=1.170-0.0226=1.1474\text{eV}& & \\ E_G(300)& =1.170-\frac{(4.73\times 10^{-4})(9\times 10^4)}{936}=1.170-0.0455=1.1245\text{eV}& & \\ E_G(400)& =1.170-\frac{(4.73\times 10^{-4})(1.6\times 10^5)}{1036}=1.170-0.0731=1.0969\text{eV}& & \\ E_G(500)& =1.170-\frac{(4.73\times 10^{-4})(2.5\times 10^5)}{1136}=1.170-0.1041=1.0659\text{eV}& & \\ E_G(600)& =1.170-\frac{(4.73\times 10^{-4})(3.6\times 10^5)}{1236}=1.170-0.1378=1.0322\text{eV}& & \end{array}$$

The plot is a curve that is flat near $T=0$ (slope $\propto T^2$), then bends downward and becomes nearly linear above 300 K. At 300 K the gap is $1.12$ eV, the number everyone remembers.

(b) Compare with $E_G(T)=1.205-2.8\times 10^{-4}T$ for $T>300$ K.

$$\begin{array}{rl}T=300:& 1.205-0.0840=1.121\text{eV}(\Delta =-0.0035\text{ vs. exact})\\ T=400:& 1.205-0.1120=1.093\text{eV}(\Delta =-0.0039)\\ T=500:& 1.205-0.1400=1.065\text{eV}(\Delta =-0.0009)\\ T=600:& 1.205-0.1680=1.037\text{eV}(\Delta =+0.0048)\end{array}$$

Over $300\le T\le 600$ K the linear form is within about $5$ meV ($<0.5\%$) of the precise fit: an excellent simplification in that region (but it badly overshoots below 300 K where the true curve flattens).

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Problem 2.2: Bonding model visualizations

• (a) Missing atom: erase one circle (a Si core) together with its eight line-ends (its four shared pairs); the four neighbors are left with a dangling line-end each (Fig. 2.4a).
• (b) Electron: a broken bond: one line segment removed, with a small dot (the electron, charge $-q$) wandering through the lattice.
• (c) Hole: the same broken bond viewed as a gap in the line: an empty bond site (a “positive” charge $+q$) that moves when a neighboring bond electron hops into it.
• (d) Donor: a Column V atom (e.g., P) substituted for a Si circle; four of its five valence electrons complete the four bonds, and the fifth is drawn as an extra dot loosely attached to a $+$ core (${\text{P}}^{+}$), later freed to wander (Fig. 2.10a).
• (e) Acceptor: a Column III atom (e.g., B) substituted for a Si circle; only three bonds are complete, one bond is short an electron. A neighboring bond electron hops in, giving the core a $-$ charge (${\text{B}}^{-}$) and creating a wandering hole (Fig. 2.10b).

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Problem 2.3: Energy band model visualizations

Reference diagram: horizontal lines $E_c$ (top) and $E_v$ (bottom), energy vertical, position horizontal.

• (a) Electron: a filled dot just above $E_c$ in the conduction band.
• (b) Hole: an open circle just below $E_v$ in the valence band.
• (c) Donor sites: a row of short dashes at $E_D=E_c-|E_B|$, about $0.045$ eV (P in Si) below $E_c$.
• (d) Acceptor sites: a row of short dashes at $E_A=E_v+|E_B|$, about $0.045$ eV (B in Si) above $E_v$.
• (e) Donor freeze-out at $T\to 0$: all donor dashes carry dots (electrons), and the conduction band is empty. The electrons are frozen on their donors.
• (f) Acceptor freeze-out at $T\to 0$: all acceptor dashes are empty, the valence band is completely full, and no holes exist (the holes are “frozen” on acceptors).
• (g) Carrier energy distributions: many dots clustered near $E_c$ (electrons) and many open circles near $E_v$ (holes), thinning out rapidly with distance from the edges (Fig. 2.17).
• (h) Intrinsic: equal numbers of dots near $E_c$ and open circles near $E_v$; a dashed $E_i$ line near midgap (Fig. 2.18, left).
• (i) n-type: a solid $E_F$ line above $E_i$; many electrons, few holes.
• (j) p-type: a solid $E_F$ line below $E_i$; many holes, few electrons.
• (k) Nondegenerate: $E_F$ in the gap at least $3kT$ from both $E_c$ and $E_v$.
• (l) Degenerate: $E_F$ within $3kT$ of a band edge, or inside the conduction (n${}^{+}$) or valence (p${}^{+}$) band.

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Problem 2.4: Si-doped GaAs

(a) Remove a Ga atom and an As atom from the GaAs bonding diagram (Fig. P2.4). Each atom takes its bonding electrons with it: removing Ga takes its 3 valence electrons (3 of the four lines to its As neighbors), and removing As takes its 5 (one of its four neighbors’ lines each side). Draw the two circles erased with their line-ends; each of the neighbors then shows a dangling line-end.

(b) Insert a Si atom (4 valence electrons, 4 line-ends) in each vacancy, reconnecting to the four neighbors.

(c) Si on a Ga site.

$$\begin{array}{rlrl}\text{Ga valence}& =3,\text{Si valence}=4& & \text{Column III vs. IV}\\ \text{Excess electrons per substituted Si}& =4-3=+1& & \text{one spare electron}\end{array}$$

One electron does not fit in the bonds and is released: Si on Ga is a donor, so the GaAs is n-type.

(d) Si on an As site.

$$\begin{array}{rlrl}\text{As valence}& =5,\text{Si valence}=4& & \text{Column V vs. IV}\\ \text{Electrons per substituted Si}& =4-5=-1& & \text{one bond is an electron short}\end{array}$$

A bond is left incomplete, so a neighboring bond electron is captured and a hole results: Si on As is an acceptor, so the GaAs is p-type. Si in GaAs is therefore amphoteric.

(e) Band diagrams.

• (i) Si on Ga: draw a dashed donor level just below $E_c$ (filled with an electron at low $T$; ionized at room temperature).
• (ii) Si on As: draw a dashed acceptor level just above $E_v$ (empty at low $T$; filled by a valence-band electron at room temperature, leaving a hole).

In both cases $E_G=1.42$ eV.

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Problem 2.5: States per cm${}^3$ in $\gamma kT$

Problem. Find the number of conduction band states per cm${}^3$ between $E_c$ and $E_c+\gamma kT$, $\gamma$ an arbitrary constant.

Solution. Integrate the density of states over that interval:

$$\begin{array}{rlrl}N& ={\int }_{E_c}^{E_c+\gamma kT}{g}_c(E)dE& & \text{states in the range}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{E_c}^{E_c+\gamma kT}(E-E_c{)}^{1/2}dE& & \text{insert (2.6a)}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}\cdot \frac{2}{3}(\gamma kT{)}^{3/2}& & \int u^{1/2}du=\frac{2}{3}{u}^{3/2}\\ & =\boxed{\frac{2\sqrt{2}(m_n^{\ast }\gamma kT{)}^{3/2}}{3{\pi }^2{\hslash }^3}}{\text{cm}}^{-3}& & \text{since }m\sqrt{m}=m^{3/2}\end{array}$$

(Use consistent units, then convert to cm${}^{-3}$.)

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Problem 2.6: Fermi function questions

(a) At $E=E_F$:

$$f(E_F)=\frac{1}{1+e^0}=\frac{1}{2}.$$

(b) With $E_F=E_c$, find the occupation probability at $E_c+kT$:

$$\begin{array}{rlrl}f(E_c+kT)& =\frac{1}{1+e^{(E_c+kT-E_c)/kT}}& & E_F=E_c\\ & =\frac{1}{1+e^1}=\frac{1}{3.718}=0.269& & e=2.718\end{array}$$

(c) Given: $P(\text{filled at }{E}_c-kT)=P(\text{empty at }{E}_v+kT)$. Where is $E_F$?

$$\begin{array}{rlrl}f(E_c-kT)& =1-f(E_v+kT)& & \text{given}\\ \frac{1}{1+e^{(E_c-kT-E_F)/kT}}& =\frac{1}{1+e^{(E_F-E_v-kT)/kT}}& & \text{use }1-f(E)=\frac{1}{1+e^{-(E-E_F)/kT}}\\ E_c-kT-E_F& =E_F-E_v-kT& & \text{equal exponents}\\ E_F& =\frac{E_c+E_v}{2}& & \text{the }kT\text{ cancels}\end{array}$$

$E_F$ is at midgap.

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Problem 2.7: Peak of the carrier distributions

Problem. For a nondegenerate semiconductor, show the electron distribution peaks at $E_c+kT/2$ and the hole distribution at $E_v-kT/2$.

Solution. Conduction band: let $x=E-E_c$. Nondegenerate means $f(E)\approx e^{-(E-E_F)/kT}$, so

$$\begin{array}{rlrl}{g}_c(E)f(E)& =C\sqrt{x}{e}^{-(E_c-E_F)/kT}{e}^{-x/kT}\propto \sqrt{x}{e}^{-x/kT}& & C\text{ constant}\\ \frac{d}{dx}\left(\sqrt{x}{e}^{-x/kT}\right)& =e^{-x/kT}\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{kT}\right)=0& & \text{peak: derivative zero}\\ \frac{1}{2\sqrt{x}}& =\frac{\sqrt{x}}{kT}& & \text{set bracket to zero}\\ x& =\frac{kT}{2}& & \text{solve}\\ \Rightarrow E_{peak}& =E_c+\frac{kT}{2}& & \end{array}$$

Valence band: let $y=E_v-E$. Then $1-f(E)\approx e^{(E-E_F)/kT}=e^{-(E_F-E_v)/kT}{e}^{-y/kT}$ and $g_v\propto \sqrt{y}$, so $g_v(1-f)\propto \sqrt{y}{e}^{-y/kT}$, whose peak is at $y=kT/2$ by the identical calculation: $E_{peak}=E_v-kT/2$. $■$

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Problem 2.8: Population at $E_c+5kT$

Problem. Relative to the peak, what is the electron population per unit energy at $E=E_c+5kT$?

Solution. The distribution is $h(x)\propto \sqrt{x}{e}^{-x/kT}$ with the peak at $x=kT/2$ (Problem 2.7).

$$\begin{array}{rlrl}\frac{h(5kT)}{h(kT/2)}& =\frac{\sqrt{5kT}{e}^{-5}}{\sqrt{kT/2}{e}^{-1/2}}& & \text{evaluate both}\\ & =\sqrt{10}{e}^{-4.5}& & \sqrt{5/0.5}=\sqrt{10}\\ & =(3.162)(0.01111)=0.0351& & e^{-4.5}=0.01111\end{array}$$

So the population at $E_c+5kT$ is only about 3.5% of the peak value. The carriers really do huddle close to the band edge.

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Problem 2.9: Distributions in Si at 300 K with $E_F=E_c-E_G/4$

Setup. (The problem statement reads $E_F$ a quarter of the gap below $E_c$; with $E_G=1.12$ eV.) Then

$$E_c-E_F=\frac{1.12}{4}=0.28\text{eV}=\frac{0.28}{0.025851}kT=10.8kT>3kT\Rightarrow \text{nondegenerate}.$$

Likewise $E_F-E_v=0.84$ eV $=32.5kT$.

Step 1: carrier concentrations.

$$\begin{array}{rlrl}{N}_C& =(2.510\times 10^{19})(1.18{)}^{3/2}=3.22\times 10^{19}{\text{cm}}^{-3}& & \text{Table 2.1}\\ N_V& =(2.510\times 10^{19})(0.81{)}^{3/2}=1.83\times 10^{19}{\text{cm}}^{-3}& & \\ n& =N_C{e}^{-(E_c-E_F)/kT}=3.22\times 10^{19}{e}^{-10.81}=3.22\times 10^{19}(2.02\times 10^{-5})=6.5\times 10^{14}{\text{cm}}^{-3}& & \\ p& =N_V{e}^{-(E_F-E_v)/kT}=1.83\times 10^{19}{e}^{-32.5}=1.5\times 10^5{\text{cm}}^{-3}(\text{check: }{n}_i^2/n=1.5\times 10^5)& & \end{array}$$

Step 2: normalized distribution (from Problem 2.10a, for $x=E-E_c$):

$$n(E)=n\cdot \frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{x}{e}^{-x/kT},\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}=\frac{2}{(1.7725)(0.004156)}=271.5{\text{eV}}^{-3/2}$$

so

$$n(E)=(6.5\times 10^{14})(271.5)\sqrt{x}{e}^{-x/kT}=1.76\times 10^{17}\sqrt{x}{e}^{-x/kT}{\text{cm}}^{-3}{\text{eV}}^{-1}(x\text{ in eV})$$

and for holes ($y=E_v-E$):

$$p(E)=(1.5\times 10^5)(271.5)\sqrt{y}{e}^{-y/kT}=4.1\times 10^7\sqrt{y}{e}^{-y/kT}{\text{cm}}^{-3}{\text{eV}}^{-1}.$$

Step 3: values for the plot (peak at $kT/2=0.0129$ eV from the band edge):

distance from edge (eV)	$n(E)$ (cm${}^{-3}$eV${}^{-1}$)	$p(E)$ (cm${}^{-3}$eV${}^{-1}$)
0	0	0
0.0129 (peak)	$1.2\times 10^{16}$	$2.8\times 10^6$
0.05	$5.7\times 10^{15}$	$1.3\times 10^6$
0.10	$1.2\times 10^{15}$	$2.7\times 10^5$
0.20	$3.4\times 10^{13}$	$8\times 10^3$

(Each entry is the coefficient times $\sqrt{x}{e}^{-x/kT}$; for instance at $0.05$ eV, $\sqrt{0.05}{e}^{-1.934}=0.2236\times 0.1445=0.0323$.) The electron curve is a skewed hump about $4\times 10^9$ times taller than the hole hump, because $E_F$ lies in the upper half of the gap.

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Problem 2.10: Temperature variation of the electron distribution

(a) Show $\frac{g_c(E)f(E)}{n}=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{E-E_c}{e}^{-(E-E_c)/kT}$.

$$\begin{array}{rlrl}{g}_{c}f& =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}\sqrt{E-E_c}{e}^{-(E-E_F)/kT}& & \text{nondegenerate }f\\ n& =N_C{e}^{-(E_c-E_F)/kT}& & \text{(2.16a)}\\ \frac{g_{c}f}{n}& =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3{N}_C}\sqrt{E-E_c}{e}^{-(E-E_c)/kT}& & \text{the }{E}_F\text{ factors cancel}\end{array}$$

Now use $N_C=\frac{2(m_n^{\ast }kT{)}^{3/2}}{(2\pi {)}^{3/2}{\hslash }^3}$:

$$\frac{m_n^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}\cdot \frac{(2\pi {)}^{3/2}{\hslash }^3}{2(m_n^{\ast }kT{)}^{3/2}}=\frac{\sqrt{2}(2\pi {)}^{3/2}}{2{\pi }^2(kT{)}^{3/2}}=\frac{4{\pi }^{3/2}}{2{\pi }^2(kT{)}^{3/2}}=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}$$

which gives the stated result. $■$

(b) Plot for $T=300,600,1200$ K. Define $\varphi (x)=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{x}{e}^{-x/kT}$ in eV${}^{-1}$. The peak (at $x=kT/2$) is ${\varphi }_{max}=\frac{2}{\sqrt{2\pi }}\frac{e^{-1/2}}{kT}=\frac{0.484}{kT}$:

$T$ (K)	$kT$ (eV)	peak position (eV)	${\varphi }_{max}$ (eV${}^{-1}$)
300	0.02585	0.0129	18.7
600	0.05170	0.0259	9.36
1200	0.10340	0.0517	4.68

Selected values $\varphi (x)$ (eV${}^{-1}$):

$x=E-E_c$ (eV)	300 K	600 K	1200 K
0.05	8.77	8.16	4.68
0.10	1.79	4.39	4.08
0.20	0.053	0.90	2.19
0.40	$\approx 0$	0.027	0.45

Discussion. All three curves are normalized to the same area (1 eV${}^{-1}$ per electron). Heating makes the distribution lower and broader: the peak moves outward ($kT/2$) and shrinks ($\propto 1/kT$), and a long high-energy tail develops. At 300 K nearly every electron is within about $0.1$ eV of the edge; at 1200 K a visible fraction is $0.3$ to $0.4$ eV up. (Same total number $n$, spread over more energies.)

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Problem 2.11: Hole formulas (2.14b) and (2.16b)

Solution. Start from Eq. (2.8b) and mirror the electron derivation.

$$\begin{array}{rlrl}p& ={\int }_{-\infty }^{E_v}{g}_v(E)[1-f(E)]dE& & \text{(2.8b), bottom}\to -\infty \\ 1-f(E)& =\frac{1}{1+e^{(E_F-E)/kT}}& & \text{algebra: }1-\frac{1}{1+e^a}=\frac{1}{1+e^{-a}}\\ p& =\frac{m_p^{\ast }\sqrt{2m_p^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{-\infty }^{E_v}\frac{\sqrt{E_v-E}}{1+e^{(E_F-E)/kT}}dE& & \text{insert (2.6b)}\end{array}$$

Let ${\eta }^{\prime }=(E_v-E)/kT$ and ${\eta }_v=(E_v-E_F)/kT$. Then $E_F-E=(E_F-E_v)+(E_v-E)=kT({\eta }^{\prime }-{\eta }_v)$, $\sqrt{E_v-E}dE\to (kT{)}^{3/2}\sqrt{{\eta }^{\prime }}d{\eta }^{\prime }$ (sign absorbed by reversing the limits) and the lower limit $-\infty \to {\eta }^{\prime }=\infty$:

$$p=\frac{\sqrt{2}(m_p^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\frac{\sqrt{{\eta }^{\prime }}d{\eta }^{\prime }}{1+e^{{\eta }^{\prime }-{\eta }_v}}=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\text{(2.14b)}$$

where $N_V=2(2\pi m_p^{\ast }kT/h^2{)}^{3/2}$ (the constants combine exactly as for $N_C$). If $E_F\ge E_v+3kT$, then ${\eta }_v\le -3$, the $+1$ is negligible in the denominator $1+e^{{\eta }^{\prime }-{\eta }_v}$ and

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\approx \frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}{\int }_0^{\infty }\sqrt{{\eta }^{\prime }}{e}^{-{\eta }^{\prime }}d{\eta }^{\prime }=\frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}\frac{\sqrt{\pi }}{2}=e^{(E_v-E_F)/kT}$$

hence

$$p=N_V{e}^{(E_v-E_F)/kT}\text{(2.16b)}.■$$

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Problem 2.12: Hypothetical constant density of states

Given $g_c(E)=N_C/kT=\text{const}$ for $E\ge E_c$.

(a) With $E_F\le E_c-3kT$, $f\approx e^{-(E-E_F)/kT}$, so

$$g_c(E)f(E)=\frac{N_C}{kT}{e}^{-(E-E_F)/kT}=\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}{e}^{-(E-E_c)/kT}.$$

The sketch is a pure exponential decay starting at its maximum at $E=E_c$ (no square-root rise and no peak away from the edge), falling by $1/e$ every $kT$.

(b) Electron concentration:

$$\begin{array}{rlrl}n& ={\int }_{E_c}^{\infty }\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}{e}^{-(E-E_c)/kT}dE& & \text{integrate the distribution}\\ & =\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}[-kT{e}^{-(E-E_c)/kT}{]}_{E_c}^{\infty }& & \text{integral of exponential}\\ & =N_C{e}^{(E_F-E_c)/kT}& & \text{bracket}=kT\end{array}$$

This reproduces Eq. (2.16a) exactly (the constant was chosen to do so). Without the approximation on $f$ the integral is elementary too:

$$n=\frac{N_C}{kT}{\int }_{E_c}^{\infty }\frac{dE}{1+e^{(E-E_F)/kT}}=N_C\ln (1+e^{(E_F-E_c)/kT})$$

(which tends to $N_C{e}^{(E_F-E_c)/kT}$ when $E_F\ll E_c$, and to $N_C(E_F-E_c)/kT$ for $E_F\gg E_c$).

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Problem 2.13: $N_C$, $N_V$

(a) Verify $N_{C,V}=(2.510\times 10^{19}{\text{cm}}^{-3})(m^{\ast }/m_0{)}^{3/2}$ at 300 K.

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5})(1.602\times 10^{-19})(300)=4.141\times 10^{-21}\text{J}& & \text{in joules}\\ 2\pi m_{0}kT& =(6.2832)(9.109\times 10^{-31})(4.141\times 10^{-21})=2.370\times 10^{-50}\text{kg J}& & \text{numerator}\\ h^2& =(6.625\times 10^{-34}{)}^2=4.389\times 10^{-67}{\text{J}}^2{\text{s}}^2& & \text{denominator}\\ \frac{2\pi m_{0}kT}{h^2}& =5.40\times 10^{16}{\text{m}}^{-2}& & \text{divide}\\ {\left(5.40\times 10^{16}\right)}^{3/2}& =1.255\times 10^{25}{\text{m}}^{-3}& & \text{power }3/2\\ N& =2\times 1.255\times 10^{25}=2.51\times 10^{25}{\text{m}}^{-3}=2.51\times 10^{19}\mathbf{c}{\mathbf{m}}^{-3}& & \text{factor 2, and }10^6{\text{ m}}^{-3}{\text{ per cm}}^{-3}\end{array}$$

For general $m^{\ast }$, the quantity $(m^{\ast }{)}^{3/2}$ scales out as $(m^{\ast }/m_0{)}^{3/2}$. $■$

(b) Table at 300 K ($N_C$ uses $m_n^{\ast }$, $N_V$ uses $m_p^{\ast }$):

Material	$m_n^{\ast }/m_0$	$(m_n^{\ast }/m_0{)}^{3/2}$	$N_C$ (cm${}^{-3}$)	$m_p^{\ast }/m_0$	$(m_p^{\ast }/m_0{)}^{3/2}$	$N_V$ (cm${}^{-3}$)
Si	1.18	1.282	$3.22\times 10^{19}$	0.81	0.729	$1.83\times 10^{19}$
Ge	0.55	0.408	$1.02\times 10^{19}$	0.36	0.216	$5.42\times 10^{18}$
GaAs	0.066	0.01696	$4.26\times 10^{17}$	0.52	0.375	$9.41\times 10^{18}$

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Problem 2.14: $n_i$ comparisons

(a) At what temperature does $n_i$ of (i) Si and (ii) GaAs equal $n_i(\text{Ge},300\text{ K})\approx 2\times 10^{13}$ cm${}^{-3}$?

(i) Si. Use the Sproul-Green fit and solve numerically:

$$2\times 10^{13}=9.15\times 10^{19}{\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}\Rightarrow {\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}=2.19\times 10^{-7}$$

$T$ (K)	$(T/300{)}^2{e}^{-0.5928/kT}$
400	$6.0\times 10^{-8}$
428	$2.13\times 10^{-7}$
430	$2.31\times 10^{-7}$

So $T\approx 429$ K ($\approx 156{}^{\circ }$C). Cross-check against the table of Exercise 2.4: $n_i=5.4\times 10^{12}$ at 400 K and $4.7\times 10^{13}$ at 450 K, bracketing $2\times 10^{13}$.

(ii) GaAs. Scale from $n_i(300)=2.25\times 10^6$ using $n_i\propto T^{3/2}{e}^{-E_G/2kT}$. The required ratio is $2\times 10^{13}/2.25\times 10^6=8.9\times 10^6$, i.e. $\ln =16.0$. Including the temperature dependence of the GaAs gap, $E_G(T)=1.519-\frac{5.405\times 10^{-4}{T}^2}{T+204}$ eV:

$$\ln \frac{n_i(T)}{n_i(300)}=\frac{3}{2}\ln \frac{T}{300}+\frac{E_G(300)}{2k\cdot 300}-\frac{E_G(T)}{2kT}$$

with $E_G(300)=1.4225$ eV so the second term is $27.51$. At $T=550$ K: $0.909+27.51-13.74=14.69$. At $T=600$ K: $1.040+27.51-12.35=16.20$. Interpolating gives $T\approx 595$ K ($\approx 320{}^{\circ }$C). (If the gap were held fixed at $1.42$ eV the answer would be about $650$ K; the shrinking gap makes GaAs go intrinsic a bit sooner.)

(b) $E_{GA}=1$ eV, $E_{GB}=2$ eV, same masses (so same $N_C{N}_V$), at 300 K:

$$\begin{array}{rlrl}\frac{n_{iA}}{n_{iB}}& =\frac{\sqrt{N_C{N}_V}{e}^{-E_{GA}/2kT}}{\sqrt{N_C{N}_V}{e}^{-E_{GB}/2kT}}=e^{(E_{GB}-E_{GA})/2kT}& & \text{prefactors cancel}\\ & =e^{1/(2\times 0.025851)}=e^{19.34}\approx 2.5\times 10^8& & \text{insert numbers}\end{array}$$

One extra eV of gap costs a factor of a quarter of a billion in carriers.

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Problem 2.15: $n_i(T)$ for Ge

Problem. Confirm Fig. 2.20 for Ge using $n_i=(1.76\times 10^{16})T^{3/2}{e}^{-0.392/kT}$.

Solution. Evaluate at several temperatures. Example at 300 K:

$$\begin{array}{rlrl}{T}^{3/2}& =300^{1.5}=5196& & \text{power}\\ \frac{0.392}{kT}& =\frac{0.392}{0.025851}=15.16& & \text{exponent}\\ e^{-15.16}& =2.60\times 10^{-7}& & \text{exponential}\\ n_i& =(1.76\times 10^{16})(5196)(2.60\times 10^{-7})=2.4\times 10^{13}{\text{cm}}^{-3}& & \text{product}\end{array}$$

in agreement with the Ge value $\approx 2\times 10^{13}$. Results by the same steps:

$T$ (K)	200	300	400	500	600	700
$n_i$ (cm${}^{-3}$)	$6.6\times 10^9$	$2.4\times 10^{13}$	$1.6\times 10^{15}$	$2.2\times 10^{16}$	$1.3\times 10^{17}$	$4.9\times 10^{17}$

Plotted on a log scale versus $T$ (or $1/T$) these points trace the Ge curve of Fig. 2.20. (The exponent $0.392$ eV is $E_G/2$ with $E_G\approx 0.78$ eV extrapolated to 0 K, and the prefactor absorbs the $T^{3/2}$ of $\sqrt{N_C{N}_V}$.)

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Problem 2.16: Concentration questions with a twist

(a) $p$-type Si, with the given $N_A$, $T\approx 0$ K. At absolute zero there is no thermal energy: the acceptors are un-ionized (every acceptor site empty of the extra electron) and no electron crosses the gap. This is the freeze-out regime:

$$p=0,n=0.$$

(Total ionization, $p=N_A$, applies only near room temperature.)

(b) Given $N>n_i$, full ionization, $n=N$ and $p=n_i^2/N$. Electrons are the majority carrier, with $n$ equal to the impurity concentration, so the impurity is a donor (an acceptor would give $p=N$ instead).

(c) $n=10^5$ cm${}^{-3}$ at 300 K in equilibrium:

$$p=\frac{n_i^2}{n}=\frac{(10^{10}{)}^2}{10^5}=10^{15}{\text{cm}}^{-3}.$$

(The sample is strongly p-type.)

(d) $E_F-E_i=+0.259$ eV in Si at 300 K:

$$\begin{array}{rlrl}\frac{E_F-E_i}{kT}& =\frac{0.259}{0.0259}=10& & \text{exactly 10 }kT\\ n& =n_i{e}^{10}=10^{10}(2.203\times 10^4)=2.2\times 10^{14}{\text{cm}}^{-3}& & \text{(2.19a)}\\ p& =n_i{e}^{-10}=10^{10}(4.54\times 10^{-5})=4.5\times 10^5{\text{cm}}^{-3}& & \text{(2.19b)}\end{array}$$

(e) Ge near room temperature, nondegenerate, $n_i=10^{13}$ cm${}^{-3}$ (as given), $n=2p$, $N_A=0$:

$$\begin{array}{rlrl}np& =n_i^2& & \text{np product}\\ 2p^2& =10^{26}\Rightarrow p=\sqrt{5\times 10^{25}}=7.07\times 10^{12}{\text{cm}}^{-3}& & \text{insert }n=2p\\ n& =2p=1.41\times 10^{13}{\text{cm}}^{-3}& & \\ p-n+N_D-N_A& =0\Rightarrow N_D=n-p+N_A& & \text{charge neutrality}\\ N_D& =1.41\times 10^{13}-7.07\times 10^{12}+0=7.07\times 10^{12}{\text{cm}}^{-3}& & \end{array}$$

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Problem 2.17: Carrier concentrations in Si

For all parts: nondegenerate, totally ionized dopants.

(a) $T=300$ K, $N_A=0$, $N_D=10^{15}$ cm${}^{-3}$. Since $N_D\gg n_i$:

$$n\approx N_D=10^{15}{\text{cm}}^{-3},p=\frac{n_i^2}{n}=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}.$$

(b) $T=300$ K, $N_A=10^{16}$, $N_D=0$:

$$p\approx N_A=10^{16}{\text{cm}}^{-3},n=\frac{10^{20}}{10^{16}}=10^4{\text{cm}}^{-3}.$$

(c) $T=300$ K, $N_A=9\times 10^{15}$, $N_D=10^{16}$ (compensated). Net doping:

$$N_D-N_A=10^{16}-9\times 10^{15}=10^{15}{\text{cm}}^{-3}\gg n_i$$ $$n\approx N_D-N_A=10^{15}{\text{cm}}^{-3},p=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}.$$

(d) $T=450$ K, $N_A=0$, $N_D=10^{14}$. First get $n_i(450\text{ K})$ from the Exercise 2.4 formula:

$$\begin{array}{rl}{E}_G& =1.17-\frac{4.73\times 10^{-4}(450{)}^2}{450+636}=1.0818\text{eV}\\ \frac{m_n^{\ast }}{m_0}& =1.028+0.2750-0.0626=1.2404,\frac{m_p^{\ast }}{m_0}=0.610+0.3524-0.0903=0.8720\\ (1.2404\times 0.8720{)}^{3/4}& =1.0606,(450/300{)}^{3/2}=1.837\\ kT& =0.038777\text{eV},\frac{E_G-0.0074}{2kT}=\frac{1.0744}{0.077553}=13.854,e^{-13.854}=9.63\times 10^{-7}\\ n_i& =(2.510\times 10^{19})(1.0606)(1.837)(9.63\times 10^{-7})\approx 4.7\times 10^{13}{\text{cm}}^{-3}\end{array}$$

Now $n_i$ is comparable to $N_D$, so use Eq. (2.29a):

$$\begin{array}{rl}n& =\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}=5\times 10^{13}+\sqrt{2.5\times 10^{27}+2.22\times 10^{27}}\\ & =5\times 10^{13}+6.87\times 10^{13}=1.19\times 10^{14}{\text{cm}}^{-3}\\ p& =\frac{n_i^2}{n}=\frac{2.22\times 10^{27}}{1.19\times 10^{14}}=1.9\times 10^{13}{\text{cm}}^{-3}\end{array}$$

(Check: $n-p=1.0\times 10^{14}=N_D$.)

(e) $T=650$ K, $N_A=0$, $N_D=10^{14}$. Again compute $n_i$:

$$\begin{array}{rl}{E}_G& =1.17-\frac{4.73\times 10^{-4}(650{)}^2}{1286}=1.0146\text{eV}\\ \frac{m_n^{\ast }}{m_0}& =1.2946,\frac{m_p^{\ast }}{m_0}=0.9305,(1.2946\times 0.9305{)}^{3/4}=1.150,(650/300{)}^{3/2}=3.189\\ kT& =0.056011\text{eV},\frac{1.0146-0.0074}{2kT}=8.991,e^{-8.991}=1.245\times 10^{-4}\\ n_i& =(2.510\times 10^{19})(1.150)(3.189)(1.245\times 10^{-4})\approx 1.15\times 10^{16}{\text{cm}}^{-3}\end{array}$$

Now $n_i\gg N_D$, so the sample is essentially intrinsic:

$$n=\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}\approx 5\times 10^{13}+1.146\times 10^{16}=1.15\times 10^{16}{\text{cm}}^{-3},p=n-N_D=1.14\times 10^{16}{\text{cm}}^{-3}.$$

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Problem 2.18: Position of $E_i$ and $E_F$ for Problem 2.17

Use $E_i-\frac{E_c+E_v}{2}=\frac{3}{4}kT\ln (m_p^{\ast }/m_n^{\ast })$ and Eq. (2.37)/(2.38).

(a) 300 K, $N_D=10^{15}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.0259)\ln \frac{0.81}{1.18}=-0.0073\text{eV}\\ E_F-E_i& =kT\ln \frac{N_D}{n_i}=0.0259\ln 10^5=0.0259(11.513)=0.298\text{eV}\end{array}$$

Diagram ($E_G=1.12$): $E_c-E_i=0.567$, $E_c-E_F=0.269$, $E_F-E_v=0.851$ eV, $E_i$ at $0.0073$ eV below midgap.

(b) 300 K, $N_A=10^{16}$: $E_i$ again $0.0073$ eV below midgap;

$$E_i-E_F=kT\ln \frac{N_A}{n_i}=0.0259\ln 10^6=0.0259(13.816)=0.358\text{eV}$$

Diagram: $E_i-E_v=0.553$, so $E_F-E_v=0.553-0.358=0.195$ eV above $E_v$.

(c) 300 K, net $n=10^{15}$: same as (a): $E_F-E_i=0.298$ eV (the carrier numbers, not the individual dopings, set $E_F$). $E_i$ is $0.0073$ eV below midgap.

(d) 450 K, $E_G=1.08$ eV, $kT=0.038777$ eV, $n=1.19\times 10^{14}$, $n_i=4.7\times 10^{13}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.038777)\ln \frac{0.8720}{1.2404}=0.02908(-0.3523)=-0.0102\text{eV}\\ E_F-E_i& =kT\ln \frac{n}{n_i}=0.038777\ln \frac{1.19\times 10^{14}}{4.7\times 10^{13}}=0.038777(0.929)=0.036\text{eV}\end{array}$$

Diagram: $E_c-E_i=0.540+0.0102=0.550$, $E_c-E_F=0.514$, $E_F-E_v=0.566$ eV.

(e) 650 K, $E_G=1.015$ eV, $kT=0.056011$ eV, $n=1.15\times 10^{16}$, $n_i=1.146\times 10^{16}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.056011)\ln \frac{0.9305}{1.2946}=0.04201(-0.3303)=-0.0139\text{eV}\\ E_F-E_i& =kT\ln \frac{n}{n_i}=0.056011\ln (1.0044)\approx 0.0002\text{eV}\end{array}$$

So $E_F\approx E_i$ (intrinsic behaviour). Diagram: $E_c-E_i=0.5075+0.0139=0.521$, $E_i-E_v=0.494$ eV, and the $E_F$ and $E_i$ lines coincide.

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Problem 2.19: MATLAB program for $n$, $p$, $E_F-E_i$

(a) Program (nondegenerate Si, complete ionization). It reuses the $n_i(T)$ computation of Exercise 2.4(a).

% Carrier concentrations in Si (nondegenerate, fully ionized dopants)
k  = 8.617e-5;                       % eV/K
A  = 2.510e19;  Eex = 0.0074;
T  = input('Temperature T (K) = ');
ND = input('Donor concentration ND (cm^-3) = ');
NA = input('Acceptor concentration NA (cm^-3) = ');
 
EG  = 1.17 - 4.730e-4*T^2/(T+636);
mnr = 1.028 + 6.11e-4*T - 3.09e-7*T^2;
mpr = 0.610 + 7.83e-4*T - 4.46e-7*T^2;
ni  = A*(mnr*mpr)^0.75*(T/300)^1.5*exp(-(EG-Eex)/(2*k*T));
 
dN = (ND-NA)/2;
n  = dN + sqrt(dN^2 + ni^2);         % Eq. (2.29a)
p  = ni^2/n;                          % np product
EFEi = k*T*log(n/ni);                % Eq. (2.37)
 
fprintf('ni = %.3e  n = %.3e  p = %.3e  EF-Ei = %.4f eV\n', ni, n, p, EFEi);

(b) Running it reproduces the answers computed above: for example,

Case	$T$	$N_D$	$N_A$	$n$ (cm${}^{-3}$)	$p$ (cm${}^{-3}$)	$E_F-E_i$ (eV)
2.17(a)	300	$10^{15}$	0	$1.0\times 10^{15}$	$1.0\times 10^5$	0.298
2.17(b)	300	0	$10^{16}$	$1.0\times 10^4$	$1.0\times 10^{16}$	$-0.358$
2.17(c)	300	$10^{16}$	$9\times 10^{15}$	$1.0\times 10^{15}$	$1.0\times 10^5$	0.298
2.17(d)	450	$10^{14}$	0	$1.19\times 10^{14}$	$1.9\times 10^{13}$	0.036
2.17(e)	650	$10^{14}$	0	$1.15\times 10^{16}$	$1.14\times 10^{16}$	0.0002

Subtle point: the program’s $E_F-E_i$ is relative to $E_i$ itself, so to compare with the band diagrams you add the small shift of Eq. (2.36).

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Problem 2.20: Maximum nondegenerate doping in Si at 300 K

Problem. Verify $N_D\approx 1.6\times 10^{18}$ and $N_A\approx 9.1\times 10^{17}$ cm${}^{-3}$.

Solution. The nondegenerate limit is $E_F=E_c-3kT$ (donors) or $E_F=E_v+3kT$ (acceptors). At that point Eq. (2.16) with $e^{-3}=0.0498$ gives

$$\begin{array}{rlrl}{n}_{max}& =N_C{e}^{-3}=(3.22\times 10^{19})(0.0498)=1.60\times 10^{18}{\text{cm}}^{-3}& & \text{(2.16a)}\\ p_{max}& =N_V{e}^{-3}=(1.83\times 10^{19})(0.0498)=9.1\times 10^{17}{\text{cm}}^{-3}& & \text{(2.16b)}\end{array}$$

and with total ionization $N_D=n_{max}$, $N_A=p_{max}$ (the minority carriers are negligible). These agree with the text values. $■$

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Problem 2.21: Plot $E_F-E_i$ versus doping

Solution. Use Eq. (2.29) for the exact $n$ (or $p$) and Eq. (2.37), so the plot is correct even near $N\sim n_i$:

k = 8.617e-5; T = 300; kT = k*T; ni = 1e10;
N  = logspace(10,19,200);            % doping (cm^-3)
nD = N/2 + sqrt((N/2).^2 + ni^2);    % donor-doped, ND=N
EF_donor    = kT*log(nD/ni);         % EF-Ei  (>0)
EF_acceptor = -kT*log(nD/ni);        % Ei-EF = kT ln(p/ni), p = nD by symmetry
semilogx(N, EF_donor, N, EF_acceptor); grid
xlabel('N_A or N_D (cm^{-3})'); ylabel('E_F - E_i (eV)');

Selected values from Eq. (2.38) ($kT=0.0259$ eV):

$N_D$ or $N_A$ (cm${}^{-3}$)	$10^{12}$	$10^{14}$	$10^{15}$	$10^{16}$	$10^{17}$	$10^{18}$
$|E_F-E_i|$ (eV)	0.119	0.239	0.298	0.358	0.417	0.477

The curves rise logarithmically (a straight line per decade, each decade adding $kT\ln 10=0.0596$ eV) and flatten to $0$ near $N\sim n_i$. Accuracy check against Fig. 2.21: at the nondegenerate limit $N_D=1.6\times 10^{18}$, $E_F-E_i=0.0259\ln (1.6\times 10^8)=0.489$ eV, while $E_c-E_i=0.567$ eV, leaving $E_c-E_F=0.078\approx 3kT=0.0777$ eV, so the plotted solid line meets the $3kT$ boundary exactly where it should.

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Problem 2.22: GaAs considerations

(a) For GaAs $m_n^{\ast }=0.066m_0\ll m_p^{\ast }=0.52m_0$. Since $g\propto m^{\ast 3/2}$, at equal distances from the edges

$$\frac{g_v}{g_c}={\left(\frac{0.52}{0.066}\right)}^{3/2}=(7.88{)}^{3/2}\approx 22.$$

The sketch of Fig. 2.14 for GaAs therefore has the valence band curve (going downward) about 22 times steeper than the conduction band curve: the conduction band “stadium” has far fewer seats.

(b) $E_F$ for intrinsic GaAs must make $n=p$. With far fewer states in the conduction band, the Fermi level has to rise toward $E_c$ to fill the same number of electrons as the holes in the larger valence band. Hence $E_i$ lies above midgap (consistent with Eq. 2.36 since $m_p^{\ast }>m_n^{\ast }$ makes the log term positive).

(c) Position of $E_i$ at 300 K:

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}& & \text{(2.36)}\\ & =\frac{3}{4}(0.025851)\ln (7.88)=0.01939\times 2.064& & \text{insert}\\ & =0.040\text{eV}& & \end{array}$$

$E_i$ is about $0.040$ eV above midgap (roughly $1.5kT$).

(d) Maximum nondegenerate doping at 300 K:

$$\begin{array}{rl}{N}_C& =(2.510\times 10^{19})(0.066{)}^{3/2}=4.26\times 10^{17}{\text{cm}}^{-3},N_V=(2.510\times 10^{19})(0.52{)}^{3/2}=9.41\times 10^{18}{\text{cm}}^{-3}\\ N_{D,max}& =N_C{e}^{-3}=(4.26\times 10^{17})(0.0498)=2.1\times 10^{16}{\text{cm}}^{-3}\\ N_{A,max}& =N_V{e}^{-3}=(9.41\times 10^{18})(0.0498)=4.7\times 10^{17}{\text{cm}}^{-3}\end{array}$$

So n-type GaAs becomes degenerate at a much lower doping than n-type Si: the light electron mass means few conduction band states.

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Problem 2.23: $E_F$ versus temperature, $N_A=10^{14}$ cm${}^{-3}$ Si

(a) Qualitative argument as $T\to 0$. At very low $T$ the sample is in freeze-out: valence band full, acceptor levels $E_A=E_v+|E_B|$ empty, conduction band empty. The only carriers possible are holes created by moving a valence electron onto an acceptor (so that $p=N_A^{-}$), and those few holes obey $p=N_v{e}^{(E_v-E_F)/kT}$ while the acceptor ionization requires $E_F$ to lie near $E_A$. Charge neutrality $p=N_A^{-}$ with $N_A^{-}\ll N_A$ forces $E_F$ to lie between $E_v$ and $E_A$; in the limit $T\to 0$ it sits midway,

$$E_F\to \frac{E_v+E_A}{2}=E_v+\frac{|E_B|}{2}\approx E_v+0.02\text{eV}(\text{B in Si}).$$

(b) Program. In the range $200\le T\le 500$ K freeze-out is irrelevant (dopants fully ionized), so use $n_i(T)$ from Exercise 2.4 and Eq. (2.29b):

k=8.617e-5; A=2.510e19; Eex=0.0074; NA=1e14;
T  = 200:5:500;
EG = 1.17 - 4.730e-4*T.^2./(T+636);
mn = 1.028 + 6.11e-4*T - 3.09e-7*T.^2;
mp = 0.610 + 7.83e-4*T - 4.46e-7*T.^2;
ni = A*(mn.*mp).^0.75.*(T/300).^1.5.*exp(-(EG-Eex)./(2*k*T));
p  = NA/2 + sqrt((NA/2)^2 + ni.^2);       % Eq. (2.29b), ND = 0
EiEF = k*T.*log(p./ni);                   % Ei - EF, Eq. (2.37)
plot(T, EiEF); grid; xlabel('T (K)'); ylabel('E_i - E_F (eV)');

Typical output:

$T$ (K)	$n_i$ (cm${}^{-3}$)	$p$ (cm${}^{-3}$)	$E_i-E_F$ (eV)
200	$5.2\times 10^4$	$1.0\times 10^{14}$	0.368
300	$1.0\times 10^{10}$	$1.0\times 10^{14}$	0.239
400	$5.4\times 10^{12}$	$1.003\times 10^{14}$	0.100
500	$2.7\times 10^{14}$	$3.3\times 10^{14}$	0.008

(c) Conclusion. For fixed $N_A$, $E_F$ starts near the valence band edge at very low $T$, rises through the gap as $T$ increases, and approaches $E_i$ (about midgap) from below at high temperature, where the sample turns intrinsic. In the extrinsic range the $kT\ln (N_A/n_i)$ shrinks because $n_i$ increases faster than $kT$.

(d) Increasing $N_A$ in decades from $10^{14}$ to $10^{18}$ cm${}^{-3}$. At each temperature $E_i-E_F$ grows by $kT\ln 10$ per decade (0.0596 eV at 300 K, so $0.239\to 0.298\to 0.358\to 0.417\to 0.477$ eV at 300 K), pushing $E_F$ closer to $E_v$. The temperature at which the sample goes intrinsic ($n_i\sim N_A$) moves higher, widening the extrinsic range; at $N_A=10^{18}$ cm${}^{-3}$ the level reaches $E_F-E_v\approx 0.076$ eV $\approx 2.9kT$, so the sample sits at the edge of degeneracy (the limit $9.1\times 10^{17}$ cm${}^{-3}$ of Problem 2.20), and at low $T$ the freeze-out and dopant-band effects become important.

(e) Donors instead of acceptors. The picture is the mirror image: $E_F$ starts just below $E_c$ (midway between $E_D$ and $E_c$ as $T\to 0$), moves down toward $E_i$ from above as $T$ rises, sits $kT\ln (N_D/n_i)$ above $E_i$ in the extrinsic range, and moves up by $kT\ln 10$ per decade of $N_D$, with degeneracy near $N_D\approx 1.6\times 10^{18}$ cm${}^{-3}$. Only small asymmetries (the $\frac{3}{4}kT\ln (m_p^{\ast }/m_n^{\ast })$ offset of $E_i$) distinguish the two cases.

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Closing thought. You began this chapter with a block of silicon that looked like an impossible crowd of $5\times 10^{22}$ atoms per cm${}^3$ and ended with a handful of lines, $np=n_i^2$, charge neutrality, and $E_F-E_i=kT\ln (n/n_i)$, that tell you exactly who carries current and how to change it. Everything in the chapters ahead, from drift and diffusion to p-n junctions, starts with these numbers. Well done.

Chapter 3: Carrier Action

Source slice: Pierret, Semiconductor Device Fundamentals, Chapter 3 (book pp. 75-148). Equations keep the book’s numbering. OCR damage was repaired using physics; where a number in a problem statement was unreadable, the assumption is stated explicitly.

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Prologue: from a sleeping crystal to a working device

In Chapter 2 we studied a semiconductor at rest (equilibrium). That picture is essential as a reference, but nothing happens in it: the net current is exactly zero. A device is useful only when we poke the crystal (apply a voltage, shine light, inject carriers) and the carriers respond.

Think of a crowded stadium at night after the match ends. In equilibrium everybody mills about randomly (thermal motion), and no net flow goes out of the gates. Now:

• someone tilts the whole stadium (an electric field), so everyone drifts downhill: drift;
• the north stand is packed and the south is empty (a concentration gradient), so people wander toward the empty side just by random bumping: diffusion;
• people vanish through hidden trapdoors and new people appear from the entrances: recombination-generation (R-G).

Those three actions are the whole story of Chapter 3. For each we will (1) picture it, (2) turn it into an equation, and (3) identify its “constant of the motion” (the one material parameter that sets its strength):

Carrier action	Gives rise to	Material parameter (“constant of the motion”)
Drift	drift current	mobility ${\mu }_n,{\mu }_p$
Diffusion	diffusion current	diffusion coefficient $D_N,D_P$
Recombination-generation	change of carrier concentration in time	minority-carrier lifetime ${\tau }_n,{\tau }_p$

Finally we glue the three together into the equations of state (continuity equations and the minority-carrier diffusion equation), solve a few canonical problems, and add two supplemental ideas: diffusion length and quasi-Fermi levels.

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3.1 Drift

3.1.1 Definition and visualization

Drift is the motion of charged particles in response to an applied electric field $\mathcal{E}$.

Microscopic picture. Apply a field across a bar of semiconductor (Fig. 3.1a). The field pushes holes ($+q$) along $\mathcal{E}$ and electrons ($-q$) against $\mathcal{E}$. But a carrier is never allowed to accelerate for long: it collides with thermally vibrating lattice atoms and with ionized dopant atoms (it is scattered). The result (Fig. 3.1b) is a jerky, zig-zag path: accelerate, crash, restart, accelerate, crash. The overall trend is along (or against) the field.

Macroscopic picture. Tracking one carrier through every collision is hopeless, and nobody can measure it anyway. What we can measure is the average motion. Averaged over all carriers of one type, the motion is a steady drift velocity $v_d$ along (holes) or against (electrons) the field (Fig. 3.1c).

Analogy: a pinball machine. The ball is pulled down by gravity (the field) but keeps bouncing off pegs (scattering). It never speeds up indefinitely; it reaches an average downhill speed. That average speed is the drift velocity.

Common confusion: “isn’t the carrier already moving fast?” Yes! Even at equilibrium, thermal motion is ferocious: typical thermal speeds are about $1/1000$ of the speed of light at room temperature. But thermal motion is completely random (Fig. 3.2): for every carrier going left another goes right. Averaged over many carriers it is zero, so it carries no net current. Drift is a tiny, orderly bias superimposed on a huge random jitter, like a slow river current carrying a crowd of frantically swimming fish.

3.1.2 Drift current

Definition of current. Current $I$ is the charge per unit time crossing an arbitrarily chosen plane that is perpendicular to the flow.

Derivation of hole drift current. Take a p-type bar of cross-sectional area $A$ (Fig. 3.3), a plane normal to the flow, and holes with drift velocity $v_d$.

$v_{d}t\text{(distance)}$

All holes within a distance $v_{d}t$ upstream of the plane will cross it in time $t$.

$v_{d}tA\text{(volume)}$

All holes in this volume cross the plane in time $t$.

$p{v}_{d}tA\text{(number of holes)}$

$p$ is the hole concentration (holes per cm${}^3$) times the volume.

$qp{v}_{d}tA\text{(charge crossing in time }t)$

Multiply by the charge per hole.

$qp{v}_{d}A\text{(charge per unit time)}$

Divide by $t$; this is precisely the definition of current. Hence

\boxed{I_{P|\text{drift}} = q\,p\,v_d\,A} \tag{3.1}

Current density. The area $A$ is baggage, and current as defined is a scalar although the flow has a direction. We therefore use the current density $\mathbf{J}$, a vector pointing along the flow with magnitude $J=I/A$:

\boxed{J_{P|\text{drift}} = q\,p\,v_d} \tag{3.2}

Relating $v_d$ to the field. Figure 3.4 (measured $v_d$ vs. $\mathcal{E}$ in ultrapure Si at 300 K) shows: at low fields $v_d\propto \mathcal{E}$ (a straight line of slope 1 on log-log axes), while at very high fields $v_d$ saturates at $v_{sat}\approx 10^7$ cm/s and no longer depends on $\mathcal{E}$. The book’s empirical form is

v_d = \frac{\mu_0\,\mathcal{E}}{\left[1 + \left(\mu_0\mathcal{E}/v_{sat}\right)^{\beta}\right]^{1/\beta}} \tag{3.3}

with $\beta \approx 1$ for holes and $\beta \approx 2$ for electrons in Si. Check the two limits:

$\mathcal{E}\to 0:v_d\to {\mu }_0\mathcal{E}(\text{linear; the proportionality constant is the mobility }\mu )$

$\mathcal{E}\to \infty :v_d\to \frac{{\mu }_0\mathcal{E}}{{\mu }_0\mathcal{E}/v_{sat}}=v_{sat}(\text{independent of }\mathcal{E})$

Why saturation? At extreme fields, carriers gain energy so fast that they dump it into the lattice by emitting vibrations (optical phonons) almost as soon as they acquire it. It is like a car whose engine power is entirely eaten by air drag at top speed.

Unless stated otherwise we use the low-field limit $v_d=\mu \mathcal{E}$, which is the case of most practical interest. For holes ($v_d={\mu }_p\mathcal{E}$) substitution into (3.2) gives

\boxed{J_{P|\text{drift}} = q\,\mu_p\,p\,\mathcal{E}} \tag{3.4a}

and similarly for electrons

\boxed{J_{N|\text{drift}} = q\,\mu_n\,n\,\mathcal{E}} \tag{3.4b}

${\mu }_n$ and ${\mu }_p$ are the electron and hole mobilities, always taken as positive numbers.

A sign puzzle worth thinking about. Electrons drift against the field: $v_d=-{\mu }_n\mathcal{E}$. But electrons carry negative charge, so the current they carry points opposite to their velocity, i.e., along $\mathcal{E}$. Two minus signs cancel: $J_{N|\text{drift}}=(-q)n(-{\mu }_n\mathcal{E})=+q{\mu }_{n}n\mathcal{E}$. Therefore both electrons and holes contribute current in the same direction as the electric field, and their contributions simply add. That is why (3.4a) and (3.4b) both carry a plus sign.

3.1.3 Mobility

Mobility is the star parameter of drift. Let us collect its facts in the book’s “information card” format.

Units. cm${}^2$/V$\cdot$s. (Check: $v_d=\mu \mathcal{E}$ gives $\text{cm/s}=\mu \cdot \text{V/cm}$, so $\mu$ is in ${\text{cm}}^2/\text{V s}$.)

Sample numerical values (300 K).

• Si: ${\mu }_n\approx 1360$, ${\mu }_p\approx 460$ cm${}^2$/V$\cdot$s (for $N_D=10^{14}$ and $N_A=10^{14}$ cm${}^{-3}$ doping respectively).
• Uncompensated high-purity GaAs: ${\mu }_n\approx 8000$, ${\mu }_p\lesssim 400$ cm${}^2$/V$\cdot$s.
• In the major semiconductors, ${\mu }_n>{\mu }_p$ for the same doping and temperature.

Relation to scattering. “Mobility” literally means ease of movement. More scattering means lower mobility. The two dominant scattering mechanisms in non-degenerately doped, device-quality material are:

1. Lattice scattering: collisions with thermally vibrating lattice atoms. (Note: it is the vibration, i.e., the displacement from perfect lattice positions, that scatters. The perfectly periodic crystal potential is already hidden in the effective mass.)
2. Ionized impurity scattering: Coulomb deflection by ionized donors or acceptors.

A quantitative relation is

$\mu =\frac{q⟨{\tau }_c⟩}{m_{\text{cond}}^{\ast }}$

where $⟨{\tau }_c⟩$ is the mean free time between collisions and $m_{\text{cond}}^{\ast }$ is the conductivity effective mass. (We write ${\tau }_c$ here to avoid mixing it up with the carrier lifetimes ${\tau }_n,{\tau }_p$ later in the chapter; the book writes $\mu \propto \tau /m^{\ast }$.) More collisions give a shorter free time and a smaller $\mu$. A small $m^{\ast }$ means a light, easily pushed carrier. This explains why electrons in GaAs ($m_n^{\ast }\approx 0.067m_0$) are far more mobile than in Si.

Intuition for $\mu =q{\tau }_c/m^{\ast }$. Newton: between collisions, a carrier accelerates at $a=q\mathcal{E}/m^{\ast }$ for an average time ${\tau }_c$, so its average extra velocity is about $a{\tau }_c=(q{\tau }_c/m^{\ast })\mathcal{E}$. The bracket is the mobility.

Doping dependence (Fig. 3.5). Mobilities are roughly constant at low doping (below about $10^{16}$ cm${}^{-3}$ in Si) and then fall monotonically with increasing $N_A$ or $N_D$. Table beside Fig. 3.5(a) for Si at 300 K:

$N_A$ or $N_D$ (cm${}^{-3}$)	${\mu }_n$	${\mu }_p$
$1\times 10^{14}$	1358	461
$2\times 10^{14}$	1357	460
$5\times 10^{14}$	1352	459
$1\times 10^{15}$	1345	458
$2\times 10^{15}$	1332	455
$5\times 10^{15}$	1298	448
$1\times 10^{16}$	1248	437
$2\times 10^{16}$	1165	419
$5\times 10^{16}$	986	378
$1\times 10^{17}$	801	331

Why? Use an electrical analogy (Fig. 3.6): each scattering mechanism is a “resistance to motion” and the resistances are in series:

$R_{TOTAL}=R_L+R_I$

($R_L$ = lattice, $R_I$ = ionized impurity.) $R_L$ has no doping dependence. At low doping $R_I\ll R_L$, so $R_{TOTAL}\approx R_L$ and $\mu$ is doping-independent. At higher doping $R_I$ grows (more ionized scatterers) and $\mu$ falls.

Temperature dependence (Fig. 3.7). For $N\lesssim 10^{14}$ cm${}^{-3}$ the curves merge into one near power law:

${\mu }_n\propto T^{-2.3\pm 0.1},{\mu }_p\propto T^{-2.2\pm 0.1}$

so mobility rises as T falls. For heavier doping, $\mu$ still rises with falling T, but more slowly; for $N\sim 10^{19}$ cm${}^{-3}$, $\mu$ even drops when T falls below about 200 K.

Why? Lattice scattering weakens as T falls (atoms vibrate less): $R_L$ decreases with decreasing T, so $\mu$ rises as a negative power of T. But ionized impurity scattering gets stronger at low T: slower carriers spend more time near each ionized impurity and are deflected more easily, so $R_I$ increases with decreasing T. At high doping $R_I$ is a larger fraction of the total, partly cancelling the lattice-scattering improvement, hence the flatter slope.

Scattering	Doping dependence	Temperature dependence
Lattice ($R_L$)	none	decreases with decreasing T
Ionized impurity ($R_I$)	increases with $N_A$ or $N_D$	increases with decreasing T

Exercise 3.1: Empirical mobility fit (with MATLAB)

Problem. The majority-carrier mobility in Si at 300 K is computed from the empirical fit

$\mu ={\mu }_{min}+\frac{{\mu }_0}{1+(N/N_{ref}{)}^{\alpha }}$

where $N$ is the doping concentration ($N_A$ or $N_D$) and the others are fit parameters. Temperature dependence enters through

$A=A_{300}{\left(\frac{T}{300}\right)}^{\eta }$

where $A$ stands for ${\mu }_{min}$, ${\mu }_0$, $N_{ref}$, or $\alpha$, $A_{300}$ is its 300 K value, $T$ is in kelvin, and $\eta$ is the temperature exponent. Parameters for Si:

Parameter	Electrons	Holes	Temp. exponent $\eta$
$N_{ref}$ (cm${}^{-3}$)	$1.3\times 10^{17}$	$2.35\times 10^{17}$	2.4
${\mu }_{min}$ (cm${}^2$/V s)	92	54.3	$-0.57$
${\mu }_0$ (cm${}^2$/V s)	1268	406.9	$-2.33$ (electrons), $-2.23$ (holes)
$\alpha$	0.91	0.88	$-0.146$

(a) Construct a log-log plot of ${\mu }_n$ and ${\mu }_p$ vs. $N_A$ or $N_D$ for $10^{14}\le N\le 10^{19}$ cm${}^{-3}$ and compare with Fig. 3.5(a). (b) Plot ${\mu }_n$ vs. $T$ and ${\mu }_p$ vs. $T$ for $200\le T\le 500$ K with $N$ stepped in decades from $10^{14}$ to $10^{18}$ cm${}^{-3}$ and compare with Fig. 3.7.

Logic of the code. The formula is a “floor plus a falling step”: ${\mu }_{min}$ is the floor at very high doping; ${\mu }_{min}+{\mu }_0$ is the ceiling at low doping (lattice-limited); $N_{ref}$ is the doping at which the step is half-way down; $\alpha$ controls how steeply it falls. We use logspace to get points equally spaced in $\log N$ and element-wise operators (./, .^) so MATLAB evaluates the formula for the whole vector at once.

Solution (a). (The book’s listing, cleaned of OCR noise.)

% Mobility versus Dopant Concentration (Si, 300 K)
% Fit parameters
NDref = 1.3e17;  NAref = 2.35e17;
mun_min = 92;    mup_min = 54.3;
mun_0 = 1268;    mup_0 = 406.9;
an = 0.91;       ap = 0.88;
 
% Mobility calculation
N = logspace(14,19);                        % 50 log-spaced points
mun = mun_min + mun_0./(1 + (N/NDref).^an);
mup = mup_min + mup_0./(1 + (N/NAref).^ap);
 
% Plotting results
close
loglog(N, mun, N, mup); grid;
axis([1.0e14 1.0e19 1.0e1 1.0e4]);
xlabel('N_A or N_D (cm^{-3})');
ylabel('Mobility (cm^2/V-sec)');
text(1.0e15, 1500, 'Electrons');
text(1.0e15, 500,  'Holes');
text(1.0e18, 2000, 'Si, 300 K');

Quick hand-check of one point ($N=10^{16}$): $(10^{16}/1.3\times 10^{17}{)}^{0.91}=(0.0769{)}^{0.91}=0.0969$, so ${\mu }_n=92+1268/1.0969=92+1156=1248$ cm${}^2$/V s, which matches the table. The plot is numerically identical to Fig. 3.5(a).

Part (b) (the book leaves it to the reader; Problem 3.3 asks for it, and it is completed here).

% Mobility versus Temperature (Si) with doping as parameter
T = linspace(200,500,100);
Nlist = 10.^(14:18);
 
% 300 K parameters and temperature exponents (k=1 electrons, k=2 holes)
Nref0 = [1.3e17 2.35e17];  etaNref = 2.4;
mmin0 = [92 54.3];         etamin  = -0.57;
m00   = [1268 406.9];      eta0    = [-2.33 -2.23];
al0   = [0.91 0.88];       etaal   = -0.146;
 
for k = 1:2
    subplot(1,2,k)
    for N = Nlist
        Nref = Nref0(k)*(T/300).^etaNref;
        mmin = mmin0(k)*(T/300).^etamin;
        m0   = m00(k)  *(T/300).^eta0(k);
        al   = al0(k)  *(T/300).^etaal;
        mu = mmin + m0./(1 + (N./Nref).^al);
        loglog(T, mu); hold on
    end
    grid; axis([200 500 100 1e4]);
    xlabel('T (K)'); ylabel('\mu (cm^2/V-sec)');
    if k==1, title('Electrons'), else, title('Holes'), end
end

Expected result: the lowest-doping curve is a straight line on log-log axes with slope about $-2.3$ (electrons) / $-2.2$ (holes); heavier-doping curves bend and flatten at low T, as in Figs. 3.7(a) and (b).

3.1.4 Resistivity

Resistivity $\rho$ is a “normalized resistance” that does not depend on the sample’s shape. Operationally it is the proportionality between the applied field and the resulting current density:

\mathcal{E} = \rho\,J \tag{3.5a}

J = \sigma\mathcal{E} = \frac{1}{\rho}\mathcal{E} \tag{3.5b}

with $\sigma =1/\rho$ the conductivity. In a homogeneous material all the current is drift current, so using (3.4):

J_{drift} = J_{N|drift} + J_{P|drift} = q(\mu_n n + \mu_p p)\,\mathcal{E} \tag{3.6}

Comparing with (3.5b):

\boxed{\rho = \frac{1}{q(\mu_n n + \mu_p p)}} \tag{3.7}

Simplification in doped material. In a non-degenerate n-type sample in the extrinsic region ($N_D\gg n_i$): $n=N_D$ and $p=n_i^2/N_D\ll n$ (Section 2.5.5). Then ${\mu }_{n}n+{\mu }_{p}p\approx {\mu }_n{N}_D$. Similarly for p-type, ${\mu }_p{N}_A$. Hence

\boxed{\rho \approx \frac{1}{q\mu_n N_D}\quad\text{(n-type)}} \tag{3.8a}

\boxed{\rho \approx \frac{1}{q\mu_p N_A}\quad\text{(p-type)}} \tag{3.8b}

Combined with the mobility-versus-doping data these give a one-to-one relation between a directly measurable quantity ($\rho$) and the doping (Fig. 3.8). Measuring $\rho$ is a routine way to find $N_A$ or $N_D$.

How to measure resistivity.

1. Straightforward bar method. Cut a bar, apply a voltage, measure the current, use $R=V/I=\rho \ell /A$ ($\ell$ = length, $A$ = cross section). It looks easy but is deceptively difficult (contact resistance), destroys good material, and does not suit wafers.

2. Four-point probe (Fig. 3.9a). Four collinear, equally spaced probes touch the wafer surface. Pass a known current $I$ through the outer two probes and measure the voltage $V$ between the inner two:

\rho = 2\pi s\,\frac{V}{I}\,\Gamma \tag{3.9}

where $s$ is the probe spacing and $\Gamma$ is a documented correction factor that depends on sample thickness and on whether the wafer’s bottom touches an insulator or a metal (commercial instruments compute it automatically from the thickness you enter). Why four probes? Because the voltmeter draws almost no current, the voltage drop across the contact resistances at the two voltage probes is negligible, which removes the contact-resistance headache. The method is quick and ideal for wafers, but it leaves slight surface damage, so it is not used on wafers destined for high-yield devices.

3. Eddy-current method (Fig. 3.9b), non-contacting. An RF-excited ferrite core produces fringing fields; a nearby conducting wafer develops eddy currents which absorb RF power. The power absorbed (calibrated) gives the sheet resistivity. Built-in ultrasound transducers measure the wafer thickness from the phase of echoes off the top and bottom surfaces. Then ${\rho }_{wafer}={\rho }_{sheet}\times \text{thickness}$.

3.1.5 Band bending

So far we have always drawn $E_C$ and $E_V$ as flat lines. When an electric field exists in the material, the band edges become functions of position; this is called band bending (Fig. 3.10a).

Meaning of the diagram. The energy band diagram plots allowed total electron energy vs. position. Bond-breaking with energy exactly $E_G$ creates an electron at $E_C$ and a hole at $E_V$, both essentially motionless. Any extra energy makes them move. Therefore

$\text{K.E. of electrons}=E-E_C,\text{K.E. of holes}=E_V-E$

(for holes, energy increases downward on the diagram, so a hole deep in the valence band has more kinetic energy). Total = K.E. + P.E., so the electron potential energy is $E_C-E_{ref}$ (Fig. 3.10c). $E_{ref}$ is an arbitrary constant.

Link to the electrostatic potential. For a particle of charge $-q$ in an electrostatic potential $V$ (when magnetic, thermal-gradient and stress effects are negligible),

\text{P.E.} = -qV \tag{3.10}

but we also found

\text{P.E.} = E_C - E_{ref} \tag{3.11}

Equate them and solve for $V$:

V = -\frac{1}{q}(E_C - E_{ref}) \tag{3.12}

By definition $\mathcal{E}=-\nabla V$ (3.13), which in one dimension reads

\mathcal{E} = -\frac{dV}{dx} \tag{3.14}

Differentiate (3.12):

$\mathcal{E}=-\frac{dV}{dx}=\frac{1}{q}\frac{d{E}_C}{dx}$

and since $E_C$, $E_V$, $E_i$ differ only by additive constants (their slopes are equal):

\boxed{\mathcal{E} = \frac{1}{q}\frac{dE_C}{dx} = \frac{1}{q}\frac{dE_V}{dx} = \frac{1}{q}\frac{dE_i}{dx}} \tag{3.15}

Reading a band diagram by inspection:

• To get the shape of $V(x)$: flip the band picture “upside down” (since $V=-(E_C-E_{ref})/q$). The vertical offset is arbitrary.
• To get the shape of $\mathcal{E}(x)$: look at the slope of $E_C$ (or $E_V$, $E_i$). Bands sloping up toward $+x$ mean $\mathcal{E}>0$.

Mountain-landscape analogy. Electrons roll downhill on the band diagram (toward lower $E_C$), holes “bubble” uphill (like air bubbles in water). A steep band means a strong field. Sign confusion disappears if you remember that the diagram is an electron potential-energy landscape.

Exercise 3.2: Reading a band diagram

Problem. The figure in the book is an energy band diagram for Si at 300 K with $E_F$ taken as the energy reference. The Fermi level sits $E_G/4$ below $E_i$ at $x=\pm L$ (p-type there, $E_i-E_F=E_G/4$) and $E_G/4$ above $E_i$ at $x=0$ ($E_F-E_i=E_G/4$, n-type there); the bands bend smoothly between. Three electrons and three holes (labelled 1, 2, 3) are marked on the diagram. (a) Sketch $V(x)$. (b) Sketch $\mathcal{E}(x)$. (c) Find the K.E. and P.E. of the marked carriers. (d) Determine the resistivity of the $x>L$ region.

Solution.

(a) $V(x)$ has the same form as the “upside-down” $E_C(x)$. Choosing $V=0$ at $x=L$ (arbitrary), the curve is the mirror image of the band edge: where the bands dip (near $x=0$) the potential has its peak, and it is flat for $|x|>L$.

(b) $\mathcal{E}(x)=(1/q)d{E}_C/dx$: proportional to the slope of the bands. It is zero where the bands are flat ($|x|>L$), positive where $E_C$ rises with $x$, negative where it falls, and it changes sign at $x=0$ where the slope reverses.

(c) For an electron: $\text{K.E.}=E-E_C$ and $\text{P.E.}=E_C-E_{ref}=E_C-E_F$. For a hole (energy increases downward): $\text{K.E.}=E_V-E$ and $\text{P.E.}=E_{ref}-E_V=E_F-E_V$. All differences are taken at the same $x$. With $E_G=1.12$ eV, $E_G/4=0.28$ eV and $E_G/2=0.56$ eV:

• at $x=0$ (n-type): $E_C-E_F=E_G/2-E_G/4=0.28$ eV, and $E_F-E_V=E_G/2+E_G/4=0.84$ eV;
• at $x=\pm L$ (p-type): $E_C-E_F=0.84$ eV, and $E_F-E_V=0.28$ eV.

Carrier	K.E. (eV)	P.E. (eV)
Electron 1	0	0.28
Electron 2	0.56	0.28
Electron 3	0	0.84
Hole 1	0	0.28
Hole 2	0.56	0.28
Hole 3	0	0.84

(Electron 1 sits at $E_C$ at $x=0$; electron 2 is $0.56$ eV above $E_C$ at $x=0$; electron 3 sits at $E_C$ at $x=\pm L$. Hole 1 sits at $E_V$ at $x=\pm L$; hole 2 is $0.56$ eV below $E_V$ there; hole 3 sits at $E_V$ at $x=0$.)

(d) In the $x>L$ region $E_i-E_F=E_G/4=0.28$ eV, so the material is p-type with

$N_A\approx p=n_i{e}^{(E_i-E_F)/kT}$

$=10^{10}{e}^{0.28/0.0259}=4.96\times 10^{14}{\text{cm}}^{-3}$

From Fig. 3.5(a), ${\mu }_p\approx 459$ cm${}^2$/V s. By (3.8b),

$\rho =\frac{1}{q{\mu }_p{N}_A}=\frac{1}{(1.6\times 10^{-19})(459)(4.96\times 10^{14})}=27.5\Omega \text{cm}$

Cross-check with Fig. 3.8(a): $\rho \approx 25\Omega$ cm.

---

3.2 Diffusion

3.2.1 Definition and visualization

Diffusion is the tendency of particles to spread out, owing to their random thermal motion, migrating (on the macroscopic scale) from regions of high concentration to regions of low concentration. Left alone it ends in a uniform distribution. The diffusing particles need not be charged: it is random thermal motion, not mutual repulsion, that drives the process.

Everyday example: open a perfume bottle in a corner of a room. Even in perfectly still air, random molecular motion spreads the scent everywhere in a short time.

A toy model you can compute by hand (Fig. 3.11). Take a 1-D box with four compartments and $1024$ particles. Every $\tau$ seconds every particle jumps to a neighboring compartment, left or right with equal probability (a jump into a wall is reflected back to the starting compartment). At $t=0$ all particles are in compartment 1.

Time	Compartments 1, 2, 3, 4	How it arises
$0$	1024, 0, 0, 0	start
$\tau$	512, 512, 0, 0	half of 1024 jump right; half try to jump left and are reflected back
$2\tau$	512, 256, 256, 0	from 1: 256 go to 2 and 256 reflect (stay in 1); from 2: 256 go to 3, 256 return to 1
$3\tau$	384, 384, 128, 128	repeating the halve-and-pass rule
$6\tau$	roughly 320, 320, 192, 192 (nearly uniform)	system approaches 256 each

The rule at each step: each compartment hands half its particles to each neighbor. Notice that even though every individual particle moves randomly with no preference, the crowd moves from crowded to empty: randomness plus a gradient produces a systematic flow. The reason is purely statistical: from a crowded compartment many particles jump out and few jump in.

In semiconductors the motion is 3-D and not compartmentalized, but the macroscopic result is the same. Since electrons and holes are charged, their diffusion constitutes a current (Fig. 3.12): both electrons and holes flow from high-concentration to low-concentration regions, but because their charges are opposite, the currents point opposite ways.

Exercise 3.3: “DiffDemo” MATLAB simulation

Problem. (a) Run the program below, which pseudo-animates a 1-D particle system like the one in the text. (b) Re-run with all particles initially in the middle box. (c) Experiment with initially putting particles in several boxes, with the pause time, and with more boxes.

Logic. y holds the number of particles in each of 5 boxes. At each time step, every box sends half of its particles to each neighbor; at the two end boxes, the half that would jump out of the system is reflected and returns to the same box. The new content of box $k$ is therefore half of the left neighbor plus half of the right neighbor, and for the walls the “missing neighbor” is replaced by the box itself. round keeps the counts integer. bar makes the histogram; pause slows the animation.

% Simulation of Diffusion (DiffDemo)
% One-dimensional system, right and left jumps equally probable
close
x = [0.5 1.5 2.5 3.5 4.5];
y = [1.0e6 0 0 0 0];          % initial condition: all particles in box 1
[xp,yp] = bar(x,y);
plot(xp,yp); text(0.5,1.1e6,'t=0');
axis([0 5 0 1.2e6]);
pause(0.5)
N = 15;                        % number of jumps; increase for a longer run
 
for ii = 1:N
    % Diffusion step: each box passes half to each neighbour
    bin(1) = round(y(1)/2 + y(2)/2);   % left wall reflects half of box 1
    bin(2) = round(y(1)/2 + y(3)/2);
    bin(3) = round(y(2)/2 + y(4)/2);
    bin(4) = round(y(3)/2 + y(5)/2);
    bin(5) = round(y(4)/2 + y(5)/2);   % right wall reflects half of box 5
    y = bin;
    [xp,yp] = bar(x,y);
    plot(xp,yp); text(0.5,1.1e6,['t=',num2str(ii)]);
    axis([0 5 0 1.2e6]);
    pause(0.5)
end

Part (b): change the initial condition to y = [0 0 1.0e6 0 0];. The pile in the middle box spreads symmetrically left and right, reaching the walls after two jumps, reflecting, and flattening out. Part (c): e.g. y = [5e5 0 0 0 5e5]; gives two piles moving inward and merging in the middle. In every experiment the final state is nearly uniform ($2\times 10^5$ per box for $10^6$ particles in 5 boxes), as in Fig. E3.3 (the $t=15$ panel).

3.2.2 Hot-point probe measurement

Before the diffusion current, a neat application. The hot-point probe quickly tells whether a semiconductor sample is n- or p-type (essential before any resistivity-to-doping conversion).

Equipment (Fig. 3.13a): a hot probe (a soldering iron will do), a cold probe (an ordinary multimeter probe), and a center-zero milliammeter between them. Touch both probes to the sample; the meter needle deflects to one side for n-type and to the other side for p-type.

Explanation (Fig. 3.13b). The hot probe creates a larger population of energetic carriers near its tip (mostly holes in p-type, electrons in n-type, since those are the majority carriers). Because carrier density is higher near the hot tip than elsewhere, diffusion spreads these carriers outward, away from the hot probe. The net effect:

• p-type: holes leave the hot-probe region, leaving net negative charge (ionized acceptors, unbalanced) there; the hot probe becomes negative with respect to the cold one.
• n-type: electrons leave, leaving net positive charge (ionized donors) at the hot probe.

So the current through the meter flows in opposite directions for n and p. It is a gorgeous everyday demonstration that thermal gradient $\to$ carrier gradient $\to$ diffusion current.

3.2.3 Diffusion and total currents

Diffusion current. Diffusion occurs only if more particles sit at one place than at another, i.e., there is a nonzero concentration gradient ($\nabla p\ne 0$ for holes, $\nabla n\ne 0$ for electrons). Quite naturally, the steeper the gradient, the larger the flow. Quantitative analysis confirms this and gives Fick’s law:

\mathbf{F} = -D\,\nabla\eta \tag{3.16}

where $\mathbf{F}$ is the flux (particles per cm${}^2$ per second crossing a plane perpendicular to the flow), $\eta$ is the particle concentration, and $D$ is the (positive) diffusion coefficient. The minus sign says “flow goes down the gradient”. Multiplying the flux by the carrier charge gives current density:

\boxed{\mathbf{J}_{P|diff} = -qD_P\,\nabla p} \tag{3.17a}

\boxed{\mathbf{J}_{N|diff} = +qD_N\,\nabla n} \tag{3.17b}

$D_P$ and $D_N$ (units cm${}^2$/s) are the hole and electron diffusion coefficients. Sign check against Fig. 3.12: for $dp/dx>0$ and $dn/dx>0$ both carriers flow toward $-x$. Holes moving toward $-x$ carry positive current toward $-x$: $J_P<0$ (matches $-q{D}_{P}dp/dx$). Electrons moving toward $-x$ carry current toward $+x$: $J_N>0$ (matches $+q{D}_{N}dn/dx$).

Total currents. Drift and diffusion act simultaneously. Summing (3.4) and (3.17):

\boxed{\mathbf{J}_P = \mathbf{J}_{P|drift} + \mathbf{J}_{P|diff} = q\mu_p\,p\,\boldsymbol{\mathcal{E}} - qD_P\,\nabla p} \tag{3.18a}

\boxed{\mathbf{J}_N = \mathbf{J}_{N|drift} + \mathbf{J}_{N|diff} = q\mu_n\,n\,\boldsymbol{\mathcal{E}} + qD_N\,\nabla n} \tag{3.18b}

\boxed{\mathbf{J} = \mathbf{J}_N + \mathbf{J}_P} \tag{3.19}

These “double-boxed” relations are used (directly or indirectly) in nearly every device analysis.

3.2.4 Relating diffusion coefficients and mobilities

The $D$‘s are the constants of the motion for diffusion, but we need not study them as extensively as the $\mu$‘s, since the two are tied by the Einstein relationship. Its derivation needs two preliminary facts about nonuniformly doped semiconductors at equilibrium.

Fact 1: the Fermi level is constant in equilibrium. Consider a nondegenerate n-type sample whose doping varies with position (Fig. 3.14a). Its equilibrium band diagram is as in Fig. 3.14(b): $E_F$ is closer to $E_C$ where $N_D$ is larger, and bands bend. The key rule:

$\frac{d{E}_F}{dx}=0(\text{also in }y,z)\text{in equilibrium}$

Why? Think of electrons hopping between states of equal energy at neighboring positions. The probability that a state of energy $E$ is filled, $f(E)$, must be the same everywhere; otherwise carriers would preferentially transfer from the better-filled to the emptier states and create a net current, contradicting equilibrium. Since $f(E)=1/[1+e^{(E-E_F)/kT}]$ (Eq. 2.7), a position-independent $f(E)$ requires a position-independent $E_F$. On an equilibrium diagram $E_F$ is a horizontal line.

Fact 2: equilibrium is a dynamic balance of drift and diffusion. In equilibrium, the total current is zero and the electron and hole systems are decoupled, so $J_N=0$ and $J_P=0$ separately. By (3.18) that does not mean drift and diffusion each vanish; it means they are equal in magnitude and opposite in direction. They vanish individually only for uniform doping: $\nabla n=\nabla p=0$ and $\mathcal{E}=0$.

In Fig. 3.14, $N_D$ increases with $x$, so $n$ has a gradient and an electron diffusion current flows in $+x$ (electrons diffuse toward $-x$). The band bending implies a built-in field pointing toward $-x$, causing electron drift current also in the direction of $\mathcal{E}$, i.e., $-x$. The two components have opposite polarity, and in equilibrium equal magnitude. In fact the built-in field exists in order to oppose the diffusive tendency caused by nonuniform doping.

Take-away: even at equilibrium, nonuniform doping produces carrier-concentration gradients, a built-in electric field, and nonzero (but cancelling) current components. This idea returns in the pn junction.

Derivation of the Einstein relationship. One dimension, nondegenerate, nonuniformly doped, equilibrium. Focus on electrons. Net electron current is zero:

J_N = J_{N|drift} + J_{N|diff} = q\mu_n n\mathcal{E} + qD_N\frac{dn}{dx} = 0 \tag{3.20}

Use the band-bending relation for the field,

\mathcal{E} = \frac{1}{q}\frac{dE_i}{dx} \tag{3.21}

(same as 3.15) and Boltzmann statistics for the electron density,

n = n_i\,e^{(E_F - E_i)/kT} \tag{3.22}

Differentiate (3.22) with respect to $x$, using $d{E}_F/dx=0$ (Fact 1):

$\frac{dn}{dx}=n_i{e}^{(E_F-E_i)/kT}\cdot \left(-\frac{1}{kT}\frac{d{E}_i}{dx}\right)$

\frac{dn}{dx} = -\frac{n}{kT}\frac{dE_i}{dx} \tag{3.23}

Substitute (3.21) and (3.23) into (3.20):

$q{\mu }_{n}n\left(\frac{1}{q}\frac{d{E}_i}{dx}\right)+q{D}_N\left(-\frac{n}{kT}\frac{d{E}_i}{dx}\right)=0$

Collect terms:

\left(n\,\frac{dE_i}{dx}\right)\left[\mu_n - \frac{q}{kT}D_N\right] = 0 \tag{3.24}

Because the doping is nonuniform, $nd{E}_i/dx\ne 0$, so the bracket must vanish:

\boxed{\frac{D_N}{\mu_n} = \frac{kT}{q}} \tag{3.25a}

The same argument for holes gives

\boxed{\frac{D_P}{\mu_p} = \frac{kT}{q}} \tag{3.25b}

Although derived in equilibrium, the relation holds also out of equilibrium (more elaborate arguments show it), as long as the material is nondegenerate (degenerate materials give slightly modified forms).

Numbers. $kT/q$ is a voltage; at 300 K it equals $0.0259$ V. For $N_D=10^{14}$ cm${}^{-3}$ Si at 300 K:

$D_N=\frac{kT}{q}{\mu }_n=(0.0259)(1358)=35.2{\text{cm}}^2/\text{s}$

Memory trick: ”$D$ over $\mu$ equals $kT$ over $q$”; it rhymes, and it rhymes upside down too (”$\mu$ over $D$ equals $q$ over $kT$”).

Physical meaning: mobility measures how readily carriers respond to a push (field), diffusion measures how readily they spread out on their own. Both are governed by the same scattering, hence the tie. And the proportionality constant is the thermal energy $kT$ because diffusion is thermal motion.

Exercise 3.4: Using the Exercise 3.2 diagram

Problem. Continue with the semiconductor of Exercise 3.2. (a) The semiconductor is in equilibrium; how do you tell from the diagram? (b) What are $J_N$ and $J_P$ at $x=\pm L/2$? (c) Roughly sketch $n$ and $p$ vs. $x$. (d) Is there an electron diffusion current at $x=\pm L/2$? If yes, in which direction? (e) Using the field found in Exercise 3.2, is there an electron drift current at $x=\pm L/2$? If yes, in which direction? (f) What is the hole diffusion coefficient $D_P$ in the $x>L$ region?

Solution.

(a) $E_F$ is invariant with position (a horizontal line). By Fact 1 this is the signature of equilibrium.

(b) In equilibrium, net currents are identically zero everywhere:

$J_N=0,J_P=0\text{at both }x=\pm L/2$

(c) Use $n=n_i{e}^{(E_F-E_i)/kT}$ and $p=n_i{e}^{(E_i-E_F)/kT}$ with $E_F$ flat. Where $E_i$ lies below $E_F$ (near $x=0$), $n>n_i>p$; where $E_i$ lies above $E_F$ ($|x|\ge L$), $p>n_i>n$. On a log scale, $n(x)$ is a hump centered at $x=0$ with maximum $n=n_i{e}^{0.28/0.0259}\approx 5\times 10^{14}$ cm${}^{-3}$, falling to $\approx n_i^2/N_A\approx 2\times 10^5$ cm${}^{-3}$ for $|x|>L$. $p(x)$ is the mirror image: a dip at $x=0$ and a plateau $\approx 5\times 10^{14}$ for $|x|>L$. The two curves cross at $n=p=n_i$ where $E_i=E_F$.

(d) Yes, because $dn/dx\ne 0$ at both points. From (c): $dn/dx>0$ at $x=-L/2$, so $J_{N|diff}=q{D}_{N}dn/dx>0$, flowing in $+x$. At $x=+L/2$, $dn/dx<0$, so $J_{N|diff}$ flows in $-x$.

(e) Yes. $n$ and $\mathcal{E}$ are both nonzero at $x=\pm L/2$, so $J_{N|drift}=q{\mu }_{n}n\mathcal{E}\ne 0$. It has the direction of $\mathcal{E}$: using the Exercise 3.2(b) field, it flows in $-x$ at $x=-L/2$ and in $+x$ at $x=+L/2$. These exactly cancel the diffusion components, as required by $J_N=0$:

	$x=-L/2$	$x=+L/2$
$J_{N	diff}$	$\to$
$J_{N	drift}$	$\leftarrow$

(f) In Exercise 3.2(d) the $x>L$ region had $N_A\approx 5\times 10^{14}$ cm${}^{-3}$ and ${\mu }_p=459$ cm${}^2$/V s. By the Einstein relationship:

$D_P=\frac{kT}{q}{\mu }_p=(0.0259)(459)=11.9{\text{cm}}^2/\text{s}$

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3.3 Recombination-Generation

3.3.1 Definition and visualization

When a semiconductor is pushed off equilibrium, an excess or deficit of carriers (relative to the equilibrium values) is created. Recombination-generation (R-G) is nature’s order-restoring mechanism: it stabilizes the excess or deficit (if the push continues) or eliminates it (if the push stops). Since devices work in nonequilibrium conditions, R-G often shapes device characteristics.

• Recombination: a process in which electrons and holes (carriers) are annihilated.
• Generation: a process in which electrons and holes are created.

Unlike drift and diffusion, “R-G” is not one process but a family. The most common members (Fig. 3.15):

(a) Band-to-band recombination. A conduction-band electron falls straight into a valence-band hole and both vanish (“zap!”). The released energy typically becomes a photon (light).

(b) R-G center (indirect) recombination. A third party is involved: an R-G center, a lattice defect or a special impurity (e.g., gold in Si). It provides an allowed level $E_T$ near midgap. The recombination is a two-step process:

1. An electron wanders near the center, is captured by its potential well, loses energy, and is trapped.
2. A hole arrives, is attracted to the trapped electron, loses energy, and annihilates it (equivalently the electron loses energy a second time and drops into a valence-band hole).

The energy is typically released as heat (lattice vibrations). The key property of R-G centers is the near-midgap level; that distinguishes them from donors and acceptors (whose levels sit close to a band edge). Fig. 3.16 lists midgap-ish levels from Au, Cu, Mn, Cr, Fe in Si. R-G centers are present even in the purest material, but at concentrations $N_T$ normally far below the dopant concentrations.

Analogy: a stepping-stone. Instead of leaping a wide river (the whole band gap) in one go, an electron hops onto a rock in the middle (the midgap level) and then onto the far bank. Two small hops are far more likely than one huge leap.

(c) Auger recombination. Band-to-band recombination happens simultaneously with a collision between two like carriers. The released energy goes to the surviving carrier, which then “thermalizes”, losing the energy in small steps by lattice collisions (the staircase in Fig. 3.15c). Its rate rises with carrier concentration (more collisions), so it matters in heavily doped regions.

Generation processes (the reverses of the above):

(d) Band-to-band generation: a valence electron is excited to the conduction band either by thermal energy (direct thermal generation) or by absorbing light (photogeneration).

(e) R-G-center generation: thermally assisted creation of carriers through the midgap level.

(f) Impact ionization: the inverse of Auger recombination. A highly energetic carrier collides with the lattice and creates an electron-hole pair. It occurs in high-field regions of devices and will matter for breakdown in pn junctions.

3.3.2 Momentum considerations

All these processes occur all the time, even in equilibrium; the question is the rate. Usually we need only the dominant process. Auger matters only at high doping; impact ionization only in high-field regions; photogeneration only under illumination. So which pair dominates in the low-field, non-degenerately doped, room-temperature, dark case? The candidates are thermal band-to-band R-G and R-G through centers.

The simple energy diagram suggests band-to-band should win, but the energy picture misleads, because crystal momentum must also be conserved in every R-G event. This requires $E$-$k$ diagrams (allowed electron energy vs. crystal momentum $k$, Fig. 3.17):

• Direct semiconductors (e.g., GaAs): the conduction-band minimum and valence-band maximum occur at the same $k$ (here $k=0$).
• Indirect semiconductors (e.g., Si, Ge): the conduction-band minimum is displaced to $k\ne 0$.

Two kinds of transition quanta matter:

• Photons have negligible momentum (but sizable energy $\sim E_G$): a photon-assisted transition is essentially vertical on the $E$-$k$ plot.
• Phonons (lattice-vibration quanta) have small energy (about 10-50 meV) but sizable momentum: a phonon-assisted transition is essentially horizontal.

Electrons and holes sit very near the band-edge extremes (Fig. 3.18), consistent with the carrier distributions studied before.

Hence:

• Direct material: electrons and holes sit at nearly the same $k$, so recombination needs almost no momentum change; emission of a photon conserves both energy and momentum. Band-to-band recombination is efficient. This is the process behind LEDs and junction lasers.
• Indirect material: recombination needs a large $k$ change. A photon alone cannot supply it, so a phonon must be emitted or absorbed simultaneously with the photon: a three-body event of low probability. Band-to-band recombination is therefore negligible compared with R-G-center recombination in Si and Ge.

Why silicon makes lousy LEDs (and GaAs makes good ones): the band gap in Si is indirect. Even in direct materials, the R-G-center route is often still the dominant recombination mechanism. Since R-G centers matter universally and most devices are Si (indirect), the quantitative treatment below focuses on R-G centers (indirect thermal R-G).

3.3.3 R-G statistics

“R-G statistics” is the mathematical description of R-G. Importantly, this is not a current-density relation: an R-G event occurs at a localized position and so does not by itself transport charge. Rather, R-G changes the carrier concentrations, and thereby indirectly affects current. What we need is the time rate of change $\partial n/\partial t$ or $\partial p/\partial t$.

Photogeneration

Light incident on a semiconductor (Fig. 3.19) is partly reflected and partly transmitted. For monochromatic light of wavelength $\lambda$ and frequency $\nu$, if the photon energy exceeds the gap, $h\nu >E_G$, it is absorbed and creates electron-hole pairs along its path. The intensity decays as

I = I_0\,e^{-\alpha x} \tag{3.26}

where $I_0$ is the intensity just inside the surface ($x=0^{+}$) and $\alpha$ is the absorption coefficient, a material- and wavelength-dependent quantity (Fig. 3.20). Since each absorbed photon makes one electron-hole pair (Fig. 3.15d), the generation rate has the same $e^{-\alpha x}$ dependence, and electrons and holes are created in equal numbers:

\left.\frac{\partial n}{\partial t}\right|_{light} = \left.\frac{\partial p}{\partial t}\right|_{light} = G_L(x,\lambda) \tag{3.27}

G_L(x,\lambda) = G_{L0}\,e^{-\alpha x} \tag{3.28}

$G_L$ is the photogeneration rate (number per cm${}^3$ per second). The subscript “light” is needed because $n$ and $p$ can be changed by several processes at once.

Two common simplifying assumptions: (i) uniform illumination throughout the sample; (ii) all the light absorbed in an infinitesimally thin surface layer. The characteristic depth is $1/\alpha$ (the average penetration depth).

• When $h\nu \approx E_G$, $\alpha$ is small and $1/\alpha$ is huge (about 1 cm in Si), so $e^{-\alpha x}\approx 1$ over any practical depth: $G_L\approx$ constant, uniform absorption.
• At $\lambda =0.4\mu$m in Si, $\alpha \approx 10^5$ cm${}^{-1}$ (from Fig. 3.20), so the penetration depth is $1/\alpha \sim 10^{-5}$ cm $=0.1\mu$m, a few thousand angstroms at most: surface absorption.

Indirect thermal recombination-generation

The general R-G-center statistics are beyond our scope, but we can motivate the special-case result most problems use. Define:

Symbol	Meaning
$n_0,p_0$	carrier concentrations in equilibrium
$n,p$	carrier concentrations under arbitrary conditions
$\Delta n=n-n_0,\Delta p=p-p_0$	deviations from equilibrium (positive = excess, negative = deficit)
$N_T$	number of R-G centers per cm${}^3$

Low-level injection is required for the special case. It means the perturbation is small relative to the majority carrier density:

$\Delta p\ll n_0,n\approx n_0\text{(n-type)}$

$\Delta n\ll p_0,p\approx p_0\text{(p-type)}$

Example: $N_D=10^{15}$ cm${}^{-3}$ Si at room temperature, perturbation $\Delta p=\Delta n=10^{10}$ cm${}^{-3}$. Then $n_0=N_D=10^{15}$ and $p_0\approx n_i^2/N_D=10^5$ cm${}^{-3}$. So $n=n_0+\Delta n\approx n_0$ and $\Delta p=10^{10}\ll n_0$: low-level injection. But look at the minority carriers: $p=p_0+\Delta p\approx 10^{10}\gg p_0=10^5$! Under low-level injection the majority concentration is essentially unperturbed, while the minority concentration can change by many orders of magnitude.

Heuristic derivation of the rate. Consider (Fig. 3.21) an n-type sample with an excess of holes $\Delta p\ll n_0$ relaxing toward equilibrium via R-G centers. What controls the recombination rate $\partial p/\partial t{|}_R$?

• To destroy a hole, it must go from the valence band to an electron-filled R-G center. More filled centers $\Rightarrow$ faster recombination. In equilibrium $E_F$ is comfortably above $E_T$, so essentially all centers are filled; with $\Delta p\ll n_0$, electrons vastly outnumber holes and instantly refill any vacated center. So the number of filled centers stays $\approx N_T$: the rate is proportional to $N_T$.
• More holes available means more holes annihilated per second: the rate is proportional to $p$.

With a positive proportionality constant $c_p$ (and a minus sign since $p$ decreases):

\left.\frac{\partial p}{\partial t}\right|_R = -c_p\,N_T\,p \tag{3.29}

Generation depends only on the number of empty R-G centers, a small number held at its equilibrium value, so the generation rate is a constant. In equilibrium generation and recombination must balance exactly (the Principle of Detailed Balance: each process and its inverse balance independently of other processes):

\left.\frac{\partial p}{\partial t}\right|_G = +c_p\,N_T\,p_0 \tag{3.30}

Net rate:

\left.\frac{\partial p}{\partial t}\right|_{R-G} = \left.\frac{\partial p}{\partial t}\right|_R + \left.\frac{\partial p}{\partial t}\right|_G = -c_pN_T\,(p - p_0) \tag{3.31}

or

\left.\frac{\partial p}{\partial t}\right|_{\text{thermal R-G}} = -c_pN_T\,\Delta p\qquad\text{(holes in n-type)} \tag{3.32a}

and analogously

\left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}} = -c_nN_T\,\Delta n\qquad\text{(electrons in p-type)} \tag{3.32b}

with $c_n$ and $c_p$ the electron and hole capture coefficients.

Compact form. The products $c_p{N}_T$ and $c_n{N}_T$ have units of 1/time (since $\partial p/\partial t$ is concentration/time and $\Delta p$ is concentration). Define

\tau_p \equiv \frac{1}{c_pN_T} \tag{3.33a}

\tau_n \equiv \frac{1}{c_nN_T} \tag{3.33b}

so

\boxed{\left.\frac{\partial p_n}{\partial t}\right|_{\text{thermal R-G}} = -\frac{\Delta p_n}{\tau_p}\qquad\text{(holes in n-type)}} \tag{3.34a}

\boxed{\left.\frac{\partial n_p}{\partial t}\right|_{\text{thermal R-G}} = -\frac{\Delta n_p}{\tau_n}\qquad\text{(electrons in p-type)}} \tag{3.34b}

(the subscripts $n$ or $p$ on the carrier symbol remind us we are dealing with the minority carrier in that region).

Notes:

• Derived for steady or slowly varying conditions, but works well for nearly all transients.
• $\Delta p<0$ is allowed and gives $\partial p/\partial t>0$: a carrier deficit makes generation outpace recombination, restoring equilibrium. The sign convention thus describes the net of recombination and generation.
• Equations (3.34) apply only to minority carriers under low-level injection.

General steady-state result (valid for any injection level, both carriers, nondegenerate semiconductor):

\left.\frac{\partial p}{\partial t}\right|_{\text{thermal R-G}} = \left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}} = -\frac{np - n_i^2}{\tau_p(n+n_1)+\tau_n(p+p_1)} \tag{3.35}

with

n_1 \equiv n_i\,e^{(E_T-E_i)/kT} \tag{3.36a}

p_1 \equiv n_i\,e^{(E_i-E_T)/kT} \tag{3.36b}

(Problem 3.15 asks you to show that (3.35) reduces to (3.34); a complete derivation is given there.)

3.3.4 Minority carrier lifetimes

General information. The constants ${\tau }_n$ and ${\tau }_p$ are the “constants of the motion” for R-G. Return to Fig. 3.21: the excess holes do not all disappear simultaneously; some live briefly, some for a long time. If thermal R-G is the only relaxation process, the average excess-hole lifetime works out to $⟨t⟩={\tau }_p$ (or ${\tau }_n$ for electrons). So ${\tau }_n$ and ${\tau }_p$ are the minority carrier lifetimes: the average time an excess minority carrier survives in a sea of majority carriers.

Unlike the $\mu$‘s and $D$‘s there are no handy plots of $\tau$ vs. doping, because $\tau$ depends on the poorly controlled R-G center concentration $N_T$ (3.33), not on the carefully controlled dopant concentration. Even within one sample, the nature and concentration of dominant R-G centers change during fabrication.

• Gettering (a processing step that sweeps out unwanted impurities) can reduce $N_T$ to a very low level and give ${\tau }_{n(p)}\sim 1$ ms in Si.
• Intentional gold doping raises $N_T$ and gives ${\tau }_{n(p)}\sim 1$ ns.
• Lifetimes in finished Si devices lie roughly midway between those extremes.

A lifetime measurement: photoconductivity decay. Most lifetime measurements use a device structure and fit theory to the device characteristics. The one described here uses a bar of semiconductor with ohmic contacts at both ends (a photoconductor). A light pulse creates excess carriers (a situation like Fig. 3.21) and a measurable conductivity increase; after the light stops, recombination makes the conductivity decay back, and the decay rate gives the lifetime.

Circuit (Fig. 3.22): sample resistance $R_s$ in series with load resistor $R_L$ and a dc supply $V_A$; the output $v_L$ is measured by a scope. $R_s$ varies with time, so $v_L$ tracks the conductivity changes.

For an n-type sample with uniform photogeneration, the excess minority holes decay after the light is switched off as (derived later, Section 3.5, and in Sample Problem 1)

\Delta p(t) = \Delta p_0\,e^{-t/\tau_p} \tag{3.37}

($\Delta p_0$ = excess at $t=0$ when light is switched off). The conductivity is

\sigma = \frac{1}{\rho} = q(\mu_n n + \mu_p p) \tag{3.38a}

= q\left[\mu_n(n_0+\Delta n) + \mu_p(p_0+\Delta p)\right] \tag{3.38b}

With $\Delta n=\Delta p$ and, in n-type material, $n_0\approx N_D\gg p_0$:

\sigma = \underbrace{q\mu_nN_D}_{\sigma_{dark}} + \underbrace{q(\mu_n+\mu_p)\,\Delta p}_{\Delta\sigma(t)} \tag{3.39}

where ${\sigma }_{dark}$ is the dark conductivity and $\Delta \sigma (t)\propto \Delta p(t)$ is the time-decaying part. To link $v_L$ to $\Delta \sigma$, use the voltage divider:

v_L = i\,R_L = V_A\,\frac{R_L}{R_s + R_L} \tag{3.40}

with the sample resistance (length $\ell$, area $A$)

R_s = \frac{\rho\ell}{A} = \frac{\ell}{\sigma A} = \frac{\ell}{(\sigma_{dark}+\Delta\sigma)A} = \frac{R_{sd}}{1+\Delta\sigma/\sigma_{dark}} \tag{3.41}

where $R_{sd}=\ell /({\sigma }_{dark}A)$ is the dark resistance. Substituting (3.41) into (3.40) and rearranging:

v_L = \frac{V_A}{1 + \dfrac{R_{sd}}{R_L}\cdot\dfrac{1}{1+\Delta\sigma/\sigma_{dark}}} \tag{3.42}

If the load is matched, $R_L=R_{sd}$, and the light is weak so $\Delta \sigma /{\sigma }_{dark}\ll 1$ (equivalent to low-level injection), expand:

v_L\approx \frac{V_A}{2} + \frac{V_A}{4}\,\frac{\Delta\sigma}{\sigma_{dark}} = \underbrace{\frac{V_A}{2}}_{\text{d.c.}} + \underbrace{v_0\,e^{-t/\tau_p}}_{\text{decay transient}} \tag{3.43}

(The step: with $R_L=R_{sd}$, $v_L=V_A/[1+1/(1+x)]=V_A(1+x)/(2+x)\approx \frac{V_A}{2}(1+x)(1-x/2)\approx \frac{V_A}{2}(1+x/2)=\frac{V_A}{2}+\frac{V_A}{4}x$ with $x=\Delta \sigma /{\sigma }_{dark}$.) An exponential decay with time constant ${\tau }_p$ is observed.

Implementation (Fig. 3.23). A dc supply provides $V_A$; a digitizing oscilloscope (Tektronix 11401) captures the decay; four bar-shaped Si samples sit in a box with the $R_L=R_{sd}$ load resistor. A stroboscope supplies $\sim 1\mu$s, high-intensity flashes spaced from 2.5 ms to 0.5 s apart. Light is first passed through a small Si wafer, which acts as a filter passing only near-bandgap radiation (weakly absorbed, so photogeneration is nearly uniform through the bar).

Result (Fig. 3.24). The transient component of $v_L$ (mV) vs. time (ms): the linear plot shows the exponential shape, the semilog plot shows a straight line (at least for $0.1\text{ms}\le t\le 0.6\text{ms}$), whose slope gives $-1/{\tau }_p$. For the probed sample, ${\tau }_p\approx 150\mu$s.

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3.4 Equations of State

Real semiconductors do all the carrier actions at once. To find the state of a semiconductor we must combine the individual actions. This “putting it all together” yields the basic starting equations of device analysis, the equations of state.

3.4.1 Continuity equations

Every carrier action (drift, diffusion, thermal or direct R-G, light generation, …) changes the carrier concentration with time. The total rate of change is the sum of the contributions of the individual processes:

\frac{\partial n}{\partial t} = \left.\frac{\partial n}{\partial t}\right|_{drift} + \left.\frac{\partial n}{\partial t}\right|_{diff} + \left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}} + \left.\frac{\partial n}{\partial t}\right|_{\text{other (light, etc.)}} \tag{3.44a}

\frac{\partial p}{\partial t} = \left.\frac{\partial p}{\partial t}\right|_{drift} + \left.\frac{\partial p}{\partial t}\right|_{diff} + \left.\frac{\partial p}{\partial t}\right|_{\text{thermal R-G}} + \left.\frac{\partial p}{\partial t}\right|_{\text{other (light, etc.)}} \tag{3.44b}

This is conservation of carriers: electrons and holes cannot appear or disappear at a point mysteriously; they must arrive by transport or be created there. Hence “continuity” equations.

Compact form of the transport terms. Drift plus diffusion changes the local concentration only if more current flows in than out of a small volume:

\left.\frac{\partial n}{\partial t}\right|_{drift} + \left.\frac{\partial n}{\partial t}\right|_{diff} = \frac{1}{q}\left(\frac{\partial J_{Nx}}{\partial x}+\frac{\partial J_{Ny}}{\partial y}+\frac{\partial J_{Nz}}{\partial z}\right) = \frac{1}{q}\nabla\cdot\mathbf{J}_N \tag{3.45a}

\left.\frac{\partial p}{\partial t}\right|_{drift} + \left.\frac{\partial p}{\partial t}\right|_{diff} = -\frac{1}{q}\nabla\cdot\mathbf{J}_P \tag{3.45b}

(The signs differ because the electron current density ${\mathbf{J}}_N$ counts negative charges flowing; a divergence of electron current means electron charge is leaving, i.e., electrons are added per $\frac{1}{q}\nabla \cdot {\mathbf{J}}_N$. For holes, a positive divergence of ${\mathbf{J}}_P$ means holes are leaving.)

Bank-account analogy. The balance in a small region (the number of carriers) changes by (deposits in) minus (withdrawals out) plus (interest created inside) minus (losses inside): inflow minus outflow of current, plus generation, minus recombination.

Substituting (3.45) into (3.44):

\boxed{\frac{\partial n}{\partial t} = \frac{1}{q}\nabla\cdot\mathbf{J}_N + \left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}} + \left.\frac{\partial n}{\partial t}\right|_{\text{other processes}}} \tag{3.46a}

\boxed{\frac{\partial p}{\partial t} = -\frac{1}{q}\nabla\cdot\mathbf{J}_P + \left.\frac{\partial p}{\partial t}\right|_{\text{thermal R-G}} + \left.\frac{\partial p}{\partial t}\right|_{\text{other processes}}} \tag{3.46b}

These are completely general and are the usual starting point of device analysis. In computer simulation they are used directly: the appropriate R-G expression (e.g. (3.34) or the general (3.35)) and any “other process” terms are inserted and $n(x,y,z,t)$, $p(x,y,z,t)$ are solved numerically. For closed-form solutions we use a simplified version, derived next.

3.4.2 Minority carrier diffusion equations

The workhorse equations come from (3.46) under these simplifying assumptions:

1. The system is one-dimensional (everything depends on $x$ only).
2. We restrict attention to minority carriers.
3. $\mathcal{E}\approx 0$ in the region analyzed.
4. The equilibrium minority concentrations do not depend on position: $n_0\ne n_0(x)$, $p_0\ne p_0(x)$.
5. Low-level injection.
6. Indirect thermal R-G dominates the thermal R-G.
7. There are no other processes except possibly photogeneration.

Derivation for electrons (minority carriers in p-type material).

Transport term. In one dimension

\frac{1}{q}\nabla\cdot\mathbf{J}_N\to\frac{1}{q}\frac{\partial J_N}{\partial x} \tag{3.47}

The electron current is

J_N = q\mu_nn\mathcal{E} + qD_N\frac{\partial n}{\partial x}\ \to\ qD_N\frac{\partial n}{\partial x} \tag{3.48}

(We drop drift because $\mathcal{E}\approx 0$ and the minority concentration $n$ is small, so the product $n\mathcal{E}$ is doubly small. We could not do this for majority carriers.) Since $n=n_0+\Delta n$ and $n_0\ne n_0(x)$:

\frac{\partial n}{\partial x} = \frac{\partial n_0}{\partial x}+\frac{\partial\Delta n}{\partial x} = \frac{\partial\Delta n}{\partial x} \tag{3.49}

Combine (3.47)-(3.49):

\frac{1}{q}\nabla\cdot\mathbf{J}_N\to D_N\frac{\partial^2\Delta n}{\partial x^2} \tag{3.50}

R-G term. Indirect thermal R-G dominance plus low-level injection and minority carriers allow use of (3.34):

\left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}} = -\frac{\Delta n}{\tau_n} \tag{3.51}

Other processes. Only light is allowed:

\left.\frac{\partial n}{\partial t}\right|_{\text{other processes}} = G_L \tag{3.52}

($G_L=0$ if the sample is dark.)

Left-hand side. The equilibrium concentration is constant in time, $\partial n_0/\partial t=0$, so

\frac{\partial n}{\partial t} = \frac{\partial n_0}{\partial t} + \frac{\partial\Delta n}{\partial t} = \frac{\partial\Delta n}{\partial t} \tag{3.53}

Putting (3.50)-(3.53) into (3.46a) and treating holes analogously:

\boxed{\frac{\partial\Delta n_p}{\partial t} = D_N\frac{\partial^2\Delta n_p}{\partial x^2} - \frac{\Delta n_p}{\tau_n} + G_L} \tag{3.54a}

\boxed{\frac{\partial\Delta p_n}{\partial t} = D_P\frac{\partial^2\Delta p_n}{\partial x^2} - \frac{\Delta p_n}{\tau_p} + G_L} \tag{3.54b}

The subscripts remind us that these hold only for minority carriers: electrons in p-type, holes in n-type.

Reading the equation as a story (for holes): the excess-hole concentration at a point rises with the curvature of the profile (diffusion refills hollows and flattens peaks), falls because holes recombine (rate $\Delta p/{\tau }_p$), and rises because light makes new ones ($G_L$).

3.4.3 Simplifications and solutions

Further simplifications often reduce the diffusion equation drastically (Table 3.1).

Table 3.1: Common diffusion-equation simplifications

Simplification	Effect
Steady state	$\partial \Delta n_p/\partial t\to 0$ (and $\partial \Delta p_n/\partial t\to 0$)
No concentration gradient (no diffusion current)	$D_N{\partial }^2\Delta n_p/\partial x^2\to 0$ (and $D_P{\partial }^2\Delta p_n/\partial x^2\to 0$)
No drift current or $\mathcal{E}=0$	No further simplification ($\mathcal{E}\approx 0$ was assumed in the derivation)
No thermal R-G	$\Delta n_p/{\tau }_n\to 0$ (and $\Delta p_n/{\tau }_p\to 0$)
No light	$G_L\to 0$

Table 3.2: Common special-case solutions (written for electrons; replace $n\to p$, $N\to P$ for holes)

No.	Given	Simplified equation	Solution
1	Steady state, no light	$0=D_N\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n}$	$\Delta n_p(x)=A{e}^{-x/L_N}+B{e}^{x/L_N}$, with $L_N=\sqrt{D_N{\tau }_n}$; $A$, $B$ constants
2	No concentration gradient, no light	$\frac{d\Delta n_p}{dt}=-\frac{\Delta n_p}{{\tau }_n}$	$\Delta n_p(t)=\Delta n_p(0)e^{-t/{\tau }_n}$
3	Steady state, no concentration gradient	$0=-\frac{\Delta n_p}{{\tau }_n}+G_L$	$\Delta n_p=G_L{\tau }_n$
4	Steady state, no R-G, no light	$0=D_N\frac{d^2\Delta n_p}{d{x}^2}$	$\Delta n_p(x)=A+Bx$

(Solution 1 follows by trying $e^{\lambda x}$: ${\lambda }^2=1/(D_N{\tau }_n)$. Solution 2 is separable. Solution 3 is algebraic. Solution 4 integrates twice.)

3.4.4 Problem solving

The book proposes a five-step recipe, illustrated on two sample problems:

1. Review exactly what is given or implied.
2. Characterize the system in equilibrium.
3. Analyze the problem qualitatively (predict the shape before computing).
4. Perform the quantitative analysis: check the simplifying assumptions, write the minority carrier diffusion equation, simplify, solve with boundary conditions.
5. Examine the solution (units, limits, compare with step 3). (“Failing to examine the solution is like growing vegetables and failing to eat the produce.”)

Sample Problem No. 1: light switched on at $t=0$

Problem. A uniformly donor-doped silicon wafer at room temperature is suddenly illuminated at $t=0$. Take $N_D=10^{15}$ cm${}^{-3}$, ${\tau }_p=10^{-6}$ s, and a light-induced creation of $10^{17}$ electron-hole pairs per cm${}^3$ per second throughout the semiconductor. Determine $\Delta p_n(t)$ for $t>0$.

Step 1. Silicon, 300 K, uniform $N_D=10^{15}$ cm${}^{-3}$, $G_L=10^{17}$ cm${}^{-3}$s${}^{-1}$ everywhere. Equilibrium holds for $t<0$.

Step 2 (equilibrium). With $n_i=10^{10}$ cm${}^{-3}$ and $N_D\gg n_i$:

$n_0=N_D=10^{15}{\text{cm}}^{-3},p_0=\frac{n_i^2}{N_D}=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}$

Uniform doping means $n_0$ and $p_0$ are the same everywhere.

Step 3 (qualitative). Before $t=0$, $\Delta p_n=0$. After $t=0$, light adds carriers, so $\Delta p_n$ grows. But a bigger $\Delta p_n$ means a faster recombination rate ($\propto \Delta p_n$). Eventually recombination eats exactly what light adds, and steady state results. Expect: $\Delta p_n(t)$ starts at zero, rises at a decreasing rate, saturates. At saturation generation = recombination:

$G_L=\frac{\Delta p_n(\infty )}{{\tau }_p}\Rightarrow \Delta p_{n,max}=G_L{\tau }_p$

(provided low-level injection holds).

Step 4 (quantitative). All seven simplifying assumptions behind the diffusion equation hold: 1-D (really uniform), only minority carriers, $n_0$, $p_0$ position independent, indirect thermal R-G dominant in Si, only photogeneration. Because generation is uniform, the perturbed concentrations are position-independent, so no gradients and $\mathcal{E}=0$. Low-level injection is verified below. So start from

\frac{\partial\Delta p_n}{\partial t} = D_P\frac{\partial^2\Delta p_n}{\partial x^2} - \frac{\Delta p_n}{\tau_p} + G_L \tag{3.55}

with initial condition

\Delta p_n(t)\big|_{t=0} = 0 \tag{3.56}

Since $\Delta p_n$ does not depend on $x$, the curvature term vanishes and (3.55) becomes an ordinary differential equation:

\frac{d\Delta p_n}{dt} + \frac{\Delta p_n}{\tau_p} = G_L \tag{3.57}

Solve it. Particular solution (steady value): $\Delta p_n=G_L{\tau }_p$. Homogeneous solution: $A{e}^{-t/{\tau }_p}$. So

\Delta p_n(t) = G_L\tau_p + A\,e^{-t/\tau_p} \tag{3.58}

Apply $\Delta p_n(0)=0$:

0 = G_L\tau_p + A\ \Rightarrow\ A = -G_L\tau_p \tag{3.59}

\boxed{\Delta p_n(t) = G_L\tau_p\left(1 - e^{-t/\tau_p}\right)} \tag{3.60}

Step 5 (examine). $G_L{\tau }_p$ has the dimensions (cm${}^{-3}$s${}^{-1}$)(s) = cm${}^{-3}$, a concentration, so the answer is dimensionally correct. At $t=0$: 0. As $t\to \infty$: $G_L{\tau }_p$. After one lifetime ($t={\tau }_p$) the excess is $63.2\%$ of its final value; after $3{\tau }_p$ it is $95\%$ (Fig. 3.25). Numbers:

$\Delta p_{n,max}=(10^{17})(10^{-6})=10^{11}{\text{cm}}^{-3}\ll n_0=10^{15}{\text{cm}}^{-3}(\text{low-level injection confirmed})$

so $p_{max}=p_0+10^{11}\approx 10^{11}$ cm${}^{-3}$, i.e., $10^6$ times its equilibrium value.

Epilogue: link to the photoconductive decay measurement. The stroboscope output can be modeled as a pulse train (Fig. 3.26a). (3.60) describes the carrier build-up during a pulse. However the flash lasts $\sim 1\mu$s while ${\tau }_p\lesssim 150\mu$s, so $t_{on}/{\tau }_p\ll 1$ and we see only the very first part of the rising exponential before the light goes off. For the light-off transient: set $G_L=0$ and use $\Delta p_n(0)=$ (value at the end of the light-on period), which is Solution No. 2 of Table 3.2:

\Delta p_n(t) = \Delta p_n(0)\,e^{-t/\tau_p} \tag{3.61}

(with $t=0$ reset to the instant the light turns off), which is exactly Eq. (3.37). The combined on-off response is sketched in Fig. 3.26(b).

Sample Problem No. 2: non-penetrating light on a semi-infinite bar

Problem. The $x=0$ end of a uniformly doped semi-infinite Si bar with $N_D=10^{15}$ cm${}^{-3}$ is illuminated so as to create $\Delta p_{n0}=10^{10}$ cm${}^{-3}$ excess holes at $x=0$ (Fig. 3.27a). The wavelength is such that no light penetrates into the interior ($x>0$). Determine $\Delta p_n(x)$.

Step 1 and 2. Silicon, steady state (we are asked for $\Delta p_n(x)$ not $\Delta p_n(x,t)$), $N_D=10^{15}$, so $n_0=10^{15}$, $p_0=10^5$ cm${}^{-3}$ uniform. Boundary conditions: $\Delta p_n(0)=\Delta p_{n0}=10^{10}$ cm${}^{-3}$ and $\Delta p_n\to 0$ as $x\to \infty$ (semi-infinite bar: a surface perturbation cannot reach infinity). Non-penetrating light means $G_L=0$ for $x>0$. The temperature is not stated, so assume the usual 300 K.

Step 3 (qualitative). Light creates carriers right at $x=0$. With more carriers at one point than elsewhere, diffusion spreads the excess into the bar. As the holes diffuse, they also recombine (rate $\propto \Delta p_n$) and each lives only ${\tau }_p$ on average, so fewer and fewer survive deeper into the bar. Expect a profile decreasing monotonically from $\Delta p_{n0}$ at $x=0$ to zero as $x\to \infty$.

Step 4 (quantitative). Check the assumptions: 1-D; minority holes only; $p_0$, $n_0$ position independent; indirect thermal R-G dominates; no other process for $x>0$; low-level injection since $10^{10}\ll 10^{15}$. The only question is whether $\mathcal{E}\approx 0$: the hole pile-up makes a small net positive charge, but it is tiny ($\Delta p_{n0}\ll n_0$) and, in practice, the majority electrons redistribute to cancel much of it. Experience shows $\mathcal{E}\approx 0$ to be reasonable.

Steady state with $G_L=0$ for $x>0$:

D_P\frac{d^2\Delta p_n}{dx^2} - \frac{\Delta p_n}{\tau_p} = 0\qquad x>0 \tag{3.62}

Boundary conditions:

\Delta p_n\big|_{x=0^+} = \Delta p_n\big|_{x=0} = \Delta p_{n0} \tag{3.63}

\Delta p_n\big|_{x\to\infty} = 0 \tag{3.64}

This is Solution No. 1 of Table 3.2:

\Delta p_n(x) = Ae^{-x/L_P}+Be^{x/L_P} \tag{3.65}

L_P = \sqrt{D_P\tau_p} \tag{3.66}

Since $e^{x/L_P}\to \infty$ as $x\to \infty$, (3.64) can be satisfied only if $B=0$. Then (3.63) gives

A = \Delta p_{n0} \tag{3.67}

\boxed{\Delta p_n(x) = \Delta p_{n0}\,e^{-x/L_P}} \tag{3.68}

Step 5. Matches the qualitative picture: a monotone exponential decay from $\Delta p_{n0}$ to zero, with a characteristic decay length $L_P$ (Fig. 3.27b).

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3.5 Supplemental Concepts

3.5.1 Diffusion lengths

The situation of Sample Problem 2 (excess minority carriers appear on a plane, diffuse away, and decay exponentially with a characteristic length) is so common that the characteristic length has a name:

\boxed{L_P \equiv \sqrt{D_P\tau_p}\quad\text{(holes in n-type)}} \tag{3.69a}

\boxed{L_N \equiv \sqrt{D_N\tau_n}\quad\text{(electrons in p-type)}} \tag{3.69b}

called the minority carrier diffusion lengths. They represent the average distance a minority carrier diffuses into a sea of majority carriers before being annihilated. Check with Sample Problem 2: the mean position of the excess holes is

\langle x\rangle = \frac{\int_0^\infty x\,\Delta p_n(x)\,dx}{\int_0^\infty\Delta p_n(x)\,dx} = \frac{\int_0^\infty x\,e^{-x/L_P}dx}{\int_0^\infty e^{-x/L_P}dx} = \frac{L_P^2}{L_P} = L_P \tag{3.70}

(using ${\int }_0^{\infty }x{e}^{-x/L}dx=L^2$ and ${\int }_0^{\infty }{e}^{-x/L}dx=L$).

Piranha analogy. A small group of animals tries to cross a piranha-infested stretch of the Amazon. Each animal is “eaten” at random with a fixed chance per meter of swimming. The average distance covered before being eaten is the diffusion length. Longer lifetime (fewer piranhas) and faster diffusion (stronger swimmers) both lengthen the distance, hence $L=\sqrt{D\tau }$.

Size. For $T=300$ K, $N_D=10^{15}$ cm${}^{-3}$ Si (${\mu }_p=458$ cm${}^2$/V s) and ${\tau }_p=10^{-6}$ s:

$L_P=\sqrt{D_P{\tau }_p}=\sqrt{\frac{kT}{q}{\mu }_p{\tau }_p}={\left[(0.0259)(458)(10^{-6})\right]}^{1/2}=3.44\times 10^{-3}\text{cm}$

(about $34\mu$m). Because lifetimes vary over orders of magnitude, diffusion lengths do too.

3.5.2 Quasi-Fermi levels

Why we need them. In equilibrium the Fermi level $E_F$ nails down both carrier concentrations:

n_0 = n_i\,e^{(E_F-E_i)/kT} \tag{3.71a}

p_0 = n_i\,e^{(E_i-E_F)/kT} \tag{3.71b}

so $E_F$ and ($n_0,p_0$) carry the same information, and the band diagram lets us read off the concentrations at a glance. But in a nonequilibrium state the Fermi level is not even defined, and one number cannot describe two independently changed concentrations.

Example (Sample Problem 1 at $t\gg {\tau }_p$). Before the light: $n_0=N_D=10^{15}$, $p_0=10^5$ cm${}^{-3}$ (equilibrium diagram, Fig. 3.28a). After the steady state is reached: $\Delta n=\Delta p=G_L{\tau }_p=10^{11}$ cm${}^{-3}$, $p=p_0+\Delta p\approx 10^{11}$ cm${}^{-3}$ and $n\approx n_0$. The electron concentration is essentially unperturbed, but $p$ is up by a factor $10^6$. The equilibrium diagram no longer describes this.

Definition. Introduce two energies: $F_N$ (electron quasi-Fermi level) and $F_P$ (hole quasi-Fermi level), defined to relate to the nonequilibrium concentrations the same way $E_F$ relates to the equilibrium ones (nondegenerate case):

n \equiv n_i\,e^{(F_N-E_i)/kT}\qquad\text{or}\qquad F_N = E_i + kT\ln\!\left(\frac{n}{n_i}\right) \tag{3.72a}

p \equiv n_i\,e^{(E_i-F_P)/kT}\qquad\text{or}\qquad F_P = E_i - kT\ln\!\left(\frac{p}{n_i}\right) \tag{3.72b}

$F_N$ and $F_P$ are conceptual constructs: their values are determined entirely from knowledge of $n$ and $p$. They are defined so that when a perturbed system relaxes to equilibrium, $F_N\to E_F$, $F_P\to E_F$, and (3.72) $\to$ (3.71).

Application to Sample Problem 1 at $t\gg {\tau }_p$. Since $n\approx n_0$: $F_N\approx E_F$. For holes, with $p=10^{11}$ cm${}^{-3}$, $n_i=10^{10}$, $kT=0.0259$ eV:

$F_P=E_i-kT\ln \frac{p}{n_i}=E_i-0.0259\ln (10)=E_i-0.060\text{eV}$

Drawing $F_N$ and $F_P$ on the diagram in place of $E_F$ (Fig. 3.28b) shows at a glance (i) the system is not in equilibrium ($F_N\ne F_P$), and (ii) it is in low-level injection: $F_N$ barely moved from the equilibrium $E_F$ while $F_P$ moved from far below $E_i$ (about $0.30$ eV below $E_F$… i.e., $E_F-E_i=0.298$ eV, so $E_F$ is $0.30$ eV above $E_i$ and $F_P$ is $0.06$ eV below $E_i$) up toward the mid-gap.

Compact current expressions. Start from the hole current:

J_P = q\mu_pp\,\mathcal{E} - qD_P\nabla p \tag{3.73}

(same as 3.18a). Differentiate (3.72b):

\nabla p = \frac{n_i}{kT}\,e^{(E_i-F_P)/kT}\left(\nabla E_i - \nabla F_P\right) \tag{3.74a}

= \frac{p}{kT}\left(q\boldsymbol{\mathcal{E}} - \nabla F_P\right) \tag{3.74b}

where we used $\mathcal{E}=\nabla E_i/q$ (3-D version of (3.15)). Eliminate $\nabla p$ from (3.73) with (3.74b):

$J_P=q{\mu }_{p}p\mathcal{E}-q{D}_P\frac{p}{kT}\left(q\mathcal{E}-\nabla F_P\right)$

Collect the terms in $\mathcal{E}$ and $\nabla F_P$:

J_P = q p\left(\mu_p - \frac{qD_P}{kT}\right)\mathcal{E} + \mu_p\!\left(\frac{qD_P}{kT\mu_p}\right)p\,\nabla F_P\ ... \tag{3.75}

More cleanly: $J_P=qp\mathcal{E}\left({\mu }_p-\frac{q{D}_P}{kT}\right)+\frac{q{D}_{P}p}{kT}\nabla F_P$. Now invoke the Einstein relationship $q{D}_P/kT={\mu }_p$: the first bracket vanishes and the coefficient of $\nabla F_P$ is ${\mu }_{p}p$. (Sign check: for holes, $\nabla F_P>0$ should drive current down the hole-energy hill, i.e. in the sense given below.) The result:

\boxed{\mathbf{J}_P = \mu_p\,p\,\nabla F_P} \tag{3.76a}

and similarly

\boxed{\mathbf{J}_N = \mu_n\,n\,\nabla F_N} \tag{3.76b}

(The book’s expressions are in this form. In the standard sign convention, with energies measured for electrons, ${\mathbf{J}}_N={\mu }_{n}n\nabla F_N$ and ${\mathbf{J}}_P={\mu }_{p}p\nabla F_P$ both point toward increasing quasi-Fermi energy; in a detailed check, one verifies these reduce to the drift-diffusion forms using the Einstein relation. The statement of consequence is what we need.)

The big interpretation. $J_P\propto \nabla F_P$ and $J_N\propto \nabla F_N$. Therefore a quasi-Fermi level that varies with position ($d{F}_P/dx\ne 0$ or $d{F}_N/dx\ne 0$) shows at a glance that a current is flowing, while a flat quasi-Fermi level means no current of that carrier type. Drift and diffusion unify into a single idea: carriers flow because their (quasi-)Fermi level is not flat, exactly like water flowing because the water level is not flat.

Analogy: two reservoirs. $F_N$ and $F_P$ are the “water levels” of the electron and hole reservoirs. In equilibrium both reservoirs sit at the same level $E_F$ and nothing flows. When we perturb the system, the two levels separate (here $F_N-F_P\ne 0$), and slopes in these levels drive flows.

Exercise 3.5: Quasi-Fermi levels in the illuminated bar

Problem. In Sample Problem 2, non-penetrating illumination of a semiconductor bar produced a steady-state excess-hole concentration $\Delta p_n(x)=\Delta p_{n0}\exp (-x/L_P)$. With low-level injection and $p=p_0+\Delta p$: $n\approx n_0$, $p=p_0+\Delta p_{n0}{e}^{-x/L_P}$. (a) Use (3.72) to establish relationships for $F_N$ and $F_P$ in the illuminated bar. (b) Simplify $F_P$ in the region where $\Delta p_n\gg p_0$. (c) Using (a), (b), sketch the energy band diagrams of the bar for equilibrium and the illuminated steady state (assume $\mathcal{E}=0$). (d) Is there a hole current in the illuminated bar in steady state? (e) Is there an electron current? ($N_D=10^{15}$, $\Delta p_{n0}=10^{10}$ cm${}^{-3}$, Si, 300 K.)

Solution.

(a) Since $n\approx n_0$, from (3.72a) $F_N\approx E_F$. For holes, substitute $p$ into (3.72b):

$F_P=E_i-kT\ln \frac{p}{n_i}=E_i-kT\ln \left[\frac{p_0}{n_i}+\frac{\Delta p_{n0}}{n_i}{e}^{-x/L_P}\right]$

(b) If $\Delta p_n\gg p_0$, then $(\Delta p_{n0}/n_i)e^{-x/L_P}\gg p_0/n_i$ and

$F_P\approx E_i-kT\ln \left[\frac{\Delta p_{n0}}{n_i}{e}^{-x/L_P}\right]$

Split the logarithm of a product:

$F_P=E_i-kT\ln \frac{\Delta p_{n0}}{n_i}+kT\frac{x}{L_P}$

so $F_P$ is a linear function of $x$ with slope $kT/L_P$.

(c) Numbers: $\Delta p_{n0}=10^{10}$, $n_i=10^{10}$, $p_0=n_i^2/N_D=10^5$ cm${}^{-3}$.

• (i) Near $x=0$, $\Delta p_n\gg p_0$ and $F_P$ is linear in $x$.
• (ii) At $x=0$: $\Delta p_{n0}=n_i$, so $\ln (\Delta p_{n0}/n_i)=0$ and $F_P(0)=E_i$.
• (iii) For large $x$, $F_P\to E_F$ (back to equilibrium).
• (iv) $F_N-E_i\approx E_F-E_i=kT\ln (N_D/n_i)=0.0259\ln (10^5)=0.30$ eV.

Sketch: $E_C$, $E_V$, $E_i$ flat (since $\mathcal{E}=0$). $F_N$ is a horizontal line $0.30$ eV above $E_i$ (coinciding with $E_F$ of the equilibrium diagram). $F_P$ starts at $E_i$ at $x=0$, rises linearly with slope $kT/L_P$ (an excess-hole pile gives $F_P$ near mid-gap), then bends over and merges into $F_N=E_F$ at large $x$ (roughly where $\Delta p_n$ falls to $p_0$, $x\approx L_P\ln (10^5)\approx 11.5L_P$).

(d) Yes. By (3.76a), $J_P={\mu }_{p}p\nabla F_P\ne 0$ wherever $d{F}_P/dx\ne 0$, which is the case near $x=0$ (and everywhere that $F_P$ is sloping).

(e) Appearances deceive: yes, $J_N\ne 0$ near $x=0$, even though $F_N$ looks flat. In steady state with no current leaving the bar at $x=0$ and no net electron or hole pile-up, the total current is the same at every point and must vanish at $x=0$ (no current leaves the bar there). Hence $J_N(x)=-J_P(x)$, and since $J_P\ne 0$ near $x=0$, $J_N\ne 0$. The apparent discrepancy is resolved by (3.76b): $J_N={\mu }_{n}n\nabla F_N$ is proportional to both $n$ and $\nabla F_N$. The electron concentration (majority) is enormously larger than the minority hole concentration, so $d{F}_N/dx$ must be correspondingly smaller than $d{F}_P/dx$ to carry the same current. The slope of $F_N$ is simply too tiny to be seen on a band diagram.

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3.6 Summary and Concluding Comments

Most of this chapter examined the three primary carrier actions:

• Drift: charged-particle motion in response to an electric field. Current: $J_{N|drift}=q{\mu }_{n}n\mathcal{E}$, $J_{P|drift}=q{\mu }_{p}p\mathcal{E}$ (Eqs. 3.4).
• Diffusion: migration from regions of high concentration to regions of low concentration due to random thermal motion. Current: Eqs. 3.17-3.19.
• Recombination-generation: annihilation and creation of carriers; changes the concentrations in time (Eqs. 3.34).

Each has its constant of the motion: mobilities $\mu$ (drift), diffusion coefficients $D$ (diffusion), minority-carrier lifetimes $\tau$ (R-G). Mobilities are well characterized versus doping and temperature (Figs. 3.5 and 3.7, and the fit in Exercise 3.1). Diffusion coefficients in nondegenerate semiconductors follow from the mobilities via the Einstein relationship (3.25). Lifetimes must be measured for the sample at hand.

Mathematically combining the actions gives the continuity equations (3.46), which simplify to the minority carrier diffusion equations (3.54) under common conditions. Table 3.1 gives additional simplifications; Table 3.2 gives the most-used solutions; Table 3.3 collects the key working relationships:

Table 3.3: Carrier action equation summary

Equations of state

$\frac{\partial n}{\partial t}=\frac{1}{q}\nabla \cdot {\mathbf{J}}_N+{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial n}{\partial t}|}_{\text{other}}$

$\frac{\partial p}{\partial t}=-\frac{1}{q}\nabla \cdot {\mathbf{J}}_P+{\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial p}{\partial t}|}_{\text{other}}$

$\frac{\partial \Delta n_p}{\partial t}=D_N\frac{{\partial }^2\Delta n_p}{\partial x^2}-\frac{\Delta n_p}{{\tau }_n}+G_L,\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L$

Current and R-G relationships

${\mathbf{J}}_N={\mathbf{J}}_{N|drift}+{\mathbf{J}}_{N|diff}=q{\mu }_{n}n\mathcal{E}+q{D}_N\nabla n,{\frac{\partial n}{\partial t}|}_{\text{i-thermal R-G}}=-\frac{\Delta n}{{\tau }_n}$

${\mathbf{J}}_P={\mathbf{J}}_{P|drift}+{\mathbf{J}}_{P|diff}=q{\mu }_{p}p\mathcal{E}-q{D}_P\nabla p,{\frac{\partial p}{\partial t}|}_{\text{i-thermal R-G}}=-\frac{\Delta p}{{\tau }_p}$

$\mathbf{J}={\mathbf{J}}_N+{\mathbf{J}}_P$

Key parametric relationships

$\frac{D_N}{{\mu }_n}=\frac{kT}{q},L_N\equiv \sqrt{D_N{\tau }_n},{\tau }_n=\frac{1}{c_n{N}_T}$

$\frac{D_P}{{\mu }_p}=\frac{kT}{q},L_P\equiv \sqrt{D_P{\tau }_p},{\tau }_p=\frac{1}{c_p{N}_T}$

Resistivity and electrostatic relationships

$\rho =\frac{1}{q({\mu }_{n}n+{\mu }_{p}p)},\rho \approx \frac{1}{q{\mu }_n{N}_D}(\text{n-type}),\rho \approx \frac{1}{q{\mu }_p{N}_A}(\text{p-type})$

$\mathcal{E}=\frac{1}{q}\frac{d{E}_C}{dx}=\frac{1}{q}\frac{d{E}_V}{dx}=\frac{1}{q}\frac{d{E}_i}{dx},V=-\frac{1}{q}(E_C-E_{ref})$

Quasi-Fermi level relationships

${\mathbf{J}}_N={\mu }_{n}n\nabla F_N,{\mathbf{J}}_P={\mu }_{p}p\nabla F_P$

Auxiliary topics met along the way: resistivity and its measurement (four-point probe, eddy current), the hot-point probe, constancy of the equilibrium Fermi level, nonuniform doping and the built-in field, $E$-$k$ diagrams (direct and indirect semiconductors), lifetime measurement by photoconductive decay, and diffusion lengths. The energy band diagram itself was developed further: (i) an electric field causes band bending, and the shape of $V(x)$ and $\mathcal{E}(x)$ can be read from the diagram; (ii) R-G centers add a near-midgap level that dominates thermal communication between the bands; (iii) quasi-Fermi levels describe nonequilibrium conditions.

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Chapter 4: Basics of Device Fabrication

Reading note. This is a “read-only” chapter in Pierret’s scheme: qualitative, a handful of equations, no computer exercises and no end-of-chapter problems, and nothing from it appears in the Review Problem Sets. So there is nothing to solve here. In place of problems I give a few small worked illustrations (marked as such) that use the chapter’s few equations, so you can feel their size. Put down your pencil, as the author says, and listen to the story of how a sand-derived crystal becomes a computer.

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Introduction: why a physics course visits the factory

Imagine you must analyse a violin without ever having seen one being built. You could write down the wave equation for a vibrating string and still be baffled that the instrument’s voice depends on the varnish, the wood’s grain and the thickness of the top plate. Devices are the same. The equations of Chapters 1-3 describe ideal crystal and carriers, but a real diode has a particular doping profile, a particular oxide thickness and a particular metal contact, and each of those is a consequence of how it was made. Knowing the fabrication steps tells you which simplifications (“the doping is uniform”, “the junction is abrupt”, “the oxide is perfect”) are honest and which are fibs.

The chapter proceeds in three moves:

1. Building-block processes (Section 4.1): oxidation, diffusion, ion implantation, lithography, thin-film deposition, epitaxy. Think of these as the six verbs of chip making: grow, spread, shoot, print, coat, extend.
2. Sequencing them (Section 4.2): first into a simple $pn$ junction diode, then into a real CPU process (Intel’s CHMOS).
3. Summary (Section 4.3).

The same verbs are repeated, in different orders, dozens of times for a modern IC. Like a cook with six techniques and thousands of recipes.

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4.1 Fabrication Processes

4.1.1 Oxidation

Why silicon won. Silicon’s dominance is partly an accident of chemistry: its native oxide, silicon dioxide ($\mathrm{Si}{\mathrm{O}}_2$), is easy to make, mechanically tough, an excellent electrical insulator, and bonds to silicon with a remarkably clean interface. Germanium’s oxide dissolves in water; GaAs has no comparable stable oxide. $\mathrm{Si}{\mathrm{O}}_2$ plays two roles in fabrication:

• an insulator inside devices (for example the gate dielectric of a MOSFET);
• a barrier (mask) against diffusion of dopants during fabrication (see 4.1.2).

The chemistry. Silicon is highly reactive. Expose a bare wafer to air at room temperature and a native oxide a few nanometres thick forms instantly. To get a controllable, thicker layer, the wafer is heated and exposed to oxygen or water vapour. The key physical point: the oxidizing species (O${}_2$ or H${}_2$O) diffuses through the oxide already present and reacts at the Si-SiO${}_2$ interface, not at the outer surface. So new oxide is made at the bottom, consuming silicon (about 44% of the final oxide thickness is silicon taken from the wafer), and the interface moves into the wafer. (That 44% figure is standard background, not stated in this text.)

The two overall reactions are

$\mathrm{Si}+{\mathrm{O}}_2\to \mathrm{Si}{\mathrm{O}}_2\text{(4.1a, dry oxidation)}$

$\mathrm{Si}+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Si}{\mathrm{O}}_2+2{\mathrm{H}}_2\text{(4.1b, wet oxidation)}$

	Dry oxidation	Wet oxidation
Oxidant	${\mathrm{O}}_2$ gas	water vapour
Growth rate	slower	faster
Interface quality	superior Si-SiO${}_2$ interface	poorer
Typical use	most critical insulators, e.g. MOSFET gate oxide	thick barrier / field oxides

Thought experiment. Water is far more soluble in the growing oxide, so many more oxidant molecules get through the glass per second than with ${\mathrm{O}}_2$. That is the intuition for “wet is fast”. Dry is slow but gentle, like carefully laying a brick floor versus pouring concrete quickly.

The furnace (Figs. 4.1 and 4.2). The wafers (typically many at once) stand on a slotted quartz boat that is pushed by an insertion rod into a long, clear fused-quartz tube inside a resistance-heated furnace. Features:

• Operating temperature $800{}^{\circ }\mathrm{C}$ to $1200{}^{\circ }\mathrm{C}$.
• Heater coils in three separately controlled zones: the two outer zones run at higher power to compensate for heat leaking out of the tube ends, leaving a long centre zone at uniform temperature (controllable to about $\pm 1{}^{\circ }\mathrm{C}$).
• Tube material: usually fused quartz (SiC and polycrystalline Si tubes also used in some cases; the OCR is damaged here).
• Dry oxidation: ${\mathrm{O}}_2$ is simply fed into the back of the tube.
• Wet oxidation: either bubble a carrier gas (Ar or N${}_2$) through heated water, or burn ${\mathrm{O}}_2$ and ${\mathrm{H}}_2$ at the tube input (called pyrogenic wet oxidation).
• Time needed depends on furnace temperature, desired thickness, and crystal orientation of the surface (other factors matter under special conditions).
• In commercial systems loading, insertion, temperature ramping and gas control are fully automated.

Figure 4.2(b) (pictorial): a wafer, ${\mathrm{O}}_2$ or ${\mathrm{H}}_2\mathrm{O}$ arrows, $800$-$1200{}^{\circ }\mathrm{C}$, and afterwards the wafer wears a layer of $\approx 0.01$-$1.0\mu \mathrm{m}$ of $\mathrm{Si}{\mathrm{O}}_2$.

Growth curves (Fig. 4.3). Figure 4.3(a) and (b) plot oxide thickness (log scale, $0.01$ to $\sim 10\mu \mathrm{m}$) against oxidation time (log scale, $0.1$ to $100$ hr) for $(100)$ and $(111)$ silicon. Each panel has dry and wet curves at $900,1000,1100,1200{}^{\circ }\mathrm{C}$. What the figures must show, and what to read from them:

• Higher temperature gives thicker oxide at a given time.
• Wet curves lie above dry curves at the same temperature.
• $(111)$ surfaces oxidize somewhat faster than $(100)$ (more atoms per unit area to react with).
• Growth is roughly linear at early times and slows (parabola-like) at long times, because the oxidant must diffuse through an ever-thicker layer.

Optional extra (not in the book): the Deal-Grove picture. The two regimes follow from $x_o^2+A{x}_o=B(t+\tau )$. For short times $x_o\approx (B/A)(t+\tau )$ (reaction limited, linear); for long times $x_o\approx \sqrt{Bt}$ (diffusion limited, parabolic). The figure’s curves are exactly this behaviour on a log-log plot.

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4.1.2 Diffusion

The idea. Solid-state diffusion is the historic and still widely used method of putting dopant atoms into a crystal. Put the wafer in contact with a solid, liquid or gaseous source of the desired impurity. A reaction at the surface keeps a supply of dopant atoms right next to the crystal. At high temperature the atoms slowly wander (diffuse) into regions of the wafer not protected by oxide, driven simply by the fact that concentration is higher outside than inside: the same random-walk “smelling perfume spread across a room” physics as the carrier diffusion in Chapter 3.

Facts (Fig. 4.4):

• Surface concentrations reach very high values (up to about $10^{21}/\mathrm{c}{\mathrm{m}}^3$), so the surface region adopts the $n$- or $p$-type character of the diffusing atoms; the Fig. 4.4 sketch shows a curve of diffused $n$-dopant concentration decaying with depth $x$ and crossing the original uniform $p$-dopant level, which marks the junction depth.
• Dopants also diffuse into $\mathrm{Si}{\mathrm{O}}_2$, but slowly for common dopants, so oxide protects underlying Si only for a limited time. The masking time depends on oxide thickness, diffusion temperature and the background doping.

The equipment. Diffusion is done in an open-tube furnace identical in construction to the oxidation furnace; temperatures are similar ($\approx 900$-$1200{}^{\circ }\mathrm{C}$). The only change is a dopant source in place of the oxidizing ambient.

Example: phosphorus from a liquid source (Fig. 4.5).

1. ${\mathrm{N}}_2$ carrier gas bubbles through liquid phosphorus oxychloride ($\mathrm{POC}{\mathrm{l}}_3$, a liquid at room temperature) and carries its vapour into the tube.
2. A small amount of ${\mathrm{O}}_2$ simultaneously admitted reacts with $\mathrm{POC}{\mathrm{l}}_3$ to deposit ${\mathrm{P}}_2{\mathrm{O}}_5$ on the wafer.
3. Silicon reduces ${\mathrm{P}}_2{\mathrm{O}}_5$, liberating P atoms that diffuse into the Si.

Two-step process. At the higher temperatures undesirable surface compounds tend to form, so the process is split:

Step	Source	Temperature	Purpose
Predeposition	present	moderate	introduce a controlled amount of dopant near the surface
Drive-in	cut off	increased	drive the already-introduced dopant deeper, faster

Mathematics (Eqs. 4.2). Neglecting second-order effects the dopant concentration $N(x,t)$ obeys Fick’s second law (the diffusion equation, same form as the minority-carrier equation of Chapter 3 without R-G):

$\frac{\partial N}{\partial t}=D\frac{{\partial }^{2}N}{\partial x^2}$

Predeposition. The surface is held at fixed concentration $N_0$ (source-limited), and the wafer is initially dopant-free:

$N(0,t)=N_0,N(x\to \infty ,t)=0,N(x,0)=0(x>0).$

The solution is

$N_1(x,t_1)=N_0\mathrm{erfc}\left(\frac{x}{2\sqrt{D_1{t}_1}}\right)\text{(4.2a)}$

Verification (every step). Let $\eta =x/(2\sqrt{Dt})$ and $N=N_0\mathrm{erfc}(\eta )$, with $\frac{d}{d\eta }\mathrm{erfc}\eta =-\frac{2}{\sqrt{\pi }}{e}^{-{\eta }^2}$.

$\frac{\partial \eta }{\partial x}=\frac{1}{2\sqrt{Dt}},\frac{\partial \eta }{\partial t}=-\frac{\eta }{2t}$

$\frac{\partial N}{\partial t}=N_0\left(-\frac{2}{\sqrt{\pi }}{e}^{-{\eta }^2}\right)\left(-\frac{\eta }{2t}\right)=\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }t}$

$\frac{\partial N}{\partial x}=-\frac{N_0}{\sqrt{\pi Dt}}{e}^{-{\eta }^2}$

$\frac{{\partial }^{2}N}{\partial x^2}=-\frac{N_0}{\sqrt{\pi Dt}}{e}^{-{\eta }^2}(-2\eta )\frac{1}{2\sqrt{Dt}}=\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }Dt}$

$D\frac{{\partial }^{2}N}{\partial x^2}=\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }t}=\frac{\partial N}{\partial t}✓$

Also $\mathrm{erfc}(0)=1$ gives $N(0,t)=N_0$ and $\mathrm{erfc}(\infty )=0$. The natural length is the diffusion length $2\sqrt{D_1{t}_1}$: the profile is essentially “one erfc, stretched”.

Dose retained after predeposition.

$Q={\int }_0^{\infty }{N}_{1}dx=N_{0}2\sqrt{D_1{t}_1}{\int }_0^{\infty }\mathrm{erfc}(\eta )d\eta =N_{0}2\sqrt{D_1{t}_1}\frac{1}{\sqrt{\pi }}=\frac{2N_0\sqrt{D_1{t}_1}}{\sqrt{\pi }}$

using ${\int }_0^{\infty }\mathrm{erfc}(\eta )d\eta =1/\sqrt{\pi }$.

Drive-in. The source is removed, so the dopant already in the wafer (dose $Q$) simply spreads, with no dopant crossing the surface: $\partial N/\partial x{|}_{x=0}=0$. The solution is a half-Gaussian whose area stays $Q$:

$N_2(x,t_2)=\frac{Q}{\sqrt{\pi D_2{t}_2}}\exp \left(-\frac{x^2}{4D_2{t}_2}\right)$

Substituting $Q$:

$N_2(x,t_2)=\frac{2}{\pi }{N}_0\sqrt{\frac{D_1{t}_1}{D_2{t}_2}}\exp \left(-\frac{x^2}{4D_2{t}_2}\right)\text{(4.2b, valid when }{D}_2{t}_2\gg D_1{t}_1\text{)}$

The OCR of Eq. 4.2b is truncated; it is printed in the book with a prefactor of the form $N_0\sqrt{D_1{t}_1/D_2{t}_2}$ times the Gaussian, and the inequality $D_2{t}_2\gg D_1{t}_1$ is the condition that the predeposited profile is thin compared with the drive-in spread (so it can be treated as a delta function at the surface). The derivation above shows the exact numerical factor is $2/\pi$. The dependence (everything scales as $\sqrt{D_1{t}_1/D_2{t}_2}{e}^{-x^2/4D_2{t}_2}$) is what matters physically.

Symbols: subscripts 1 and 2 denote predeposition and drive-in; $x$ is depth from the surface; $N_0$ is the surface concentration during predeposition (set by solid solubility); $D$ is the diffusion coefficient for the given impurity and furnace temperature (it rises roughly exponentially with temperature); $t$ is the duration. $\mathrm{erfc}$ is the complementary error function found in math handbooks.

Fig. 4.6 (sample computation). Phosphorus predeposition at $1000{}^{\circ }\mathrm{C}$ for $10$ min, then drive-in at $1200{}^{\circ }\mathrm{C}$ for $30$ min, with $N_0\sim 10^{21}\mathrm{c}{\mathrm{m}}^{-3}$ and $D\approx 2.58\times 10^{-14}\mathrm{c}{\mathrm{m}}^2/s$ for the predeposition (the remaining parameters are cut off in the OCR). The figure is a semilog plot of phosphorus concentration ($10^{15}$ to $10^{21}\mathrm{c}{\mathrm{m}}^{-3}$ roughly) versus $x$ from $0$ to $5\mu \mathrm{m}$ showing two curves: a steep erfc “predeposition” curve that is very shallow (a fraction of a micron), and a much flatter, deeper Gaussian “drive-in” curve with a lower surface concentration. The moral: drive-in trades surface concentration for depth, conserving the total dose.

Worked illustration (mine). Predeposition: $D_1{t}_1=(2.58\times 10^{-14})(600\mathrm{s})=1.55\times 10^{-11}\mathrm{c}{\mathrm{m}}^2$, so $2\sqrt{D_1{t}_1}=2\sqrt{1.55\times 10^{-11}}=7.9\times 10^{-6}\mathrm{cm}=0.079\mu \mathrm{m}.$ Dose (with $N_0=10^{21}$): $Q=2N_0\sqrt{D_1{t}_1/\pi }=2\times 10^{21}\times 2.22\times 10^{-6}\approx 4.4\times 10^{15}\mathrm{c}{\mathrm{m}}^{-2}$. The drive-in then lowers the surface concentration to $N_2(0)=Q/\sqrt{\pi D_2{t}_2}$, and the larger $D_2{t}_2$ is, the lower and deeper the profile.

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4.1.3 Ion Implantation

The idea. Instead of coaxing dopants in with heat, shoot them in. Create ions of the impurity, accelerate them to energies from $5\mathrm{keV}$ to $1\mathrm{MeV}$, and literally fire them into the semiconductor. Picture a microscopic shotgun; the ions slow by collisions and stop at a depth that depends on their energy. Because of the need for shallower junctions, lower-temperature processing and more precise control, implantation has become a workhorse process.

Damage and anneal. Implanted ions knock semiconductor atoms off their lattice sites along the path, and the ions themselves do not necessarily stop on lattice sites (so they are not yet electrically active). A follow-up anneal (heating) is therefore an integral part of the process: it repairs crystal damage and activates the dopants by letting them settle on substitutional sites.

The machine (Figs. 4.7 and 4.8). Left to right in Fig. 4.7:

Component	Job
Ion source	ionizes the desired impurity (also produces unwanted ions)
Analyzer magnet / mass analyzer (with resolving aperture)	bends the beam; only the desired mass/charge passes, removing unwanted ions
Acceleration tube (high-voltage terminal)	accelerates ions to the preset energy
Lens (focus)	focuses the beam
x-scan and y-scan plates	electrostatically sweep the beam over the wafer (or the wafer is moved mechanically, or both)
Wafer (target) and beam mask	the target
Faraday cup	collects charge to measure the beam current

An electrical contact to the wafer lets electrons flow in to neutralize the implanted ions. Integrating the target current over the implant time counts the ions exactly. The total number of implanted ions per unit area is the dose $\Phi$ (ions/cm${}^2$). Fig. 4.8 photographs show the commercial end-station: a wafer carousel in two ports and robot arms that automatically load and unload it.

The profile (Eq. 4.3). The implanted concentration has a Gaussian shape:

$N(x)=\frac{\Phi }{\sqrt{2\pi }\Delta R_p}\exp \left\{-\frac{1}{2}{\left[\frac{x-R_p}{\Delta R_p}\right]}^2\right\}\text{(4.3)}$

For readers who know statistics: $R_p$ is the mean (the projected range) and $\Delta R_p$ the standard deviation (the straggle). Both depend on ion species and substrate, and are roughly proportional to ion energy. Key consequences:

${\int }_{-\infty }^{\infty }Ndx=\Phi (\text{normalized}),N_{\max }=N(R_p)=\frac{\Phi }{\sqrt{2\pi }\Delta R_p}\approx \frac{0.4\Phi }{\Delta R_p}.$

(The integral is $\Phi$ because a Gaussian pdf of unit area times $\Phi$; the small part with $x<0$ is neglected when $R_p\gg \Delta R_p$.) Unlike diffusion, the peak is buried at depth $R_p$ below the surface, not at it.

Fig. 4.9. Computed P implants into Si at constant dose $10^{14}\mathrm{c}{\mathrm{m}}^{-2}$ for energies $25,50,75,100\mathrm{keV}$: semilog curves from $x=0$ to $0.4\mu \mathrm{m}$. Each is a bell (parabola on a log axis) peaking at $R_p$ with a peak concentration near $10^{19}$ to $10^{20}\mathrm{c}{\mathrm{m}}^{-3}$; higher energy means deeper $R_p$, broader curve and lower peak (same area).

Worked illustration (mine; approximate range statistics). A $50$ keV phosphorus implant into Si has $R_p\approx 0.06\mu \mathrm{m}$ and $\Delta R_p\approx 0.024\mu \mathrm{m}$. For $\Phi =10^{14}\mathrm{c}{\mathrm{m}}^{-2}$: $N_{\max }=\frac{10^{14}}{\sqrt{2\pi }(2.4\times 10^{-6}\mathrm{cm})}=\frac{10^{14}}{6.0\times 10^{-6}}\approx 1.7\times 10^{19}\mathrm{c}{\mathrm{m}}^{-3}.$

Masking. As with diffusion, parts of a wafer can be masked: thin films of $\mathrm{Si}{\mathrm{O}}_2$, $\mathrm{S}{\mathrm{i}}_3{\mathrm{N}}_4$, and less frequently photoresist and Al stop the ions before they reach the silicon.

Advantages over diffusion.

1. Intrinsically low temperature (implant itself at room temperature). The anneal is often some other high-temperature step already in the flow (such as a later oxidation); a dedicated anneal can be as low as $600{}^{\circ }\mathrm{C}$ under optimum conditions.
2. Minimal spreading of the profile.
3. Very precise control of the impurity amount (dose from current integration).
4. Virtually any atom into any substrate. A striking recent use: implanting high doses of oxygen to produce a buried $\mathrm{Si}{\mathrm{O}}_2$ layer, giving silicon-on-insulator (SOI) structures.
5. Shallow profiles (Fig. 4.9). Implants have therefore replaced the predeposition step of diffusion and are ideal for devices needing extremely shallow junctions.

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4.1.4 Lithography

Purpose. Masks of $\mathrm{Si}{\mathrm{O}}_2$ and other materials must have openings in exactly the right places; metals must be patterned into wires. Lithography (literally “stone writing”) selectively removes a thin film at prescribed positions across the wafer surface. It is the printing press of chip making. Fig. 4.10 shows a production line: a coat-and-develop system (the SSI150) next to a Nikon stepper, with wafers moving between them automatically on tracks.

The steps (Fig. 4.11), using patterning of an $\mathrm{Si}{\mathrm{O}}_2$ film as the example.

1. Resist coating. Coat the oxide-covered wafer with photoresist, a UV-sensitive liquid. A few drops are placed on the wafer, which is spun at high speed to make a thin uniform coat.
2. Prebake at $80$-$100{}^{\circ }\mathrm{C}$ to drive out solvent and improve adhesion. The hardened resist resembles a photographic emulsion.
3. Exposure through a mask with UV light (Fig. 4.11b). The mask is a carefully made glass or quartz photoplate carrying a copy of the pattern; darkened emulsion regions block UV. Exposed resist changes chemically, in a way depending on the resist type:

	Negative resist	Positive resist
Effect of light	exposed areas polymerize, harder to remove	sensitizer breaks down, exposed regions dissolve faster
After developer	exposed regions remain, unexposed wash away	exposed regions are removed, unexposed remain
Remark	widely used in early IC processing	now the main type: better small-geometry control

For a positive resist the sensitizer (present in large amounts) dramatically slows dissolution in an alkaline developer until light destroys it. Fig. 4.11(c): left half (positive) has a window where the mask was clear; right half (negative) has resist where the mask was clear.

4. Develop (Fig. 4.11c).
5. Hard bake, $30$ min at $120$-$180{}^{\circ }\mathrm{C}$: improves adhesion and resistance to the etch.
6. Etch the unprotected oxide with an acid, buffered HF for $\mathrm{Si}{\mathrm{O}}_2$ (Fig. 4.11d).
7. Strip the resist, either with a chemical solution that swells and lifts it, or by oxidizing (burning) it in an oxygen plasma system called an asher.

Limits. Ever-shrinking features push UV lithography: shorter-wavelength sources and special compensating techniques extended the minimum feature size, and x-ray lithography was under development and expected soon.

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4.1.5 Thin-Film Deposition

Devices must be wired to the outside world: metal layers must be deposited and patterned. Complex ICs have three, sometimes four, electrically isolated metal layers, which needs intervening dielectric layers. Thin films also act as interdiffusion barriers and protect circuits from contamination.

Evaporation

An old, direct method. Place the source (for Al, a short piece of wire on a tungsten filament or boat) on a resistance-heated holder inside a vacuum chamber (Fig. 4.12) with the substrate facing it. Pump down, heat the source, and vapour travels unimpeded (low pressure means few collisions) to condense as a thin film on the wafer.

• Hot-filament evaporation: moderately high contamination (filament material).
• Electron-beam evaporation (source heated by e-beam): no contamination, but it generates x-rays that degrade devices.
• Therefore seldom used in production-line modern ICs, but still common for simple devices.

Sputtering

Also in vacuum, but the source (“target”) sits on the cathode of a parallel-plate system and the wafer on the anode (Fig. 4.13).

1. Evacuate, then admit low-pressure argon.
2. Apply an interelectrode voltage; Ar ionizes, forming a plasma.
3. The source plate is negative, so $\mathrm{A}{\mathrm{r}}^{+}$ ions accelerate into it.
4. The impacts knock out source atoms/molecules (momentum transfer, like a cue ball breaking a rack).
5. Being neutral, they travel to the substrate and form the film.

A d.c. supply suffices for metals; RF is needed for insulating films (otherwise charge builds up on the insulator). For compounds, a gas containing one component may be added to keep the film near stoichiometric. Low-temperature, low-contamination, acceptable throughput: sputtering is the chief commercial method for Al and other metals.

Chemical Vapor Deposition (CVD)

The film forms from gas-phase components: a compound decomposes, or gases react. The reactions are almost always surface catalyzed, so film grows on the wafer surfaces placed in the gas stream.

Type	Features
APCVD (atmospheric pressure)	simple systems
LPCVD (low pressure)	comparable kinetics, better uniformity, less gas consumption
PECVD (plasma-enhanced)	plasma electrons energize reaction gases, allowing very low deposition temperatures

Fig. 4.14 is a cutaway barrel reactor: wafers stand on a central susceptor heated radiantly by surrounding quartz lamps (infrared sensors monitor temperature), process gases enter at the top, flow across the wafers and exit coaxially at the bottom. CVD makes the masking and inter-metal dielectric films of complex ICs, and polycrystalline Si (a pseudo-metal when heavily doped). All three CVD types appear somewhere in a typical IC flow.

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4.1.6 Epitaxy

Evaporation, sputtering and CVD give amorphous or polycrystalline films. Epitaxy (Greek epi “upon” + taxis “arrangement”) yields a single-crystal layer that is an extension of the substrate lattice: new atoms obey the pattern of the crystal beneath, like children’s blocks added neatly to an existing perfect stack.

For silicon: vapour-phase decomposition of $\mathrm{SiC}{\mathrm{l}}_4$ or silanes ($\mathrm{Si}{\mathrm{H}}_4,\mathrm{Si}{\mathrm{H}}_2\mathrm{C}{\mathrm{l}}_2,\mathrm{SiHC}{\mathrm{l}}_3$) in a reactor like the CVD one (Fig. 4.14) lets Si literally grow on the existing crystal. The epi-layer doping, intentionally different from the substrate’s, is set by adding a dopant gas: phosphine ($\mathrm{P}{\mathrm{H}}_3$), diborane (${\mathrm{B}}_2{\mathrm{H}}_6$) or arsine ($\mathrm{As}{\mathrm{H}}_3$). Whole-wafer epitaxy must be done early in the flow, before any permanent surface films are grown or deposited. Wafers are sometimes purchased with the epi-layer already in place.

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4.2 Device Fabrication Examples

The purpose: to show how individual processes are combined and sequenced. Two examples, both qualitative.

4.2.1 $pn$ Junction Diode Fabrication

Starting point: a flat, damage-free, single-crystal Si wafer, precleaned to remove particulates, organic films and adsorbed metal ions. Here it is $p$-type, uniformly boron-doped during crystal growth. Fig. 4.15 steps:

Step	Process	What happens
(0)	Start	$p$-Si wafer
(1)	Oxidation	grow thermal $\mathrm{Si}{\mathrm{O}}_2$, a diffusion barrier. Its thickness must be comfortably greater than the projected masking thickness
(2)	**Lithography 1 (positive resist assumed)	open “diffusion holes” in the oxide where the diodes will be
(3)	Phosphorus diffusion (predeposition then drive furnaces)	forms $n^{+}$-$p$ junctions where oxide does not protect the surface. The ”+” in $n^{+}$ means very heavy doping
(4)	Metallization	sputter (or evaporate) Al over the entire wafer
(5)	**Lithography 2	remove excess metal outside the junction areas

Practical notes: some oxygen is needed in the P predeposition (as in 4.1.2); the drive-in may be in an oxidizing atmosphere to minimize out-diffusion, so after diffusion a subsidiary lithography step is usually needed to reopen the oxide holes (not shown in Fig. 4.15). A low-temperature anneal ($\le 500{}^{\circ }\mathrm{C}$) promotes a low-resistance metal-Si contact.

Diodes now work across the wafer. For commercial parts a diamond-edged saw cuts the wafer into dice (a die is a wafer piece with a single device or IC); a large-area metal contact is made on the die back, a lead attached to the top contact, and the device encapsulated in plastic or hermetically sealed in a metal package.

4.2.2 Computer CPU Process Flow

This is a direct excerpt (with permission) of Intel’s Components Quality and Reliability 1991/92. Fig. 4.16 is the block diagram of the single poly, single metal, CHMOS flow. The blocks, in order, include: starting silicon wafer; initial oxidation; well ion implant plus drive; silicon-nitride deposition and patterning; field threshold ion implant; field oxidation; silicon-nitride removal; gate oxidation; transistor threshold ion implant; polycide deposition plus impurity doping; polycide patterning plus oxidation; n-channel source-drain implant; p-channel source-drain implant; poly re-ox; doped glass deposition; contact etch; metal deposition; metal patterning; passivation; alloy; wafer backside preparation; process control/electrical test; wafer sort; assembly; class test; outgoing quality monitor; customer. (Between most blocks are lithography “MASK” steps.)

Vocabulary.

• Poly: polycrystalline Si.
• CHMOS: Complementary High-density Metal-Oxide-Semiconductor, a fabrication technology placing both $n$-channel and $p$-channel MOS transistors on one chip. “Single poly, single metal” means one level of heavily doped polysilicon and one level of metal contact and interconnect the devices.
• Wet etch: dissolving a material (such as $\mathrm{Si}{\mathrm{O}}_2$) in a liquid chemical bath.
• Dry etch: removing material by a plasma-enhanced gas-phase reaction.

The author asks that the description be “looked over, not digested”. Here are the 11 stages (Fig. 4.17 a-l), with a one-line meaning.

1. Starting silicon and well definition (4.17a). $150$ mm (6 in.) high-resistivity $(100)$ boron-doped $p$-type wafers. $p$-type is needed for $n$-channel transistors. For $p$-channel transistors, $n$-type regions ($n$-wells) are implanted (As or P) through a mask. The $n$-well is the background for $p$-channel devices; the protected $p$-type substrate is the background for $n$-channel ones. A high-temperature drive finishes the wells. Analogy: neighbouring plots of land, one planted for rice and one for wheat, on the same field.
2. Field threshold implant and field oxidation (4.17b,c). Silicon nitride ($\mathrm{S}{\mathrm{i}}_3{\mathrm{N}}_4$) is deposited, masked and etched. Etched-away nitride defines where the field threshold implant goes and where oxide may grow in field oxidation; remaining nitride blocks oxide growth (these are future transistor sites). The thick (about $6000\text{A˚}$) field oxide electrically isolates adjacent transistors. Nitride is then removed.
3. Gate oxidation (4.17d). Grow a thin, dense thermal oxide everywhere; what remains after processing is the MOS gate oxide. Performance depends critically on its quality (hence dry oxidation).
4. Transistor threshold implant (4.17d). A boron implant adjusts the threshold voltage $V_T$ of both device types. Thick field oxide keeps the boron out of isolation regions.
5. Polysilicon-to-diffusion (buried) contacts (4.17e). Openings are defined in the thin gate oxide where poly must contact diffused silicon regions.
6. Polycide deposition, doping, patterning, oxidation (4.17f,g). About $1500\text{A˚}$ of poly deposited by CVD, doped with phosphorus by gas-source diffusion (which also diffuses through poly into the substrate, lowering buried-contact resistance). About $1500\text{A˚}$ of silicide deposited conformally on top to lower resistance and increase speed in small transistors. The polycide stack (poly + silicide) is defined by plasma dry etch using resist as mask, then oxidized to protect the gates.
7. Source and drain implants and poly re-ox (4.17h). The polycide gate stack, thin oxide, field oxide and resist form the mask for the $p$-channel and $n$-channel source/drain implants. This is the self-aligned source/drain process: the source and drain are aligned directly to the gate, which fixes their location relative to the channel. (Intuition: the gate is its own stencil, so the alignment cannot be wrong, regardless of mask-alignment errors.) After removing remaining poly oxide, a re-ox high-temperature oxidation (a) drives in the S/D implants, (b) grows a high-integrity dielectric on poly, (c) caps exposed S/D regions to prevent out-diffusion during later glass deposition.
8. Doped glass deposition (4.17i). $5000$-$10,000\text{A˚}$ of boron- and/or phosphorus-doped silicate glass (BPSG) deposited by low-temperature CVD for electrical isolation between the metal lines and the underlying gates and devices; then prepared for masking.
9. Contact mask and etch (4.17j). Open windows for metal-to-polycide and metal-to-silicon contact holes, by wet, dry or wet+dry etch. In a wet/dry combination the wet etch gives a sloped top for better metal step coverage, and an isotropic dry etch removes the remaining glass leaving vertical sidewalls.
10. Metal deposition and definition (4.17k). Sputter metal; pattern it. These lines connect transistors to the outside world and apply the voltages that turn transistors on and off.
11. Passivation (4.17l). PECVD deposits a dual film: oxynitride (moisture and contamination barrier, vital for plastic packages since plastic is a poor moisture barrier) under plasma oxide (handling protection).

The remaining flow steps (Fig. 4.16, right column) prepare for die separation and packaging. Fig. 4.18 is a photograph of an Intel Pentium processor die about $1.5$ cm square containing $3.1$ million transistors, with labelled functional units: instruction fetch, branch prediction, code cache, instruction decode, code TLB, clock driver, bus interface, superscalar integer execution units, pipelined floating-point unit, data TLB, and data cache. (Fabricated by an upgraded version of the described process.)

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4.3 Summary

The tour of silicon device fabrication gave a feel for the construction and physical nature of devices:

• Component processes: oxidation, diffusion, ion implantation, lithography, evaporation, sputtering, chemical vapor deposition and epitaxy.
• Sequencing: a simple $pn$ diode (oxidize, open windows, diffuse, metallize, pattern) and a moderately complex IC (CHMOS) show how the same processes are combined, again and again, to build real structures.

Key formulas of the chapter

Quantity	Formula
Wet / dry oxidation	$\mathrm{Si}+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Si}{\mathrm{O}}_2+2{\mathrm{H}}_2$ ; $\mathrm{Si}+{\mathrm{O}}_2\to \mathrm{Si}{\mathrm{O}}_2$
Predeposition profile	$N_1=N_0\mathrm{erfc}(x/2\sqrt{D_1{t}_1})$
Drive-in profile	Gaussian $\propto \exp (-x^2/4D_2{t}_2)$, with $D_2{t}_2\gg D_1{t}_1$
Implant profile	$N=\frac{\Phi }{\sqrt{2\pi }\Delta R_p}\exp \{-\frac{1}{2}[(x-R_p)/\Delta R_p{]}^2\}$

End-of-chapter problems / computer exercises: none (read-only chapter).

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Part I Supplement and Review

Part I covered Chapters 1-3 (crystals, bands, carriers, transport, recombination-generation) plus the read-only Chapter 4. This supplement collects the reading list, figure sources, a self-test vocabulary list and two review problem sets with answer keys. Chapter 4 itself is not examined in the problem sets.

R1.1 Alternative / Supplemental Reading List

Author(s)	Type (A = alternative, S = supplemental)	Level	Relevant chapters
For semiconductor physics
Ferendeci	A/S	Advanced undergrad / intro grad	1-6
Neamen	A	Undergraduate	1-6
Pierret	S	Introductory graduate	1-6
Streetman	A	Undergraduate	1-4
Tyagi	A/S	Advanced undergrad / intro grad	1-5
For device fabrication
Jaeger	S	Undergraduate	All
Runyan and Bean	S	Professional	All
Sze (editor)	S	Professional	All

Full citations:

1. A. M. Ferendeci, Physical Foundations of Solid State and Electron Devices, McGraw-Hill, 1991.
2. R. C. Jaeger, Introduction to Microelectronic Fabrication, Vol. V, Modular Series on Solid State Devices (Neudeck and Pierret, eds.), Addison-Wesley, 1988.
3. D. A. Neamen, Semiconductor Physics and Devices, Basic Principles, Irwin, 1992.
4. R. F. Pierret, Advanced Semiconductor Fundamentals, Vol. VI, Modular Series, Addison-Wesley, 1987.
5. W. R. Runyan and K. E. Bean, Semiconductor Integrated Circuit Processing Technology, Addison-Wesley, 1990.
6. B. G. Streetman, Solid State Electronic Devices, 4th ed., Prentice Hall, 1995.
7. S. M. Sze (ed.), VLSI Technology, 2nd ed., McGraw-Hill, 1988.
8. M. S. Tyagi, Introduction to Semiconductor Materials and Devices, Wiley, 1991.

Other recent texts: Bar-Lev (Prentice Hall, 1993); Navon (Holt, Rinehart and Winston, 1986); Pulfrey and Tarr (Prentice Hall, 1989); Sah (World Scientific, 1991); Singh (McGraw-Hill); Sze, Semiconductor Devices: Physics and Technology (Wiley, 1985); Yang, Microelectronic Devices (McGraw-Hill); Zambuto, Semiconductor Devices (McGraw-Hill, 1989). (Years partly garbled in OCR.)

R1.2 Figure Sources / Cited References

(1) Shockley, Electrons and Holes in Semiconductors (1950); (2) Sze, Physics of Semiconductor Devices, 2nd ed. (Wiley, 1981); (3) a Springer-Verlag reference (1982, title lost in OCR); (4) Jacoboni, Canali, Ottaviani, Quaranta, “A Review of Some Charge Transport Properties of Silicon,” Solid-State Electronics 20, 77 (1977); (5) Smith, Frey, Chatterjee, “High-Field Transport of Holes in Silicon,” Appl. Phys. Lett. 39, 332 (1981); (6) Schroder, Semiconductor Material and Device Characterization (Wiley, 1990); (7) Pierret, Advanced Semiconductor Fundamentals; (8) Jaeger, Introduction to Microelectronic Fabrication; (9) Runyan and Bean; (10) T. Deacon, “Silicon Epitaxy: An Overview,” Microelectronic Manufacturing and Testing, Sept. 1984.

R1.3 Review List of Terms

The book asks you to define these in your own words. Here are compact definitions to check yourself against.

#	Term	Meaning
1	amorphous	no long-range atomic order
2	polycrystalline	many small single-crystal grains with random orientations
3	crystalline	atoms in a periodic array throughout
4	lattice	the periodic set of points on which the atom pattern repeats
5	unit cell	smallest volume that, translated, fills space and reproduces the crystal
6	ingot	large single-crystal boule pulled from the melt, sliced into wafers
7	carrier	mobile charge: electron or hole
8	electron	negative carrier in the conduction band
9	hole	missing valence electron; behaves as a positive carrier
10	conduction band	lowest band of allowed states that is empty at 0 K
11	valence band	highest band filled at 0 K
12	band gap	$E_G=E_C-E_V$, the forbidden energy range
13	effective mass	$m^{\ast }$ absorbing lattice forces so $F=m^{\ast }a$
14	intrinsic semiconductor	pure; $n=p=n_i$
15	extrinsic semiconductor	doped so one carrier type dominates
16	dopant	deliberately added impurity
17	donor	impurity giving an electron to the conduction band
18	acceptor	impurity capturing a valence electron, creating a hole
19	n-type	electrons are majority
20	p-type	holes are majority
21	$n^{+}$ ($p^{+}$)	very heavily doped
22	majority carrier	the more abundant carrier
23	minority carrier	the less abundant carrier
24	density of states	number of allowed states per unit energy per unit volume, $g(E)$
25	Fermi function	$f(E)=1/[1+e^{(E-E_F)/kT}]$, occupation probability
26	Fermi energy (level)	energy where $f=1/2$
27	nondegenerate	$E_F$ more than a few $kT$ inside the gap; Boltzmann statistics hold
28	degenerate	$E_F$ within about $3kT$ of, or inside, a band
29	charge neutrality	$p-n+N_D^{+}-N_A^{-}=0$ in a uniformly doped sample
30	extrinsic temperature region	$n\approx N_D$ (or $p\approx N_A$), all dopants ionized
31	intrinsic temperature region	$n_i\gg$ doping, so $n\approx p\approx n_i$
32	freeze-out	at very low $T$ carriers fall back onto donors/acceptors
33	drift	carrier motion caused by an electric field
34	scattering	collisions that randomize carrier velocity
35	thermal motion	random motion with $\frac{1}{2}{m}^{\ast }{v}_{th}^2\sim \frac{3}{2}kT$
36	drift velocity	average velocity from an applied field
37	saturation velocity	limiting drift velocity at high field ($\sim 10^7$ cm/s in Si)
38	current density	charge crossing unit area per unit time
39	mobility	$\mu =v_d/\mathcal{E}$
40	resistivity	$\rho =1/\sigma$
41	conductivity	$\sigma =q({\mu }_{n}n+{\mu }_{p}p)$
42	band bending	position-dependent band edges caused by an electrostatic potential
43	diffusion	net flow from high to low concentration
44	diffusion coefficient	$D$ in $J=qDdn/dx$; $D=(kT/q)\mu$
45	recombination	an electron-hole pair annihilates
46	generation	an electron-hole pair is created
47	band-to-band R-G	direct electron transition across the gap
48	R-G centers	deep-level traps giving two-step R-G
49	E-k plots	energy vs wavevector band diagrams
50	direct semiconductor	conduction minimum and valence maximum at the same $k$ (GaAs)
51	indirect semiconductor	extrema at different $k$ (Si, Ge)
52	photogeneration	pair creation by absorbed photons ($h\nu \ge E_G$)
53	absorption coefficient	$\alpha$; light intensity decays as $e^{-\alpha x}$
54	low-level injection	excess minority carriers $\ll$ majority concentration
55	equilibrium	no external perturbation; $np=n_i^2$, constant $E_F$
56	perturbation	any external influence driving the system from equilibrium
57	steady state	time-independent, but not necessarily equilibrium
58	minority carrier lifetime	$\tau$, mean time before recombination
59	minority carrier diffusion length	$L=\sqrt{D\tau }$
60	quasi-Fermi level	$F_N$, $F_P$ describing non-equilibrium carrier populations
61	dry oxidation	oxidizing Si with ${\mathrm{O}}_2$
62	wet oxidation	oxidizing Si with water vapour
63	predeposition	first diffusion step introducing dopant at fixed surface concentration
64	drive-in	second step redistributing dopant deeper, source off
65	diffusion mask	film (usually $\mathrm{Si}{\mathrm{O}}_2$) blocking dopant diffusion
66	ion implantation	shooting accelerated dopant ions into the wafer
67	dose	implanted ions per cm${}^2$, $\Phi$
68	projected range	mean implant depth $R_p$
69	photoresist	light-sensitive coating used to pattern films
70	negative resist	exposed regions remain after development
71	positive resist	exposed regions are removed
72	sputtering	ion bombardment ejects target atoms that coat the wafer
73	CVD	film grown from gas-phase reactions
74	LPCVD	CVD at low pressure
75	PECVD	plasma-enhanced CVD
76	epitaxy	single-crystal layer extending the substrate lattice
77	die	wafer piece holding one device or IC
78	poly	polycrystalline silicon
79	wet etch	etching in a liquid chemical bath
80	dry etch	plasma/gas-phase etching

R1.4 Part I Review Problem Sets and Answers

Problem Set A was adapted from a one-hour open-book exam; Set B from a one-hour closed-book exam. Use $T=300$ K, $n_i=10^{10}\mathrm{c}{\mathrm{m}}^{-3}$, $kT=0.0259$ eV.

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Problem Set A (open book; use accurate values)

Problem A1: Crystal geometry

Problem. For the Si diamond unit cell (figure: a cube with the origin at the lower back corner; points A, B, C are on the three axes, at distance $a$ from the origin; point D is on the front face):

(a) Miller indices of the plane through A, B, C. (b) Miller indices of the direction from the origin to D. (c) Nearest-neighbour distance in terms of $a$. (d) How to visualize a donor in the cell.

(The OCR lost the figure. Consistent with the book’s answers: A, B, C are the corners at $(a,0,0)$, $(0,a,0)$, $(0,0,a)$, and D is the face-centre atom at $(a,a/2,a/2)$.)

Solution.

(a) Miller procedure:

$\text{intercepts}=(1,1,1)(\text{in units of }a)$ $\text{reciprocals}=(1,1,1)$ $\text{no fractions to clear}\Rightarrow (111).$

(b) Direction components from the origin to D, in units of $a/2$:

$(a,a/2,a/2)=\frac{a}{2}(2,1,1)\Rightarrow [211].$

(c) In the diamond lattice each atom sits at a quarter of the body diagonal from its nearest neighbour. Body diagonal $=\sqrt{3}a$, so

$d_{nn}=\frac{\sqrt{3}}{4}a.$

(d) A donor is a group-V atom (five valence electrons) replacing one Si atom. Four electrons complete the four bonds; the fifth is a weakly bound extra electron (draw an extra bond or electron on that atom), easily freed into the conduction band.

Problem A2: n-type wafer table

Problem. A Si wafer is uniformly doped $N_D=10^{17}\mathrm{c}{\mathrm{m}}^{-3}$ under equilibrium at 300 K, with ${\tau }_p=10^{-6}$ s. Complete the table of $E_G$, $n_i$, $n_0$, $p_0$, $E_F-E_i$, charge density, ${\mu }_n$, $\rho$, $L_p$, $D_p$.

Solution.

Parameter	Value	Reasoning
$E_G$	1.12 eV	Si band gap at 300 K
$n_i$	$10^{10}\mathrm{c}{\mathrm{m}}^{-3}$	given
$n_0$	$10^{17}\mathrm{c}{\mathrm{m}}^{-3}$	$n_0=N_D$ since $N_D\gg n_i$
$p_0$	$10^3\mathrm{c}{\mathrm{m}}^{-3}$	see below
$E_F-E_i$	0.417 eV	see below
charge density	0	charge neutrality holds in a uniform equilibrium sample
${\mu }_n$	801 cm${}^2$/V s	from Fig. 3.5 at $10^{17}$
$\rho$	0.078 $\Omega$ cm	see below
$L_p$	$2.93\times 10^{-3}$ cm	see below
$D_p$	8.57 cm${}^2$/s	see below

$p_0=\frac{n_i^2}{n_0}=\frac{10^{20}}{10^{17}}=10^3\mathrm{c}{\mathrm{m}}^{-3}$

$E_F-E_i=kT\ln \frac{n_0}{n_i}=0.0259\ln (10^7)=0.0259\times 16.12=0.417\mathrm{eV}$

$\rho =\frac{1}{q{\mu }_n{N}_D}=\frac{1}{(1.6\times 10^{-19})(801)(10^{17})}=\frac{1}{12.8}=0.078\Omega \mathrm{cm}$

(holes are negligible in $\sigma$.) Minority-hole diffusion coefficient from the Einstein relation, with ${\mu }_p\approx 331$ cm${}^2$/V s at this doping (Fig. 3.5):

$D_p=\frac{kT}{q}{\mu }_p=0.0259\times 331=8.57\mathrm{c}{\mathrm{m}}^2/s$

$L_p=\sqrt{D_p{\tau }_p}=\sqrt{(8.57)(10^{-6})}=2.93\times 10^{-3}\mathrm{cm}$

Problem A3: Band diagrams

Problem. (a) On a band diagram, show the usual positions of $E_i$, $E_D$, $E_A$, $E_T$ (trap/R-G centre) and $E_F$ for degenerately doped $p$-type. (b) A Si device in equilibrium at 300 K has the electric-field profile shown (zero for $0\le x\le x_a$, constant negative for $x_a\le x\le x_b$, zero for $x_b\le x\le x_c$), with $N_A=10^{16}\mathrm{c}{\mathrm{m}}^{-3}$ on the left and $N_D=10^{16}\mathrm{c}{\mathrm{m}}^{-3}$ on the right (junction inside the field region). Draw the bands with $E_C$, $E_i$, $E_V$ and $E_F$ and justify. (c) Find $V(x_c)-V(0)$.

Solution.

(a) $E_i$ lies near midgap; $E_D$ lies just below $E_C$; $E_A$ just above $E_V$; $E_T$ lies near midgap (deep level, most effective R-G centre); $E_F$ for degenerate $p$-type lies below $E_V+3kT$, i.e. inside the valence band or within a few $kT$ of its edge. Comment: $E_T$ is a localized level, drawn as a dashed line, not a band.

(b) In equilibrium $E_F$ is position independent. Since

$\mathcal{E}=\frac{1}{q}\frac{d{E}_C}{dx}=\frac{1}{q}\frac{d{E}_i}{dx}=\frac{1}{q}\frac{d{E}_V}{dx},$

zero field means flat bands for $0\le x\le x_a$ and $x_b\le x\le x_c$; a constant negative field means all bands have a constant negative slope in $x_a\le x\le x_b$.

Far from the junction:

$E_i-E_F=kT\ln \frac{N_A}{n_i}=0.0259\ln \frac{10^{16}}{10^{10}}=0.0259(13.82)=0.358\mathrm{eV}(0\le x\le x_a)$

$E_F-E_i=kT\ln \frac{N_D}{n_i}=0.358\mathrm{eV}(x_b\le x\le x_c)$

So the diagram: horizontal $E_F$; on the left $E_i$ is 0.358 eV above $E_F$; the bands slope down linearly across $x_a$-$x_b$; on the right $E_i$ is 0.358 eV below $E_F$. $E_C$ and $E_V$ stay $E_G/2$ above and below $E_i$ everywhere.

(c) Total band drop:

$\Delta E=0.358+0.358=0.716\mathrm{eV}$

Electron energy is $E=-qV$ (plus constant), so

$V(x_c)-V(0)=-\frac{1}{q}\left[E_i(x_c)-E_i(0)\right]=-\frac{-0.716\mathrm{eV}}{q}=+0.716\mathrm{V}.$

Problem A4: Two light flashes

Problem. An $N_D=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ Si sample at room temperature receives two flashes of light. Each creates $\Delta p_f=10^{10}\mathrm{c}{\mathrm{m}}^{-3}$ holes uniformly. The first is at $t=0$, the second at $t=t_f$, with $t_f$ comparable to ${\tau }_p$; flashes are infinitely short. (a) Sketch $\Delta p(t)$. (b) Is low-level injection valid? (c) Solve for $\Delta p(t)$, $t\ge 0$. (d) Fractional conductivity change right after the first flash, symbolically and numerically (use ${\mu }_n=1350$, ${\mu }_p=458$ cm${}^2$/V s at this doping; the numbers are recovered from the printed answer).

Solution.

(a) $\Delta p$ jumps to $\Delta p_f$ at $t=0$, decays exponentially with time constant ${\tau }_p$ to $\Delta p_f{e}^{-t_f/{\tau }_p}$ at $t_f^{-}$, jumps up by another $\Delta p_f$ to $\Delta p_f(1+e^{-t_f/{\tau }_p})$ at $t_f^{+}$ (below $2\Delta p_f$), then decays exponentially again toward zero.

(b) Maximum excess is below $2\Delta p_f$:

$\Delta p_{max}<2\Delta p_f=2\times 10^{10}\ll n_0=N_D=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$

So yes, low-level injection holds at all times.

(c) For a uniform sample with no field, no gradient and no generation after the flash, the minority-carrier diffusion equation reduces to

$\frac{d\Delta p}{dt}+\frac{\Delta p}{{\tau }_p}=0.$

For $0\le t\le t_f$: general solution $\Delta p=A{e}^{-t/{\tau }_p}$; condition $\Delta p(0)=\Delta p_f\Rightarrow A=\Delta p_f$:

$\Delta p(t)=\Delta p_f{e}^{-t/{\tau }_p}.$

For $t\ge t_f$: general solution $\Delta p=A{e}^{-(t-t_f)/{\tau }_p}$; boundary condition (jump by $\Delta p_f$ at $t_f$):

$\Delta p(t_f^{+})=\Delta p(t_f^{-})+\Delta p_f=\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)\Rightarrow A=\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)$

$\Delta p(t)=\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)e^{-(t-t_f)/{\tau }_p}.$

(d) $\sigma =q({\mu }_{n}n+{\mu }_{p}p)$.

$\sigma (0^{-})=q{\mu }_n{N}_D(n_0=N_D\gg p_0)$

$\sigma (0^{+})=q\left[{\mu }_n(N_D+\Delta p_f)+{\mu }_p(p_0+\Delta p_f)\right]\approx q{\mu }_n{N}_D+q({\mu }_n+{\mu }_p)\Delta p_f$

$\Delta \sigma =q({\mu }_n+{\mu }_p)\Delta p_f$

$\frac{\sigma (0^{+})-\sigma (0^{-})}{\sigma (0^{-})}=\frac{q({\mu }_n+{\mu }_p)\Delta p_f}{q{\mu }_n{N}_D}=\left(1+\frac{{\mu }_p}{{\mu }_n}\right)\frac{\Delta p_f}{N_D}$

$=\left(1+\frac{458}{1350}\right)\frac{10^{10}}{10^{15}}=1.34\times 10^{-5}.$

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Problem Set B (closed book)

Use $n_i=10^{10}$ cm${}^{-3}$, $kT=0.0259$ eV.

I. General information

(1) A cubic cell has one atom centred on each vertical edge and one in the middle of the top and bottom faces. Atoms in the cell? Options: (a) 4, (b) 2, (c) 1, (d) 1/2.

Edge atoms are shared by four cells; face atoms by two:

$N=4\times \frac{1}{4}+2\times \frac{1}{2}=1+1=2.$

Answer: (b).

(2) Lattice name? Options: simple cubic, bcc, fcc, diamond.

Shift the origin to one edge atom: the other atoms are then at the corners and centres of cells of a displaced cube, i.e. every point is a corner or body centre of a cubic cell. One atom at each corner and one at the centre is bcc. Answer: (b).

(3) Miller indices of the plane with intercepts $x=a$, $y=-2a$, $z=3a$.

$\text{intercepts}=1,-2,3$ $\text{reciprocals}=1,-\frac{1}{2},\frac{1}{3}$ $\times 6(\text{lowest common denominator})\Rightarrow 6,-3,2$

So $(6\bar{3}2)$. Answer: (e).

(4) Which direction is not perpendicular to $[100]$? Options: $[0\bar{1}1]$, $[032]$, $[001]$, $[\bar{1}00]$.

Perpendicular means dot product zero; $[100]\cdot [uvw]=u$. Only $[\bar{1}00]$ has nonzero first index ($-1$). Answer: (d).

(5) The occupation probability at $E_C+kT$ is $e^{-10}$. Find $E_F$.

$\frac{1}{1+e^{(E_C+kT-E_F)/kT}}=e^{-10}$ $e^{(E_C+kT-E_F)/kT}=e^{10}-1\approx e^{10}$ $E_C+kT-E_F\approx 10kT$ $E_C-E_F\approx 9kT.$

Answer: (b).

(6) Dominant scattering in lightly doped Si at room temperature: (b) lattice scattering (phonons); impurity scattering matters only at high doping or low $T$.

(7) $D_N$ versus $N_D$ in Si at 300 K. For nondegenerate doping $D_N=(kT/q){\mu }_n$, so the curve has the same shape as ${\mu }_n$ vs $N_D$ (Fig. 3.5): flat at low doping (lattice-scattering limited), then falling as impurity scattering grows. Answer: sketch (c).

(8) Donor in the bonding model: a group-V atom (five valence electrons) substituting for Si; four electrons form bonds, the fifth is a weakly bound extra electron orbiting the ion core (drawn as an extra, unpaired electron on that atom).

(9) Freeze-out in band-diagram language: as $T\to 0$ K the electrons in the conduction band (that came from the donors at $E_D$, just below $E_C$) lose thermal energy and drop back onto the donor level $E_D$; no electrons remain in $E_C$. (Sketch: $E_C$ empty, $E_D$ filled with electrons, $E_V$ filled.)

(10) Resistivities:

${\rho }_n=\frac{1}{q{\mu }_n{N}_D},{\rho }_p=\frac{1}{q{\mu }_p{N}_A}.$

Because ${\mu }_n>{\mu }_p$ in a given semiconductor at a given temperature, for $N_D=N_A$ we get ${\rho }_p>{\rho }_n$: the $n$-type resistivity is smaller.

(11) Recombination via R-G centres: an electron from $E_C$ drops to the trap level $E_T$ (near midgap), then falls from $E_T$ into $E_V$ (equivalently, the trap captures a hole). Sketch: $E_C$, dashed $E_T$, $E_V$ with two downward arrows.

(12) Drift velocity is the average electron or hole velocity resulting from an applied electric field. At low fields $v_d=\mu \mathcal{E}$, with $\mu$ the mobility.

II. Interpretation of an energy band diagram

The diagram (figure lost in OCR) is for a Si device at 300 K with a single flat $E_F$. From the answer key: $E_C$ and $E_V$ are flat on the ends and drop linearly with slope $E_G/L$ over the middle region $L/3\le x\le 2L/3$ (total drop $E_G/3$), and $E_F$ approaches a band edge near $x=L$. Where numeric values are needed I follow the answer key’s stated assumptions.

(13) $V=-E_i/q$: the potential has the shape of the band diagram turned upside down. Answer: (d).

(14) $\mathcal{E}=\frac{1}{q}\frac{d{E}_i}{dx}$ is proportional to the band slope: zero, constant, zero. Answer: (e).

(15) Equilibrium prevails because $E_F$ is invariant with position. Answer: (a).

(16) Degenerate where $E_F$ is at or within a few $kT$ of a band edge: near $x=L$. Answer: (c).

(17) At $x=x_2$ the Fermi level is $E_G/3$ from the band edge, hence $E_G/2-E_G/3=E_G/6$ from midgap:

$|E_i-E_F|=\frac{E_G}{2}-\frac{E_G}{3}=\frac{E_G}{6}$

$\text{concentration}=n_i{e}^{E_G/(6kT)}=10^{10}{e}^{1.12/(6\times 0.0259)}=10^{10}{e}^{7.21}=1.35\times 10^{13}\mathrm{c}{\mathrm{m}}^{-3}.$

Answer: (b). (The key is applied as printed; the minority density would be about $n_i^2/1.35\times 10^{13}\approx 7.4\times 10^6\mathrm{c}{\mathrm{m}}^{-3}$, which is the distractor (a).)

(18) Equilibrium implies no net current: $J_N=0$. Answer: (a).

(19) At the point where $E_i=E_F$ we have $p=n_i$. The field in the sloped region:

$\mathcal{E}=\frac{1}{q}\frac{\Delta E_i}{\Delta x}=\frac{1}{q}\frac{E_G/3}{L/3}=\frac{E_G}{qL}$

$J_P{|}_{drift}=q{\mu }_{p}p\mathcal{E}=q{\mu }_p{n}_i\frac{E_G}{qL}=\frac{{\mu }_p{n}_i{E}_G}{L}.$

Answer: (b).

(20) Hole kinetic energy is the distance from the hole’s energy (drawn on the diagram) to the valence-band edge, $E_G/3$ per the key. Answer: (c).

III. Problem solving

An $n$-type Si bar of length $L$ is maintained at room temperature; boundary conditions $\Delta p_n(0)=\Delta p_{n0}$ and $\Delta p_n(L)=0$; steady state, no photogeneration, $\Delta p_{n0}\le n_0$; $N_D=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ (recovered from the answer key).

(21) Equilibrium holes:

$n_0\approx N_D=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ $p_0=\frac{n_i^2}{n_0}=\frac{10^{20}}{10^{15}}=10^5\mathrm{c}{\mathrm{m}}^{-3}.$

Answer: (c).

(22) $\Delta p_n{|}_{max}=\Delta p_{n0}\le n_0$, so low-level injection holds. Answer: (a).

(23) The minority-carrier diffusion equation is

$\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L.$

Steady state ($\partial /\partial t=0$) and no photogeneration ($G_L=0$) leave

$0=D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}⟺0=\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{L_P^2}.$

Answer: (e).

(24) Second-order linear ODE with characteristic roots $\pm 1/L_P$:

$\Delta p_n(x)=A{e}^{-x/L_P}+B{e}^{x/L_P},L_P^2=D_P{\tau }_p.$

Answer: (a).

(25) The boundary conditions are those stated for the bar: $\Delta p_n(0)=\Delta p_{n0}$ and $\Delta p_n(L)=0$. Answer: (c).

Completing the problem (bonus). Applying them:

$A+B=\Delta p_{n0},A{e}^{-L/L_P}+B{e}^{L/L_P}=0$

$\Delta p_n(x)=\Delta p_{n0}\frac{\sinh [(L-x)/L_P]}{\sinh (L/L_P)}.$

For $L\ll L_P$ this becomes the straight line $\Delta p_{n0}(1-x/L)$, and for $L\gg L_P$ it becomes $\Delta p_{n0}{e}^{-x/L_P}$.

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End of Part I supplement. Part II (Chapter 5, pn junctions) begins on the next page of the source.

Part II: pn Junction Diodes and Chapter 5: pn Junction Electrostatics

Source slice. Pierret, Semiconductor Device Fundamentals, Part II opener (Part IIA: Junction Diodes) and Chapter 5, book pages 195-234. Equation numbers (5.x) follow the book. OCR damage has been repaired using physics; wherever the original text or a figure was unreadable, the assumption made is stated explicitly.

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Part II Opener: Where We Are Going

Part I taught us the materials: how electrons and holes are born, how they drift and diffuse. Part II is where we build machines from them. The machines are built from one magical structure: the pn junction, the place where p-type meets n-type material.

Think of Part II as a family tree.

Family member	Where it appears	What it is
pn junction diode	Part IIA (the “charter member”)	One junction, two terminals
Optoelectronic diodes	End of Part IIA (read-only chapter)	Solar cells, LEDs, photodiodes, lasers
Bipolar junction transistor (BJT)	Part IIB	Two junctions, three terminals
PNPN devices	Part IIB	More than two junctions (thyristors)
Schottky diode	Part IIB (last chapter)	Metal-semiconductor “cousin” of the pn diode

The analysis in Part IIA matters beyond diodes: it establishes concepts and procedures of universal utility for every device in the family.

The four-step recipe of every device analysis

A complete, systematic device analysis is broken into four segments:

1. Electrostatics: the charge density $\rho$, electric field $\mathcal{E}$, and electrostatic potential $V$ inside the device under equilibrium and steady state. (This chapter.)
2. Steady-state (d.c.) response: the I-V characteristic.
3. Small-signal (a.c.) response: how it behaves for small wiggles on top of a d.c. bias.
4. Transient (pulsed) response: how it switches.

Analogy. Before you can predict how traffic flows through a city, you must first know the map: where the hills are, where the roads slope. Electrostatics is the map of potential hills and valleys that carriers must navigate. Chapters 6-8 will use this map to compute the traffic (current).

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Chapter 5: pn Junction Electrostatics

5.1 Preliminaries

5.1.1 Junction Terminology and Idealized Profiles

Thought experiment: making a junction

Take a wafer that is uniformly n-type with donor concentration $N_D$. Now diffuse a p-type dopant (acceptors, $N_A$) in from the surface. Near the surface the acceptor concentration is huge; deeper inside it decays away (Fig. 5.1(a): $N_A$ (diffused) falling with $x$, $N_D$ (original) flat).

• Near the surface where $N_A>N_D$ the semiconductor is p-type.
• Deep inside where $N_D>N_A$ it is n-type.
• The dividing plane is the metallurgical junction, defined by

$N_D-N_A=0.$

Key point. Only the net doping $N_D-N_A$ matters, both for locating the metallurgical junction and for finding the electrostatic variables. (A donor and an acceptor next to each other “cancel” in effect: the electron from the donor falls into the acceptor.) So instead of two superimposed plots of $N_A(x)$ and $N_D(x)$, we plot the single curve $N_D-N_A$ versus $x$ (Fig. 5.1(b)). That plot is the doping profile.

Idealized profiles

Real diffusion and ion-implantation profiles make the mathematics miserable. Luckily only the doping very near the metallurgical junction matters, so surprisingly accurate results come from idealizations (Fig. 5.2):

Idealization	Profile	When appropriate
Step junction	$N_D-N_A=-N_A$ for $x<0$, $+N_D$ for $x>0$ (abrupt jump)	Ion implantation or shallow diffusion into a lightly doped wafer
Linearly graded junction	$N_D-N_A=ax$	Deep diffusion into a moderately to heavily doped wafer

Which is better depends on the slope of the real profile at the metallurgical junction and the background doping of the starting wafer. In most of the book we use the step junction purely to keep the algebra manageable.

5.1.2 Poisson’s Equation

From electromagnetism, the 3-D form is

\nabla\cdot\boldsymbol{\mathcal{E}} = \frac{\rho}{K_S\varepsilon_0} \tag{5.1}

In one dimension, $\mathcal{E}={\mathcal{E}}_x$:

\frac{d\mathcal{E}}{dx} = \frac{\rho}{K_S\varepsilon_0} \tag{5.2}

• $K_S$ is the semiconductor dielectric constant (11.8 for Si) and ${\epsilon }_0=8.85\times 10^{-14}$ F/cm.
• Beware the symbol: earlier, $\rho$ meant resistivity. In the electrostatics context $\rho$ is the charge density in C/cm${}^3$.

With totally ionized dopants,

\rho = q\,(p - n + N_D - N_A) \tag{5.3}

(this is the first part of Eq. (2.23), used earlier to derive charge neutrality). Far from any surface in a uniformly doped semiconductor in equilibrium, $\rho \equiv 0$. Under less restrictive conditions $\rho$ can be nonzero and position dependent.

A handy visual trick. Since $\rho \propto d\mathcal{E}/dx$, you can read the shape of $\rho (x)$ off an $\mathcal{E}$-vs-$x$ plot simply by looking at its slope: flat field means zero charge, steeply rising field means large positive charge, and so on. Conversely, the area under $\rho$ gives the change in $\mathcal{E}$. This slope/area dance between $\rho$, $\mathcal{E}$, $V$ will carry us through the whole chapter.

5.1.3 Qualitative Solution

Before any algebra: what should the answer look like? Experienced physicists always sketch first.

From the Einstein-relationship discussion (Subsection 3.2.4) we already know that non-uniform doping produces band bending and an internal field. Assume a 1-D step junction in equilibrium.

Building the equilibrium band diagram in three moves (Fig. 5.3)

1. (a) Far from the metallurgical junction the material is just uniform p-type on the left and uniform n-type on the right. Draw each region’s own band diagram: on the p-side $E_F$ sits near $E_v$; on the n-side $E_F$ sits near $E_c$.
2. (b) In equilibrium the Fermi level is a constant, independent of position. So slide the two diagrams vertically until the $E_F$ lines coincide. Now the n-side bands sit lower than the p-side bands.
3. (c) Connect $E_c$, $E_i$, $E_v$ across the middle with a smooth monotonic curve that is flat (zero slope) at both ends of the central region.

The exact shape in the middle is not known yet, only that it is monotonic and joins smoothly. That is Fig. 5.3(c) and Fig. 5.4(a).

A footnote in the text: from the band diagram alone one can only conclude that the magnitude of the field (slope of $E_c$) rises from zero on the p-side, peaks near the metallurgical boundary, and drops to zero on the n-side. The pseudo-linear shape sketched in Fig. 5.4(c) anticipates the quantitative result.

Reading the diagram for $V$, $\mathcal{E}$, $\rho$ (Fig. 5.4)

• Potential: from Subsection 3.1.5 on band bending, $V(x)$ has the same shape as the upside-down $E_c$ (or $E_i$ or $E_v$), because electron energy $=-qV$. Setting $V=0$ on the far p-side, $V$ rises to $V_{bi}$ on the n-side (Fig. 5.4(b)).
• Field: $\mathcal{E}=-dV/dx$, i.e. proportional to the slope of $E_c$ (Fig. 5.4(c)). It is negative (pointing from n to p) and nonzero only in the central region.
• Charge: $\rho =K_S{\epsilon }_{0}d\mathcal{E}/dx$: negative on the p-side (field becoming more negative as $x$ increases), positive on the n-side (Fig. 5.4(d)).

The big question: where does that charge come from?

Perform a thought experiment (Fig. 5.5). Hold a p-block and an n-block apart.

• In the p-block, mobile holes ($+$) exactly balance fixed ionized acceptor ions ($-$). In the n-block, mobile electrons ($-$) exactly balance fixed ionized donor ions ($+$). Both blocks are neutral.
• Now press them together with a perfect connection. There are enormously more holes on the p-side than on the n-side, so holes diffuse to the n-side; electrons diffuse to the p-side. Think of two rooms, one crowded with red balloons, one with blue; open the door and each color spills into the other room.
• But the dopant ions are bolted into the crystal. When the mobile carriers leave the neighborhood of the junction, they leave behind uncovered ions: negative acceptor ions on the p-side, positive donor ions on the n-side.

This is the source of the junction charge. The region around the metallurgical junction where this significant charge exists is the space charge region or depletion region (the latter because the mobile carriers there are greatly reduced, “depleted”).

The charge build-up does not continue forever. The growing field pushes carriers back (drift) while concentration gradients push them forward (diffusion). The process halts when diffusion and drift precisely balance. Individually $J_N=0$ and $J_P=0$ in equilibrium: each carrier type’s drift cancels its own diffusion.

Analogy. Think of a hiker climbing a hill while a crowd pushes from behind. The hill (potential barrier) grows until the push from the crowd (diffusion) is exactly matched by the pull back down the slope (drift).

5.1.4 The Built-in Potential ($V_{bi}$)

The voltage drop across the depletion region in equilibrium is the built-in potential $V_{bi}$. We want a computational formula.

Take a nondegenerately doped pn junction in equilibrium with $x=0$ at the metallurgical boundary, and depletion region edges at $-x_p$ (p-side) and $x_n$ (n-side).

Derivation (Eqs. 5.4-5.10)

Start from the definition of potential:

\mathcal{E} = -\frac{dV}{dx} \tag{5.4}

Integrate across the depletion region:

${\int }_{-x_p}^{x_n}\mathcal{E}dx=-{\int }_{V(-x_p)}^{V(x_n)}dV=-\left[V(x_n)-V(-x_p)\right]=-V_{bi}$

(the book writes this with the sign convention $V_{bi}=V(x_n)-V(-x_p)$, i.e. $-\int \mathcal{E}dx=V_{bi}$):

-\int_{-x_p}^{x_n}\mathcal{E}\,dx = \int_{V(-x_p)}^{V(x_n)} dV = V(x_n) - V(-x_p) = V_{bi} \tag{5.5}

In equilibrium the electron current is zero:

J_N = q\mu_n n\mathcal{E} + qD_N\frac{dn}{dx} = 0 \tag{5.6}

Solve for $\mathcal{E}$ and use the Einstein relation $D_N/{\mu }_n=kT/q$:

\mathcal{E} = -\frac{D_N}{\mu_n}\frac{dn/dx}{n} = -\frac{kT}{q}\frac{dn/dx}{n} \tag{5.7}

Substitute (5.7) into (5.5) and integrate:

V_{bi} = -\int_{-x_p}^{x_n}\mathcal{E}\,dx = \frac{kT}{q}\int_{n(-x_p)}^{n(x_n)}\frac{dn}{n} = \frac{kT}{q}\ln\frac{n(x_n)}{n(-x_p)} \tag{5.8}

Notice that nothing here required a step junction, only nondegenerate doping. Now specialize to a nondegenerate step junction:

n(x_n) = N_D \tag{5.9a}

n(-x_p) = \frac{n_i^2}{p(-x_p)} = \frac{n_i^2}{N_A} \tag{5.9b}

(on the p-side, $p=N_A$ and $n=n_i^2/N_A$). Therefore

\boxed{V_{bi} = \frac{kT}{q}\ln\!\left(\frac{N_AN_D}{n_i^2}\right)} \tag{5.10}

Sample computation. $N_A=N_D=10^{15}$/cm${}^3$, Si at 300 K ($n_i=10^{10}$/cm${}^3$, $kT/q=0.0259$ V):

$V_{bi}=0.0259\ln \left(\frac{10^{15}\cdot 10^{15}}{10^{20}}\right)=0.0259\ln (10^{10})=0.0259(23.03)\approx 0.6\text{V}$

This is typical. In nondegenerately doped diodes $V_{bi}<E_G/q$. Nondegenerately doped Ge, Si, GaAs diodes at room temperature have $V_{bi}$ below 0.66 V, 1.12 V, 1.42 V respectively.

Alternative derivation from the band diagram

From Figs. 5.4(a) and (b), $q{V}_{bi}$ is the total band bending. Since $E_i$ (the reference for electron energy) drops by $q{V}_{bi}$ from the p-side to the n-side:

V_{bi} = V(x_n) - V(-x_p) \tag{5.11a}

qV_{bi} = (E_i)_{p\text{-side}} - (E_i)_{n\text{-side}} \tag{5.11b}

Since $E_F$ is constant, add and subtract $E_F$ on the right:

$q{V}_{bi}=(E_i-E_F{)}_{p\text{-side}}+(E_F-E_i{)}_{n\text{-side}}$

or

V_{bi} = \frac{1}{q}\left[(E_i - E_F)_{p\text{-side}} + (E_F - E_i)_{n\text{-side}}\right] \tag{5.12}

In a nondegenerate diode each term is less than $E_G/2$, so $q{V}_{bi}<E_G$, which explains the bound above. For a nondegenerate step junction,

(E_i - E_F)_{p\text{-side}} = kT\ln(N_A/n_i) \tag{5.13a}

(E_F - E_i)_{n\text{-side}} = kT\ln(N_D/n_i) \tag{5.13b}

Inserting (5.13) into (5.12):

$V_{bi}=\frac{kT}{q}\left[\ln \frac{N_A}{n_i}+\ln \frac{N_D}{n_i}\right]=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2},$

which is (5.10). Important: (5.13), hence (5.10), need nondegeneracy, but (5.12) has no doping restrictions.

Common confusion. “If $V_{bi}\approx 0.6$ V exists inside the diode, why can’t I measure it with a voltmeter or light a bulb?” Because every metal-semiconductor contact has its own built-in potential, and these exactly cancel $V_{bi}$ around the closed loop (a diode in equilibrium cannot deliver free energy; the Fermi level is flat everywhere).

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Exercise 5.1: $V_{bi}$ for $p^{+}$-n and $n^{+}$-p junctions (computer-based)

Problem. Most real diodes are heavily doped on one side. In computing $V_{bi}$ of $p^{+}$-n and $n^{+}$-p step junctions, it is common practice to assume that the Fermi level on the heavily doped side is at the band edge: $E_F=E_v$ in a $p^{+}$ material and $E_F=E_c$ in an $n^{+}$ material. With this assumption, compute and plot $V_{bi}$ versus the doping ($N_A$ or $N_D$) on the lightly doped side of Si $p^{+}$-n and $n^{+}$-p step junctions at 300 K, for $10^{14}/{\text{cm}}^3\le N_A$ or $N_D\le 10^{17}/{\text{cm}}^3$.

Solution logic.

• On the $p^{+}$ side, $E_F=E_v$ so $(E_i-E_F{)}_{p\text{-side}}=E_i-E_v=E_G/2$ (midgap).
• On the lightly doped n-side, $(E_F-E_i{)}_{n\text{-side}}=kT\ln (N_D/n_i)$.
• Use (5.12), valid for arbitrary doping:

$V_{bi}=\frac{1}{q}\left[\frac{E_G}{2}+kT\ln \frac{N_D}{n_i}\right]=\frac{E_G}{2q}+\frac{kT}{q}\ln \frac{N_D}{n_i}$

For $n^{+}$-p junctions replace $N_D$ by $N_A$ (same computation by symmetry).

Hand check. At $N_D=10^{14}$: $V_{bi}=0.56+0.0259\ln (10^4)=0.56+0.2385=0.80$ V. At $N_D=10^{17}$: $0.56+0.0259\ln (10^7)=0.56+0.417=0.98$ V. This matches the plot’s range (0.8 to 0.98 V, gently rising, almost a straight line on the semilog axis).

MATLAB. logspace(14,17) gives 50 log-spaced doping values; the formula is vectorized with .* and ./; semilogx gives a logarithmic horizontal axis.

% Vbi computation (p+/n and n+/p junctions)
% Constants
EG = 1.12;      % band gap, eV
kT = 0.0259;    % thermal energy, eV
ni = 1.0e10;    % intrinsic concentration, cm^-3
% Computation
ND = logspace(14,17);
Vbi = EG/2 + kT.*log(ND./ni);
% Plotting
close
semilogx(ND,Vbi); grid
axis([1.0e14 1.0e17 0.75 1])
xlabel('NA or ND (cm^-3)'); ylabel('Vbi (volts)')
text(1e16,0.8,'Si, 300K')
text(1e16,0.78,'p+/n and n+/p diodes')

Result (Fig. E5.1). A slowly rising, nearly straight line on a semilog plot from about 0.80 V at $10^{14}$ to about 0.98 V at $10^{17}$. Intuition: each decade of doping on the light side moves $E_F$ by $kT\ln 10=59.6$ meV, so $V_{bi}$ rises by about 60 mV per decade.

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5.1.5 The Depletion Approximation

Now we need to actually solve Poisson’s equation. Rewrite it:

\frac{d\mathcal{E}}{dx} = \frac{\rho}{K_S\varepsilon_0} = \frac{q}{K_S\varepsilon_0}\left(p - n + N_D - N_A\right) \tag{5.14}

The doping profile $N_D-N_A$ is known. But $p(x)$ and $n(x)$ in the region where $\rho \ne 0$ are not known in advance; they depend on the potential, which is what we are solving for (see the band diagram, Fig. 5.4(a)). This is a chicken-and-egg problem. Exact closed-form solutions exist in certain cases (they are complicated; see Problem 5.14), but the depletion approximation sidesteps the issue and is “far and away the most important and most widely encountered of the simplifying approximations used in the modeling of devices.”

Statement of the approximation (Fig. 5.6)

1. In a region $-x_p\le x\le x_n$ straddling the metallurgical junction, the carrier concentrations are assumed negligible compared to the net doping: $p,n\ll |N_D-N_A|$.
2. Outside that region, the charge density is taken to be identically zero (quasineutral).

Within the depletion region this gives $\rho =q(N_D-N_A)$, and Poisson’s equation becomes

\frac{d\mathcal{E}}{dx} = \frac{q(N_D - N_A)}{K_S\varepsilon_0}, \quad -x_p\le x\le x_n \tag{5.15a}

\frac{d\mathcal{E}}{dx} = 0, \quad x\le -x_p \ \text{and}\ x \ge x_n \tag{5.15b}

Analogy. The real charge distribution has soft, fuzzy edges: carrier numbers drop gradually. The approximation replaces the fuzzy edge by a razor-sharp cliff: “Inside this box, no carriers. Outside it, perfect neutrality.” Fig. 5.6(b) shows the exact $\rho$ (smooth, rounded) and the depletion-approximation $\rho$ (sharp rectangles) for a step junction. The picture is crude but it captures the area (total charge) and location of the real charge, which is what matters for $\mathcal{E}$ and $V$.

Everything about $\rho$ is now specified except the unknown edges $-x_p$, $x_n$. Moreover, inside the depletion region $\rho$ has precisely the same functional form as $N_D-N_A$ (Fig. 5.7 shows two moderately complicated examples): to get $\rho (x)$ just copy the doping profile shape (times $q$) inside the box.

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5.2 Quantitative Electrostatic Relationships

Now the main event: quantitative relationships. We emphasize the step junction, and give every derivation in detail, partly to get the formulas but also to establish procedures that apply to any doping profile. The plan: set up the system, solve the step junction at $V_A=0$, extend to $V_A\ne 0$, and finally treat the linearly graded junction.

5.2.1 Assumptions and Definitions

The diode (Fig. 5.8):

• All variables depend only on $x$, the coordinate normal to the surface: the device is one-dimensional (edge effects at lateral ends are ignored).
• $x=0$ sits at the metallurgical boundary.
• The end contacts are ohmic: by definition a negligible portion of an applied voltage appears across an ohmic contact.
• $V_A$ denotes the applied voltage (subscript A distinguishes it from internal junction voltages). It is positive when the p-terminal is positive relative to the n-terminal.
• To start, $V_A=0$ (equilibrium).

5.2.2 Step Junction with $V_A=0$

Take $N_A>N_D$ for illustration (Fig. 5.9(a)).

Solution for $\rho$

By the depletion approximation:

\rho = \begin{cases} -qN_A, & -x_p\le x\le 0 \\ +qN_D, & 0\le x\le x_n \\ 0, & x\le -x_p \text{ and } x\ge x_n \end{cases} \tag{5.16a-c}

($x_n$, $x_p$ unknown for now.)

Solution for $\mathcal{E}$

Poisson’s equation gives

\frac{d\mathcal{E}}{dx} = \begin{cases} -qN_A/(K_S\varepsilon_0), & -x_p\le x\le 0 \\ +qN_D/(K_S\varepsilon_0), & 0\le x\le x_n \\ 0, & x\le -x_p \text{ and } x\ge x_n\end{cases} \tag{5.17a-c}

Boundary conditions: $\mathcal{E}=0$ far from the junction, thus $\mathcal{E}=0$ everywhere outside the depletion region; and since $\mathcal{E}$ can’t jump, $\mathcal{E}(-x_p)=0$ and $\mathcal{E}(x_n)=0$.

p-side. Separate variables and integrate from the edge $-x_p$ to a point $x$:

\int_0^{\mathcal{E}(x)} d\mathcal{E}' = -\int_{-x_p}^{x}\frac{qN_A}{K_S\varepsilon_0}\,dx' \tag{5.18}

\mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}\left[x' \right]_{-x_p}^{x} \;\Rightarrow\; \mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}(x + x_p), \quad -x_p\le x\le 0 \tag{5.19}

n-side. Integrate from $x$ to the edge $x_n$:

\int_{\mathcal{E}(x)}^{0} d\mathcal{E}' = \int_{x}^{x_n}\frac{qN_D}{K_S\varepsilon_0}\,dx' \tag{5.20}

-\mathcal{E}(x) = \frac{qN_D}{K_S\varepsilon_0}(x_n - x) \;\Rightarrow\; \mathcal{E}(x) = -\frac{qN_D}{K_S\varepsilon_0}(x_n - x), \quad 0\le x\le x_n \tag{5.21}

So the field is always negative inside the depletion region, zero outside, and linear in position with a “peak” (largest magnitude) at $x=0$. It looks like a tent/triangle (Fig. 5.9(c)). This agrees with the qualitative sketch of Fig. 5.4(c).

Continuity at $x=0$. The field is continuous across a boundary unless a sheet of charge lies on it, and there is none. Equating (5.19) and (5.21) at $x=0$:

-\frac{qN_A}{K_S\varepsilon_0}x_p = -\frac{qN_D}{K_S\varepsilon_0}x_n \;\Rightarrow\; \boxed{N_Ax_p = N_Dx_n} \tag{5.22}

Gauss’ law view. $\mathcal{E}=0$ outside means the total charge in the depletion region is zero: $-q{N}_A{x}_p\cdot (\text{area})+q{N}_D{x}_n\cdot (\text{area})=0$, i.e. the rectangular areas in Fig. 5.9(b) are equal. So (5.22) is just charge balance. A see-saw: the more heavily doped side has the shorter arm.

Solution for $V$

Since $\mathcal{E}=-dV/dx$, with (5.19), (5.21):

\frac{dV}{dx} = \begin{cases}\dfrac{qN_A}{K_S\varepsilon_0}(x + x_p), & -x_p\le x\le 0 \\[2mm] \dfrac{qN_D}{K_S\varepsilon_0}(x_n - x), & 0\le x\le x_n\end{cases} \tag{5.23a,b}

Take the arbitrary reference $V(-x_p)=0$; the total drop across the depletion region in equilibrium is $V_{bi}$:

V(-x_p) = 0, \qquad V(x_n) = V_{bi} \tag{5.24a,b}

p-side. Integrate from $-x_p$ to $x$:

\int_0^{V(x)}dV' = \frac{qN_A}{K_S\varepsilon_0}\int_{-x_p}^{x}(x'+x_p)\,dx' \tag{5.25}

V(x) = \frac{qN_A}{K_S\varepsilon_0}\left[\frac{(x'+x_p)^2}{2}\right]_{-x_p}^{x} = \frac{qN_A}{2K_S\varepsilon_0}(x+x_p)^2, \quad -x_p\le x\le 0 \tag{5.26}

n-side. Integrate from $x$ to $x_n$:

\int_{V(x)}^{V_{bi}}dV' = \frac{qN_D}{K_S\varepsilon_0}\int_x^{x_n}(x_n - x')\,dx' \tag{5.27}

$V_{bi}-V(x)=\frac{q{N}_D}{K_S{\epsilon }_0}{\left[-\frac{(x_n-x^{\prime }{)}^2}{2}\right]}_x^{x_n}=\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2$

V(x) = V_{bi} - \frac{qN_D}{2K_S\varepsilon_0}(x_n - x)^2, \quad 0\le x\le x_n \tag{5.28}

$V(x)$ is quadratic: it rises with an upward-opening (concave-up, “smile”) curvature on the p-side and an upside-down (concave-down) curvature on the n-side, giving the S-shaped hill of Fig. 5.9(d). Continuity of $V$ at $x=0$ is justified because there is no dipole layer (closely spaced sheets of plus and minus charge) at the boundary. Equating (5.26) and (5.28) at $x=0$:

\frac{qN_A}{2K_S\varepsilon_0}x_p^2 = V_{bi} - \frac{qN_D}{2K_S\varepsilon_0}x_n^2 \tag{5.29}

Solving for $x_n$ and $x_p$

The only unknowns in (5.22) and (5.29) are $x_n$, $x_p$. Rewrite (5.29) as

$V_{bi}=\frac{q}{2K_S{\epsilon }_0}\left(N_A{x}_p^2+N_D{x}_n^2\right)$

Eliminate $x_p$ with (5.22), $x_p=(N_D/N_A)x_n$:

$V_{bi}=\frac{q}{2K_S{\epsilon }_0}\left(\frac{N_D^2}{N_A}{x}_n^2+N_D{x}_n^2\right)=\frac{q{N}_D}{2K_S{\epsilon }_0}\frac{N_D+N_A}{N_A}{x}_n^2$

Solve for $x_n$, and do the same for $x_p$ (using $x_n=(N_A/N_D)x_p$):

\boxed{x_n = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\,\frac{N_A}{N_D(N_A+N_D)}\right]^{1/2}} \tag{5.30a}

\boxed{x_p = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\,\frac{N_D}{N_A(N_A+N_D)}\right]^{1/2}} \tag{5.30b}

The total depletion width $W$:

$W=x_n+x_p=x_n\left(1+\frac{N_D}{N_A}\right)=x_n\frac{N_A+N_D}{N_A}$

\boxed{W = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\left(\frac{N_A+N_D}{N_AN_D}\right)\right]^{1/2}} \tag{5.31}

$W$ is what you usually need in practical device computations. Another quick result: peak field, from (5.21) at $x=0$:

$|\mathcal{E}{|}_{max}=|\mathcal{E}(0)|=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}$

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Exercise 5.2: How big are $W$ and $|\mathcal{E}{|}_{max}$?

Problem. Si step junction at 300 K with $N_A=10^{17}$/cm${}^3$ and $N_D=10^{14}$/cm${}^3$. Compute $W$ and $|\mathcal{E}{|}_{max}$ at equilibrium.

Solution.

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}=0.0259\ln \frac{(10^{17})(10^{14})}{10^{20}}=0.0259\ln (10^{11})=0.656\text{V}$

Use $K_S{\epsilon }_0=11.8(8.85\times 10^{-14})=1.044\times 10^{-12}$ F/cm. By (5.30a), since $N_A\gg N_D$, $N_A/(N_A+N_D)\approx 1$:

$x_n={\left[\frac{2(11.8)(8.85\times 10^{-14})(0.656)}{(1.6\times 10^{-19})(10^{14})}\right]}^{1/2}=2.93\times 10^{-4}\text{cm}=2.93\mu \text{m}$

$x_p=\frac{N_D}{N_A}{x}_n=10^{-3}{x}_n=2.93\times 10^{-7}\text{cm}$

$W=x_n+x_p\approx x_n=2.93\mu \text{m}$

$|\mathcal{E}{|}_{max}=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{(1.6\times 10^{-19})(10^{14})(2.93\times 10^{-4})}{(11.8)(8.85\times 10^{-14})}=4.49\times 10^3\text{V/cm}$

(One could have used (5.31) directly. The purpose of computing $x_n$, $x_p$ separately is to make a point.)

Lesson. In an asymmetrically doped junction ($N_A\gg N_D$ or $N_D\gg N_A$) the depletion region lies almost entirely on the lightly doped side. Here the p-side depletion is 0.3 nm worth of … precisely, 2.93 nm versus 2.93 micrometers on the n-side. Intuition: the lightly doped side has few dopant ions per unit length, so it must expose a long stretch to supply the same total charge.

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5.2.3 Step Junction with $V_A\ne 0$

To be practical we must handle $V_A\ne 0$. We could repeat everything, but there is a shortcut.

Apply $V_A$ to the diode terminals (Fig. 5.10). The voltage must drop somewhere inside:

• A negligible portion drops across the ohmic contacts (by definition).
• Under low-level injection (reasonable currents) the resistive drop across the quasineutral p- and n-regions between the contacts and the depletion edges is also negligible.
• Therefore all of $V_A$ is dropped across the depletion region.

When $V_A>0$ (forward bias, p-terminal positive) the potential of the n-side is lowered relative to the p-side, reducing the hill. When $V_A<0$ (reverse bias), the n-side potential rises relative to the p-side, increasing the hill. So the voltage drop across the depletion region becomes

$V(x_n)-V(-x_p)=V_{bi}-V_A$

The only change in the mathematical problem is one boundary condition (5.24b). Hence the $V_A\ne 0$ solution follows from the $V_A=0$ solution by replacing every explicit $V_{bi}$ with $V_{bi}-V_A$. Summary:

For $-x_p\le x\le 0$:

\mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}(x_p + x) \tag{5.32}

V(x) = \frac{qN_A}{2K_S\varepsilon_0}(x_p + x)^2 \tag{5.33}

x_p = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_D}{N_A(N_A+N_D)}\right]^{1/2} \tag{5.34}

For $0\le x\le x_n$:

\mathcal{E}(x) = -\frac{qN_D}{K_S\varepsilon_0}(x_n - x) \tag{5.35}

V(x) = V_{bi} - V_A - \frac{qN_D}{2K_S\varepsilon_0}(x_n - x)^2 \tag{5.36}

x_n = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_A}{N_D(N_A+N_D)}\right]^{1/2} \tag{5.37}

and

\boxed{W = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_A+N_D}{N_AN_D}\right]^{1/2}} \tag{5.38}

To avoid an imaginary square root, $V_A\le V_{bi}$ is obviously required in (5.34), (5.37), (5.38). The formulation actually fails before this: as $V_A\to V_{bi}$ a large current flows, and the voltage drops across the quasineutral regions can no longer be neglected.

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Exercise 5.3: $W$ versus doping at several biases (computer-based)

Problem. Construct a log-log plot of the depletion width $W$ versus the impurity concentration ($N_A$ or $N_D$) on the lightly doped side of Si $p^{+}$-n and $n^{+}$-p step junctions at 300 K. Include curves for $V_A=0.5$ V, 0 V, $-10$ V over $10^{14}\le N_A$ or $N_D\le 10^{17}$/cm${}^3$.

Solution logic.

1. $V_{bi}$ comes from the Exercise 5.1 relation, $V_{bi}=E_G/2q+(kT/q)\ln (N_B/n_i)$, where $N_B$ is the light-side doping.
2. For a strongly one-sided junction the doping factor in (5.38) simplifies:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{1}{N_D}+\frac{1}{N_A}\approx \frac{1}{N_B}$

(the heavily doped term is negligible).

3. So $W\approx {\left[\frac{2K_S{\epsilon }_0}{q{N}_B}(V_{bi}-V_A)\right]}^{1/2}$.

Hand check at $N_B=10^{15}$: $V_{bi}=0.56+0.0259\ln (10^5)=0.858$ V.

• $V_A=0$: $W=\sqrt{(1.305\times 10^7)(0.858)/10^{15}}=1.06\mu$m.
• $V_A=0.5$ V: $(V_{bi}-V_A)=0.358$, $W=0.68\mu$m.
• $V_A=-10$ V: $(V_{bi}-V_A)=10.86$, $W=3.77\mu$m.

(Here $2K_S{\epsilon }_0/q=1.305\times 10^7$ cm${}^{-1}\cdot$V${}^{-1}\cdot$cm${}^{-2}$ units, i.e. $W^2=(1.305\times 10^7)(V)/N_B$ in cm${}^2$.)

MATLAB logic. Build vectors of doping (logspace), a three-element bias vector, and a $3\times 50$ matrix W whose rows are the three bias curves. Multiply by $10^4$ to convert cm to micrometers. loglog(NB,W,'-') plots all three rows at once. (The OCR rendering of VA=[0.5 0 -10] and of the e0 variable name was garbled; restored below.)

% This program calculates and plots the depletion width vs impurity
% concentration in Silicon p+/n and n+/p step junctions at 300K.
% Three plots are generated corresponding to VA=0.5V, 0.0V and -10V
% The Vbi relationship employed is Vbi=(EG/2q)+(kT/q)ln(NB/ni)
% where NB is the impurity concentration on the lightly doped side.
 
% Constants and Parameters
T  = 300;        % Temperature in Kelvin
k  = 8.617e-5;   % Boltzmann constant (eV/K)
e0 = 8.85e-14;   % permittivity of free space (F/cm)
q  = 1.602e-19;  % charge on an electron (coul)
KS = 11.8;       % dielectric constant of Si at 300K
ni = 1e10;       % intrinsic carrier conc. in Silicon at 300K (cm^-3)
EG = 1.12;       % band gap of Silicon (eV)
 
% Choose variable values
NB = logspace(14,17);   % doping ranges from 1e14 to 1e17
VA = [0.5 0 -10];       % VA values set
 
% Depletion width calculation
Vbi = EG/2 + k*T.*log(NB./ni);
W(1,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(1))./NB);
W(2,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(2))./NB);
W(3,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(3))./NB);
 
% Plot
close
loglog(NB,W,'-'); grid
axis([1.0e14 1.0e17 1.0e-1 1.0e1])
xlabel('NA or ND (cm^-3)')
ylabel('W (micrometers)')
set(gca,'DefaultTextUnits','normalized')
text(.38,.26,'VA=0.5V')
text(.38,.50,'VA=0')
text(.38,.76,'VA=-10V')
text(.77,.82,'Si, 300K')
text(.77,.79,'p+/n and n+/p')
set(gca,'DefaultTextUnits','data')

Result (Fig. E5.3). Three nearly parallel straight lines of slope about $-1/2$ on log-log axes (because $W\propto N_B^{-1/2}$; the slight curvature comes from the weak $\ln$ dependence of $V_{bi}$). The reverse-bias curve ($-10$ V) lies highest, forward bias ($0.5$ V) lowest. Depletion widths span roughly 0.1 to 10 micrometers over this range.

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5.2.4 Examination and Extrapolation of Results

How do the electrostatic variables respond to bias?

• Depletion width. From (5.34) and (5.37), $x_p$ and $x_n$ shrink under forward bias ($V_A>0$) and grow under reverse bias ($V_A<0$), consistent with Exercise 5.3.
• Field. From (5.32), (5.35), smaller $x_p$, $x_n$ under forward bias lower $|\mathcal{E}|$ everywhere in the depletion region; reverse bias gives larger $x_p$, $x_n$ and a larger $|\mathcal{E}|$. Physically: less space charge means weaker field, more space charge means stronger field.
• Potential. From (5.33), (5.36) the potential decreases everywhere for $V_A>0$ and increases everywhere for $V_A<0$. The potential hill shrinks in both height and width under forward bias, and grows taller and wider under reverse bias (Fig. 5.11).

Analogy. Reverse bias is like heightening a dam: more water (carriers) are held back and the dam wall must be wider and thicker (more exposed charge). Forward bias is like lowering the dam gate.

Energy band diagrams under bias (Fig. 5.12)

Run the Subsection 3.1.5 procedure in reverse: take the “upside-down” potential plots of Fig. 5.11(d) and modify the equilibrium band diagram: smaller width and smaller hill for forward bias, larger width and larger hill for reverse bias.

Comments on the $V_A\ne 0$ diagrams:

1. $E_F$ is omitted from the depletion region, because the device is no longer in equilibrium and one level cannot describe both carrier populations there. The levels labeled $E_{Fp}$ and $E_{Fn}$ on the quasineutral sides are actually majority-carrier quasi-Fermi levels; the deviation from equilibrium in the neutral regions is small (especially far from the junction), so we keep calling them $E_F$.
2. By inspection of the diagrams,

E_{Fp} - E_{Fn} = -qV_A \tag{5.39}

(For $V_A>0$, $E_{Fn}$ lies above $E_{Fp}$ by $q{V}_A$.)

Visualization. Equation (5.39) says the terminals give you direct handles on the two ends of the equilibrium Fermi level. Forward bias: grab the n-side and lift it up by $q{V}_A$ while holding the p-side fixed, which shrinks the hill from $q{V}_{bi}$ to $q(V_{bi}-V_A)$. Reverse bias: pull the n-side down.

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Exercise 5.4: Equilibrium Energy Band Diagram Generator (computer-based)

Problem. Given a quantitative expression for the electrostatic potential, a fully dimensioned band diagram can be drawn. Run the “Diagram Generator” program, which prompts for $N_A$ and $N_D$. Try combinations with $N_A>N_D$, $N_A=N_D$, $N_A<N_D$. Asymmetric junctions give “one-sided” diagrams unlike typical textbooks. Consider modifying the program to draw an arbitrary applied bias.

Program logic.

1. $V_{bi}=(kT/q)\ln (N_A{N}_D/n_i^2)$ and $x_n$, $x_p$ from (5.30) are computed.
2. An $x$ array from $-3.5\mu$m to $+3.5\mu$m is built.
3. $V(x)$ is assembled piecewise: $0$ for $x<-x_p$; (5.26) for $-x_p\le x<0$; (5.28) for $0\le x\le x_n$; $V_{bi}$ for $x>x_n$. In MATLAB the logical masks (x>=0 & x<=xN) multiply each piece, turning it on only in its range.
4. Band edges are $E_i=-V+V_{MAX}/2$ (energy in eV, since $E=-qV$ and $q$ cancels), $E_c=E_i+E_G/2$, $E_v=E_i-E_G/2$, and the Fermi level is placed $kT\ln (N_A/n_i)$ below $E_i$ at the far p-side.

(The OCR damaged the line defining the n-side potential; it has been reconstructed from (5.28), and an x>xN term has been added so that $V$ stays at $V_{bi}$ beyond the depletion edge.)

% Equilibrium Energy Band Diagram Generator
% (Si, 300K, nondegenerately doped step junction)
% Constants
T  = 300;        % Temperature in Kelvin
k  = 8.617e-5;   % Boltzmann constant (eV/K)
e0 = 8.85e-14;   % permittivity of free space (F/cm)
q  = 1.602e-19;  % charge on an electron (coul)
KS = 11.8;       % Dielectric constant of Si
ni = 1.0e10;     % intrinsic carrier conc. in Silicon at 300K (cm^-3)
EG = 1.12;       % Silicon band gap (eV)
% Control constants
xleft  = -3.5e-4;  % Leftmost x position (cm)
xright = -xleft;   % Rightmost x position (cm)
NA = input('Please enter p-side doping (cm^-3), NA=');
ND = input('Please enter n-side doping (cm^-3), ND=');
% Computations
Vbi = k*T*log((NA*ND)/ni^2);
xN = sqrt(2*KS*e0/q*NA*Vbi/(ND*(NA+ND)));   % Depletion width n-side
xP = sqrt(2*KS*e0/q*ND*Vbi/(NA*(NA+ND)));   % Depletion width p-side
x = linspace(xleft,xright,200);
Vx1 = (Vbi - 0.5*q*ND.*(xN-x).^2/(KS*e0)).*(x>=0 & x<=xN) + Vbi*(x>xN);
Vx2 = 0.5*q*NA.*(xP+x).^2/(KS*e0).*(x>=-xP & x<0);
Vx = Vx1 + Vx2;                              % V as a function of x
VMAX = 3;                                    % Maximum Plot Voltage
EF = Vx(1) + VMAX/2 - k*T*log(NA/ni);        % Fermi level
% Plot Diagram
close
plot(x,-Vx+EG/2+VMAX/2);                     % Ec
axis([xleft xright 0 VMAX]);
axis('off'); hold on
plot(x,-Vx-EG/2+VMAX/2);                     % Ev
plot(x,-Vx+VMAX/2,'w:');                     % Ei (dotted)
plot([xleft xright],[EF EF],'w--');          % EF
plot([0 0],[0.15*VMAX 0.5*VMAX],'w-');       % metallurgical junction marker
text(xleft*1.08,(-Vx(1)+EG/2+VMAX/2-.05),'Ec');
text(xright*1.02,(-Vx(200)+EG/2+VMAX/2-.05),'Ec');
text(xleft*1.08,(-Vx(1)-EG/2+VMAX/2-.05),'Ev');
text(xright*1.02,(-Vx(200)-EG/2+VMAX/2-.05),'Ev');
text(xleft*1.08,(-Vx(1)+VMAX/2-.05),'Ei');
text(xright*1.02,EF-.05,'EF');
set(gca,'DefaultTextUnits','normalized')
text(.18,0,'p-side'); text(.47,0,'x=0'); text(.75,0,'n-side');
set(gca,'DefaultTextUnits','data')
hold off

What to expect. For $N_A=N_D$ you get a symmetric S-shaped band bending. For $N_A\gg N_D$ nearly all the bending occurs on the n-side (the p-side is nearly flat, a “one-sided” diagram, with total drop $q{V}_{bi}$). To include bias, replace $V_{bi}$ by $V_{bi}-V_A$ in $x_n$, $x_p$, $V(x)$, and draw the n-side bands raised by $q{V}_A$ relative to the p-side, as in (5.39).

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5.2.5 Linearly Graded Junctions

Step junctions are the “straight-line” idealization. But real junctions made by a deep diffusion into a moderately or heavily doped wafer look very different: the doping does not jump, it slides smoothly from p-type to n-type. Picture a ramp, not a cliff. The linearly graded junction models that ramp (Fig. 5.13a) as

N_D - N_A = a\,x \tag{5.40}

where $a$ (units ${\text{cm}}^{-4}$) is the grading constant. The metallurgical junction sits at $x=0$, where the net doping is exactly zero.

Treat this subsection as an unnumbered exercise: a chance to see the procedure you learned for the step junction applied to a new profile. A few things make it pleasantly different.

1. The profile is continuous through $x=0$. So there is no need to solve the p-side and n-side separately and then stitch them together at $x=0$. There is one $\rho$, one $\mathcal{E}$, one $V$ for the whole depletion region.
2. The profile is symmetric about $x=0$ (it is an odd function), so the depletion region extends equally far to both sides, $x_n=x_p=W/2$, and all electrostatic variables inherit a symmetry.
3. A complication we choose to ignore. The doping outside the depletion region is nonuniform too, so strictly there is a small residual charge, field, and potential drop out there. We ignore this and take $\rho =0$, $\mathcal{E}=0$, $V=\text{constant}$ outside the depletion region.
4. $V_{bi}$ cannot be taken from the step-junction formula (5.10). It still has to be recomputed, and this is done at the end.

Analogy. In a step junction the two sides are like two countries with a sharp border wall. In a linearly graded junction it is a mountain slope: the “no-man’s-land” is the whole gentle gradient, and the point where it crosses zero is the middle of the pass.

Depletion approximation applied

With (5.40) and the depletion approximation, the charge density is just $q$ times the net doping inside the depletion region:

\rho(x) = \begin{cases} q\,a\,x, & -W/2 \le x \le W/2 \\[2pt] 0, & x \le -W/2 \ \text{and}\ x \ge W/2\end{cases} \tag{5.41}

Because $\rho$ is odd in $x$, the net charge is zero only if the two edges are symmetric:

x_n = x_p = \frac{W}{2} \tag{5.42}

Thought experiment. The positive charge on the right ($qax$ for $x>0$) and the negative charge on the left must cancel exactly (neutrality). Because the ramp is a straight line through the origin, the only way two triangles can have equal area is for them to have equal base. That is the whole argument for (5.42).

Electric field

Poisson’s equation, $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$, with the boundary condition $\mathcal{E}(-W/2)=0$ (integrate from the p-edge):

$\mathcal{E}(x)=\frac{qa}{K_S{\epsilon }_0}{\left[\frac{x^{\prime 2}}{2}\right]}_{-W/2}^x=\frac{qa}{2K_S{\epsilon }_0}\left(x^2-\frac{W^2}{4}\right)$

\boxed{\mathcal{E}(x) = -\frac{qa}{2K_S\varepsilon_0}\left(\frac{W^2}{4} - x^2\right), \qquad -\frac{W}{2}\le x\le\frac{W}{2}} \tag{5.43}

A linear $\rho$ gives a parabolic $\mathcal{E}$. The field is negative everywhere (pointing from n toward p, as always), zero at both edges, and most negative at the center:

$|\mathcal{E}{|}_{max}=|\mathcal{E}(0)|=\frac{qa{W}^2}{8K_S{\epsilon }_0}$

Electrostatic potential

Use $\mathcal{E}=-dV/dx$ and integrate from the p-edge, where we set $V=0$:

$V(x)=-{\int }_{-W/2}^x\mathcal{E}d{x}^{\prime }=\frac{qa}{2K_S{\epsilon }_0}{\int }_{-W/2}^x\left(\frac{W^2}{4}-x^{\prime 2}\right)d{x}^{\prime }$

$V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^2}{4}\left(x+\frac{W}{2}\right)-\frac{1}{3}\left(x^3+\frac{W^3}{8}\right)\right]$

$V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^{2}x}{4}-\frac{x^3}{3}+\frac{W^3}{8}-\frac{W^3}{24}\right]$

\boxed{V(x) = \frac{qa}{6K_S\varepsilon_0}\left(\frac{3W^2}{4}\,x - x^3 + \frac{W^3}{4}\right), \qquad -\frac W2\le x\le\frac W2} \tag{5.44}

A parabolic field gives a cubic potential. Sanity check at the n-edge, $x=W/2$:

$V\left(\frac{W}{2}\right)=\frac{qa}{6K_S{\epsilon }_0}\left(\frac{3W^3}{8}-\frac{W^3}{8}+\frac{W^3}{4}\right)=\frac{qa}{6K_S{\epsilon }_0}\cdot \frac{W^3}{2}=\frac{qa{W}^3}{12K_S{\epsilon }_0}$

Depletion width

The total potential drop across the depletion region must be $V_{bi}-V_A$. Setting $V(W/2)=V_{bi}-V_A$:

$\frac{qa{W}^3}{12K_S{\epsilon }_0}=V_{bi}-V_A$

\boxed{W = \left[\frac{12K_S\varepsilon_0}{qa}\,(V_{bi}-V_A)\right]^{1/3}} \tag{5.45}

Remember this contrast. The linearly graded depletion width varies as the cube root of $V_{bi}-V_A$, whereas the step-junction width varies as the square root. Why? In a step junction $\rho$ is uniform, so exposed charge grows linearly with $W$ and $V\propto W^2$. In a graded junction the charge density is zero at the center and grows outward, so the exposed charge grows like $W^2$ and the potential like $W^3$.

The built-in voltage of a linearly graded junction

Equation (5.10) was derived for a step junction. But the derivation of (5.8) placed no restriction on the profile except nondegeneracy, so we just re-evaluate the carrier concentrations at the two edges of the equilibrium depletion region ($W_0=W{|}_{V_A=0}$). At the edges the net doping magnitude is $a{W}_0/2$, so

n(x_n) = (N_D - N_A)\big|_{x=W_0/2} = \frac{aW_0}{2} \tag{5.46a}

p(-x_p) = -(N_D-N_A)\big|_{x=-W_0/2} = \frac{aW_0}{2} \tag{5.46b}

Substituting into (5.8), $V_{bi}=(kT/q)\ln [n(x_n)p(-x_p)/n_i^2]$:

$V_{bi}=\frac{kT}{q}\ln \left[\frac{(a{W}_0/2)(a{W}_0/2)}{n_i^2}\right]$

\boxed{V_{bi} = \frac{2kT}{q}\ln\!\left(\frac{aW_0}{2n_i}\right)} \tag{5.47}

Now insert $W_0$ from (5.45) with $V_A=0$:

\boxed{V_{bi} = \frac{2kT}{q}\ln\!\left\{\frac{a}{2n_i}\left[\frac{12K_S\varepsilon_0V_{bi}}{qa}\right]^{1/3}\right\}} \tag{5.48}

$V_{bi}$ appears on both sides. Equation (5.48) cannot be solved explicitly and must be iterated numerically: guess $V_{bi}$, compute the right-hand side, feed the result back in, and repeat. Because the dependence on $V_{bi}$ is only through the logarithm of a cube root, the iteration converges in two or three passes. (Problem 5.13 does exactly this.)

Summary table, step versus linearly graded.

Quantity	Step junction	Linearly graded
$\rho (x)$	constant on each side	linear in $x$
$\mathcal{E}(x)$	piecewise linear (triangle)	parabolic
$V(x)$	piecewise parabolic	cubic
$W\propto$	$(V_{bi}-V_A{)}^{1/2}$	$(V_{bi}-V_A{)}^{1/3}$
$V_{bi}$	closed form (5.10)	iterate (5.48)

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5.3 Summary

Take a breath and look back at how far we have come. The electrostatics of the pn junction is the foundation for the operational model of the diode and of every other device containing a pn junction.

What we did, in order.

1. Vocabulary. We defined the metallurgical junction and doping profile, and introduced the idealized step and linearly graded profiles used throughout the analysis.
2. Poisson’s equation, the usual starting point for quantitative solutions of $\rho$, $\mathcal{E}$, $V$.
3. A qualitative solution read directly from the energy band diagram: bands bend, a potential hill forms, and a space charge region appears.
4. The built-in voltage, $V_{bi}=(kT/q)\ln (N_A{N}_D/n_i^2)$ for a step junction, the height of the potential hill.
5. The depletion approximation, the most important and most widely used simplification in device modeling: the transition region is completely empty of mobile carriers and everything beyond it is perfectly neutral.
6. Quantitative solutions for $\rho$, $\mathcal{E}$, $V$ in a step junction at equilibrium, then extended to $V_A\ne 0$ by replacing $V_{bi}\to V_{bi}-V_A$.
7. Effect of bias on the electrostatic variables and the resulting band diagrams: forward bias shrinks the hill and the depletion region, reverse bias grows both.
8. A linearly graded junction, which showed that the same procedure gives parabolic fields and cubic potentials.

What you should now be able to do. Feel (not just compute) what is happening electrostatically inside a pn junction, and, given any other doping profile, set up and carry out the solution: write $\rho$ from the depletion approximation, integrate Poisson once for $\mathcal{E}$, again for $V$, and force the total potential drop to equal $V_{bi}-V_A$.

Encouragement. The ritual ”$\rho \to \mathcal{E}\to V\to$ match $V_{bi}-V_A$” is one of the most reusable ideas in this course. You will meet it again in the MOS capacitor, the MOSFET, and the Schottky diode. Master it here and later chapters will feel like old friends.

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Chapter 5 Problems

A note on the source. The OCR of the problem statements is damaged in a few places (garbled exponents, a missing figure label). Wherever a value or a figure could not be read, the reconstruction I used is stated explicitly at the start of that solution, chosen to be consistent with the problem-information table (difficulty level, topic, and “NA ~ ND” or “NA $\ge$ ND” descriptions). The method is the point; if your textbook shows different numbers, just substitute them into the same chain of steps.

Constants used throughout (Si, 300 K). $kT/q=0.0259$ V, $n_i=10^{10}{\text{cm}}^{-3}$, $K_S=11.8$, ${\epsilon }_0=8.85\times 10^{-14}$ F/cm so $K_S{\epsilon }_0=1.044\times 10^{-12}$ F/cm, $q=1.6\times 10^{-19}$ C, $E_G=1.12$ eV.

Problem information table (from the book).

Problem	After Section	Level	Description
5.1	5.2.5	1	True-or-false quiz
5.2	5.1.4	1	$V_{bi}$ from energy band
5.3	5.1.4	2	Isotype step junction
5.4	5.2.2	2	Step junction, $N_A\sim N_D$
5.5	5.2.2	2	Step junction, $N_A\ge N_D$
5.6	5.2.3	2	Compute using diary
5.7	5.2.4	3	Step junction program
5.8	5.2.5	2	Combination profile
5.9	5.2.5	2	Exponential profile
5.10	5.2.5	2-3	Modified-step profile
5.11	5.2.5	3-4	PIN diode
5.12	5.2.5	2-3	Given $V$, find $\mathcal{E}$, $\rho$, $N_D-N_A$
5.13	5.2.5	3-4	Linearly graded program
5.14	5.2.5	5	Exact solution

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Problem 5.1: True or false

Problem. Decide whether each statement is true or false (and, as a good habit, say why).

Solution. One statement at a time.

(a) The step junction is an idealized doping profile used to model $p^{+}$-n and $n^{+}$-p junctions. $\text{Heavy doping on one side and light on the other, with an abrupt switch, is exactly a step profile.}$ TRUE.

(b) The $\rho$ in Poisson’s equation is the charge density, with units of C/cm${}^3$. $[\rho ]=\frac{\text{C}}{{\text{cm}}^3},\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}\Rightarrow \frac{\text{V/cm}}{\text{cm}}=\frac{{\text{C/cm}}^3}{\text{F/cm}}✓$ TRUE.

(c) The space charge region about the metallurgical junction is due to a pile-up of electrons on the p-side and holes on the n-side. $\text{Actually, the carriers diffuse across and recombine/are swept away, leaving behind uncovered ionized dopants }(N_A^{-},N_D^{+}).$ Mobile carriers are depleted, not piled up. FALSE.

(d) The built-in potential is typically less than the band gap energy converted to volts. $q{V}_{bi}=kT\ln \frac{N_A{N}_D}{n_i^2}<E_G\text{(it approaches }{E}_G\text{ only for very heavy doping)}$ For Si, $V_{bi}\sim 0.6$-$0.9$ V $<1.12$ V. TRUE.

(e) Invoking the depletion approximation makes the charge density inside the depletion region directly proportional to the net doping concentration. $\rho \approx q(N_D-N_A)(p=n=0)$ TRUE.

(f) Ohmic contacts reduce the built-in voltage drop across a junction. $\text{Contact potentials at the two metal-semiconductor contacts exactly cancel the difference, so the total equilibrium voltage around the loop is zero; the voltage across the junction itself is unchanged.}$ FALSE.

(g) In solutions based on the depletion approximation, the magnitude of the electric field reaches a maximum right at the metallurgical boundary. $\mathcal{E}(x)\text{ is a triangle that peaks where }\rho \text{ changes sign, i.e., at }x=0.$ TRUE.

(h) If one has a $p^{+}$-n step junction, with $N_A\gg N_D$, it follows that $x_p\le x_n$. $N_A{x}_p=N_D{x}_n\Rightarrow x_p=\frac{N_D}{N_A}{x}_n\le x_n$ TRUE.

(i) The potential hill between the n-side and the p-side of a junction increases with forward biasing. $\text{Hill height}=V_{bi}-V_A\text{ decreases for }{V}_A>0.$ FALSE.

(j) The depletion width in a linearly graded junction varies as the square root of $V_{bi}-V_A$. (The OCR of this item is garbled; this is the statement it matches, consistent with the chapter’s emphasis.) $W={\left[\frac{12K_S{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}$ It varies as the cube root. FALSE.

Answer key. (a) T, (b) T, (c) F, (d) T, (e) T, (f) F, (g) T, (h) T, (i) F, (j) F.

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Problem 5.2: $V_{bi}$ from the energy band diagram

Problem. (OCR-damaged; reconstruction.) A pn junction has its Fermi level at $E_F=E_v+2kT$ on the p-side and $E_F=E_c-kT/4$ on the n-side. (a) Draw the equilibrium energy band diagram. (b) Determine the built-in voltage, symbolically and numerically.

Assumption note. The garbled text reads ”$E_F=E-2kT$ on the p-side and $E_F=E-/4$ on the n-side.” I interpret it as above, with the reference levels being the band edges on each side. Note that a Fermi level only $kT/4$ below $E_c$ means the n-side is just on the edge of degeneracy, so we are using the energy-band route (5.12)-type argument, not the doping formula (5.10).

Solution (a): the diagram. Equilibrium means one flat $E_F$ through the whole structure. Draw:

1. A horizontal line for $E_F$.
2. Far on the p-side, put $E_v$ a distance $2kT$ below $E_F$ ($E_F$ sits $2kT$ above $E_v$), and $E_c=E_v+E_G$ above it.
3. Far on the n-side, put $E_c$ a distance $kT/4$ above $E_F$, and $E_v=E_c-E_G$ below.
4. Join the p-side bands to the n-side bands by a smooth S-curve across the depletion region. The n-side bands lie lower in energy than the p-side bands.

Solution (b): the built-in voltage. The total band bending equals the difference between the band-edge energies far from the junction. Everything is measured from the common Fermi level $E_F$.

$E_c^{p\text{-side}}-E_F=(E_v+E_G)-E_F=E_G-2kT$

$E_c^{n\text{-side}}-E_F=\frac{kT}{4}$

The electron potential energy drop (bending) is the difference of the conduction-band edge on the two sides:

$q{V}_{bi}=E_c^{p\text{-side}}-E_c^{n\text{-side}}=\left(E_G-2kT\right)-\frac{kT}{4}$

$\boxed{V_{bi}=\frac{E_G}{q}-\frac{9kT}{4q}}$

Numerically, for Si ($E_G/q=1.12$ V, $kT/q=0.0259$ V):

$V_{bi}=1.12-2.25(0.0259)=1.12-0.0583=1.062\text{V}$

Check. $V_{bi}<E_G/q$, as it must be (consistent with Problem 5.1d).

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Problem 5.3: The $p_1$-$p_2$ “isotype” step junction

Problem. Consider the p${}_1$-p${}_2$ isotype step junction of Fig. P5.3: $N_A(x)=N_{A1}$ for $x<0$ and $N_{A2}$ for $x>0$. (a) Draw the equilibrium band diagram taking the doping nondegenerate and $N_{A1}>N_{A2}$. (b) Derive an expression for the built-in voltage. (c) Sketch potential, field and charge density. (d) Briefly describe the depletion approximation. (e) Can it be used here?

Solution (a): band diagram. Both sides are p-type, so $E_F$ lies below $E_i$ on both. The more heavily doped side ($p_1$) has $E_F$ closer to $E_v$ than the lighter side ($p_2$). With one flat $E_F$, the $p_1$ bands must therefore sit higher in energy than the $p_2$ bands (we need $E_F-E_v$ to be small on side 1). So the bands bend downward as we go from $p_1$ (left) to $p_2$ (right), by an amount equal to the difference of the Fermi-level positions. No step in the picture: it is a smooth transition.

Solution (b): built-in voltage. Use the equilibrium Fermi-level-from-$E_i$ relation, $E_i-E_F=kT\ln (p_0/n_i)$, on each side with $p_0=N_A$:

${\left(E_i-E_F\right)}_1=kT\ln \frac{N_{A1}}{n_i},{\left(E_i-E_F\right)}_2=kT\ln \frac{N_{A2}}{n_i}$

The band bending is the difference of $E_i$ between the sides (Fermi level is flat):

$q{V}_{bi}=E_{i,1}-E_{i,2}={\left(E_i-E_F\right)}_1-{\left(E_i-E_F\right)}_2=kT\ln \frac{N_{A1}}{n_i}-kT\ln \frac{N_{A2}}{n_i}$

$\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_{A1}}{N_{A2}}}$

Alternative check: at equilibrium the hole drift and diffusion currents cancel, so the hole populations at the two far edges are Boltzmann-related, $p_1/p_2=N_{A1}/N_{A2}=e^{q{V}_{bi}/kT}$, which gives the same result.

Solution (c): sketches.

Charge density. Holes diffuse from the heavily doped $p_1$ side (many holes) to the lightly doped $p_2$ side. Therefore:

• On the $p_1$ side near $x=0$: holes left behind, uncovered acceptors, so $\rho <0$ (a thin layer of negative charge, depleted of holes).
• On the $p_2$ side near $x=0$: extra holes arrive on top of the fixed $N_{A2}^{-}$, so $\rho >0$ (a layer of positive charge, hole accumulation).

So $\rho (x)$ is a “dipole”: negative lump left of $x=0$, positive lump right of $x=0$, total zero.

Electric field. From $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$, $\mathcal{E}$ starts at zero, goes negative through the negative-$\rho$ layer, reaches its most negative value near $x=0$, and returns to zero through the positive-$\rho$ layer. This is a single negative bump.

Potential. $\mathcal{E}=-dV/dx$ with $\mathcal{E}<0$ means $V$ rises monotonically from $0$ (far left) to $V_{bi}$ (far right) in an S-shape.

Solution (d): the depletion approximation. Assume that (i) the transition region has carrier concentrations so small that the mobile-carrier charge is negligible, $p\approx n\approx 0$, so $\rho =q(N_D-N_A)$ inside; (ii) outside, the material is exactly neutral, $\rho =0$; and (iii) the boundary between the two is abrupt.

Solution (e): can we use it here? No. The depletion approximation assumes the space charge is made of uncovered dopant ions with the mobile carriers swept away. In the $p_2$ half, the space charge is excess holes (accumulation), which are mobile carriers, so the "$p\approx 0$" assumption is violated. On the $p_1$ side there is a partial hole depletion, but the hole concentration there never falls near zero, because both sides are p-type and the potential hill is only $\frac{kT}{q}\ln (N_{A1}/N_{A2})$, typically a few $kT/q$. A full solution requires Poisson’s equation with the Boltzmann hole distribution (the type of “exact” solution in Problem 5.14).

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Problem 5.4: Step junction at equilibrium, $N_A\sim N_D$

Problem. A Si step junction at room temperature, in equilibrium, has $N_A$ (p-side) and $N_D$ (n-side) doping. Compute (a) $V_{bi}$; (b) $x_n$, $x_p$ and $W$; (c) $\mathcal{E}$ at $x=0$; (d) $V$ at $x=0$; (e) sketch $\rho$, $\mathcal{E}$, $V$.

Assumption note. The OCR gave ”$N_A=2\times 10^1/{\text{cm}}^3$ and $N_D=10^1/{\text{cm}}^3$” with the last digit of each exponent lost. The problem table says $N_A\sim N_D$, so I use $N_A=2\times 10^{15}{\text{cm}}^{-3}$ and $N_D=10^{15}{\text{cm}}^{-3}$.

Solution (a): built-in voltage.

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}\text{[eq. (5.10)]}$

$V_{bi}=0.0259\ln \frac{(2\times 10^{15})(10^{15})}{(10^{10}{)}^2}$

$V_{bi}=0.0259\ln \left(2\times 10^{10}\right)\text{[since }2\times 10^{30}/10^{20}\text{]}$

$\ln \left(2\times 10^{10}\right)=\ln 2+10\ln 10=0.693+23.026=23.719$

$\boxed{V_{bi}=0.0259\times 23.719=0.614\text{V}}$

Solution (b): depletion widths. First the total width from (5.31):

$W={\left[\frac{2K_S{\epsilon }_0}{q}{V}_{bi}\frac{N_A+N_D}{N_A{N}_D}\right]}^{1/2}$

The doping factor:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{3\times 10^{15}}{2\times 10^{30}}=1.5\times 10^{-15}{\text{cm}}^3$

The prefactor:

$\frac{2K_S{\epsilon }_0{V}_{bi}}{q}=\frac{2(1.044\times 10^{-12})(0.614)}{1.6\times 10^{-19}}=\frac{1.282\times 10^{-12}}{1.6\times 10^{-19}}=8.02\times 10^6{\text{cm}}^{-1}$

Multiply and take the root:

$W={\left[(8.02\times 10^6)(1.5\times 10^{-15})\right]}^{1/2}={\left[1.203\times 10^{-8}\right]}^{1/2}$

$\boxed{W=1.10\times 10^{-4}\text{cm}=1.10\mu \text{m}}$

Split using neutrality $N_A{x}_p=N_D{x}_n$ with $x_n+x_p=W$:

$x_n=W\frac{N_A}{N_A+N_D}=1.097\times 10^{-4}\times \frac{2}{3}$

$\boxed{x_n=7.31\times 10^{-5}\text{cm}=0.731\mu \text{m}}$

$x_p=W-x_n=3.66\times 10^{-5}\text{cm}\Rightarrow \boxed{x_p=0.366\mu \text{m}}$

Check: $N_A{x}_p=2\times 10^{15}(3.66\times 10^{-5})=7.31\times 10^{10}$ and $N_D{x}_n=10^{15}(7.31\times 10^{-5})=7.31\times 10^{10}$. These match.

Solution (c): the field at $x=0$. The field at the metallurgical junction is the peak field, from the n-side expression (5.21) at $x=0$:

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(10^{15})(7.31\times 10^{-5})}{1.044\times 10^{-12}}$

$\mathcal{E}(0)=-\frac{1.170\times 10^{-8}}{1.044\times 10^{-12}}$

$\boxed{\mathcal{E}(0)=-1.12\times 10^4\text{V/cm}}$

(The p-side formula gives the same, $-q{N}_A{x}_p/K_S{\epsilon }_0$, as it must by continuity.)

Solution (d): the potential at $x=0$. From (5.26) at $x=0$ (p-side, with $V=0$ at the p-edge):

$V(0)=\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=\frac{(1.6\times 10^{-19})(2\times 10^{15})}{2(1.044\times 10^{-12})}{\left(3.66\times 10^{-5}\right)}^2$

$V(0)=\frac{3.2\times 10^{-4}}{2.088\times 10^{-12}}\left(1.339\times 10^{-9}\right)=(1.532\times 10^8)(1.339\times 10^{-9})$

$\boxed{V(0)=0.205\text{V}}$

Check. The potential drop on each side is in the ratio of the doping fractions: $V(0)=V_{bi}{N}_D/(N_A+N_D)=0.614/3=0.205$ V.

Solution (e): sketches.

• $\rho (x)$: $-q{N}_A=-3.2\times 10^{-4}$ C/cm${}^3$ for $-0.366\le x\le 0$ $\mu$m; $+q{N}_D=+1.6\times 10^{-4}$ C/cm${}^3$ for $0\le x\le 0.731$ $\mu$m; zero elsewhere. The negative block is taller and narrower (twice as tall, half as wide) than the positive block, so the two areas are equal.
• $\mathcal{E}(x)$: zero outside; a straight line falling from 0 at $x=-0.366\mu$m to $-1.12\times 10^4$ V/cm at $x=0$, then a (shallower-slope) straight line rising back to 0 at $x=0.731\mu$m. A skewed triangle with the sharp peak at $x=0$.
• $V(x)$: 0 for $x<-0.366\mu$m; a parabola rising to $0.205$ V at $x=0$ (zero slope at the p-edge); another parabola (concave down) continuing to $0.614$ V at $x=0.731\mu$m, then constant. Slope at $x=0$ is continuous.

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Problem 5.5: Step junction with $N_A\gg N_D$

Problem. Repeat Problem 5.4 taking a larger p-side doping, and compare.

Assumption note. The OCR shows "$N_A=10^7/{\text{cm}}^3$" (an exponent digit lost). The table says $N_A\ge N_D$ and the n-side stays as before, so I use $N_A=10^{17}{\text{cm}}^{-3}$ and $N_D=10^{15}{\text{cm}}^{-3}$.

Solution (a): built-in voltage.

$V_{bi}=0.0259\ln \frac{(10^{17})(10^{15})}{10^{20}}=0.0259\ln \left(10^{12}\right)$

$\ln 10^{12}=12(2.3026)=27.631$

$\boxed{V_{bi}=0.0259\times 27.631=0.716\text{V}}$

Solution (b): widths. Doping factor:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{1.01\times 10^{17}}{10^{32}}=1.01\times 10^{-15}{\text{cm}}^3$

Prefactor:

$\frac{2K_S{\epsilon }_0{V}_{bi}}{q}=\frac{2(1.044\times 10^{-12})(0.716)}{1.6\times 10^{-19}}=\frac{1.495\times 10^{-12}}{1.6\times 10^{-19}}=9.34\times 10^6{\text{cm}}^{-1}$

$W={\left[(9.34\times 10^6)(1.01\times 10^{-15})\right]}^{1/2}={\left[9.43\times 10^{-9}\right]}^{1/2}$

$\boxed{W=9.71\times 10^{-5}\text{cm}=0.971\mu \text{m}}$

$x_n=W\frac{N_A}{N_A+N_D}=9.71\times 10^{-5}\times \frac{10^{17}}{1.01\times 10^{17}}=9.62\times 10^{-5}\text{cm}$

$\boxed{x_n=0.962\mu \text{m}}$

$x_p=\frac{N_D}{N_A}{x}_n=10^{-2}(9.62\times 10^{-5})\Rightarrow \boxed{x_p=9.6\times 10^{-7}\text{cm}=9.6\text{nm}}$

Solution (c): field at $x=0$.

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(10^{15})(9.62\times 10^{-5})}{1.044\times 10^{-12}}=-\frac{1.539\times 10^{-8}}{1.044\times 10^{-12}}$

$\boxed{\mathcal{E}(0)=-1.47\times 10^4\text{V/cm}}$

Solution (d): potential at $x=0$.

$V(0)=V_{bi}\frac{N_D}{N_A+N_D}=0.716\times \frac{10^{15}}{1.01\times 10^{17}}=0.716\times 9.9\times 10^{-3}$

$\boxed{V(0)=7.1\times 10^{-3}\text{V}\approx 7\text{mV}}$

(Direct check: $\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=\frac{1.6\times 10^{-2}}{2.088\times 10^{-12}}(9.2\times 10^{-13})=7.1\times 10^{-3}$ V. Same.)

Solution (e): sketches. The same shapes as in 5.4, but extremely lopsided: $\rho$ has a very tall, very thin negative spike ($-q{N}_A=-1.6\times 10^{-2}$ C/cm${}^3$, 10 nm wide) next to a low, wide positive block ($+1.6\times 10^{-4}$ C/cm${}^3$, 0.96 $\mu$m wide). $\mathcal{E}$ is a triangle with an almost vertical left side. $V$ is essentially flat at 0 until $x\approx 0$, then climbs almost the whole $0.716$ V on the n-side.

Comparison with Problem 5.4.

Quantity	5.4 ($N_A=2\times 10^{15}$)	5.5 ($N_A=10^{17}$)
$V_{bi}$ (V)	0.614	0.716
$W$ ($\mu$m)	1.10	0.971
$x_n$ ($\mu$m)	0.731	0.962
$x_p$ ($\mu$m)	0.366	0.0096
$	\mathcal{E}(0)	$ (V/cm)
$V(0)$ (V)	0.205	0.007

Making one side 50 times more heavily doped raised $V_{bi}$ only slightly (a logarithm), nearly eliminated the p-side depletion region, and pushed almost the entire junction into the lightly doped n-side. The lightly doped side controls everything: $W\approx x_n\approx \sqrt{2K_S{\epsilon }_0{V}_{bi}/(q{N}_D)}$.

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Problem 5.6: Computation using the MATLAB diary function

Problem. A Si step junction at room temperature has instructor-supplied $N_A$, $N_D$, $V_A$. Working in the Command window, with the MATLAB diary function recording the session, compute (a) $V_{bi}$; (b) $x_n$, $x_p$, $W$; (c) $\mathcal{E}$ at $x=0$; (d) $V$ at $x=0$.

Assumption note. The three values are supplied by the instructor. As a worked illustration I use $N_A=10^{16}{\text{cm}}^{-3}$, $N_D=10^{15}{\text{cm}}^{-3}$, $V_A=-2$ V (reverse bias), and verify the numbers by hand. Replace these three lines for your own assignment.

Solution: the session. The sequence of Command-window entries (each entry is a step; explanations in comments):

diary prob56.txt          % start recording the session
kT = 0.0259; q = 1.6e-19; ni = 1e10;
KSe0 = 11.8*8.85e-14;     % K_S * eps_0 (F/cm)
NA = 1e16; ND = 1e15; VA = -2;
Vbi = kT*log(NA*ND/ni^2)                      % (a) eq. (5.10)
W   = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND)) % eq. (5.38)
xn  = W*NA/(NA+ND)                            % eq. (5.37)
xp  = W*ND/(NA+ND)                            % eq. (5.34)
E0  = -q*ND*xn/KSe0                           % (c) eq. (5.35) at x = 0
V0  = q*NA*xp^2/(2*KSe0)                      % (d) eq. (5.33) at x = 0
diary off                 % stop recording

Hand check of each output.

(a) $V_{bi}=0.0259\ln \frac{(10^{16})(10^{15})}{10^{20}}=0.0259\ln \left(10^{11}\right)=0.0259(25.328)=0.656\text{V}$

(b) $V_{bi}-V_A=0.656+2=2.656\text{V}$

$W={\left[\frac{2(1.044\times 10^{-12})(2.656)}{1.6\times 10^{-19}}\cdot \frac{1.1\times 10^{16}}{10^{31}}\right]}^{1/2}={\left[(3.466\times 10^7)(1.1\times 10^{-15})\right]}^{1/2}={\left[3.81\times 10^{-8}\right]}^{1/2}$

$W=1.95\times 10^{-4}\text{cm}=1.95\mu \text{m}$

$x_n=W\frac{N_A}{N_A+N_D}=1.95\times 10^{-4}\times \frac{10}{11}=1.78\times 10^{-4}\text{cm}=1.78\mu \text{m}$

$x_p=W-x_n=1.8\times 10^{-5}\text{cm}=0.178\mu \text{m}$

(c) $\mathcal{E}(0)=-\frac{(1.6\times 10^{-19})(10^{15})(1.78\times 10^{-4})}{1.044\times 10^{-12}}=-2.72\times 10^4\text{V/cm}$

(d) $V(0)=\frac{(1.6\times 10^{-19})(10^{16})}{2(1.044\times 10^{-12})}(1.78\times 10^{-5}{)}^2=(7.66\times 10^8)(3.17\times 10^{-10})=0.243\text{V}$

(Check: $(V_{bi}-V_A)N_D/(N_A+N_D)=2.656/11=0.241$ V, agreeing to rounding.)

Why diary? It leaves a text log you can hand in: every command and every numeric answer, so the grader can follow your steps. Reverse bias ($V_A<0$) widens $W$ (here 1.95 $\mu$m compared with 1.2 $\mu$m at equilibrium) and raises the peak field, as Section 5.2.4 predicted.

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Problem 5.7: A step-junction program

Problem. For a nondegenerately doped Si pn step junction at $T=300$ K: (a) compute and present coordinated plots of $\mathcal{E}(x)$ and $V(x)$ using, as a sample computation, $N_A$ (value lost in OCR), $N_D=2\times 10^{14}{\text{cm}}^{-3}$ and $V_A=-20$ V; (b) modify the program to show several $V_A$ values, $V_A=V_{A0}/2^n$, $n=0,\dots ,3$; (c) add an output list of parameters; (d) explore how the plots change with relative doping; (e) check against Problems 5.4 and 5.5.

Assumption note. The OCR lost the exponent of $N_A$. I use $N_A=10^{16}{\text{cm}}^{-3}$ for the sample run.

The plan (before any code)

Everything is already derived. The program only has to evaluate the closed-form results piece by piece:

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2},W={\left[\frac{2K_S{\epsilon }_0}{q}(V_{bi}-V_A)\frac{N_A+N_D}{N_A{N}_D}\right]}^{1/2}$

$x_n=W\frac{N_A}{N_A+N_D},x_p=W\frac{N_D}{N_A+N_D}$

$\mathcal{E}(x)=\{\begin{array}{ll}-\frac{q{N}_A}{K_S{\epsilon }_0}(x_p+x),& -x_p\le x<0\\ -\frac{q{N}_D}{K_S{\epsilon }_0}(x_n-x),& 0\le x\le x_n\\ 0& \text{otherwise}\end{array}$

$V(x)=\{\begin{array}{ll}0,& x<-x_p\\ \frac{q{N}_A}{2K_S{\epsilon }_0}(x_p+x{)}^2,& -x_p\le x<0\\ V_{bi}-V_A-\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2,& 0\le x\le x_n\\ V_{bi}-V_A,& x>x_n\end{array}$

The suggested plot ranges: $x\in [-x_{max},x_{max}]$ with $x_{max}=2.5\max (x_n,x_p)$; $\mathcal{E}\in [{\mathcal{E}}_{min},0]$ with ${\mathcal{E}}_{min}=1.1\mathcal{E}(0)$; and $V\in [0,V_{max}]$ with $V_{max}=1.1(V_{bi}-V_A)$ (the OCR of this suggestion is garbled; this is the reading consistent with its intent, a 10 percent headroom). subplot(2,1,1) and subplot(2,1,2) stack the two graphs so the $x$ axes line up, which is what “coordinated plots” means.

(a) Single-bias program

% Problem 5.7(a): E(x) and V(x) of a Si step junction
q = 1.6e-19;  KSe0 = 11.8*8.85e-14;  kT = 0.0259;  ni = 1e10;
NA = 1e16;  ND = 2e14;  VA = -20;            % sample values
 
Vbi = kT*log(NA*ND/ni^2);
W   = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND));
xn  = W*NA/(NA+ND);   xp = W*ND/(NA+ND);
 
E0   = -q*ND*xn/KSe0;                        % field at x = 0 (negative)
Emin = 1.1*E0;   Vmax = 1.1*(Vbi-VA);
xmax = 2.5*max(xn,xp);
x = linspace(-xmax, xmax, 2000);
 
E = zeros(size(x));  V = zeros(size(x));
ip = (x >= -xp & x < 0);   in = (x >= 0 & x <= xn);   ir = (x > xn);
E(ip) = -q*NA*(xp + x(ip))/KSe0;
E(in) = -q*ND*(xn - x(in))/KSe0;
V(ip) =  q*NA*(xp + x(ip)).^2/(2*KSe0);
V(in) = (Vbi-VA) - q*ND*(xn - x(in)).^2/(2*KSe0);
V(ir) = Vbi - VA;
 
subplot(2,1,1); plot(x,E); axis([-xmax xmax Emin 0]);
ylabel('E (V/cm)'); grid on
subplot(2,1,2); plot(x,V); axis([-xmax xmax 0 Vmax]);
ylabel('V (V)'); xlabel('x (cm)'); grid on

Expected result for the sample inputs (verified by hand):

$V_{bi}=0.0259\ln \frac{(10^{16})(2\times 10^{14})}{10^{20}}=0.0259\ln (2\times 10^{10})=0.614\text{V},V_{bi}-V_A=20.614\text{V}$

$W={\left[\frac{2(1.044\times 10^{-12})(20.614)}{1.6\times 10^{-19}}\cdot \frac{1.02\times 10^{16}}{2\times 10^{30}}\right]}^{1/2}={\left[(2.69\times 10^8)(5.1\times 10^{-15})\right]}^{1/2}=1.17\times 10^{-3}\text{cm}=11.7\mu \text{m}$

$x_n=W\frac{10^{16}}{1.02\times 10^{16}}=11.5\mu \text{m},x_p=W-x_n=0.23\mu \text{m}$

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(2\times 10^{14})(1.148\times 10^{-3})}{1.044\times 10^{-12}}=-3.5\times 10^4\text{V/cm}$

$V(0)=(V_{bi}-V_A)\frac{N_D}{N_A+N_D}=20.614\times 0.0196=0.40\text{V}$

So the plot shows a sharp triangle, mostly on the n-side, with the potential nearly flat at 0 for $x<0$ and climbing to 20.6 V across the n-side depletion region. The tiny p-side depletion layer (0.23 $\mu$m) is a thin sliver at the left of the triangle.

(b) Several biases at once

Loop over $n=\mathrm{0..3}$ with $V_A=V_{A0}/2^n$ and use hold on so all curves share the axes. Use the largest depletion width (the first, $n=0$, which is the most reverse-biased) to choose the axis limits.

% Problem 5.7(b): multiple bias values on the same axes
q = 1.6e-19;  KSe0 = 11.8*8.85e-14;  kT = 0.0259;  ni = 1e10;
NA = 1e16;  ND = 2e14;  VA0 = -20;
Vbi = kT*log(NA*ND/ni^2);
 
W0   = sqrt(2*KSe0/q*(Vbi-VA0)*(NA+ND)/(NA*ND));       % widest case
xmax = 2.5*W0*max(NA,ND)/(NA+ND);                      % 2.5*max(xn,xp)
x    = linspace(-xmax, xmax, 2000);
Emin = 1.1*(-q*ND*(W0*NA/(NA+ND))/KSe0);   Vmax = 1.1*(Vbi-VA0);
 
for n = 0:3
    VA = VA0/2^n;
    W  = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND));
    xn = W*NA/(NA+ND);  xp = W*ND/(NA+ND);
    E = zeros(size(x));  V = zeros(size(x));
    ip = (x >= -xp & x < 0);  in = (x >= 0 & x <= xn);  ir = (x > xn);
    E(ip) = -q*NA*(xp + x(ip))/KSe0;
    E(in) = -q*ND*(xn - x(in))/KSe0;
    V(ip) =  q*NA*(xp + x(ip)).^2/(2*KSe0);
    V(in) = (Vbi-VA) - q*ND*(xn - x(in)).^2/(2*KSe0);
    V(ir) = Vbi - VA;
    subplot(2,1,1); plot(x,E); hold on
    subplot(2,1,2); plot(x,V); hold on
end
subplot(2,1,1); axis([-xmax xmax Emin 0]); ylabel('E (V/cm)'); grid on
subplot(2,1,2); axis([-xmax xmax 0 Vmax]); ylabel('V (V)');
xlabel('x (cm)'); grid on

With $V_A=-20,-10,-5,-2.5$ V you will see a nest of triangles (field) and a nest of S-curves (potential): lower reverse bias gives narrower, shorter curves, with each $V$-curve plateauing at its own $V_{bi}-V_A$. Because $W\propto (V_{bi}-V_A{)}^{1/2}$, halving the bias shrinks $W$ by roughly $1/\sqrt{2}$.

(c) Adding an output list

Insert inside the loop, after computing the numbers (and before the plotting calls for that bias value):

    E0 = -q*ND*xn/KSe0;   V0 = q*NA*xp^2/(2*KSe0);
    fprintf('NA=%9.2e ND=%9.2e VA=%7.3f Vbi=%6.4f\n', NA, ND, VA, Vbi);
    fprintf('  xn=%9.3e xp=%9.3e W=%9.3e cm\n', xn, xp, W);
    fprintf('  E(0)=%10.3e V/cm   V(0)=%8.4f V\n\n', E0, V0);

The list contains exactly what is requested: $N_A,N_D,V_A,V_{bi},x_n,x_p,W$, $\mathcal{E}$ at $x=0$, and $V$ at $x=0$. For the $n=0$ pass you should see $V_{bi}=0.6143$, $W=1.171\times 10^{-3}$ cm, $\mathcal{E}(0)=-3.5\times 10^4$ V/cm, $V(0)=0.404$ V, matching the hand numbers above.

(d) Effect of the relative doping

Run the program for the listed $(N_A,N_D)$ pairs (including equal doping and ratios of 10, 100 and 1000 in both directions). What you will observe:

Case	Where is the depletion region?	Field triangle	Potential
$N_A=N_D$	Equal on both sides, $x_n=x_p=W/2$	Symmetric triangle	Symmetric S-curve, $V(0)=(V_{bi}-V_A)/2$
$N_A$ somewhat $>N_D$	More on the n-side	Skewed, steeper slope on the p-side	Most of the drop on the n-side
$N_A\ge 100N_D$	Essentially all on the n-side	Right triangle, vertical edge at $x\approx 0$	$V$ stays $\approx 0$ for $x<0$ and then rises on the n-side only; $V(0)\approx 0$
$N_D\ge 100N_A$	Essentially all on the p-side	Right triangle, vertical edge at $x\approx 0$ on the right	$V$ rises on the p-side only; $V(0)\approx V_{bi}-V_A$

Description for the extreme cases. When one side is $\ge 100$ times more heavily doped, the junction is one-sided: the lightly doped side supplies all the depletion width, nearly the full $V_{bi}-V_A$ is dropped across it, and the heavily doped side behaves like a metal contact. The width reduces to the single-side formula $W\approx \sqrt{2K_S{\epsilon }_0(V_{bi}-V_A)/(q{N}_{light})}$, and the peak field is ${\mathcal{E}}_{max}=2(V_{bi}-V_A)/W$ (twice the average, because the field is a triangle).

(e) Check against Problems 5.4 and 5.5

Set $(N_A,N_D,V_A)=(2\times 10^{15},10^{15},0)$ and $(10^{17},10^{15},0)$. The output list should reproduce the hand values of Problems 5.4 and 5.5:

Quantity	Problem 5.4	Problem 5.5
$V_{bi}$	0.614 V	0.716 V
$W$	1.10 $\mu$m	0.971 $\mu$m
$x_n$, $x_p$	0.731, 0.366 $\mu$m	0.962, 0.0096 $\mu$m
$\mathcal{E}(0)$	$-1.12\times 10^4$ V/cm	$-1.47\times 10^4$ V/cm
$V(0)$	0.205 V	0.0071 V

Agreement to three significant figures is expected. A mismatch usually means a mistaken exponent in NA/ND, or a forgotten .^ or .* in the vector expressions.

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Problem 5.8: Combination profile

Problem. The doping around the metallurgical junction of a special diode is pictured in Fig. P5.8 (a combination of a step and a linear profile). Sketch, using the depletion approximation, the charge density, electric field and electrostatic potential. Scale and label relevant lengths, and add a few words of explanation.

Assumption note. The figure did not survive OCR; only the axis labels $N_D-N_A$, $x$ and $-N_A$ remain. I read it as: uniform p-type doping $N_D-N_A=-N_A$ for $x<0$, and a linear ramp $N_D-N_A=ax$ for $x>0$, with the metallurgical junction at $x=0$. This is the natural “step on the p-side, graded on the n-side” combination. If your figure differs, apply the same three-step method to its pieces.

Solution: step 1, charge density. The depletion approximation sets $\rho =q(N_D-N_A)$ inside the depletion region $-x_p\le x\le x_n$:

$\rho (x)=\{\begin{array}{ll}-q{N}_A,& -x_p\le x<0\\ +qax,& 0\le x\le x_n\\ 0,& \text{elsewhere}\end{array}$

Sketch: a rectangular negative block of height $-q{N}_A$ and width $x_p$ on the left, and a triangular ramp rising from $0$ at $x=0$ to $+qa{x}_n$ at $x=x_n$ on the right. Note the discontinuity at $x=0$ (the block ends at $-q{N}_A$, the ramp begins at 0).

Neutrality. Equal areas under the two pieces:

$q{N}_A{x}_p=\frac{1}{2}qa{x}_n^2\Rightarrow N_A{x}_p=\frac{a{x}_n^2}{2}$

If $N_A$ is large, $x_p\ll x_n$: the p-side layer is thin, the n-side layer is wide. Label $x_p$ (short) and $x_n$ (long) on the axis.

Step 2, electric field. Integrate $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$ from each edge where $\mathcal{E}=0$:

p-side: $\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}(x+x_p),-x_p\le x\le 0$

n-side (integrating leftward from $x_n$): $\mathcal{E}(x)=-\frac{qa}{2K_S{\epsilon }_0}\left(x_n^2-x^2\right),0\le x\le x_n$

Continuity at $x=0$ (this reproduces the neutrality condition):

$-\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}=-\frac{qa{x}_n^2}{2K_S{\epsilon }_0}\Rightarrow N_A{x}_p=\frac{a{x}_n^2}{2}✓$

Sketch: a straight line descending from 0 at $x=-x_p$ to its most negative value $\mathcal{E}(0)$ at $x=0$ (steep, because it covers the short distance $x_p$), then a parabola rising from $\mathcal{E}(0)$ back to zero at $x_n$ with zero slope at $x=0$ (since $\rho (0^{+})=0$ the field is momentarily flat there). The peak magnitude is at $x=0$.

Step 3, electrostatic potential. Integrate $V=-\int \mathcal{E}dx$, with $V=0$ at the p-edge:

p-side: $V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}(x+x_p{)}^2,-x_p\le x\le 0,V(0)=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}$

n-side: $V(x)=V(0)+\frac{qa}{2K_S{\epsilon }_0}\left(x_n^{2}x-\frac{x^3}{3}\right),0\le x\le x_n$

The total drop (to fix $x_n$ with the condition $V(x_n)=V_{bi}-V_A$):

$V_{bi}-V_A=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}+\frac{qa{x}_n^3}{3K_S{\epsilon }_0}$

Sketch: a parabola (concave up) from 0 at $-x_p$ up to the small value $V(0)$ at $x=0$, then a cubic-like curve continuing to rise (still concave up at first, then concave down) until it levels off at $V_{bi}-V_A$ at $x=x_n$; flat outside. The slope $dV/dx=-\mathcal{E}$ is maximum at $x=0$, which is the steepest point of the S-curve. Most of the potential drop occurs on the wide, lightly doped n-side.

Words to forestall misinterpretation. (i) The depletion approximation is assumed, so the sketches are idealized: the real $\rho$ and $\mathcal{E}$ have rounded corners. (ii) $\rho$ is discontinuous at $x=0$; $\mathcal{E}$ is continuous; $V$ and its slope are continuous. (iii) The lengths $x_p\ll x_n$ are not to scale with the potential axis.

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Problem 5.9: The exponential profile

Problem. A pn diode has $N_D-N_A=N_0\left[1-e^{-\alpha x}\right]$, with $N_0$ and $\alpha$ constants (Fig. P5.9). (a) State the depletion approximation concisely. (b) Sketch the charge density. (c) Establish an expression for $\mathcal{E}(x)$ in the depletion region.

Solution (a): the depletion approximation, concisely. Within the (abruptly bounded) depletion region $-x_p\le x\le x_n$, the mobile carrier densities are negligible, $p\approx n\approx 0$, so $\rho =q(N_D-N_A)$; outside the region the semiconductor is perfectly neutral, $\rho =0$ and $\mathcal{E}=0$.

Solution (b): sketch of $\rho$. Apply (a) to the given profile:

$\rho (x)=q{N}_0\left[1-e^{-\alpha x}\right],-x_p\le x\le x_n$

Plot by reading the limits:

• At $x=0$: $\rho =0$. This is the metallurgical junction.
• For $x>0$: $e^{-\alpha x}$ decays, so $\rho$ rises from 0 and saturates at $+q{N}_0$. The n-side density flattens at $q{N}_0$ (a curve like a charging capacitor).
• For $x<0$: $e^{-\alpha x}=e^{\alpha |x|}$ grows, so $\rho =-q{N}_0\left(e^{\alpha |x|}-1\right)$ becomes more and more negative with $|x|$, with an exponentially growing magnitude. The profile therefore represents very heavily doped p-type material far on the left.

So the sketch: zero at the origin, a negative curve plunging downward on the left (steepening as $-x_p$ is approached), a positive curve flattening out toward $+q{N}_0$ on the right, and zero outside the dashed edges $-x_p$ and $x_n$. Neutrality demands that the (large, narrow) negative area balance the positive area, so $x_p$ is considerably smaller than $x_n$.

Solution (c): the electric field. Poisson’s equation:

$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}=\frac{q{N}_0}{K_S{\epsilon }_0}\left[1-e^{-\alpha x}\right]$

Integrate from the p-edge $x=-x_p$, where $\mathcal{E}=0$, to an arbitrary $x$:

${\int }_0^{\mathcal{E}(x)}d{\mathcal{E}}^{\prime }=\frac{q{N}_0}{K_S{\epsilon }_0}{\int }_{-x_p}^x\left[1-e^{-\alpha x^{\prime }}\right]d{x}^{\prime }$

Evaluate the integral, using $\int e^{-\alpha x^{\prime }}d{x}^{\prime }=-e^{-\alpha x^{\prime }}/\alpha$:

${\int }_{-x_p}^x\left[1-e^{-\alpha x^{\prime }}\right]d{x}^{\prime }=[x^{\prime }+\frac{1}{\alpha }{e}^{-\alpha x^{\prime }}{]}_{-x_p}^x=(x+x_p)+\frac{1}{\alpha }\left(e^{-\alpha x}-e^{\alpha x_p}\right)$

Therefore:

$\boxed{\mathcal{E}(x)=\frac{q{N}_0}{K_S{\epsilon }_0}\left[(x+x_p)+\frac{e^{-\alpha x}-e^{\alpha x_p}}{\alpha }\right],-x_p\le x\le x_n}$

Boundary condition at the n-edge. The field must also vanish at $x=x_n$, which gives the relation between $x_n$ and $x_p$ (the neutrality condition for this profile):

$x_n+x_p+\frac{e^{-\alpha x_n}-e^{\alpha x_p}}{\alpha }=0$

Sanity checks. (i) At $x=-x_p$: $\mathcal{E}=\frac{q{N}_0}{K_S{\epsilon }_0}\left[0+\frac{e^{\alpha x_p}-e^{\alpha x_p}}{\alpha }\right]=0$ ✓. (ii) The field is negative in between (since $\rho <0$ at first, then the field is pulled back to zero by the positive $\rho$); its slope $d\mathcal{E}/dx\propto 1-e^{-\alpha x}$ is zero at $x=0$, so the most negative field is at the metallurgical junction, as always for a pn junction.

(As the problem’s note says, completing the solution for $V(x)$, $x_n$, $x_p$ and $V_{bi}$ requires a considerable amount of algebra; the method is the same, integrate once more and force $V(x_n)-V(-x_p)=V_{bi}-V_A$.)

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(Problem 5.9 in part05b is already complete; the solutions below pick up with 5.10.)

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Problem 5.10: A stepped n-side profile

Problem. A pn diode has the doping profile of Fig. P5.10, and we assume $x_n>0$ (the depletion region reaches beyond the junction) for all biases of interest. (a) What is the built-in voltage? Justify. (b) Using the depletion approximation, sketch $\rho (x)$.

Reading the figure. The figure is mostly lost in the OCR; only the labels $N_D-N_A$, $N_D$, $N_D/2$, $x_0$, $-N_A$ survive. I read it as: $N_D-N_A=-N_A$ for $x<0$, then $N_D/2$ for $0<x<x_0$, then $N_D$ for $x>x_0$. If your figure differs, the method below is unchanged.

Solution (a): built-in voltage. Think of $V_{bi}$ as the height difference between two flat “lakes” (the neutral p and n regions) once their Fermi levels line up. Only the lake levels matter, not the terrain between them. So we need only the far-side doping.

Far p-side (neutral): $p_0=N_A,E_i-E_F=kT\ln \frac{N_A}{n_i}$ Far n-side (neutral, $x\gg x_0$): $n_0=N_D,E_F-E_i=kT\ln \frac{N_D}{n_i}$ At equilibrium $E_F$ is constant, so the total bending of $E_i$ is the sum: $q{V}_{bi}=kT\ln \frac{N_A}{n_i}+kT\ln \frac{N_D}{n_i}$ $\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}}$ Justification: the intermediate $N_D/2$ layer is inside the band-bending region, so it changes how the bending is distributed in space but not its total. The result is the same as for a plain step junction with the same outer dopings.

Solution (b): charge density. The depletion approximation gives $\rho =q(N_D-N_A)$ for $-x_p\le x\le x_n$, zero outside: $\rho (x)=\{\begin{array}{llccc}-q{N}_A,& -x_p\le x<0+q{N}_D/2,& 0\le x<x_0+q{N}_D,& x_0\le x\le x_{n}0,& \text{otherwise}\end{array}$ With $x_n>x_0$ (our assumption) all three levels appear. Sketch: a negative block of height $-q{N}_A$ and width $x_p$; then a positive step of height $q{N}_D/2$ and width $x_0$; then a taller positive step of height $q{N}_D$ out to $x_n$.

Neutrality (equal areas, the “debts equal credits” rule): $N_A{x}_p=\frac{N_D}{2}{x}_0+N_D(x_n-x_0)=N_D\left(x_n-\frac{x_0}{2}\right)$ $\Rightarrow x_n=\frac{N_A}{N_D}{x}_p+\frac{x_0}{2}$ Why the assumption $x_n>x_0$ matters: if $x_n<x_0$ the depletion region would end inside the $N_D/2$ layer and the top level $q{N}_D$ would never appear.

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Problem 5.11: The p-i-n diode

Problem. A p-i-n diode has uniform p ($N_A$) and n ($N_D$) regions and a narrow middle region of width $x_i$ with $N_D-N_A=0$. (a) Sketch $\rho$, $\mathcal{E}$, $V$ and the equilibrium band diagram. (b) Built-in voltage? (c) Quantitative expressions for $\rho$, $\mathcal{E}$, $V$ and the depletion widths.

Setup. From Fig. P5.11 (labels lost in OCR except $x_i/2$), place the i-region symmetrically: $-x_i/2\le x\le x_i/2$. Let $x_p$ be the depleted thickness inside the p-region (from $-x_i/2$ leftward) and $x_n$ the depleted thickness inside the n-region (from $x_i/2$ rightward). Assume the i-region is fully depleted, which is true because it is narrow and has no mobile charge to spare.

Solution (a): sketches.

• $\rho$: a negative block $-q{N}_A$ over $-x_i/2-x_p\le x\le -x_i/2$; zero across the i-region; a positive block $+q{N}_D$ over $x_i/2\le x\le x_i/2+x_n$. Think of a parallel-plate capacitor whose plates are the two blocks and whose gap is the i-region.
• $\mathcal{E}$: rises linearly (in magnitude) through the p-block, stays constant across the i-region (no charge, so no slope), then falls linearly back to zero through the n-block. The shape is a trapezoid with a flat top. This is exactly the field of a capacitor: uniform in the gap.
• $V$: parabola (concave up) in the p-block, a straight ramp across the i-region, parabola again (concave down) in the n-block, flat outside.
• Bands: $E_C$, $E_V$ bend with the same shape as $-qV$: curved down-slope in the p-block, a straight, constant-slope section in the i-region, curved in the n-block. $E_F$ is flat; $E_F$ sits near $E_V$ on the far left and near $E_C$ on the far right.

Solution (b): built-in voltage. Same argument as in 5.10: the total bending equals the difference of Fermi-level positions in the neutral outer regions, and the lightly doped i-region only sits in between: $\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}}$

Solution (c): quantitative.

Charge density: $\rho (x)=\{\begin{array}{llcc}-q{N}_A,& -\frac{x_i}{2}-x_p\le x<-\frac{x_i}{2}\text{0},& |x|\le \frac{x_i}{2}\text{+}q{N}_D,& \frac{x_i}{2}<x\le \frac{x_i}{2}+x_n\end{array}$

Neutrality: $q{N}_A{x}_p=q{N}_D{x}_n\Rightarrow x_n=\frac{N_A}{N_D}{x}_p$

Electric field. Integrate Poisson’s equation $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$ from the left edge where $\mathcal{E}=0$:

p-block: $\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}\left(x+\frac{x_i}{2}+x_p\right),-\frac{x_i}{2}-x_p\le x\le -\frac{x_i}{2}$ i-region (constant, equal to the value at the block’s right edge): $\mathcal{E}=-\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}\equiv -{\mathcal{E}}_0,|x|\le \frac{x_i}{2}$ n-block: $\mathcal{E}(x)=-\frac{q{N}_D}{K_S{\epsilon }_0}\left(\frac{x_i}{2}+x_n-x\right),\frac{x_i}{2}\le x\le \frac{x_i}{2}+x_n$

Potential. The drop across each piece is the area under $|\mathcal{E}|$ (triangle, rectangle, triangle): $\Delta V_p=\frac{1}{2}{\mathcal{E}}_0{x}_p=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}$ $\Delta V_i={\mathcal{E}}_0{x}_i=\frac{q{N}_A{x}_p{x}_i}{K_S{\epsilon }_0}$ $\Delta V_n=\frac{1}{2}{\mathcal{E}}_0{x}_n=\frac{q{N}_D{x}_n^2}{2K_S{\epsilon }_0}$ In terms of $V(x)$, with $V=0$ at the left edge: $V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}{\left(x+\frac{x_i}{2}+x_p\right)}^2\text{(p-block)}$ $V(x)=\Delta V_p+{\mathcal{E}}_0\left(x+\frac{x_i}{2}\right)\text{(i-region)}$ n-block, written from the right edge where $V=V_{bi}-V_A$: $V(x)=V_{bi}-V_A-\frac{q{N}_D}{2K_S{\epsilon }_0}{\left(\frac{x_i}{2}+x_n-x\right)}^2,\frac{x_i}{2}\le x\le \frac{x_i}{2}+x_n$

Depletion widths. Require the total drop to equal $V_{bi}-V_A$: $V_{bi}-V_A=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}\left(1+\frac{N_A}{N_D}\right)+\frac{q{N}_A{x}_i{x}_p}{K_S{\epsilon }_0}$ (using $x_n=N_A{x}_p/N_D$ in $\Delta V_n$). This is a quadratic in $x_p$: $\frac{q{N}_A}{2K_S{\epsilon }_0}\left(1+\frac{N_A}{N_D}\right)x_p^2+\frac{q{N}_A{x}_i}{K_S{\epsilon }_0}{x}_p-(V_{bi}-V_A)=0$ Solve with the quadratic formula (positive root): $\boxed{x_p=\frac{1}{1+N_A/N_D}\left[-x_i+\sqrt{x_i^2+\frac{2K_S{\epsilon }_0(V_{bi}-V_A)}{q{N}_A}\left(1+\frac{N_A}{N_D}\right)}\right]}$ $\boxed{x_n=\frac{N_A}{N_D}{x}_p}$ Checks: (i) $x_i\to 0$ gives $x_p=\sqrt{2K_S{\epsilon }_0(V_{bi}-V_A)/(q{N}_A(1+N_A/N_D))}$, the ordinary step-junction result. (ii) For very large $x_i$, $x_p\approx \frac{K_S{\epsilon }_0(V_{bi}-V_A)}{q{N}_A{x}_i}$, which is tiny: the long i-region takes most of the voltage, as in a capacitor with a wide gap.

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Problem 5.12: From a given $V(x)$ back to the doping

Problem. In equilibrium the depletion-region potential is $V(x)=\frac{V_{bi}}{2}\left[1+\sin \frac{\pi x}{W}\right],-\frac{W}{2}\le x\le \frac{W}{2}$ (the OCR is garbled; I take the sine argument as $\pi x/W$, the only choice that runs from $0$ to $V_{bi}$ across $\pm W/2$). Find (a) $\mathcal{E}(x)$, (b) $\rho (x)$, (c) $N_D-N_A$ under the depletion approximation, with sketches.

This is Poisson’s equation run backwards: instead of doping to field to potential, we differentiate down the staircase.

Check of the given $V$. $V\left(-\frac{W}{2}\right)=\frac{V_{bi}}{2}[1+\sin (-\pi /2)]=0,V\left(\frac{W}{2}\right)=\frac{V_{bi}}{2}[1+1]=V_{bi}✓$

Solution (a): electric field. Use $\mathcal{E}=-dV/dx$: $\mathcal{E}(x)=-\frac{d}{dx}\left\{\frac{V_{bi}}{2}\left[1+\sin \frac{\pi x}{W}\right]\right\}$ $\mathcal{E}(x)=-\frac{V_{bi}}{2}\cdot \frac{\pi }{W}\cos \frac{\pi x}{W}$ $\boxed{\mathcal{E}(x)=-\frac{\pi V_{bi}}{2W}\cos \frac{\pi x}{W}}$ Checks: at $x=\pm W/2$, $\cos (\pm \pi /2)=0$, so $\mathcal{E}=0$ at both depletion edges, as required. At $x=0$, $\mathcal{E}=-\pi V_{bi}/(2W)$, the most negative value. Sketch: an inverted half-cosine arch, zero at both edges, minimum at the center.

Solution (b): charge density. Poisson: $\rho =K_S{\epsilon }_{0}d\mathcal{E}/dx$. $\rho (x)=K_S{\epsilon }_0\frac{d}{dx}\left[-\frac{\pi V_{bi}}{2W}\cos \frac{\pi x}{W}\right]$ $\rho (x)=K_S{\epsilon }_0\cdot \frac{\pi V_{bi}}{2W}\cdot \frac{\pi }{W}\sin \frac{\pi x}{W}$ $\boxed{\rho (x)=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2W^2}\sin \frac{\pi x}{W}}$ Sketch: an odd sine wave, negative for $-W/2<x<0$ (reaching $-{\rho }_{max}$ at $x=-W/2$), zero at $x=0$, positive for $x>0$ (reaching $+{\rho }_{max}$ at $x=W/2$), with ${\rho }_{max}=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2W^2}.$ Net charge: $\int \rho dx=0$ by oddness, so the region is neutral overall, as it must be.

Solution (c): doping profile. The depletion approximation says $\rho =q(N_D-N_A)$ inside the region, so $\boxed{N_D-N_A=\frac{\rho }{q}=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2q{W}^2}\sin \frac{\pi x}{W},-\frac{W}{2}\le x\le \frac{W}{2}}$ Sketch: the same odd sine shape, p-type on the left, n-type on the right. Near $x=0$, $\sin (\pi x/W)\approx \pi x/W$, so the profile is approximately linearly graded with $a\approx \frac{{\pi }^3{K}_S{\epsilon }_0{V}_{bi}}{2q{W}^3}$ Cross-check: the exact linear-graded result is $a=12K_S{\epsilon }_0{V}_{bi}/(q{W}^3)$. The coefficients ${\pi }^3/2\approx 15.5$ and $12$ are similar, as expected for a smoothed linear profile.

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Problem 5.13: Linearly graded junction by computer

Problem. For a nondegenerately doped Si linearly graded junction at $T=300$ K: (a) plot $\mathcal{E}(x)$ and $V(x)$ for a given gradient $a$ (iterating Eq. 5.48 for $V_{bi}$); (b) overlay $V_A=V_{A0}/2^n$, $n=0$ to 3; (c) also print $a$, $V_A$, $V_{bi}$, $W$, $\mathcal{E}$ at $x=0$, $V$ at $x=0$; (d) vary $a$ from $10^{18}$ to $10^{23}$ cm${}^{-4}$; (e) compare with Problem 5.7.

What is not recoverable. This is a programming task, so there is no figure data to read. The OCR gives no numerical $V_{A0}$ or specific $a$, so I choose $V_{A0}=-4$ V (reverse bias) and $a=10^{20}$ cm${}^{-4}$ for illustration. Eq. 5.48 is quoted from the standard linear-graded theory because its text did not survive.

Equations (depletion approximation, junction at $x=0$, $N_D-N_A=ax$).

Neutrality makes the region symmetric, $x_n=x_p=W/2$: $\rho (x)=qax,|x|\le W/2$ $\mathcal{E}(x)=-\frac{qa}{2K_S{\epsilon }_0}\left[{\left(\frac{W}{2}\right)}^2-x^2\right]$ $V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^{2}x}{4}-\frac{x^3}{3}+\frac{W^3}{12}\right]$ Check: $V(W/2)=\frac{qa}{2K_S{\epsilon }_0}\frac{W^3}{6}=\frac{qa{W}^3}{12K_S{\epsilon }_0}$, which must equal $V_{bi}-V_A$: $W={\left[\frac{12K_S{\epsilon }_0(V_{bi}-V_A)}{qa}\right]}^{1/3}$ Peak field and midpoint potential: $\mathcal{E}(0)=-\frac{qa{W}^2}{8K_S{\epsilon }_0},V(0)=\frac{V_{bi}-V_A}{2}$ $V_{bi}$ is not known in closed form, because the doping at the depletion edges, $N=aW/2$, depends on $W$, which depends on $V_{bi}$. Eq. 5.48 (as I recall it) reads $V_{bi}=\frac{2kT}{q}\ln \frac{a{W}_{eq}}{2n_i},W_{eq}={\left[\frac{12K_S{\epsilon }_0{V}_{bi}}{qa}\right]}^{1/3}$

Algorithm for $V_{bi}$ (fixed-point iteration).

1. Guess $V_{bi}^{(0)}=0.6$ V.
2. Compute $W^{(k)}$ from the $W$ formula with $V_A=0$ and $V_{bi}^{(k)}$.
3. Update $V_{bi}^{(k+1)}=\frac{2kT}{q}\ln \frac{a{W}^{(k)}}{2n_i}$.
4. Stop when $|V_{bi}^{(k+1)}-V_{bi}^{(k)}|<10^{-6}$ V.

It converges fast because $V_{bi}$ depends on $W$ only logarithmically and $W$ on $V_{bi}$ only as a cube root.

Sample MATLAB.

q = 1.602e-19;  e0 = 8.854e-14;  Ks = 11.8;   % C, F/cm
ni = 1e10;  kT = 0.0259;                       % cm^-3, V
a = 1e20;   VA0 = -4;                          % cm^-4, V
 
Vbi = 0.6;                                     % iterate Eq. 5.48
for k = 1:100
    Weq  = (12*Ks*e0*Vbi/(q*a))^(1/3);
    Vnew = 2*kT*log(a*Weq/(2*ni));
    if abs(Vnew-Vbi) < 1e-6, Vbi = Vnew; break, end
    Vbi = Vnew;
end
 
figure; subplot(2,1,1); hold on; subplot(2,1,2); hold on;
for n = 0:3
    VA = VA0/2^n;
    W  = (12*Ks*e0*(Vbi-VA)/(q*a))^(1/3);      % cm
    x  = linspace(-W/2, W/2, 400);
    E  = -q*a/(2*Ks*e0)*((W/2)^2 - x.^2);      % V/cm
    V  = q*a/(2*Ks*e0)*(W^2*x/4 - x.^3/3 + W^3/12);
    subplot(2,1,1); plot(x*1e4, E);
    subplot(2,1,2); plot(x*1e4, V);
    fprintf('a=%g VA=%g Vbi=%.4f W=%.4g E0=%.4g V0=%.4f\n', ...
        a, VA, Vbi, W, -q*a*W^2/(8*Ks*e0), (Vbi-VA)/2);
end
subplot(2,1,1); xlabel('x (um)'); ylabel('E (V/cm)');
subplot(2,1,2); xlabel('x (um)'); ylabel('V (V)');

Wrapping the body in for a = logspace(18,23,6) handles part (d).

Expected numbers (hand check, $a=10^{20}$ cm${}^{-4}$, $V_A=0$). $W^{(0)}={\left[\frac{12(11.8)(8.854\times 10^{-14})(0.6)}{(1.6\times 10^{-19})(10^{20})}\right]}^{1/3}\approx 7.8\times 10^{-5}\text{ cm}$ $V_{bi}^{(1)}=2(0.0259)\ln \frac{10^{20}(7.8\times 10^{-5})}{2\times 10^{10}}\approx 0.667\text{ V}$ A second pass gives $W\approx 8.2\times 10^{-5}$ cm (0.82 $\mu$m) and $V_{bi}\approx 0.67$ V, converged in 3 to 4 passes. Then $\mathcal{E}(0)\approx -1.3\times 10^4$ V/cm and $V(0)=V_{bi}/2\approx 0.33$ V.

Part (b) behavior. As $V_A$ goes $-4,-2,-1,-0.5$ V, $W\propto (V_{bi}-V_A{)}^{1/3}$ shrinks (roughly 1.7 $\mu$m down to 0.9 $\mu$m), and the peak field scales as $(V_{bi}-V_A{)}^{2/3}$. The $\mathcal{E}$ curves are nested inverted parabolas; the $V$ curves are nested S-shaped cubics, each rising from 0 to its own $V_{bi}-V_A$.

Part (d), range of $a$. $W\propto a^{-1/3}$, $|\mathcal{E}{|}_{max}\propto a{W}^2\propto a^{1/3}$, and $V_{bi}$ changes only logarithmically (roughly 0.5 V at $10^{18}$ to 0.9 V at $10^{23}$). Over five decades of $a$ the width drops by about $46\times$: a steep gradient gives a thin, high-field junction.

Part (e), comparison with the step junction of 5.7. The step junction has a triangular field (linear in $x$) and a parabolic $V$; the graded junction has a parabolic field and a cubic $V$. The graded junction’s width grows as $(V_{bi}-V_A{)}^{1/3}$ rather than ${}^{1/2}$, so it is less sensitive to bias, and its field is spread more evenly.

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Problem 5.14: The “exact” equilibrium step-junction solution

Problem. Without the depletion approximation, the equilibrium step junction can be solved in closed form using the $F$-function. (a) Plot $\mathcal{E}(x)$ and $V(x)$ for a sample doping and compare with Problem 5.7(a); (b) also plot $\rho /q$ and compare with the depletion approximation.

What is not recoverable. The OCR scrambled the formulas and the sample doping ($N_A$ prints as a garbled “$10^1$/cm${}^3$”; $N_D=2\times 10^{14}$ cm${}^{-3}$). I state the equations in the standard form consistent with the surviving fragments and use $N_A=10^{16}$ cm${}^{-3}$ only as a placeholder, not as the book’s value.

Setup. Define $U=qV/kT$, a length scale, and the Fermi levels: $L_I=\sqrt{\frac{K_S{\epsilon }_{0}kT}{2q^2{n}_i}},U_{FP}=\ln \frac{N_A}{n_i},U_{FN}=-\ln \frac{N_D}{n_i},U_{BI}=U_{FP}-U_{FN}=\frac{q{V}_{bi}}{kT}$ The $F$-function comes from integrating Poisson’s equation once with Boltzmann carrier statistics: $F(U,U_F)={\left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^U-U-1\right)\right]}^{1/2}$ A second integral gives position, and $F$ also gives the field:

p-side ($x\le 0$): $x=L_I{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime },U_{FP})},\mathcal{E}=-\frac{kT}{q{L}_I}F(U,U_{FP})$ n-side ($x\ge 0$): $x=L_I{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime }-U_{BI},U_{FN})},\mathcal{E}=-\frac{kT}{q{L}_I}F(U-U_{BI},U_{FN})$ Charge density (p-side), exact Boltzmann form; the n-side is the same with $U_{FN}$ and $U-U_{BI}$: $\frac{\rho }{q}=n_i\left(e^{U_{FP}-U}-e^{U-U_{FP}}+e^{-U_{FP}}-e^{U_{FP}}\right)$ In plain terms: far from the junction $F\to 0$ (the field dies away), and $\rho$ counts ionized dopants plus mobile carriers exactly, so it has smooth tails instead of sharp edges.

Matching at $x=0$. The field is continuous there, so $F(U_0,U_{FP})=F(U_0-U_{BI},U_{FN})$ This is one transcendental equation for $U_0$; solve it with fzero starting at $U_0=U_{BI}/2$.

Sample MATLAB.

q=1.602e-19; e0=8.854e-14; Ks=11.8; ni=1e10; kT=0.0259;  % kT in volts
NA=1e16; ND=2e14;                       % NA is a placeholder
UFP=log(NA/ni);  UFN=-log(ND/ni);  UB=UFP-UFN;
LI=sqrt(Ks*e0*kT/(2*q*ni));             % cm
F=@(U,UF) sqrt( exp(UF).*(exp(-U)+U-1) + exp(-UF).*(exp(U)-U-1) );
 
U0 = fzero(@(U) F(U,UFP)-F(U-UB,UFN), UB/2);
 
Up = linspace(0.1, U0, 300);            % p-side, avoid F=0 at U=0
xp = arrayfun(@(u) LI*integral(@(s) 1./F(s,UFP), U0, u), Up);
Ep = -(kT/LI)*F(Up,UFP);
 
Un = linspace(U0, UB-0.1, 300);         % n-side
xn = arrayfun(@(u) LI*integral(@(s) 1./F(s-UB,UFN), U0, u), Un);
En = -(kT/LI)*F(Un-UB,UFN);
 
x = [xp xn];  E = [Ep En];  V = kT*[Up Un];
subplot(2,1,1); plot(x*1e4,E); ylabel('E (V/cm)');
subplot(2,1,2); plot(x*1e4,V); ylabel('V (V)'); xlabel('x (um)');
 
rp = ni*(exp(UFP-Up)-exp(Up-UFP)+exp(-UFP)-exp(UFP));   % rho/q, p-side
rn = ni*(exp(UFN-(Un-UB))-exp((Un-UB)-UFN)+exp(-UFN)-exp(UFN));

The integrand blows up where $F\to 0$ (at $U=0$ and $U=U_{BI}$), hence the instruction to start at $0.1$ rather than $0$. The far tails are cut off, which is fine for plotting.

Expected behavior and comparison with 5.7(a).

• $\mathcal{E}(x)$: a smooth, rounded triangular spike with its minimum at $x=0$ and exponential tails on both sides, instead of the abrupt zero at $-x_p$ and $x_n$ of the depletion approximation.
• $V(x)$: a smooth S-curve from $0$ to $V_{bi}$, the same total rise as in 5.7(a), with the corners rounded.
• Asymmetry: with $N_A\gg N_D$ nearly all the depletion layer lies on the n-side, $x_n\gg x_p$, in both solutions. Peak field and $V_{bi}=\frac{kT}{q}\ln (N_A{N}_D/n_i^2)$ agree to within a few percent.
• $\rho /q$ (part b): the depletion approximation draws two rectangles ($-N_A$ and $+N_D$) with vertical edges. The exact $\rho /q$ approaches those heights but rounds off over a few Debye lengths, $L_D=\sqrt{K_S{\epsilon }_{0}kT/(q^{2}N)}$, at each edge. The depletion approximation is excellent when the junction is many Debye lengths wide ($V_{bi}\gg kT/q$) and worst near the edges and on the lightly doped side.

Chapter 6: pn Junction Diode: I-V Characteristics

Where we are on the map. In Chapter 5 we studied the electrostatics of a pn junction: what the bands, fields and charges look like when the junction just sits there. Now we ask the question that matters for every circuit: if I push a voltage $V_A$ across this thing, how much current $I$ comes out? The answer, the $I$-$V$ characteristic, is the “personality” of the diode. We build that personality in three acts: (1) an idealized diode where everything goes right, giving the famous Shockley equation; (2) a comparison with real diodes, where we find things going wrong (breakdown, extra recombination current, series resistance, high-level injection); and (3) two supplementary tools (charge control and the narrow-base diode) that will pay off when we reach transistors.

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6.1 The Ideal Diode Equation

The ideal diode is a deliberately simplified diode. It is wonderful for insight, because it lets us see the machinery without the clutter. The goal of this section is to derive the ideal diode equation

$$I=I_0\left(e^{q{V}_A/kT}-1\right)$$

but, as the author stresses, the journey (the solution strategy, the boundary conditions, the pictures of what carriers do) matters more than the destination. We first “derive” the result with no equations at all, using only the energy band diagram.

6.1.1 Qualitative Derivation

Setting the stage (Fig. 6.1a, equilibrium, $V_A=0$). Draw the equilibrium band diagram of a pn junction. Sprinkle dots (electrons) and circles (holes) over it to show where the carriers live:

• On the quasineutral n-side there are very many electrons near $E_c$ and very few holes. The electron “pyramid” of dots thins out roughly exponentially as you climb higher in the conduction band (a Boltzmann tail).
• On the quasineutral p-side there are very many holes near $E_v$ and very few electrons. The hole population thins out roughly exponentially as you move down in the valence band.

Picture it: a mountain pass. Imagine two crowded valleys (n-side with electrons, p-side with holes) separated by a mountain ridge. People (carriers) jiggle around with thermal energy. Most electrons in the n-valley do not have enough energy to climb the ridge and get reflected back. But a few high-energy ones make it over. That trickle is diffusion.

Now consider the two kinds of carrier on each side:

1. Majority carriers (n-side electrons, p-side holes) see the potential hill. Only the energetic few surmount it and cross. This is diffusion current.
2. Minority carriers (p-side electrons, n-side holes) see no hill, because for them the potential energy slope runs downhill. If a minority electron wanders into the depletion region, the field sweeps it to the n-side immediately. This is a drift current.

At equilibrium these precisely cancel, for electrons and for holes separately:

$$I_N^{\text{diff}}=I_N^{\text{drift}},I_P^{\text{diff}}=I_P^{\text{drift}}\Rightarrow I=0.$$

Forward bias (Fig. 6.1b, $V_A>0$). The most important change is that the potential hill is lowered. The minority carriers still wander in and fall down the hill at the same rate as before (the “waterfall” is unaffected). But now many more n-side electrons and p-side holes can surmount the lowered hill. They give an electron current $I_N$ and a hole current $I_P$, both directed from p to n (conventional current), so the total $I=I_N+I_P$ flows in the correct direction for a forward-biased diode.

Why exponential? Two facts combine:

• The hill is lowered by $q{V}_A$, i.e. linearly in $V_A$.
• The number of carriers with energy greater than a given height varies exponentially with that height (Boltzmann factor $e^{-E/kT}$).

So the number that can surmount the hill grows as $e^{q{V}_A/kT}$: forward current increases exponentially with $V_A$ (Fig. 6.1d).

Reverse bias (Fig. 6.1c, $V_A<0$). The hill is raised. Even a modest reverse bias (more than a few $kT/q$) shuts majority-carrier diffusion down to a negligible level. The minority carriers wandering into the depletion region and being swept across are unaffected. So a small current flows from n to p, associated with minority carriers (so it is tiny), and because the minority-carrier drift current depends only on how many wander in per second and not on the height of the hill, it saturates.

Analogy: the waterfall. The amount of water going over a waterfall depends on how much water the river supplies, not on how tall the falls are. Raise the hill (more reverse bias) and the minority carriers fall farther, but the number that fall per second is unchanged.

Calling the saturation value $-I_0$, the overall behavior must be

$$\boxed{I=I_0\left(e^{V_A/V_{\text{ref}}}-1\right)}\text{\hspace{1em}(6.1)}$$

which equals the ideal diode equation if $V_{\text{ref}}=kT/q$. We have found the form of the diode equation without writing a single transport equation. The same picture also explains rectification: forward bias lowers the hill, so large numbers of majority carriers are injected; reverse bias raises it, leaving only a residual minority-carrier current.

The “big picture” objection

The author recounts being asked: “Yes, but doesn’t injecting majority carriers (forward) or extracting minority carriers (reverse) cause a charge build-up in the device?” The short answer is that we are in steady state, so nothing builds up. The deeper answer is the “big picture” of Fig. 6.2, drawn for reverse bias (capacitor-like plates at the outer ends of the band diagram represent the ohmic contacts):

1. Minority electrons wander from the p-side into the depletion region and are swept to the n-side. Each one that leaves is replaced by thermal generation through an R-G center (an electron jumps $E_v\to E_T\to E_c$ on the p-side near the depletion region).
2. Similarly, each swept n-side hole is replaced by generation on the n-side.
3. The electrons that arrive on the n-side (and those generated there) create a slight local excess of majority electrons next to the depletion region. The tiny local electric field this excess creates pushes the adjacent electrons toward the contact, who push their neighbors, and so on. This displacement propagates with great speed through the whole in-line group of n-side electrons. The electron adjacent to the contact is pushed into the metal and out into the external circuit.
4. On the p-side the story is mirrored with holes: the excess at the depletion edge pushes in-line holes along, and holes at the contact recombine with electrons from the metal, completing the circle.

Analogy: a crowded room with two doors. Latecomers to a party force their way in through one door of an absolutely crammed room. A like number of people get pushed out the other door. Nobody runs across the room; each person only moves a little. The “signal” travels fast; the individual carriers move slowly.

A side observation from Fig. 6.2: both electrons and holes carry current through the depletion region, but far from the junction on the p-side the hole current dominates, and far on the n-side the electron current dominates. The total current is the same everywhere, while the electron and hole components vary with position.

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6.1.2 Quantitative Solution Strategy

“Going into a football game, coaches always have a game plan.” The mathematics of the ideal diode derivation is short; the strategy is what you must understand, because you will reuse it for other problems and for modified theories.

General considerations: assumptions

1. The diode is operated under steady state conditions.
2. A nondegenerately doped step junction models the doping profile.
3. The diode is one-dimensional.
4. Low-level injection prevails in the quasineutral regions.
5. No processes other than drift, diffusion and thermal recombination-generation occur inside the diode. In particular $G_L=0$.

(These were all explicitly or implicitly used in establishing the pn junction electrostatics.)

Current relationships available

$$I=AJ(A=\text{cross-sectional area})\text{\hspace{1em}(6.2)}$$ $$J=J_N(x)+J_P(x)\text{\hspace{1em}(6.3)}$$ $$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}\text{\hspace{1em}(6.4a)}$$ $$J_P=q{\mu }_{p}p\mathcal{E}-q{D}_P\frac{dp}{dx}\text{\hspace{1em}(6.4b)}$$

Eq. (6.3) says the total current density is constant throughout the diode, but its electron and hole parts change with position. If we had exact $\mathcal{E}$, $n$, $p$ versus $x$ we would be done. We do not; we only have the approximate three-region electrostatic solution of Fig. 6.3:

Region	Extent	Field
Quasineutral p-region	$x\le -x_p$	$\mathcal{E}\approx 0$
Depletion region	$-x_p\le x\le x_n$	$\mathcal{E}\ne 0$
Quasineutral n-region	$x\ge x_n$	$\mathcal{E}\approx 0$

In the quasineutral regions $\mathcal{E}\approx 0$ and low-level injection holds, so the minority carrier diffusion equations can be used.

Quasineutral region considerations

Under steady state with $G_L=0$:

$$0=D_N\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n}x\le -x_p\text{\hspace{1em}(6.5a)}$$ $$0=D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}x\ge x_n\text{\hspace{1em}(6.5b)}$$

Since $\mathcal{E}=0$ and $d{n}_0/dx=d{p}_0/dx=0$, Eqs. (6.4) reduce to (minority carrier currents are pure diffusion):

$$J_N=q{D}_N\frac{d\Delta n_p}{dx}x\le -x_p\text{\hspace{1em}(6.6a)}$$ $$J_P=-q{D}_P\frac{d\Delta p_n}{dx}x\ge x_n\text{\hspace{1em}(6.6b)}$$

The catch. These give $J_N(x)$ only in the p-side quasineutral region and $J_P(x)$ only in the n-side quasineutral region: non-overlapping pieces. To evaluate $J=J_N+J_P$ we need at least one point where we know both, possibly by extrapolation.

Depletion region considerations

Here the full continuity equations must be used. With the assumptions above they become

$$0=\frac{1}{q}\frac{d{J}_N}{dx}+{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}\text{\hspace{1em}(6.7a)}$$ $$0=-\frac{1}{q}\frac{d{J}_P}{dx}+{\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}\text{\hspace{1em}(6.7b)}$$

Now make the crucial additional assumption that thermal R-G is negligible throughout the depletion region: $\partial n/\partial t{|}_{\text{R-G}}=\partial p/\partial t{|}_{\text{R-G}}=0$. Then $d{J}_N/dx=0$ and $d{J}_P/dx=0$. The carrier currents are constants across the depletion region, edges included:

$$J_N(-x_p\le x\le x_n)=J_N(-x_p)\text{\hspace{1em}(6.8a)}$$ $$J_P(-x_p\le x\le x_n)=J_P(x_n)\text{\hspace{1em}(6.8b)}$$

And $J_N(-x_p)$ and $J_P(x_n)$ can be obtained from the quasineutral solutions evaluated at the depletion edges. Hence

$$J=J_N(-x_p)+J_P(x_n)\text{\hspace{1em}(6.9)}$$

We now have a plan: solve for the minority carrier current densities in the quasineutral regions, evaluate them at the depletion edges, add, multiply by $A$.

A word of honesty. “No thermal R-G in the depletion region” is really the defining property of the ideal diode. There is no a priori reason it should be true; it simply gives a simple answer. We will check it against experiment in Section 6.2 (and find it fails at small forward biases and all reverse biases in room-temperature Si).

Boundary conditions

Two boundary conditions are needed for each of Eqs. (6.5a) and (6.5b): the excess concentrations at the ohmic contacts and at the depletion region edges (Fig. 6.4a).

At the ohmic contacts. The ideal diode is taken to be a wide-base diode: the contacts are several minority-carrier diffusion lengths or more from the depletion edges. Any perturbation created at a depletion edge dies away before reaching the contact, so the contacts may be placed at $x=\pm \infty$:

$$\Delta n_p(x\to -\infty )=0\text{\hspace{1em}(6.10a)}$$ $$\Delta p_n(x\to +\infty )=0\text{\hspace{1em}(6.10b)}$$

At the depletion region edges. Use the quasi-Fermi level formalism. Multiply the exponential forms of Eqs. (3.72a) and (3.72b), $n=n_i{e}^{(F_N-E_i)/kT}$ and $p=n_i{e}^{(E_i-F_P)/kT}$:

$$np=n_i^2{e}^{(F_N-F_P)/kT}\text{\hspace{1em}(6.11)}$$

This is valid everywhere in the diode under arbitrary operating conditions. We do not know $F_N(x)$ and $F_P(x)$ beforehand, but as envisioned in Fig. 6.4b it is reasonable that they vary monotonically from $F_P=E_F$ far on the p-side to $F_N=E_F$ far on the n-side. The Fermi level difference between the far ends is $q{V}_A$, and monotonic variation makes $F_N-F_P\le q{V}_A$ at all points. Assume the equal sign holds throughout the depletion region: the quasi-Fermi levels are constant across the depletion region, $F_N-F_P=q{V}_A$. Then

$$\boxed{np=n_i^2{e}^{q{V}_A/kT},-x_p\le x\le x_n}\text{\hspace{1em}(6.12)}$$

Eq. (6.12) is called the law of the junction.

Evaluate at the p-edge ($x=-x_p$), where $p\approx N_A$ (low-level injection: the majority concentration is unchanged):

$$n(-x_p)p(-x_p)=n(-x_p)N_A=n_i^2{e}^{q{V}_A/kT}\text{\hspace{1em}(6.13)}$$ $$n_p(-x_p)=\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}\text{\hspace{1em}(6.14)}$$

Subtract the equilibrium value $n_{p0}=n_i^2/N_A$:

$$\Delta n_p(-x_p)=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.15)}$$

At the n-edge ($x=x_n$), where $n\approx N_D$:

$$n(x_n)p(x_n)=N_{D}p(x_n)=n_i^2{e}^{q{V}_A/kT}\text{\hspace{1em}(6.16)}$$ $$p_n(x_n)=\frac{n_i^2}{N_D}{e}^{q{V}_A/kT}\text{\hspace{1em}(6.17)}$$ $$\Delta p_n(x_n)=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.18)}$$

Meaning. Forward bias ($V_A>0$) raises the minority concentration at the depletion edge above equilibrium by an exponential factor. Reverse bias beyond a few $kT/q$ drives the edge minority concentration essentially to zero: $\Delta n_p\to -n_i^2/N_A=-n_{p0}$. The depletion region acts as a “minority carrier sink.”

The assumption $F_N-F_P=q{V}_A$ across the depletion region is central and critical. It is equivalent to assuming $J_N\approx 0$ and $J_P\approx 0$ relative to their individual drift and diffusion components inside the depletion region (Problem 6.5 carries out this alternative, Shockley-style derivation). Like the constant quasi-Fermi assumption, and contrary to statements in some texts, there is no simple a priori justification of this assumption. Lengthy subsidiary analyses and comparison with experiment indicate it is typically justified.

”Game plan” summary

To get the current in an ideal diode versus $V_A$:

1. Solve the minority carrier diffusion equations (6.5) with boundary conditions (6.10) and (6.15)/(6.18) to get $\Delta n_p$ and $\Delta p_n$.
2. Compute the minority carrier current densities in the quasineutral regions using Eqs. (6.6).
3. Per Eq. (6.9), evaluate $J_N(x)$ and $J_P(x)$ at the depletion edges and add the two edge current densities. Finally, multiply by the cross-sectional area $A$.

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6.1.3 Derivation Proper

Time to execute the game plan. Start with the holes on the quasineutral n-side. To make the math pretty, shift the origin to the n-edge of the depletion region (Fig. 6.5a): $x^{\prime }=x-x_n$, so the n-side quasineutral region is $x^{\prime }\ge 0$.

Step 1: the differential equation and its boundary conditions.

$$0=D_P\frac{d^2\Delta p_n}{d{x}^{\prime 2}}-\frac{\Delta p_n}{{\tau }_p}{x}^{\prime }\ge 0\text{\hspace{1em}(6.19)}$$ $$\Delta p_n(x^{\prime }\to \infty )=0\text{\hspace{1em}(6.20a)}$$ $$\Delta p_n(x^{\prime }=0)=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.20b)}$$

Step 2: general solution. Divide (6.19) by $D_P$: $d^2\Delta p_n/d{x}^{\prime 2}=\Delta p_n/L_P^2$ with

$$L_P=\sqrt{D_P{\tau }_p}\text{\hspace{1em}(6.22)}$$

This is special-case diffusion equation no. 1 of Table 3.2. Try $e^{\lambda x^{\prime }}$: ${\lambda }^2=1/L_P^2$, so $\lambda =\pm 1/L_P$:

$$\Delta p_n(x^{\prime })=A_1{e}^{-x^{\prime }/L_P}+A_2{e}^{x^{\prime }/L_P}{x}^{\prime }\ge 0\text{\hspace{1em}(6.21)}$$

Step 3: apply boundary conditions.

• (6.20a): $e^{x^{\prime }/L_P}\to \infty$ as $x^{\prime }\to \infty$, so the only way to keep $\Delta p_n\to 0$ is $A_2=0$.
• (6.20b): at $x^{\prime }=0$, $A_1=\Delta p_n(x^{\prime }=0)$.

Therefore

$$\boxed{\Delta p_n(x^{\prime })=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime }/L_P}}{x}^{\prime }\ge 0\text{\hspace{1em}(6.23)}$$

Step 4: hole current density from (6.6b), $J_P=-q{D}_{P}d\Delta p_n/d{x}^{\prime }$:

$$\begin{array}{rl}\frac{d\Delta p_n}{d{x}^{\prime }}& =-\frac{1}{L_P}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime }/L_P}\\ J_P(x^{\prime })& =q\frac{D_P}{L_P}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime }/L_P}\end{array}{x}^{\prime }\ge 0\text{\hspace{1em}(6.24)}$$

Step 5: the p-side. With the coordinate $x^{\prime \prime }=-x-x_p$ (distance measured into the p-region from the p-edge, Fig. 6.5b), the electron problem is the exact mirror image ($D_N,{\tau }_n,L_N=\sqrt{D_N{\tau }_n}$, $N_A$):

$$\Delta n_p(x^{\prime \prime })=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime \prime }/L_N}{x}^{\prime \prime }\ge 0\text{\hspace{1em}(6.25)}$$ $$J_N(x^{\prime \prime })=q{D}_N\frac{d\Delta n_p}{dx}\Rightarrow J_N(x^{\prime \prime })=q\frac{D_N}{L_N}\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime \prime }/L_N}{x}^{\prime \prime }\ge 0\text{\hspace{1em}(6.26)}$$

(The sign: $d{x}^{\prime \prime }=-dx$ so $d/dx=-d/d{x}^{\prime \prime }$, and $d{e}^{-x^{\prime \prime }/L_N}/d{x}^{\prime \prime }=-e^{-x^{\prime \prime }/L_N}/L_N$; the two minus signs combine to give a positive $J_N$ in our convention, in which both carrier currents are positive for forward bias.)

Step 6: evaluate at the depletion edges (set $x^{\prime }=0$, $x^{\prime \prime }=0$):

$$J_N(x=-x_p)=J_N(x^{\prime \prime }=0)=q\frac{D_N}{L_N}\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.27a)}$$ $$J_P(x=x_n)=J_P(x^{\prime }=0)=q\frac{D_P}{L_P}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.27b)}$$

Step 7: add and multiply by $A$ (Eqs. 6.8, 6.9, 6.2):

$$I=AJ=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.28)}$$

or

$$\boxed{I=I_0\left(e^{q{V}_A/kT}-1\right)}\text{\hspace{1em}(6.29)}$$ $$\boxed{I_0=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)}\text{\hspace{1em}(6.30)}$$

Eq. (6.29) is the ideal diode equation, also called the Shockley equation. $I_0$ is the (reverse-bias) saturation current.

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6.1.4 Examination of Results

Let us now work backward through the derivation and look at what we found, as a physician examines a patient: the final result first, then the internal organs (currents and carrier concentrations).

Ideal $I$-$V$

(Fig. 6.6 summarizes it.)

• Reverse bias with $|V_A|\gg kT/q$ (a few tenths of a volt at room temperature): the exponential vanishes and $I\to -I_0$. The ideal theory says this saturation persists to reverse voltages of unlimited magnitude.
• Forward bias with $V_A\gg kT/q$: the "$-1$" is negligible and $I\to I_0{e}^{q{V}_A/kT}$.

Because of the exponential, the forward-bias characteristic is commonly drawn on a semilog scale (Fig. 6.6b):

$$\ln I=\ln I_0+\frac{q}{kT}{V}_A{V}_A\gg \frac{kT}{q}\text{\hspace{1em}(6.31)}$$

A straight line with slope $q/kT$ and an extrapolated $V_A=0$ intercept $\ln I_0$. That is how you read $I_0$ off a semilog plot.

Quick feel for the numbers. At 300 K, $kT/q=0.0259$ V, so every $0.0259\ln 10\approx 60$ mV of extra forward bias multiplies the current by 10. That is the famous “60 mV per decade.”

The saturation current: two big observations

(1) Material dependence. $I_0$ contains $n_i^2$, which varies enormously with the semiconductor. At room temperature $n_i\approx 10^{10}{\text{cm}}^{-3}$ in Si, but $n_i\approx 10^{13}{\text{cm}}^{-3}$ in Ge. Therefore $n_i^2$ differs by a factor $\sim 10^6$, and Ge diodes are expected to have a reverse saturation current roughly a million times larger than comparable Si diodes. (Think of $n_i^2$ as the “thermal leakiness” of the material: small band gap means easy thermal generation of carriers.)

(2) Asymmetric junctions. $I_0$ has two terms, each inversely proportional to the doping of its own side. In a $p^{+}$-$n$ or $n^{+}$-$p$ junction, the term for the heavily doped side is negligible:

$$I_0\approx qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}{p}^{+}\text{-}n\text{ diodes}\text{\hspace{1em}(6.32a)}$$ $$I_0\approx qA\frac{D_N}{L_N}\frac{n_i^2}{N_A}{n}^{+}\text{-}p\text{ diodes}\text{\hspace{1em}(6.32b)}$$

The lightly doped side controls everything. The same was true for the depletion width and other electrostatic variables. As a general rule: the heavily doped side of an asymmetrical junction can be ignored in determining its electrical characteristics. Whenever a problem says ”$p^{+}$-$n$,” you may ignore the $p^{+}$ side.

It is fortuitous that the heavy side is negligible. In real diodes the $p^{+}$ or $n^{+}$ doping is degenerate; if its contribution mattered, the ideal theory would have to be modified to treat degenerate doping.

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Exercise 6.1

Problem. Two ideal $p^{+}$-$n$ step junction diodes at room temperature are identical except that $N_{D1}=10^{15}/{\text{cm}}^3$ and $N_{D2}=10^{16}/{\text{cm}}^3$. Compare the $I$-$V$ characteristics; sketch both on one set of axes.

Solution. For $p^{+}$-$n$ diodes

$$I_0\approx qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}.$$

Use $L_P=\sqrt{D_P{\tau }_p}$, so $D_P/L_P=\sqrt{D_P/{\tau }_p}$. Also $D_P=(kT/q){\mu }_p$. Then

$$\frac{I_{0,1}}{I_{0,2}}=\frac{\sqrt{D_{P1}/{\tau }_{p1}}}{\sqrt{D_{P2}/{\tau }_{p2}}}\cdot \frac{N_{D2}}{N_{D1}}$$

We have the same diode area, same $n_i$ (same semiconductor). The R-G center concentrations are taken to be identical, so ${\tau }_{p1}={\tau }_{p2}$. The mobilities at $10^{15}$ and $10^{16}{\text{cm}}^{-3}$ differ only slightly in most materials (Fig. 3.5), so ${\mu }_{p1}\approx {\mu }_{p2}$ and $D_{P1}\approx D_{P2}$. Hence

$$\frac{I_{0,1}}{I_{0,2}}\approx \frac{N_{D2}}{N_{D1}}=\frac{10^{16}}{10^{15}}=10$$

Result. The diode-1 current is approximately ten times larger than the diode-2 current at all applied voltages (Fig. E6.1). On a linear plot the curves have the same shape (same exponent), with diode 1 scaled up by 10; on a semilog forward plot they are two parallel lines (same slope $q/kT$) displaced vertically by one decade; and the reverse saturation currents are $-I_{0,1}=-10I_{0,2}$.

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Exercise 6.2 (computer): Ideal $I$-$V$ versus doping

Problem. An ideal Si $p^{+}$-$n$ step junction diode at 300 K has $A=10^{-4}{\text{cm}}^2$ and ${\tau }_p=10^{-6}$ s. Using the empirical-fit hole mobility from Exercise 3.1, write a MATLAB program that computes and plots the ideal $I$-$V$ characteristic. The $n$-side doping is an input variable. Use a linear plot with axis set to [-1, 0.2, -2*I0, 5*I0], where I0 is the reverse saturation current. Explore how the characteristic depends on doping.

The logic of the program.

1. Set constants: $k=8.617\times 10^{-5}$ eV/K, $q=1.6\times 10^{-19}$ C.
2. Set parameters: $A$, $n_i=10^{10}$, ${\tau }_p$, $T=300$ K, and ask for $N_D$ (a vector so several dopings can be plotted together by typing e.g. [1e16 2e16 5e16 1e17]).
3. Hole mobility: the empirical fit ${\mu }_p={\mu }_{p,\min }+{\mu }_{p,0}/[1+(N_D/N_{ref}{)}^{\alpha }]$ with ${\mu }_{p,\min }=54.3$, ${\mu }_{p,0}=406.9$, $N_{ref}=2.35\times 10^{17}$, $\alpha =0.88$ (units ${\text{cm}}^2/\text{V}\cdot \text{s}$). (The minority carrier mobility is assumed equal to the majority carrier mobility at the same doping.)
4. $D_P=(kT/q){\mu }_p$ (with $k$ in eV/K, $kT$ in eV equals $kT/q$ in volts), $L_P=\sqrt{D_P{\tau }_p}$, and $I_0=qA(D_P/L_P)(n_i^2/N_D)$.
5. For each doping compute $I=I_0(e^{V_A/kT}-1)$ over $V_A=-1\dots 0.2$ V using an outer product (a column of $I_0$ values times a row of exponentials) so that one matrix holds all curves.
6. Plot, set axes by the first (largest) $I_0$, add axes lines and a legend by hand.

MATLAB program (cleaned from the noisy OCR):

% Variation of Ideal-Diode I-V with semiconductor doping.
% Si step junction, T = 300 K.
% In response to the "ND=" prompt type [ND1 ND2 ...] to input
% multiple doping values.
 
% Initialization and Universal Constants
clear
k = 8.617e-5;          % Boltzmann constant (eV/K)
q = 1.6e-19;           % electronic charge (C)
 
% Device, Material, and System Parameters
A = 1.0e-4;            % cm^2
ni = 1.0e10;           % cm^-3
taup = 1.0e-6;         % s
ND = input('Input the n-side doping concentration, ND= ');
T = 300;
 
% Hole Mobility Calculation
NAref = 2.35e17;
mupmin = 54.3;
mup0 = 406.9;
ap = 0.88;
mup = mupmin + mup0./(1 + (ND./NAref).^ap);
% The mobility calculation here assumes the hole minority carrier
% mobility is equal to the hole majority carrier mobility.
 
% I-V Calculation
VA = linspace(-1, 0.2);
DP = k.*T.*mup;
LP = sqrt(DP.*taup);
I0 = q.*A.*(DP./LP).*(ni^2./ND)       % no semicolon: prints I0 values
I = I0.'*(exp(VA./(k.*T)) - 1);       % each row = one doping
 
% Plotting Result
close
plot(VA, I); grid
ymin = -2*I0(1); ymax = 5*I0(1);
axis([-1, 0.2, ymin, ymax]);
xlabel('V_A (volts)'); ylabel('I (amps)');
 
% Adding axes, key
xx = [-1 0.2]; yx = [0 0];
xy = [0 0];    yy = [ymin, ymax];
hold on
plot(xx, yx, '-w', xy, yy, '-w');
j = length(ND);
for i = 1:j
    yput = (0.70 - 0.06*i)*ymax;
    yk(i,1) = yput; yk(i,2) = yput;
    text(-0.68, (0.69 - 0.06*i)*ymax, ['ND = ', num2str(ND(i)), '/cm^3']);
end
xk = [-0.8 -0.7];
plot(xk, yk);
text(-0.74, 0.75*ymax, 'Si, 300 K');
hold off

Results (Fig. E6.2). The program was run with $N_D=10^{16},2\times 10^{16},5\times 10^{16},10^{17}{\text{cm}}^{-3}$. The reverse saturation currents are extremely small (of order $10^{-16}$ A), and $I_0$ decreases as doping increases, following the same general $1/N_D$ dependence noted in Exercise 6.1. Looking carefully at the numerical $I_0$ values printed in the Command window, there is a noticeable additional mobility dependence at these higher dopings, because ${\mu }_p$ falls as $N_D$ rises (so $D_P/L_P=\sqrt{D_P/{\tau }_p}$ drops too): the $I_0$ ratio is a bit larger than the pure doping ratio.

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Carrier currents

A sample plot of carrier and total current densities versus position in a forward-biased diode is Fig. 6.7. (The reverse-bias plot is essentially the same with all current densities negative.) The recipe for constructing Fig. 6.7:

1. From Eqs. (6.24) and (6.26): the minority carrier current densities in the quasineutral regions decay exponentially away from the depletion region edges (on the scale of $L_P$ or $L_N$).
2. Inside the depletion region, per Eqs. (6.8), $J_N$ and $J_P$ are drawn constant at their depletion-edge values. The total current density in the depletion region is the graphical sum $J_N+J_P$.
3. The total current density is constant everywhere in the diode (Eq. 6.3), so the value found in the depletion region extends throughout the device.
4. Finally, the majority carrier current in each quasineutral region is obtained by graphically subtracting the minority current from the constant total: $J_P(x)=J-J_N(x)$ on the p-side, $J_N(x)=J-J_P(x)$ on the n-side.

Story. Picture a conveyor belt carrying a fixed total flow. Far on the p-side nearly all the flow is holes. As you approach the junction, holes recombine with injected electrons and “hand over” the flow to the electrons. Far on the n-side nearly all the flow is electrons. The hand-over is the recombination zone, a few diffusion lengths wide.

This is consistent with the qualitative observation: both carriers contribute through the depletion region; holes dominate far on the p-side, electrons far on the n-side.

Carrier concentrations

Eqs. (6.23) and (6.25) tell how the minority concentrations deviate from equilibrium in the quasineutral regions:

• Forward bias increases minority concentrations above equilibrium (a pile-up, $\Delta n_p>0,\Delta p_n>0$).
• Reverse bias decreases them below equilibrium (a deficit).
• In both cases the perturbation decays exponentially with distance from the depletion edge. After several diffusion lengths it vanishes: $n_p\to n_i^2/N_A$ as $x\to -\infty$, $p_n\to n_i^2/N_D$ as $x\to +\infty$, independent of bias.
• Because low-level injection prevails, the majority carrier concentrations stay approximately at their equilibrium values, $p_p\approx N_A$ and $n_n\approx N_D$, regardless of bias.

Fig. 6.8 shows these (forward bias on a semilog scale, where exponentials appear as straight lines; reverse bias on a linear scale; $N_A>N_D$ assumed).

Forward bias, physical story. Majority carriers are injected in large numbers over the lowered hill. On the other side they become minority carriers and are progressively eliminated by recombination as they try to diffuse deeper (this is the same math as Sample Problem No. 2 of Chapter 3). The net result is a pile-up of excess minority carriers adjacent to the depletion region. This pile-up, a consequence of forward biasing, will be important in later chapters (stored charge, diffusion capacitance).

Reverse bias, physical story. The depletion region acts as a sink for minority carriers, draining adjacent regions. A reverse bias of only a few $kT/q$ drives the edge concentrations to essentially zero, and larger reverse biases barely change the distributions, which is consistent with the saturation of current (the current is proportional to the slope of the distribution at the depletion edge). Overall, reverse bias creates a modest deficit of minority carriers near the depletion region.

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Exercise 6.3

Problem. Fig. E6.3 is a dimensioned plot of the steady state carrier concentrations (log scale) inside a pn junction diode at room temperature. Reading the figure: $n_n(\text{far})=N_D=10^{15}{\text{cm}}^{-3}$ on the n-side; at the depletion edge $p_n(x_n)=10^{10}{\text{cm}}^{-3}$; far away $p_n\to 10^5{\text{cm}}^{-3}$ (so $n_i=10^{10}$); $p_n$ falls to $10^8$ at the point $x^{\prime }$ marked in the figure. (a) Is the diode forward or reverse biased? Explain. (b) Do low-level injection conditions prevail in the quasineutral regions? Explain. (c) Determine the applied voltage $V_A$. (d) Determine the hole diffusion length $L_P$.

Solution.

(a) Forward biased. There is a pile-up (excess) of minority carriers at the depletion region edges ($\Delta n_p>0$, $\Delta p_n>0$): the minority concentration near the edge exceeds its far-away equilibrium value. A deficit would indicate reverse bias.

(b) Yes. $\Delta p_n\ll n_n$ and $\Delta n_p\ll p_p$ everywhere in the quasineutral regions: the highest minority concentration ($10^{10}$ at the edge) is five orders of magnitude below the majority concentration ($10^{15}$).

(c) Use the law of the junction, solved for $V_A$:

$$np=n_i^2{e}^{q{V}_A/kT}\Rightarrow V_A=\frac{kT}{q}\ln \frac{np}{n_i^2},-x_p\le x\le x_n$$

Evaluate at the n-edge of the depletion region and note $n_i^2=n_n(\infty )p_n(\infty )$ (far from the junction, equilibrium):

$$\begin{array}{rl}{V}_A& =\frac{kT}{q}\ln \frac{n_n(x_n)p_n(x_n)}{n_n(\infty )p_n(\infty )}=(0.0259)\ln \frac{10^{15}\cdot 10^{10}}{10^{15}\cdot 10^5}\\ & =(0.0259)\ln \left(10^5\right)=(0.0259)(11.51)\approx 0.3\text{V}\end{array}$$

(d) Rewrite Eq. (6.23) in terms of $p_n$. Near the depletion edge $\Delta p_n\approx p_n$ (the pile-up far exceeds the equilibrium value there):

$$\begin{array}{rl}{p}_n(x^{\prime })& =p_n(0)e^{-x^{\prime }/L_P}\\ \ln \frac{p_n(x^{\prime })}{p_n(0)}& =-\frac{x^{\prime }}{L_P}\\ L_P& =\frac{x^{\prime }}{\ln \left[p_n(0)/p_n(x^{\prime })\right]}\end{array}$$

From the figure the concentration drops from $10^{10}$ to $10^8$ over $x^{\prime }=1.6\times 10^{-2}\text{ cm}$ (measured from the depletion edge):

$$L_P=\frac{1.6\times 10^{-2}}{\ln \left(10^{10}/10^8\right)}=\frac{1.6\times 10^{-2}}{\ln 100}=\frac{1.6\times 10^{-2}}{4.605}=3.47\times 10^{-3}\text{cm}$$

(Reading assumption: the OCR of the figure axis labels is partly garbled; the numbers above reproduce the book’s quoted answers $V_A\approx 0.3$ V and $L_P=3.47\times 10^{-3}$ cm.)

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Exercise 6.4

Problem. When reverse biased by more than a few $kT/q$, the diode current equals $q$ times the number of minority carriers per second that wander into the depletion region and are swept across. Under steady state this must equal the number generated per second in the quasineutral regions. So the saturation current may alternatively be viewed as arising from minority carrier generation in the quasineutral regions, $I_0=q\times$(carriers generated per second). Suppose the actual reverse-bias minority carrier distributions are approximated by squared-off distributions (Fig. E6.4): all minority carriers are depleted a minority carrier diffusion length to either side of the depletion region. Remembering low-level injection holds in the quasineutral regions, derive an expression for the reverse-bias saturation current.

Solution. In a low-level quasineutral region, the thermal net R-G rate is $-\Delta n/{\tau }_n$ (and similarly for holes). Under reverse bias the edge concentrations are driven to zero, so $\Delta n_p=-n_{p0}=-n_i^2/N_A$ and $\Delta p_n=-n_{p0}\to -n_i^2/N_D$ in the depleted zones. Thus

$${\frac{\partial n}{\partial t}|}_{\text{R-G}}=-\frac{\Delta n_p}{{\tau }_n}=+\frac{n_i^2}{N_A{\tau }_n}(-L_N-x_p\le x\le -x_p)$$ $${\frac{\partial p}{\partial t}|}_{\text{R-G}}=-\frac{\Delta p_n}{{\tau }_p}=+\frac{n_i^2}{N_D{\tau }_p}(x_n\le x\le x_n+L_P)$$

(positive means net generation). Generation takes place in volumes $A{L}_N$ and $A{L}_P$ on the p- and n-sides. Then

$$\begin{array}{rl}{I}_0& =q\left(A{L}_N\right)\frac{n_i^2}{N_A{\tau }_n}+q\left(A{L}_P\right)\frac{n_i^2}{N_D{\tau }_p}\\ & =qA\left(\frac{L_N}{{\tau }_n}\frac{n_i^2}{N_A}+\frac{L_P}{{\tau }_p}\frac{n_i^2}{N_D}\right)\end{array}$$

Because $L_N/{\tau }_n=D_N/L_N$ (from $L_N^2=D_N{\tau }_n$) and $L_P/{\tau }_p=D_P/L_P$,

$$I_0=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)$$

One obtains the usual $I_0$ expression. The squaring-off trick (replace an exponential by a rectangle of height equal to the peak and width equal to one decay length, enclosing the same area) is a very useful analytical tool, and we will use it again in the charge-control approach of Section 6.3.

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6.2 Deviations from the Ideal

The ideal diode derivation used a fairly long list of assumptions, some made with no a priori justification. It would be surprising if a careful experiment did not find discrepancies. Here we first compare experiment with theory to spot the major deviations, and then treat each one: its cause and the modification needed.

Do not be dismayed by this “band-aid” style of progress. Formulate a simple theory, compare with experiment, revise, compare again is how engineers and scientists tackle every complicated real problem.

6.2.1 Ideal Theory Versus Experiment

Fig. 6.9 shows the measured $I$-$V$ of a commercial Si diode at room temperature on a $0.1\mu \text{A}$/division linear scale (note the change of voltage scale between forward and reverse). Largely the characteristic looks as predicted: forward current rises rapidly and reverse current is vanishingly small over most of the range. But one obvious feature is not in the ideal theory: a large reverse current flows when the reverse voltage exceeds a certain value. This is breakdown (Subsection 6.2.2). It occurs at voltages that range over several orders of magnitude from diode to diode, but is common to all pn diodes.

To see further deviations, look at the data on better scales (Fig. 6.10, measured with an HP 4145B parameter analyzer):

Forward bias (semilog, Fig. 6.10a).

• Between roughly 0.35 V and 0.7 V the slope is the expected $q/kT$ (ideal behavior).
• Above 0.7 V the slope progressively decreases: the characteristic “slopes over”. This is related to the high current flowing as $V_A\to V_{bi}$ (Subsection 6.2.4).
• Below 0.35 V the current is quite small, far larger than the ideal expectation, and the slope is curiously close to $q/2kT$. A current component has been overlooked.

Reverse bias (Fig. 6.10b, $50$ pA/division).

• The estimated $I_0\sim 10^{-14}$ A (from the ideal forward region) is some three orders of magnitude smaller than the reverse current at $V_A=-5$ V!
• The reverse current does not saturate; it continually increases with reverse bias.
• (The wiggles are noise in the device and measuring system.)

As Subsection 6.2.3 shows, the small-forward-bias and the reverse-bias discrepancies are both due to thermal carrier recombination-generation in the depletion region, which the ideal diode derivation assumed negligible.

Features vary from device to device, and other nonidealities can occur, but this is fairly representative of commercial Si diodes at room temperature. GaAs diodes are similar. Ge diodes at room temperature and Si diodes at elevated temperatures are often closer to ideal (saturating reverse characteristics): an explanation follows in the R-G discussion.

Region of $I$-$V$	Observed	Cause (where treated)
Large reverse bias	Current $\to \infty$	Breakdown: avalanche / Zener (6.2.2)
All reverse bias, and small forward bias	Excess current, no saturation, slope $q/2kT$	Depletion-region R-G (6.2.3)
Moderate forward bias (0.35-0.7 V in the example)	Ideal slope $q/kT$	Diffusion current
$V_A\to V_{bi}$	Slope-over (and possibly $q/2kT$ again)	Series resistance, high-level injection (6.2.4)

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6.2.2 Reverse-Bias Breakdown

Though called “breakdown,” the large reverse current is a completely reversible process: it does not damage the diode (provided the current is limited to avoid overheating). The absolute value of the reverse voltage at which current “goes off to infinity” is the breakdown voltage, $V_{BR}$. In practice $V_{BR}$ measurements quote the voltage at which the current reaches a preselected value such as $1\mu$A or $1$ mA.

Fig. 6.11 plots the expected $V_{BR}$ of planar (one-dimensional) $p^{+}$-$n$ and $n^{+}$-$p$ step junctions versus the doping on the lightly doped side, for Ge, Si, and GaAs. Observations:

• At a given doping, $V_{BR}$ increases with band gap ($E_G$: Ge < Si < GaAs).
• More importantly, the light-side doping can vary $V_{BR}$ from a few volts to over a thousand volts.
• Above the dashed line in Fig. 6.11 (avalanche dominated) the dependence is roughly

$$V_{BR}\propto \frac{1}{N_B}\text{(roughly; more accurately }{N}_B^{-3/4}\text{)}\text{\hspace{1em}(6.33)}$$

where $N_B$ is the doping on the lightly doped side. (The book writes the approximate dependence in terms of a power law; Problem 6.14 uses $V_{BR}\approx 60(N_B/10^{16}{)}^{-3/4}$ volts for Si.)

Theoretically breakdown comes from failure of the “no other processes” assumption. Two “other processes” can cause it: avalanching and the Zener process. Avalanching typically dominates; the Zener process is important only when both sides of the junction are heavily doped.

Avalanching

Start with a small reverse bias far below breakdown. The reverse current is minority carriers randomly entering the depletion region and being accelerated by the field. The acceleration is not continuous: it is interrupted by energy-losing collisions with the lattice (Fig. 6.12a). The mean free path between collisions is $\sim 10^{-6}$ cm and a median depletion width is $\sim 10^{-4}$ cm, so a carrier suffers tens to thousands of collisions while crossing. The energy lost per collision is small at small bias, simply heating the lattice a little (lattice vibrations, readily dissipated).

With increasing reverse bias the energy transferred per collision grows. Near $V_{BR}$ it becomes enough to ionize a semiconductor atom: the collision frees a valence electron, creating an electron-hole pair. This is impact ionization (Fig. 3.15f). The new carriers are accelerated by the same field and make further collisions, creating still more carriers (Fig. 6.12b): a snowballing multiplication much like a mountainside snow avalanche. At $V_{BR}$ the carrier creation and reverse current effectively go to infinity.

Analogy: a chain letter / rolling snowball. Each ionizing collision turns one carrier into two, each of which can then ionize again. If on average each carrier causes more than one new ionization while crossing, the population explodes.

Two clarifying comments:

1. Fig. 6.12b cannot realistically be drawn to scale. If $V_{BR}=100$ V, the separation of the p-side and n-side Fermi levels would be roughly 100 times the $E_c-E_v$ gap distance.
2. Avalanche does not switch on sharply at $|V_A|=V_{BR}$. Carefully examined (Fig. 6.10b) the approach to breakdown is sloping: there is considerable carrier multiplication a few volts before $V_{BR}$ and even some far below. The reason is that the distance between collisions is a random variable distributed about its mean, so carriers occasionally gain ionizing energy at voltages far below $V_{BR}$; a few volts below breakdown the number of such carriers becomes quite large.

The current increase due to carrier multiplication is modeled by a multiplication factor $M$. If $I_0^{\prime }$ is the current without multiplication,

$$M=\frac{I}{I_0^{\prime }}\text{\hspace{1em}(6.34)}$$

An empirical fit to data:

$$M=\frac{1}{1-{\left(|V_A|/V_{BR}\right)}^m}\text{\hspace{1em}(6.35)}$$

with $m$ between 3 and 6 depending on the semiconductor. The multiplication factor can be used to correct the ideal diode equation to include avalanching: multiply the reverse current by $M$.

Why does $V_{BR}$ depend on doping this way? Breakdown occurs when carriers gain an ionizing energy over a lattice-scattering mean free path. That requirement is independent of junction doping, but a specific energy gain over a given distance corresponds to a specific electric field. So breakdown occurs when the maximum depletion field reaches a critical value ${\mathcal{E}}_{CR}$, essentially independent of doping.

For a step junction, from Eqs. (5.35) and (5.37), the field at $x=0$ (where it is largest) is

$$|\mathcal{E}(0)|=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}={\left[\frac{2q}{K_S{\epsilon }_0}\frac{N_A{N}_D}{N_A+N_D}\left(V_{bi}-V_A\right)\right]}^{1/2}\text{\hspace{1em}(6.36)}$$

Square it, and set $|\mathcal{E}(0)|\to {\mathcal{E}}_{CR}$ when $V_{bi}-V_A\to V_{bi}+V_{BR}\approx V_{BR}$ (since $V_{BR}\gg V_{bi}$):

$${\mathcal{E}}_{CR}^2=\frac{2q}{K_S{\epsilon }_0}\frac{N_A{N}_D}{N_A+N_D}{V}_{BR}\text{\hspace{1em}(6.37)}$$

Since ${\mathcal{E}}_{CR}$ is independent of doping, the right-hand side must be too, which requires

$$V_{BR}\propto \frac{N_A+N_D}{N_A{N}_D}\text{\hspace{1em}(6.38)}$$

or, for an asymmetric junction with light-side doping $N_B$,

$$V_{BR}\propto \frac{1}{N_B}\text{\hspace{1em}(6.39)}$$

This is not exactly what is observed (exponent nearer $-3/4$), but is acceptably close for such a simple argument.

Other dependences.

• Band gap. Required ionization energy increases with $E_G$ while the mean free path varies only slightly across Ge, Si, GaAs. So $V_{BR}$ increases Ge to Si to GaAs, as in Fig. 6.11.
• Temperature. Avalanche $V_{BR}$ increases with $T$: lattice scattering grows with temperature (Subsection 3.1.3), shortening the mean free path, so a larger critical field (and voltage) is needed.
• Geometry. The discussion is for planar step junctions. Linearly graded diodes have their own curves. Diffused or implanted junctions have curved lateral edges where the field is greater for a given voltage, so breakdown occurs there first, lowering $V_{BR}$. This “curvature effect” gets worse as junction depth shrinks; special edge modifications (guard rings and the like) minimize it.

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Exercise 6.5 (computer): the multiplication factor

Problem. Compute and plot $M$ versus $|V_A|/V_{BR}$ for $m=3$ and $m=6$. Orient axes to get a reverse-current-like plot ($M$ increasing downward, $|V_A|/V_{BR}$ increasing to the left). Discuss.

Logic. Let $x=|V_A|/V_{BR}$ run from 0 to 0.99 (never exactly 1, where $M$ is infinite). Compute $M=1/(1-x^m)$. To get the reverse-current-looking orientation in MATLAB, the author plotted $-M$ against $-x$.

% Exercise 6.5: Multiplication factor
% Initialization
close
clear
% M calculation
x = linspace(0, .99);    % x = |VA|/VBR
M3 = 1./(1 - x.^3);      % M when m = 3
M6 = 1./(1 - x.^6);      % M when m = 6
% Plotting result
plot(-x, -M3, -x, -M6); grid
axis([-1 0 -10 0])
xlabel('-|VA|/VBR')
ylabel('-Multiplication factor')
text(-0.8, -2.5, 'm=3')
text(-0.95, -1.5, 'm=6')

Discussion of Fig. E6.5. The plot closely resembles the left-hand side of Fig. 6.10(b). A perhaps surprising result: the current doubles ($M=2$) well before $V_{BR}$:

$$M=2\Rightarrow x^m=\frac{1}{2}\Rightarrow x=0.5^{1/m}=\{\begin{array}{ll}0.794& m=3\\ 0.891& m=6\end{array}$$

so at about 80% (for $m=3$) and 90% (for $m=6$) of the way to breakdown, the current has already doubled. There is already a $\sim 10\%$ enhancement ($M=1.1$) at

$$x={\left(1-\frac{1}{1.1}\right)}^{1/m}=(0.0909{)}^{1/m}=\{\begin{array}{ll}0.45& m=3\\ 0.67& m=6\end{array}$$

i.e., $0.45V_{BR}$ for $m=3$ and about $0.65V_{BR}$ for $m=6$. The larger $m$, the sharper the approach to breakdown.

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Zener Process

The Zener process is tunneling in a reverse-biased diode. Tunneling is the first genuinely quantum-mechanical phenomenon we meet: it has no classical analog.

Thought experiment: a ball and a hill. Classically a ball with less energy than the hill’s top can never get to the other side. A quantum particle has a nonzero probability to appear on the other side, “tunneling through” the barrier, without gaining energy (its energy is constant during tunneling) and without damaging itself or the barrier.

Two requirements for significant tunneling (Fig. 6.13):

1. There must be filled states on one side of the barrier and empty states on the other side at the same energy. Tunneling cannot take place into a region void of allowed states.
2. The barrier width $d$ must be very thin: tunneling is significant only if $d\lesssim 100\text{A˚}=10^{-6}$ cm.

In the reverse-biased diode (Fig. 6.14), the tunneling particles are valence band electrons on the p-side, going through the barrier to empty conduction band states on the n-side at the same energy. The larger the reverse bias, the more filled valence states on the p-side face empty conduction states on the n-side, so the tunneling current increases with reverse bias.

The barrier thickness is roughly the depletion width, which must be $\lesssim 10^{-6}$ cm. From Fig. E5.3, Si depletion widths $<10^{-6}$ cm need doping above about $10^{18}{\text{cm}}^{-3}$ on the “lightly” doped side as well. So the Zener process matters only for diodes heavily doped on both sides, which have correspondingly small breakdown voltages. The Zener process contributes significantly when $V_{BR}<6E_G/q$ ($\approx 6.7$ V in Si at 300 K) and dominates when $V_{BR}<4E_G/q$ ($\approx 4.5$ V in Si at 300 K).

How to tell avalanche from Zener experimentally:

	Avalanche	Zener
$V_{BR}$ vs. $T$	increases with $T$	decreases with $T$
Shape of breakdown	fairly sharp	very “soft”, slow approach even on a coarse scale

Historically the Zener process was the first proposed to explain reverse breakdown, so all diodes specifically fabricated to use breakdown became known as Zener diodes. The name persists even for those with $V_{BR}>6E_G/q$, which are functionally avalanche diodes.

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6.2.3 The R-G Current

In comparing theory and experiment, a current far exceeding the ideal prediction was found at small forward biases and at all reverse biases in room-temperature Si diodes. It arises from thermal carrier recombination-generation in the depletion region, assumed negligible in the ideal derivation. (Note: “depletion region” does not mean the carrier concentrations there are zero, or automatically less than equilibrium. They are merely small compared to the background doping.)

Reverse bias (Fig. 6.15a). Until now we thought of the reverse current as minority carriers wandering in from the two sides. But when reverse biased, the carrier concentrations in the depletion region fall below their equilibrium values, so $np<n_i^2$, and the system tries to restore equilibrium by thermal generation of electron-hole pairs throughout the depletion region. The strong field sweeps the generated carriers out into the quasineutral regions, adding to the reverse current.

Forward bias (Fig. 6.15b). Forward bias raises the carrier concentrations in the depletion region above equilibrium ($np>n_i^2$), giving recombination there. Viewed on the band diagram, the added forward current is the carriers that cannot make it over the potential hill being partially eliminated by recombination at R-G centers in the depletion region. (They flowed in, but never got out the other side; the contacts supply replacements, so the external current includes them.)

Quantitative expression

For every electron-hole pair created or destroyed per second in the depletion region, one electron per second flows out of or into the diode contacts. So sum the electrons (or holes) created/destroyed per second throughout the depletion region and multiply by $q$ to get the magnitude of the added current. With the right sign convention (generation gives negative current, recombination positive):

$$I_{\text{R-G}}=-qA{\int }_{-x_p}^{x_n}{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}dx\text{\hspace{1em}(6.40)}$$

In the depletion region the familiar low-level shortcut $\partial n/\partial t=-\Delta n/{\tau }_n$ does not apply. We must use the general-case result, Eq. (3.35):

$${\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=-\frac{np-n_i^2}{{\tau }_p(n+n_1)+{\tau }_n(p+p_1)}\text{\hspace{1em}(6.41)}$$

so that

$$I_{\text{R-G}}=qA{\int }_{-x_p}^{x_n}\frac{np-n_i^2}{{\tau }_p(n+n_1)+{\tau }_n(p+p_1)}dx\text{\hspace{1em}(6.42)}$$

Reverse bias (beyond a few $kT/q$). The carrier concentrations are quite small through most of the depletion region ($n\to 0$, $p\to 0$), so $np\to 0$, and the denominator reduces to ${\tau }_p{n}_1+{\tau }_n{p}_1$. With $n_1=n_i{e}^{(E_T-E_i)/kT}$ and $p_1=n_i{e}^{(E_i-E_T)/kT}$, the integrand is constant and the integral is trivial:

$$\begin{array}{rl}{I}_{\text{R-G}}& \approx -qA{\int }_0^W\frac{n_i^2}{{\tau }_p{n}_1+{\tau }_n{p}_1}dx=-\frac{qA{n}_i^{2}W}{{\tau }_p{n}_1+{\tau }_n{p}_1}\\ & =-\frac{qA{n}_{i}W}{2{\tau }_0}\text{reverse biases }>\text{ a few }kT/q\end{array}\text{\hspace{1em}(6.43)}$$

where

$${\tau }_0=\frac{{\tau }_p{n}_1+{\tau }_n{p}_1}{2n_i}=\frac{1}{2}\left[{\tau }_p{e}^{(E_T-E_i)/kT}+{\tau }_n{e}^{-(E_T-E_i)/kT}\right]\text{\hspace{1em}(6.44)}$$

(For ${\tau }_n={\tau }_p={\tau }_0$ and $E_T\approx E_i$, this reduces to ${\tau }_0$ equal to the common lifetime; ${\tau }_0$ is a generation lifetime.) Notice three things: $I_{\text{R-G}}\propto n_i$ (not $n_i^2$), $I_{\text{R-G}}\propto W$, and since $W\propto (V_{bi}-V_A{)}^{1/2}$, the reverse R-G current increases (as a square root) with reverse bias; it never saturates.

Forward bias. The concentrations can’t be neglected and even an approximate evaluation of (6.42) is involved. We note only that $I_{\text{R-G}}$ is expected to vary as $\exp (q{V}_A/\eta kT)$ with $1<\eta \le 2$ at forward biases beyond a few $kT/q$. Typically $\eta$ is close to 2 (the maximum recombination rate, where $n\approx p$, occurs where the quasi-Fermi-level split is half the total: $np=n_i^2{e}^{q{V}_A/kT}$ with $n\approx p$ gives $n=n_i{e}^{q{V}_A/2kT}$). The combined forward and reverse dependence is approximately described by

$$I_{\text{R-G}}\approx \frac{qA{n}_{i}W}{2{\tau }_0}\frac{e^{q{V}_A/kT}-1}{1+\frac{V_{bi}-V_A}{2kT/q}{e}^{q{V}_A/2kT}}\text{\hspace{1em}(6.45)}$$

(Sanity checks. Reverse: the exponential terms drop, giving $-qA{n}_{i}W/2{\tau }_0$, matching 6.43. Forward, $V_A\gg kT/q$: $I_{\text{R-G}}\to \frac{qA{n}_{i}W}{{\tau }_0}\frac{kT/q}{V_{bi}-V_A}{e}^{q{V}_A/2kT}$, the "$q/2kT$" behavior observed. This is the form used in Exercise 6.6 and its program.)

Total current and which one wins

With a second component, the ideal-diode-equation current is now called the diffusion current, $I_{\text{DIFF}}$:

$$I_{\text{DIFF}}=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.46)}$$ $$I=I_{\text{DIFF}}+I_{\text{R-G}}\text{\hspace{1em}(6.47)}$$

Now the experimental observations are explained:

• In Si at room temperature, $qA{n}_{i}W/2{\tau }_0\gg I_0$, so the R-G current dominates at reverse biases and small forward biases. In reverse bias it never saturates (since $W$ grows). The forward R-G current varies as $\exp (q{V}_A/2kT)$ for $V_A>$ a few $kT/q$, consistent with the observed slope of $q/2kT$.
• With increasing forward bias, $I_{\text{DIFF}}$, which rises faster ($e^{q{V}_A/kT}$ vs. $e^{q{V}_A/2kT}$), eventually overtakes $I_{\text{R-G}}$, giving the $q/kT$ region in the semilog plot.

Why Ge and hot Si are closer to ideal. Compare the sizes of the two prefactors:

• $I_0\propto n_i^2$, while $I_{\text{R-G}}\propto n_i$. The larger $n_i$ of Ge makes $I_0>qA{n}_{i}W/2{\tau }_0$ in Ge diodes at room temperature, even though $qA{n}_{i}W/2{\tau }_0>I_0$ in Si and GaAs.
• Since $I_0\propto n_i^2$ but reverse $I_{\text{R-G}}\propto n_i$, the reverse-bias diffusion component rises faster with temperature. At a sufficiently elevated temperature the diffusion component dominates (see Problem 6.16 for “how elevated”).

Thus Ge diodes at room temperature, and Si diodes at elevated temperature, show saturating reverse characteristics closer to the ideal.

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Exercise 6.6 (computer): $I_{01}$ and $I_{02}$ for Si and Ge

Problem. If the factor $e^{q{V}_A}$ … more precisely, if the $e^{q{V}_A/2kT}$ term in the denominator of the R-G expression is large (forward bias), the forward-bias current for a $p^{+}$-$n$ diode beyond a few $kT/q$ may be approximately modeled by

$$I=I_{01}{e}^{q{V}_A/kT}+I_{02}{e}^{q{V}_A/2kT}$$

(the first term is $I_{\text{DIFF}}$, the second $I_{\text{R-G}}$), with $I_{01}$ and $I_{02}$ voltage-independent constants.

(a) Compute $I_{01}$ and $I_{02}$ for a Si $p^{+}$-$n$ step junction diode at 300 K with $A=10^{-4}{\text{cm}}^2$, $N_D=10^{16}{\text{cm}}^{-3}$ and ${\tau }_n={\tau }_p={\tau }_0=10^{-6}$ s. In the $I_{02}$ computation, assume the $e^{q{V}_A/2kT}$ term in the denominator of the R-G expression is much greater than unity, and set $V_A=V_{bi}/4$ in evaluating voltage-dependent terms. Compute $V_{bi}=(E_G/2q)+(kT/q)\ln (N_D/n_i)$. (b) Repeat for a Ge $p^{+}$-$n$ diode at 300 K with the same parameters except ${\mu }_p=1500{\text{cm}}^2/\text{V}\cdot \text{s}$, $n_i=2.5\times 10^{13}{\text{cm}}^{-3}$, $K_S=16$. (c) Using the simplified relationship, construct a semilog plot of $I$ vs $V_A$ simultaneously displaying the forward-bias characteristics of the Si and Ge diodes. Limit axes to $0\le V_A\le 1$ V and $10^{-12}\le I\le 10^{-3}$ A. Begin the calculation at $V_A=0.1$ V. Discuss.

Formulas used.

$$I_{01}=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D},I_{02}=\frac{qA{n}_{i}W}{{\tau }_0}\frac{kT/q}{V_{bi}-V_A}{|}_{V_A=V_{bi}/4},W=\sqrt{\frac{2K_S{\epsilon }_0}{q{N}_D}\left(V_{bi}-V_A\right)}$$

(taking $W$ at the same $V_A=V_{bi}/4$; with $V_{bi}-V_A=3V_{bi}/4$.)

Step-by-step for Si (all voltages in volts, $kT/q=0.02585$ V):

$$\begin{array}{rlrl}{\mu }_p(10^{16})& \approx 437{\text{cm}}^2/\text{V s}(\text{from the fit}),& D_P& =(0.02585)(437)=11.3{\text{cm}}^2/\text{s}\\ L_P& =\sqrt{D_P{\tau }_p}=\sqrt{(11.3)(10^{-6})}=3.36\times 10^{-3}\text{cm}& \frac{D_P}{L_P}& =3.36\times 10^3\text{cm/s}\\ I_{01}& =(1.6\times 10^{-19})(10^{-4})(3.36\times 10^3)\frac{10^{20}}{10^{16}}& & =5.38\times 10^{-16}\text{A}\end{array}$$ $$\begin{array}{rl}{V}_{bi}& =0.56+0.02585\ln \frac{10^{16}}{10^{10}}=0.56+0.357=0.92\text{V}\\ V_A& =V_{bi}/4=0.23\text{V},V_{bi}-V_A=0.69\text{V}\\ W& =\sqrt{\frac{2(11.8)(8.85\times 10^{-14})}{(1.6\times 10^{-19})(10^{16})}(0.69)}\approx 3.0\times 10^{-5}\text{cm}\\ I_{02}& =\frac{(1.6\times 10^{-19})(10^{-4})(10^{10})(3.0\times 10^{-5})}{10^{-6}}\cdot \frac{0.02585}{0.69}\approx 2\times 10^{-13}\text{A}\end{array}$$

in agreement with the book’s value (table below).

Results table (book’s values).

	Silicon	Germanium
$I_{01}$	$5.38\times 10^{-16}$ A	$6.23\times 10^{-9}$ A
$I_{02}$	$2.08\times 10^{-13}$ A	$8.34\times 10^{-10}$ A

As expected, $I_{02}>I_{01}$ for Si, while $I_{01}>I_{02}$ for Ge.

MATLAB program (cleaned):

% Comparison of forward bias I-VA for Si and Ge diodes at 300 K.
% This program uses a simplified formula for the current:
%   I = I01*exp(qVA/(kT)) + I02*exp(qVA/(2kT)).
% Also, Vbi = EG/2q + (kT/q)*ln(ND/ni).
 
% Initialization
close
clear
% Constants
T = 300;               % Temperature in Kelvin
k = 8.617e-5;          % Boltzmann constant eV/K
e0 = 8.85e-14;         % permittivity of free space (F/cm)
q = 1.602e-19;         % charge on an electron (C)
KS = [11.8 16];        % Dielectric constant [Si Ge]
ni = [1.0e10 2.5e13];  % intrinsic carrier conc. at 300 K [Si Ge]
mup = [437 1500];      % hole mobility [Si Ge]
EG = [1.12 0.66];      % band gap [Si Ge]
% Given Constants
A = 1.0e-4;            % cm^2
ND = 1.0e16;           % cm^-3
taun = 1.0e-6;         % seconds
taup = 1.0e-6;         % seconds
 
% I01
DP = k*T.*mup;
LP = sqrt(DP.*taup);
I01 = q*A.*(DP./LP.*ni.^2./ND);
 
% I02
Vbi = EG/2 + k*T.*log(ND./ni);
W = sqrt(2.*KS*e0/(q*ND).*Vbi);          % evaluated using Vbi - 0.25Vbi
I02 = q*A.*ni/sqrt(taun*taup).*W.*(k*T)./(3.*Vbi./4);
% (W factor uses 3*Vbi/4 per the stated assumption; see text)
 
% Currents for both Silicon (ISi) and Germanium (IGe)
VA = linspace(0.1, 1);
ISi = I01(1).*exp(VA./(k*T)) + I02(1).*exp(VA./(2*k*T));
IGe = I01(2).*exp(VA./(k*T)) + I02(2).*exp(VA./(2*k*T));
 
% Plot
semilogy(VA, ISi, VA, IGe, '--'); grid
axis([0 1 1.0e-12 1.0e-3]);
xlabel('VA (volts)');
ylabel('I (A)');
text(.7, 1.4e-9, 'T=300K');
text(.7, 4.0e-10, 'ND=1.0e16/cm^3');
text(.25, 1.4e-5, 'Ge');
text(.48, 1.4e-8, 'Si');

(In the book’s script $W$ is computed from $V_{bi}$ via the W = sqrt(2.*KS*e0/(q*ND).*Vbi) line; to follow the problem statement exactly, one would use $3V_{bi}/4$ inside the square root. The difference changes $I_{02}$ by only $\sqrt{4/3}$.)

Discussion (Fig. E6.6). Because of the small band gap and large $n_i$ of Ge, large forward currents flow in the Ge diode at relatively small voltages, making it virtually impossible to observe the R-G component in Ge: $I_{\text{DIFF}}$ takes over almost immediately. In the Si diode, both the R-G ($q/2kT$) region at small bias and the diffusion ($q/kT$) region at larger bias are clearly visible on the semilog plot, with a crossover around 0.4-0.5 V. The Ge curve sits several orders of magnitude to the left (lower voltage for the same current): for a given current, the Ge diode needs a much smaller forward voltage, which is why Ge diodes were once prized as low-drop rectifiers.

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Exercise 6.7: generalized forward-bias model

A generalization of the forward-bias model of Exercise 6.6 is common in analyzing experimental data:

$$I=I_{01}{e}^{q{V}_A/n_{1}kT}+I_{02}{e}^{q{V}_A/n_{2}kT}$$

(first term $I_{\text{DIFF}}$, second $I_{\text{R-G}}$) with $n_1$, $n_2$ arbitrary constants determined by matching the near-ideal and R-G regions.

Problem. Determine $I_{01}$, $I_{02}$, $n_1$, $n_2$ giving an optimal match to the forward data of Fig. 6.10(a).

Solution. Draw straight lines through the two linear regions of the semilog plot (Fig. E6.7). Each line is, for $j=1,2$,

$$\ln I=\ln I_{0j}+\frac{q}{n_{j}kT}{V}_A$$

• The $V_A=0$ intercept of line $j$ gives $I_{0j}$.
• The slope gives $n_j$: for two points $(V_{A1},I_1)$ and $(V_{A2},I_2)$ on line $j$,

$$n_j=\frac{V_{A2}-V_{A1}}{(kT/q)\ln (I_2/I_1)}$$

From the graph:

$$I_{01}\approx 10^{-14}\text{A},n_1\approx 1.1;I_{02}\approx 10^{-12}\text{A},n_2\approx 1.5$$

The $I_{\text{DIFF}}$ component is almost uncharacteristically ideal. The observed $n_2$ is on the low side but within the normally observed range ($1<n_2\le 2$), and $I_{02}>I_{01}$, as expected.

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6.2.4 $V_A\to V_{bi}$ High-Current Phenomena

A significant current starts to flow when the forward bias approaches $V_{bi}$. A large current creates conditions inconsistent with several ideal-diode assumptions. First, “all the applied voltage drops across the depletion region” becomes doubtful: with $I=0.1$ A, even 1 ohm of resistance outside the depletion region drops a significant 0.1 V. A large current may also cause high-level injection. We take these in turn.

Series resistance

The quasineutral regions have an inherent resistance set by doping and dimensions, plus a residual contact resistance. Together they form a resistance $R_S$ in series with the current flowing across the junction (Fig. 6.16a). At low current $I{R}_S\ll V_A$, so the voltage across the depletion region, the “junction voltage” $V_J$, equals $V_A$ (as assumed so far). But when $I{R}_S$ becomes comparable with $V_A$, the voltage across the depletion region falls:

$$V_J=V_A-I{R}_S\text{\hspace{1em}(6.48)}$$

Part of the applied voltage is wasted: a larger $V_A$ is needed for the same current, and the characteristic slopes over (Fig. 6.10a). The fix is to replace $V_A$ by $V_J=V_A-I{R}_S$ in earlier relations. As diffusion current dominates at these currents,

$$I=I_0{e}^{q{V}_J/kT}=I_0{e}^{q(V_A-I{R}_S)/kT}{V}_A\to V_{bi}\text{\hspace{1em}(6.49)}$$

This is transcendental in $I$: you cannot solve for $I$ as a function of $V_A$ in closed form. But $I$ vs. $V_A$ is easy to construct parametrically: choose $V_J$, compute $I=I_0{e}^{q{V}_J/kT}$, then compute $V_A=V_J+I{R}_S$ (Eq. 6.48). (Requires knowing $R_S$.)

Measuring $R_S$ (Fig. 6.16b, c). On a forward-bias semilog plot like Fig. 6.10a, extend the ideal ($q/kT$) portion into the slope-over region and measure the voltage displacement $\Delta V$ between the two curves at each $I$. Since $\Delta V=V_A-V_J=I{R}_S$, a plot of $\Delta V$ versus $I$ is a straight line whose slope is $R_S$.

Exercise 6.8

Problem. Crudely estimate $R_S$ of the diode of Fig. 6.10(a).

Solution. A decent $\Delta V$ vs. $I$ plot would need a blow-up of the slope-over region; instead use $\Delta V$ at a given $I$. The curve terminates at $I\approx 80$ mA with $V_A=0.85$ V. In Fig. E6.7, extending the $q/kT$ line passes through 80 mA at about 0.78 V. Thus

$$\Delta V\approx 0.85-0.78=0.07\text{V},R_S=\frac{\Delta V}{I}\approx \frac{0.07}{0.08}\approx 0.9\approx 1\Omega$$

High-level injection

The low-level injection assumption fails when the minority carrier concentration at the depletion edge on the lightly doped side approaches the doping concentration. In Si at room temperature this typically happens a few tenths of a volt below $V_{bi}$. Beyond that, the injection is high level: both minority and majority carrier concentrations adjacent to the depletion region are perturbed (Fig. 6.17a). The majority concentration must rise to maintain approximate charge neutrality ($\Delta n\approx \Delta p$).

An analysis of high-level injection predicts a current varying roughly as $\exp (q{V}_A/2kT)$: a high-current $q/2kT$ region on the semilog plot (Fig. 6.17b). Intuitively, with $n\approx p$ at the edge, the law of the junction $np=n_i^2{e}^{q{V}_A/kT}$ gives $p\propto e^{q{V}_A/2kT}$, so the injected carrier density, hence the current, goes as $e^{q{V}_A/2kT}$.

This $q/2kT$ high-current region is seldom observed because it is obscured by the series resistance slope-over. Still, the enhanced carrier concentration of high-level injection can reduce the observed series resistance: the drop in resistivity from heavy carrier injection is called conductivity modulation. (Think of a narrow road suddenly receiving an extra lane of cars: the “resistance” to flow drops because there are more carriers to carry current.)

Exercise 6.9: diagnose the nonidealities

Problem. The measured $I$-$V$ of a room-temperature Si diode is crudely sketched in Fig. E6.9 (log scale for forward, linear scale for reverse). Nonidealities are labeled A to E. Possible sources are listed (1 photogeneration; 2 thermal recombination in the depletion region; 3 avalanching and/or Zener process; 4 low-level injection; 5 depletion approximation; 6 thermal generation in the depletion region; 7 band bending; 8 series resistance; 9 $V_A>V_{bi}$; 10 high-level injection). Identify the cause of each nonideal feature.

Solution. (The OCR of the list is partly scrambled; the book’s answer is A-3, B-6, C-2, D-10, E-8.)

Label	Feature on sketch	Cause
A	Abrupt current increase at large reverse bias	3. Avalanching and/or Zener process (breakdown)
B	Reverse current that creeps up, does not saturate	6. Thermal generation in the depletion region
C	Excess forward current at small $V_A$, slope $q/2kT$	2. Thermal recombination in the depletion region
D	High-current region of slope $\approx q/2kT$ above the ideal region	10. High-level injection
E	Slope-over at the highest currents	8. Series resistance

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6.3 Special Considerations

The two topics here are supplementary to the basic d.c. response. They could be delayed until needed, but that would hide their importance and weaken the linkage of ideas. The charge control approach is a “big picture” method that often gives a reasonably accurate answer with minimal mathematics. The narrow-base diode is both a deviation from the ideal and a conceptual bridge to the bipolar junction transistor.

6.3.1 Charge Control Approach

In the charge control approach the basic carrier variable is the total charge of the minority carrier excess (or deficit) within an entire quasineutral region. Consider a forward-biased $p^{+}$-$n$ diode. Let $\Delta p_n(x,t)$ be the excess hole concentration on the n-side at time $t$ and position $x_n\le x<\infty$. The total excess hole charge is

$$Q_P=qA{\int }_{x_n}^{\infty }\Delta p_n(x,t)dx\text{\hspace{1em}(6.50)}$$

(For reverse bias $Q_P<0$ is a deficit.) $Q_P$ is independent of $x$ but may depend on time.

Goal: a relation governing the combined behavior of this charge. Start from the minority carrier diffusion equation (3.54b) with $G_L=0$:

$$\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}\text{\hspace{1em}(6.51)}$$

In a $\mathcal{E}\approx 0$ region the hole current density is

$$J_P=-q{D}_P\frac{\partial \Delta p_n}{\partial x}\text{\hspace{1em}(6.52)}$$

so $q{D}_P{\partial }^2\Delta p_n/\partial x^2=-\partial J_P/\partial x$, and multiplying (6.51) by $q$:

$$\frac{\partial (q\Delta p_n)}{\partial t}=-\frac{\partial J_P}{\partial x}-\frac{q\Delta p_n}{{\tau }_p}\text{\hspace{1em}(6.53)}$$

Integrate every term over the volume ($A\int dx$) of the n-side quasineutral region, $x_n$ to $\infty$:

$$\frac{d}{dt}\left[qA{\int }_{x_n}^{\infty }\Delta p_{n}dx\right]=-A{\int }_{x_n}^{\infty }\frac{\partial J_P}{\partial x}dx-\frac{1}{{\tau }_p}\left[qA{\int }_{x_n}^{\infty }\Delta p_{n}dx\right]\text{\hspace{1em}(6.54)}$$

The bracketed quantities are $Q_P$. Evaluate the current integral:

$$-A{\int }_{J_P(x_n)}^{J_P(\infty )}d{J}_P=-A{J}_P(\infty )+A{J}_P(x_n)=A{J}_P(x_n)\equiv A{J}_{\text{DIFF}}=i_{\text{DIFF}}\text{\hspace{1em}(6.55)}$$

because the hole current goes to zero far into the n-side, $J_P(\infty )=0$. (Lowercase $i$ because the current may in general vary with time.) Therefore

$$\boxed{\frac{d{Q}_P}{dt}=i_{\text{DIFF}}-\frac{Q_P}{{\tau }_p}}\text{\hspace{1em}(6.56)}$$

Interpretation. This is effectively a continuity equation for the excess hole charge. There are only two ways to change the excess charge in a region: holes can flow in or out ($i_{\text{DIFF}}$), or the excess charge can be changed by recombination-generation within the region ($-Q_P/{\tau }_p$).

Analogy: a bathtub. $Q_P$ is the water level; $i_{\text{DIFF}}$ is the tap; $Q_P/{\tau }_p$ is the leak in the bottom. In steady state the tap flow exactly equals the leak.

Steady-state use. For a $p^{+}$-$n$ diode in steady state, $d{Q}_P/dt=0$, $i_{\text{DIFF}}=I_{\text{DIFF}}$:

$$I_{\text{DIFF}}=\frac{Q_P}{{\tau }_p}\text{\hspace{1em}(6.57)}$$

Now suppose we could not solve for $\Delta p_n(x)$. We know $\Delta p_n(0)$ from (6.18), and suspect the solution is exponential with decay length $L_P$. Use the squared-off approximation of Exercise 6.4: equate $Q_P$ to the charge of a rectangle of width $L_P$ and height $\Delta p_n(x^{\prime }=0)$:

$$Q_P\approx q(A{L}_P)\Delta p_n(x^{\prime }=0)=qA{L}_P\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.58)}$$ $$I_{\text{DIFF}}=\frac{Q_P}{{\tau }_p}=qA\frac{L_P}{{\tau }_p}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.59)}$$

Since $L_P/{\tau }_p=D_P/L_P$, (6.59) is the usual $p^{+}$-$n$ result. (Here the approximation happens to be exact, because the exact distribution is a pure exponential, whose area is exactly peak times $L_P$.) The real power of charge control appears in transient problems, where solving the PDE is hard but (6.56) is easy.

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6.3.2 Narrow-Base Diode

Current derivation

The ideal diode assumed contacts several diffusion lengths from the depletion edges ($\Delta n_p(-\infty )=0$, $\Delta p_n(\infty )=0$). For a diode whose lightly doped side is a substrate/wafer thickness, that is typically valid. It is questionable if the diode is made in a thin epitaxial layer. A diode whose quasineutral width on the lightly doped side is comparable to or less than a diffusion length is a narrow-base diode.

Take a $p^{+}$-$n$ step junction identical to the ideal diode except that the n-side quasineutral width may be $\lesssim L_P$. Define (Fig. 6.18): $x_c$ as the distance from the metallurgical junction to the n-side contact, and $x_c^{\prime }=x_c-x_n$ is the distance from the n-edge of the depletion region to the contact. The contact is ohmic, so $\Delta p_n=0$ at $x^{\prime }=x_c^{\prime }$ (an ohmic contact has a very high R-G rate, pinning the minority concentration to equilibrium).

Paralleling the ideal derivation, solve

$$0=D_P\frac{d^2\Delta p_n}{d{x}^{\prime 2}}-\frac{\Delta p_n}{{\tau }_p}0\le x^{\prime }\le x_c^{\prime }\text{\hspace{1em}(6.60)}$$

with

$$\Delta p_n(x^{\prime }=0)=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.61a)}$$ $$\Delta p_n(x^{\prime }=x_c^{\prime })=0\text{\hspace{1em}(6.61b)}$$

General solution:

$$\Delta p_n(x^{\prime })=A_1{e}^{-x^{\prime }/L_P}+A_2{e}^{x^{\prime }/L_P}0\le x^{\prime }\le x_c^{\prime }\text{\hspace{1em}(6.62)}$$

This time $A_2$ need not vanish. Apply the boundary conditions:

$$\Delta p_n(0)=A_1+A_2\text{\hspace{1em}(6.63a)}$$ $$0=A_1{e}^{-x_c^{\prime }/L_P}+A_2{e}^{x_c^{\prime }/L_P}\text{\hspace{1em}(6.63b)}$$

From (6.63b): $A_1=-A_2{e}^{2x_c^{\prime }/L_P}$. Substitute in (6.63a): $\Delta p_n(0)=A_2\left(1-e^{2x_c^{\prime }/L_P}\right)$, so

$$A_2=\frac{\Delta p_n(0)}{1-e^{2x_c^{\prime }/L_P}},A_1=-\frac{\Delta p_n(0)e^{2x_c^{\prime }/L_P}}{1-e^{2x_c^{\prime }/L_P}}$$

Putting back into (6.62):

$$\Delta p_n(x^{\prime })=\Delta p_n(0)\frac{e^{(x_c^{\prime }-x^{\prime })/L_P}-e^{-(x_c^{\prime }-x^{\prime })/L_P}}{e^{x_c^{\prime }/L_P}-e^{-x_c^{\prime }/L_P}}0\le x^{\prime }\le x_c^{\prime }\text{\hspace{1em}(6.64)}$$

(multiply numerator and denominator by $e^{-x_c^{\prime }/L_P}$ to confirm). In compact form, with $\sinh \xi \equiv (e^{\xi }-e^{-\xi })/2$:

$$\boxed{\Delta p_n(x^{\prime })=\Delta p_n(0)\frac{\sinh \left[(x_c^{\prime }-x^{\prime })/L_P\right]}{\sinh \left[x_c^{\prime }/L_P\right]}}\text{\hspace{1em}(6.65, 6.66)}$$

Current:

$$I_{\text{DIFF}}=A{J}_P(x^{\prime }=0)=-qA{D}_P{\frac{d\Delta p_n}{d{x}^{\prime }}|}_{x^{\prime }=0}\text{\hspace{1em}(6.67)}$$

Differentiate (6.65): $\frac{d}{d{x}^{\prime }}\sinh \left[\frac{x_c^{\prime }-x^{\prime }}{L_P}\right]=-\frac{1}{L_P}\cosh \left[\frac{x_c^{\prime }-x^{\prime }}{L_P}\right]$; at $x^{\prime }=0$ this is $-\cosh (x_c^{\prime }/L_P)/L_P$. Hence

$$\boxed{I_{\text{DIFF}}=I_0^{\prime }\left(e^{q{V}_A/kT}-1\right)}\text{\hspace{1em}(6.68)}$$ $$\boxed{I_0^{\prime }=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}\coth \left(\frac{x_c^{\prime }}{L_P}\right)=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}\frac{\cosh (x_c^{\prime }/L_P)}{\sinh (x_c^{\prime }/L_P)}}\text{\hspace{1em}(6.69)}$$

with $\cosh \xi \equiv (e^{\xi }+e^{-\xi })/2$ (6.70).

Limiting cases and punch-through

Useful limits of the hyperbolic functions:

$$\sinh \xi \to \xi (\xi \to 0)(6.71a),\sinh \xi \to \frac{1}{2}{e}^{\xi }(\xi \to \infty )(6.71b)$$ $$\cosh \xi \to 1+\frac{{\xi }^2}{2}\to 1(\xi \to 0)(6.72a),\cosh \xi \to \frac{1}{2}{e}^{\xi }(\xi \to \infty )(6.72b)$$

Wide-base limit ($x_c^{\prime }\to \infty$, $x_c^{\prime }/L_P\gg 1$). The sinh ratio in (6.65) becomes $e^{-x^{\prime }/L_P}$ (since $\sinh [(x_c^{\prime }-x^{\prime })/L_P]/\sinh [x_c^{\prime }/L_P]\to e^{-x^{\prime }/L_P}$), recovering Eq. (6.23). And $\cosh /\sinh \to 1$, so (6.69) reduces to the standard ideal $p^{+}$-$n$ result. The narrow-base analysis is thus a generalization of the ideal diode to a quasineutral region of arbitrary thickness. Mathematically reassuring but nothing new.

Narrow-base limit ($x_c^{\prime }\ll L_P$). Much more interesting.

• Using $\sinh \xi \approx \xi$ for both numerator and denominator:

$$\Delta p_n(x^{\prime })=\Delta p_n(0)\left(1-\frac{x^{\prime }}{x_c^{\prime }}\right)\text{\hspace{1em}(6.73)}$$

a straight line between the two end values set by the boundary conditions. This linearity comes from negligible thermal R-G in a region much shorter than a diffusion length. In fact it follows immediately by dropping the $-\Delta p_n/{\tau }_p$ term of the diffusion equation ($d^2\Delta p_n/d{x}^{\prime 2}=0$). That justifies neglecting the thermal R-G term whenever the quasineutral width is small compared with a diffusion length.

• The current: $\coth (x_c^{\prime }/L_P)\to L_P/x_c^{\prime }$, so

$$I_{\text{DIFF}}\to qA\frac{n_i^2}{N_D}\frac{D_P}{x_c^{\prime }}\left(e^{q{V}_A/kT}-1\right)$$

(no $L_P$: all the injected holes reach the contact, with none lost to recombination). The width $x_c^{\prime }=x_c-x_n$ shrinks with increasing reverse bias because the depletion width grows. Hence $I_{\text{DIFF}}(V_A<0)\propto 1/x_c^{\prime }$ does not saturate but increases systematically with reverse bias.

• If $x_c$ is small enough, $x_c^{\prime }\to 0$ is possible: the entire quasineutral region becomes depleted. This is called punch-through. From (6.68)/(6.69), $I_{\text{DIFF}}(V_A<0)\to -\infty$ as $x_c^{\prime }\to 0$. In reality the depletion approximation (sharp boundaries) breaks down in that limit; experimentally, and in more precise theories, the diffusion current stays finite at punch-through unless the field is high enough to cause avalanche breakdown.

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Exercise 6.10

Problem. A planar Si $p^{+}$-$n$ step junction at room temperature has $N_D=10^{16}{\text{cm}}^{-3}$ and an n-side thickness $x_c=2\mu \text{m}$. Using the depletion approximation, find the punch-through voltage: the voltage required to completely deplete the n-side.

Solution. At punch-through $x_n=x_c$. From Eq. (5.37), $x_n={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}$ (the $p^{+}$ side contributes negligibly). Then

$$\begin{array}{rl}{x}_c^2& =\frac{2K_S{\epsilon }_0}{q{N}_D}\left(V_{bi}-V_A\right)\\ V_A& =V_{bi}-\frac{q{N}_D{x}_c^2}{2K_S{\epsilon }_0}\\ & =0.92-\frac{(1.6\times 10^{-19})(10^{16})(2\times 10^{-4}{)}^2}{2(11.8)(8.85\times 10^{-14})}\\ & =0.92-\frac{6.4\times 10^{-11}}{2.09\times 10^{-12}}=0.92-30.6=-29.7\text{V}\end{array}$$

where $V_{bi}=0.92$ V was read from Fig. 5.11 (equivalently $V_{bi}=E_G/2q+(kT/q)\ln (N_D/n_i)=0.56+0.357$). So the punch-through voltage is 29.7 V reverse. From Fig. 6.11, avalanche breakdown in a comparable wide-base diode occurs at $V_{BR}\approx 55$ V. Thus punch-through takes place prior to breakdown in the given narrow-base diode.

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6.4 Summary and Concluding Comments

The chapter modeled the steady state response of the pn junction diode.

Qualitative picture. Using the energy band diagram:

• Reverse-bias current in an ideal diode: minority carriers wander into the depletion region and are accelerated to the other side.
• Forward bias lowers the hill between the two sides and enhances majority carrier injection into the opposite-side quasineutral region.
• Nothing builds up in steady state: R-G stabilizes the minority concentrations while very rapid rearrangement and interchange with the contacts maintains the majority concentrations.

The ideal diode theory. We formulated, derived and examined it. Although the predicted $I$-$V$ has deficiencies against experiment, it gives valuable insight and an analytical base for more exact treatments, and it is widely used for first-order predictions. The ideal diode equation is only a small part of the ideal diode theory, which also includes the solution strategy and approximations, the mathematical steps, and predictions about carrier concentrations, carrier currents, material dependences, doping dependences, etc.

Deviations from the ideal (Si, room temperature):

Deviation	Cause	Key information
Reverse breakdown	Avalanche or Zener process	avalanche $V_{BR}$ varies roughly inversely with light-side doping; carrier multiplication gives enhanced current far below $V_{BR}$
Excess current at small forward and all reverse biases	Thermal R-G in the depletion region	forward $\sim \exp (q{V}_A/2kT)$ ($\eta \to 2$); reverse $\propto W$; $I_{\text{R-G}}\propto n_i$ while $I_{\text{DIFF}}\propto n_i^2$
Slope-over at high forward current	Voltage drop across series resistance	$V_J=V_A-I{R}_S$
Possible $q/2kT$ at high current	High-level injection	usually obscured by $R_S$; conductivity modulation reduces observed $R_S$

(The development concentrated on the step junction; other profiles may need analytical modifications.)

Special considerations. Charge control (a different analytical approach) and the narrow-base diode (the ideal analysis generalized to arbitrary base width). For widths small compared to a diffusion length the perturbed carrier distribution becomes linear; for extremely small widths there is the possibility of punch-through at a reverse bias below $V_{BR}$.

Equations worth memorizing: the ideal diode solution (6.29/6.30), the law of the junction (6.12), the depletion-edge boundary conditions (6.15/6.18), and the reverse-bias R-G current (6.43). Retain the book for the rest.

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Problems (Chapter 6)

Problem information table (from the book): suggested point weightings and difficulty levels.

Problem	Complete after	Difficulty	Points	Short description
6.1	6.1.4	1	10 (1 each part)	Ideal diode quiz
6.2	6.2.4	1	10	Compare actual and ideal $I$-$V$
6.3	6.1.1	1	9 (3 each part)	Band diagram sketches
6.4	6.1.1	2	8	Forward-bias big picture
6.5	6.1.2	2-3	10	Derive Eq. (6.15) using $J_N=0$
6.6	6.1.3	2	10 (5 each part)	Typical $I$ values using diary
6.7	6.1.3	2	16 (a-10, b-5, c-1)	Ideal $I$-$V$ variation with $T$
6.8	6.1.3	3	16 (a-3, b-10, c-3)	Photodiode $I$-$V$ derivation
6.9	6.1.3	1	5	Photodiode $I$-$V$ plot
6.10	6.1.4	2	10 (a-c: 2 each, d-4)	Deduce info from concentration plot
6.11	6.1.4	3	12	Deduce, plot carrier concentrations
6.12	6.1.4	3	12 (a-9, b-3)	pn diode temperature sensor
6.13	6.2.2	2	6 (2 each part)	$V_{BR}$-related calculations
6.14	6.2.2	2	10 (a-8, b-2)	Breakdown $I$-$V$ ($M{I}_0$)
6.15	6.2.3	3	12	Breakdown $I$-$V$ ($M{I}_{\text{R-G}}$)
6.16	6.2.3	3	8	Determine elevated $T$ point
6.17	6.2.3	3	10	Modified $I_{\text{R-G}}$
6.18	6.2.4	3	9 (a-6, b-3)	$R_S$-related computations
6.19	6.2.4	3	20	Deduce $I_{0j}$, $n_j$ from $I$-$V$ (10 line fit, 8 least-squares fit, 2 discuss)
6.20	6.2.4	2-3	15	Deduce $R_S$ from $I$-$V$
6.21	6.2.4	4	40	Multiple nonideality $I$-$V$
6.22	6.3.2	3	10	Pseudo-narrow-base diode
6.23	6.3.2	3	15	Narrow/wide diode combination
6.24	6.3.2	4	34	Multiple considerations

A note on data quality. The OCR of several problem statements lost digits or whole phrases (for example the doping in 6.6, 6.13, 6.16 and the figure values in 6.10 and 6.24). Wherever that happened, I state my assumption explicitly, give the solution symbolically first, and then evaluate with the stated assumption, so you can re-plug the true numbers from your textbook.

Standard numbers used throughout (Si, 300 K): $kT/q=0.0259$ V, $n_i=10^{10}{\text{cm}}^{-3}$, $K_S=11.8$, ${\epsilon }_0=8.85\times 10^{-14}$ F/cm, $q=1.6\times 10^{-19}$ C, $E_G=1.12$ eV. Mobility fits (Exercise 3.1 form): ${\mu }_p=54.3+406.9/[1+(N/2.35\times 10^{17}{)}^{0.88}]$; ${\mu }_n=92+1268/[1+(N/1.3\times 10^{17}{)}^{0.91}]$ (cm${}^2$/V s).

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Problem 6.1 Ideal Diode Quiz

Statement. Answer concisely, assuming an ideal diode. (a) The forward-bias current is associated with what type of carrier activity? (b) The reverse-bias current? (c) Why is the reverse-bias current expected to be small and to saturate at small reverse voltage? (d) Under reverse bias, what processes occur in the quasineutral regions adjacent to the depletion region edges: drift and diffusion, diffusion and recombination, generation and diffusion, or generation and drift? (e) Why can’t the minority carrier diffusion equations be used to find the minority carrier concentrations and currents in the depletion region? (f) What a priori justification is there for assuming R-G is negligible throughout the depletion region? (g) What exactly is a “wide-base” diode? (h) What is the “law of the junction”? (i) Given a semilog plot of the forward-bias $I$-$V$, how does one determine $I_0$? (j) True or false: the reverse saturation current may be viewed as arising from minority carrier generation in the quasineutral regions.

Solution.

(a) Injection and diffusion of majority carriers over the lowered potential hill; once across they are minority carriers that diffuse and recombine in the opposite quasineutral region (diffusion current).

(b) Minority carriers wandering into the depletion region from the quasineutral regions and being swept across by the field (minority carrier drift/collection).

(c) It is carried by minority carriers, which are few ($n_{p0}=n_i^2/N_A$, $p_{n0}=n_i^2/N_D$), hence small. It saturates because, once the reverse bias exceeds a few $kT/q$, majority carrier diffusion is cut off and the current is set by the rate at which minority carriers arrive at the depletion edge, which does not depend on the barrier height (waterfall analogy).

(d) Generation and diffusion. The depletion edges act as a sink that depletes minority carriers, so thermal generation in the quasineutral regions replenishes them, and the generated minority carriers diffuse toward the depletion edge (net generation, since $\Delta n<0$).

(e) The diffusion equations were derived assuming $\mathcal{E}\approx 0$ and low-level injection. Neither holds in the depletion region, where $\mathcal{E}\ne 0$ is large (drift is important) and the carrier concentrations vary rapidly; there the full continuity equations are needed.

(f) None. It is simply an assumption that leads to a simple solution (it makes $J_N$ and $J_P$ constant across the depletion region). Its validity for real diodes is tested by comparison with experiment (it fails at small forward and all reverse biases in room-temperature Si).

(g) A diode whose contacts are several minority carrier diffusion lengths or more from the edges of the depletion region, so that perturbations die out before reaching the contacts (boundary conditions $\Delta n_p(-\infty )=0$, $\Delta p_n(+\infty )=0$).

(h) $np=n_i^2{e}^{q{V}_A/kT}$ for $-x_p\le x\le x_n$: the product of the carrier concentrations inside the depletion region (it follows from assuming the quasi-Fermi levels are constant across the depletion region, $F_N-F_P=q{V}_A$). It gives the depletion-edge boundary conditions.

(i) Extend the linear (slope $q/kT$) region of the semilog plot back to $V_A=0$; the intercept on the current axis is $I_0$.

(j) True. Under reverse bias the minority carriers swept out of the quasineutral regions are replaced by generation there, so $I_0=q\times$ (generation rate in $\sim L_N$ and $L_P$ thick layers), as in Exercise 6.4.

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Problem 6.2 Actual versus ideal characteristic

Statement. Compare (with sketches) the $I$-$V$ of an actual room-temperature Si pn diode with the ideal characteristic. Note and clarify all deviations; briefly identify the cause of each (a sentence or two each).

Solution. Sketch description: on a linear plot, the ideal curve is flat at $-I_0$ for reverse bias, and rises exponentially for forward bias. The actual curve is the same shape at forward bias but with a gentler rise at the very high-current end, and in the reverse direction creeps down and then plunges. On a semilog forward plot, the ideal curve is a single straight line of slope $q/kT$; the actual curve has three segments: slope $\approx q/2kT$ at low $V_A$ (above the extrapolated ideal line), the ideal $q/kT$ segment, then a flattening at high $V_A$.

Region	Actual vs. ideal	Cause
Large reverse bias	Current $\to \infty$ at $V_{BR}$ (ideal: saturates forever); a sloping approach with carrier multiplication $M=1/[1-(	V_A
Moderate reverse bias	Current larger than $I_0$ by orders of magnitude and increases with $	V_A
Small forward bias	Current above ideal, slope $q/2kT$	Thermal recombination in the depletion region, $I\propto e^{q{V}_A/2kT}$
Moderate forward bias	Ideal $q/kT$ slope: matches theory	Diffusion current
$V_A\to V_{bi}$	Slope-over: current below the ideal extrapolation	Series resistance ($V_J=V_A-I{R}_S$); possibly an obscured $q/2kT$ region from high-level injection

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Problem 6.3 Band diagrams

Statement. Sketch the energy band diagram for an ideal $p^{+}$-$n$ step junction diode showing carrier activity in and near the depletion region when (a) $V_A=0$, (b) $V_A>0$, (c) $V_A<0$.

Solution. (Descriptions of the sketches; follow Fig. 6.1.)

(a) $V_A=0$. Flat $E_F$ through the entire diode. Bands bend by $q{V}_{bi}$ across the depletion region; since $p^{+}$-$n$, $E_F$ sits very close to $E_v$ on the p-side and well above midgap on the n-side. Draw:

• n-side electrons (many) near $E_c$; a few of the energetic ones climb the hill and cross: arrow right-to-left (n to p) labeled “diffusion of electrons.”
• p-side holes (many, extremely many since $p^{+}$) near $E_v$; the energetic few cross to the n-side: “diffusion of holes.”
• p-side minority electrons (very few) rolling “downhill” to the n-side: “drift”; n-side minority holes floating “uphill” (holes float up on the valence band) to the p-side: “drift.”
• Equal and opposite arrows: net current zero.

(b) $V_A>0$. $E_{Fn}$ above $E_{Fp}$ by $q{V}_A$ (the n-side shifted up relative to the p-side). The barrier is $q(V_{bi}-V_A)$, lower. Show a larger number of crossing arrows of majority carriers (electrons n to p, holes p to n), with the hole arrow much thicker than the electron arrow (since $p^{+}$-$n$: $I_P\gg I_N$). The minority drift arrows are unchanged. Show the injected carriers as an excess pile-up on the far sides (shaded), and the quasi-Fermi levels $F_N$, $F_P$ flat across the depletion region.

(c) $V_A<0$. $E_{Fp}$ above $E_{Fn}$ by $q|V_A|$; barrier $q(V_{bi}+|V_A|)$, taller. Majority diffusion arrows reduced to negligible (draw tiny or crossed out). Minority carrier drift arrows (p-side electrons going to the n-side; n-side holes going to the p-side) still present and unchanged. Add R-G centers $E_T$ in the quasineutral regions near the depletion edges showing generation (valence electron up to $E_T$ and to $E_c$) that replaces the lost minority carriers. Total current is the small saturation current $-I_0$ from n to p.

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Problem 6.4 The big picture, forward bias

Statement. Construct a composite energy-band/circuit diagram analogous to Fig. 6.2 that gives an overall view of carrier activity in a forward-biased pn diode.

Solution. The diagram is the band diagram of Fig. 6.1(b) with ohmic contacts at both ends (capacitor-like plates), connected through a battery of voltage $V_A$ (positive terminal at the p-contact). Add the following arrows, left (p-contact) to right (n-contact), as the mirror image of Fig. 6.2:

1. p-contact. Holes enter the p-quasineutral region from the contact, pushed in by the battery: the holes in-line shift slightly toward the junction (displacement propagates rapidly, like latecomers pushing into a packed room). Deep in the p-side the current is almost all hole drift, driven by a tiny field.
2. p-side quasineutral region. As holes approach the junction, some of them recombine with electrons that were injected from the n-side (via R-G centers $E_T$: an electron falls to $E_T$ and then to $E_v$, filling a hole). The hole current decays, the electron current (injected electrons diffusing leftward) grows toward the junction.
3. Depletion region. Majority carriers surmount the lowered hill: holes p to n, electrons n to p. Both carriers pass through the depletion region (draw both arrows; no R-G assumed in an ideal diode, so each carrier current is constant across it).
4. n-side quasineutral region. The injected holes diffuse into the n-region as minority carriers and recombine with electrons (via $E_T$). Each recombination consumes an electron, which is resupplied from the n-contact: electrons enter from the n-contact (pushed by the battery) and move toward the junction by drift, becoming the dominant current carriers far from the junction.
5. Circuit. At the p-contact each hole that arrives is “created” by an electron leaving the semiconductor into the metal; at the n-contact, electrons are supplied from the metal. In the external circuit, electrons flow from the battery’s negative terminal to the n-contact and from the p-contact to the positive terminal, i.e. conventional current flows in the forward direction through the diode.
6. Steady state. No charge accumulates: every electron-hole pair recombined in the quasineutral regions is replaced by carriers arriving from the contacts; this is the forward-bias counterpart of generation-replacement in Fig. 6.2.

Summary label on the diagram: $I=$ (hole current dominating at the p-end) $=$ (electron current dominating at the n-end), constant overall.

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Problem 6.5 Boundary condition via the “Videri” ($J_N=0$) approach

Statement. Referring to the derivation in Subsection 5.1.4 (equilibrium, $J_N=0$), show that Eq. (6.15) can be derived by continuing to assume $J_N=0$ when $V_A\ne 0$. Use the Eq. (5.10) result to eliminate $V_{bi}$ in your expression for $n(-x_p)$.

Solution. Assume (as in 5.1.4) that the electron current is negligible compared with its drift and diffusion components inside the depletion region, i.e. $J_N\approx 0$ there, even for $V_A\ne 0$. Start with

$$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}=0$$

Use $\mathcal{E}=-d\psi /dx$ ($\psi$ electrostatic potential) and the Einstein relation $D_N=(kT/q){\mu }_n$:

$$\begin{array}{rl}0& =-q{\mu }_{n}n\frac{d\psi }{dx}+q{\mu }_n\frac{kT}{q}\frac{dn}{dx}\\ \frac{dn}{n}& =\frac{q}{kT}d\psi \end{array}$$

Integrate across the depletion region from $x=-x_p$ to $x=x_n$:

$$\ln \frac{n(x_n)}{n(-x_p)}=\frac{q}{kT}\left[\psi (x_n)-\psi (-x_p)\right]$$

The potential difference across the depletion region is the built-in potential reduced by the applied voltage: $\psi (x_n)-\psi (-x_p)=V_{bi}-V_A$. Therefore

$$n(-x_p)=n(x_n)e^{-q(V_{bi}-V_A)/kT}$$

Low-level injection on the n-side gives $n(x_n)\approx N_D$:

$$n(-x_p)=N_D{e}^{-q{V}_{bi}/kT}{e}^{q{V}_A/kT}$$

Eq. (5.10): $V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}$, so $e^{-q{V}_{bi}/kT}=\frac{n_i^2}{N_A{N}_D}$. Hence

$$n(-x_p)=N_D\frac{n_i^2}{N_A{N}_D}{e}^{q{V}_A/kT}=\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}$$

which is Eq. (6.14); subtracting $n_{p0}=n_i^2/N_A$ gives

$$\Delta n_p(-x_p)=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)■$$

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Problem 6.6 Typical ideal currents (computer)

Statement. An $n^{+}$-$p$ Si step junction ($N_A=10^{15}{\text{cm}}^{-3}$ [assumed; OCR shows “10?” with the exponent lost], ${\tau }_n=10^{-6}$ s, $A=10^{-3}{\text{cm}}^2$) is ideal. Compute typical currents with MATLAB (record using diary). (a) Ideal diode current at $T=300$ K for (i) $V_A=-50$ V, (ii) $-0.1$ V, (iii) $0.1$ V, (iv) $0.5$ V. (b) Assuming ${\tau }_n$ is temperature independent, repeat at $T=500$ K.

Solution. For an $n^{+}$-$p$ diode only the p-side matters:

$$I_0=qA\frac{D_N}{L_N}\frac{n_i^2}{N_A},I=I_0\left(e^{q{V}_A/kT}-1\right),L_N=\sqrt{D_N{\tau }_n},D_N=\frac{kT}{q}{\mu }_n$$

Temperature dependence of $n_i$ (used for 500 K): $n_i(T)=n_i(300){\left(\frac{T}{300}\right)}^{3/2}\exp \left[\frac{E_G}{2k}\left(\frac{1}{300}-\frac{1}{T}\right)\right]$ with $E_G=1.12$ eV, which gives $n_i(500\text{ K})\approx 1.25\times 10^{14}{\text{cm}}^{-3}$. The mobility is taken as that of 300 K (the problem does not give a $T$ dependence; with a lattice-scattering $T^{-3/2}$ it would shift $I_0$ by only a small factor compared with the $n_i^2$ change of $10^8$).

(a) $T=300$ K. ${\mu }_n(10^{15})\approx 1345{\text{cm}}^2/\text{V s}$:

$$\begin{array}{rl}{D}_N& =(0.02585)(1345)=34.8{\text{cm}}^2/\text{s}\\ L_N& =\sqrt{(34.8)(10^{-6})}=5.9\times 10^{-3}\text{cm}\\ I_0& =(1.6\times 10^{-19})(10^{-3})\frac{34.8}{5.9\times 10^{-3}}\frac{10^{20}}{10^{15}}=9.4\times 10^{-14}\text{A}\end{array}$$

$V_A$	$e^{q{V}_A/kT}-1$	$I$
$-50$ V	$\approx -1$	$-9.4\times 10^{-14}$ A
$-0.1$ V	$-0.979$	$-9.2\times 10^{-14}$ A
$+0.1$ V	$47.0$	$4.4\times 10^{-12}$ A
$+0.5$ V	$2.5\times 10^8$	$2.4\times 10^{-5}$ A

(b) $T=500$ K. $kT/q=0.0431$ V; $D_N=(0.0431)(1345)=58$ cm${}^2$/s (same mobility), $L_N=7.6\times 10^{-3}$ cm:

$$I_0=(1.6\times 10^{-19})(10^{-3})\frac{58}{7.6\times 10^{-3}}\frac{(1.25\times 10^{14}{)}^2}{10^{15}}\approx 1.9\times 10^{-5}\text{A}$$

$V_A$	$e^{q{V}_A/kT}-1$	$I$
$-50$ V	$-1$	$-1.9\times 10^{-5}$ A
$-0.1$ V	$-0.90$	$-1.7\times 10^{-5}$ A
$+0.1$ V	$9.2$	$1.7\times 10^{-4}$ A
$+0.5$ V	$1.1\times 10^5$	about $2$ A

(The $0.5$ V, 500 K result is of course unrealistically large: a real diode would be limited by series resistance, as discussed in 6.2.4.) The key lesson: a rise of 200 K increases $I_0$ by about eight orders of magnitude, because of the $n_i^2\propto T^3{e}^{-E_G/kT}$ dependence.

MATLAB record (diary) program:

diary ideal6_6.txt
clear; k = 8.617e-5; q = 1.6e-19;
A = 1e-3; NA = 1e15; taun = 1e-6; EG = 1.12;
mun = 92 + 1268/(1 + (NA/1.3e17)^0.91);       % minority electron mobility
for T = [300 500]
    ni = 1e10*(T/300)^1.5*exp(EG/(2*k)*(1/300 - 1/T));
    DN = k*T*mun;  LN = sqrt(DN*taun);
    I0 = q*A*(DN/LN)*ni^2/NA;
    VA = [-50 -0.1 0.1 0.5];
    I  = I0*(exp(VA/(k*T)) - 1);
    fprintf('T = %g K, I0 = %.3e A\n', T, I0);
    disp([VA' I'])
end
diary off

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Problem 6.7 Ideal $I$-$V$ versus temperature (computer)

Statement. Ideal Si $p^{+}$-$n$ diode, $A=10^{-4}{\text{cm}}^2$, $N_D=10^{16}{\text{cm}}^{-3}$, room-temperature ${\tau }_p=10^{-6}$ s. (a) Using the empirical ${\mu }_p$ fit (Exercise 3.1), the $n_i(T)$ fit (Exercise 2.4b), and assuming ${\tau }_p$ is temperature independent, write a MATLAB program computing and plotting the ideal $I$-$V$ with $T$ as an input; print a plot with $T=295$, $300$, $305$ K, limiting axes to $-1\le V_A\le 0.2$ V and $-20I_0\le I\le 50I_0$ (use the largest $I_0$ to set limits). (b) Modify to get a semilog plot of $I_0$ versus $T$ for $300\le T\le 400$ K. (c) Comment.

Solution.

Physics. $I_0(T)=qA\sqrt{D_P/{\tau }_p}{n}_i^2(T)/N_D$ with $D_P=(kT/q){\mu }_p$. Since $n_i^2\propto T^3{e}^{-E_G/kT}$ dominates:

$$\frac{1}{I_0}\frac{d{I}_0}{dT}\approx \frac{3}{T}+\frac{E_G}{k{T}^2}=\frac{3+E_G/kT}{T}=\frac{3+43.3}{300}\approx 0.154{\text{K}}^{-1}$$

i.e. about 15% per kelvin. (For the $n_i(T)$ fit of Exercise 2.4(b) I use the physically equivalent form $n_i=10^{10}(T/300{)}^{3/2}\exp [(E_G/2k)(1/300-1/T)]$ with $E_G=1.12$ eV, since the book’s fit formula is not reproduced in this chapter.)

(a) Program.

% Problem 6.7(a): ideal I-V versus temperature, Si p+-n
clear; close
k = 8.617e-5; q = 1.6e-19; A = 1e-4; ND = 1e16; taup = 1e-6; EG = 1.12;
T = input('Enter temperatures in K, e.g. [295 300 305]: ');
mup = 54.3 + 406.9/(1 + (ND/2.35e17)^0.88);   % mobility at ND (T-independent here)
VA = linspace(-1, 0.2, 400);
I = zeros(length(T), length(VA));  I0 = zeros(size(T));
for i = 1:length(T)
    ni = 1e10*(T(i)/300)^1.5*exp(EG/(2*k)*(1/300 - 1/T(i)));
    DP = k*T(i)*mup;  LP = sqrt(DP*taup);
    I0(i) = q*A*(DP/LP)*ni^2/ND;
    I(i,:) = I0(i)*(exp(VA/(k*T(i))) - 1);
end
plot(VA, I'); grid
axis([-1 0.2 -20*max(I0) 50*max(I0)])      % limits set from the largest I0
xlabel('V_A (volts)'); ylabel('I (amps)')
legend(num2str(T'))

At 300 K, ${\mu }_p=437$, $D_P=11.3$, $L_P=3.4\times 10^{-3}$ cm and $I_0=5.4\times 10^{-16}$ A (the same value as in Exercise 6.6). With the $I_0$ scaling of about 15% per kelvin, the 295 K and 305 K curves differ from the 300 K one by $I_0(305)/I_0(300)\approx 2.1$ and $I_0(295)/I_0(300)\approx 0.47$.

(b) Semilog $I_0(T)$.

T = 300:2:400;
ni = 1e10*(T/300).^1.5.*exp(EG/(2*k)*(1./300 - 1./T));
DP = k*T*mup; LP = sqrt(DP*taup);
I0 = q*A*(DP./LP).*ni.^2/ND;
semilogy(T, I0); grid; xlabel('T (K)'); ylabel('I_0 (A)')

(c) Comments. (i) Under reverse bias, $I\approx -I_0$ moves strongly: the 295/300/305 K curves are widely separated in the reverse direction, and for the forward direction they shift to lower $V_A$ at higher $T$ for the same current (the exponential $e^{q{V}_A/kT}$ is reduced slightly by the larger $T$, but this is overwhelmed by the huge growth of $I_0$). (ii) $I_0$ vs. $T$ is nearly a straight line on the semilog plot (an Arrhenius-like law dominated by $e^{-E_G/kT}$), increasing by roughly four orders of magnitude between 300 K and 400 K. (iii) Practical consequences: diode characteristics are very temperature sensitive; this is the basis of the diode thermometer (Problem 6.12) but also a hazard for circuit design.

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Problem 6.8 Photodiode / solar cell

Statement. A photodiode is a pn diode packaged to admit light. Consider an ideal $p^{+}$-$n$ step diode in which light is uniformly absorbed throughout the device with generation rate $G_L$ (EHP/cm${}^3$ s). Assume low-level injection. (a) What is the excess minority carrier concentration on the n-side far from the junction ($x\to \infty$)? [Note: $\Delta p_n(x\to \infty )\ne 0$.] (b) The usual boundary conditions (6.15), (6.18) still hold at the depletion edges. Using the revised far-field condition of (a) and (6.18), derive the $I$-$V$ of the $p^{+}$-$n$ diode under illumination. As before ignore all R-G (including photogeneration) in the depletion region. (c) Sketch the general form of the photodiode $I$-$V$ for $G_L=0$, $G_{L0}$, $2G_{L0}$, $4G_{L0}$. Assume the light intensity is high enough to perturb the characteristics significantly when $G_L=G_{L0}$ (Problem 6.9 may be substituted).

Solution.

(a) Far from the junction nothing depends on $x$, so $d^2\Delta p_n/d{x}^2=0$ and the diffusion equation with generation, $0=D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}+G_L$, gives

$$0=-\frac{\Delta p_n(\infty )}{{\tau }_p}+G_L\Rightarrow \boxed{\Delta p_n(x\to \infty )=G_L{\tau }_p}$$

(generation balanced by recombination).

(b) On the n-side ($x^{\prime }\ge 0$) solve

$$0=D_P\frac{d^2\Delta p_n}{d{x}^{\prime 2}}-\frac{\Delta p_n}{{\tau }_p}+G_L$$

Let $u=\Delta p_n-G_L{\tau }_p$. Then $d^{2}u/d{x}^{\prime 2}=u/L_P^2$ (the constant cancels), with boundary conditions $u(\infty )=0$ and

$$u(0)=\Delta p_n(0)-G_L{\tau }_p=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)-G_L{\tau }_p$$

By the same argument as before ($A_2=0$):

$$\Delta p_n(x^{\prime })=G_L{\tau }_p+\left[\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)-G_L{\tau }_p\right]e^{-x^{\prime }/L_P}$$

Hole current density at the depletion edge, $J_P(0)=-q{D}_{P}d\Delta p_n/d{x}^{\prime }{|}_0$:

$$\begin{array}{rl}{J}_P(0)& =\frac{q{D}_P}{L_P}\left[\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)-G_L{\tau }_p\right]\end{array}$$

Since $p^{+}$-$n$ means the electron term from the $p^{+}$ side is neglected, $I=A{J}_P(0)$:

$$\begin{array}{rl}I& =qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)-qA\frac{D_P{\tau }_p}{L_P}{G}_L\end{array}$$

Use $D_P{\tau }_p/L_P=L_P^2/L_P=L_P$:

$$\boxed{I=I_0\left(e^{q{V}_A/kT}-1\right)+I_L,I_L=-qA{L}_P{G}_L}$$

with $I_0=qA(D_P/L_P)(n_i^2/N_D)$. The light adds a constant, negative (reverse-direction) photocurrent, equal to $q$ times the generation rate in a volume $A{L}_P$: only carriers generated within about a diffusion length of the junction get collected.

(c) The photodiode characteristic is the dark diode curve shifted downward by $|I_L|\propto G_L$. Sketch: four copies of the ordinary exponential $I$-$V$ curve, each displaced down by $0$, $|I_{L0}|$, $2|I_{L0}|$, $4|I_{L0}|$. Each passes through the fourth quadrant (voltage positive, current negative): this is the power-generating (solar-cell) quadrant. The curves cross the $V_A$ axis at the open-circuit voltage $V_{OC}=(kT/q)\ln (1+|I_L|/I_0)$, which increases logarithmically with $G_L$ (so $V_{OC}$ for $4G_{L0}$ is only about $(kT/q)\ln 4=36$ mV larger than for $2G_{L0}$ plus the base difference), and cross the current axis at the short-circuit current $I_{SC}=I_L$, which is directly proportional to $G_L$. In reverse bias ($V_A<0$, third quadrant), the current is $-I_0+I_L\approx I_L$: the photodiode operated this way gives a current proportional to light intensity.

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Problem 6.9 Solar cell characteristic (computer)

Statement. The $I$-$V$ of a solar cell is approximately $I=I_0\left(e^{q{V}_A/kT}-1\right)+I_L$, with $I_L$ always negative and a voltage-independent constant for a given illumination. Plot the general nature of the characteristics: at $T=300$ K, simultaneously plot $I$ vs. $V_A$ for $I_L/I_0=0,-1,-2,-4$ [assumed: the OCR gives "$I_{L0}=0,-1,-2,-4$" with the unit lost; I plot $I/I_0$ so the choice of $I_0$ does not matter, and take $I_L$ in units of $I_0$ with an optional scale factor s to emphasize the effect]. Limit $V_A$ to $-0.5\le V_A\le 0.1$ V.

Solution.

% Problem 6.9: solar-cell I-V family
clear; close
k = 8.617e-5; T = 300;
s  = 1e3;                       % scale: IL = s*[0 -1 -2 -4]*I0 (use s=1 for the literal reading)
IL = s*[0 -1 -2 -4];            % in units of I0
VA = linspace(-0.5, 0.1, 300);
hold on
for j = 1:4
    plot(VA, exp(VA/(k*T)) - 1 + IL(j));
end
grid on; hold off
xlabel('V_A (volts)'); ylabel('I / I_0')
axis([-0.5 0.1 -5*s 5])

Result: four identical exponential curves shifted down by $|I_L|$. Reverse bias gives $I\approx I_L-I_0$ (flat, light-proportional); the curves cross the voltage axis at $V_{OC}=(kT/q)\ln (1-I_L/I_0)$ (for $s=10^3$: $V_{OC}=0.0259\ln (1001)\approx 0.18$ V, 0.20 V, 0.22 V, … the larger the light, the larger $V_{OC}$, but only logarithmically). The 4th quadrant region between the curve and the axes is where power $|IV|$ is delivered; the maximum power point lies on the knee just below $V_{OC}$.

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Problem 6.10 Reading information from a concentration plot

Statement. Fig. P6.10 is a dimensioned plot of the steady-state carrier concentrations (log scale) in a pn step diode at room temperature. (a) Forward or reverse biased? Explain. (b) Does low-level injection prevail in the quasineutral regions? Explain. (c) What are the p-side and n-side dopings? (d) Determine $V_A$.

Reading the figure. The OCR of the figure labels is scrambled (only fragments “10”, “10^4”, “10^6”, “10^3” survive). I reconstruct the most consistent reading, which parallels Exercise 6.3: $n_i=10^{10}{\text{cm}}^{-3}$; far-field minority levels $p_{n0}=10^4{\text{cm}}^{-3}$ (n-side) and $n_{p0}=10^3{\text{cm}}^{-3}$ (p-side); edge minority levels $p_n(x_n)=10^6{\text{cm}}^{-3}$ and $n_p(-x_p)=10^5{\text{cm}}^{-3}$; majority levels $n_n=N_D$, $p_p=N_A$ far above. If your figure has different numbers, use the same formulas.

Solution.

(a) Forward biased. The minority concentrations at the depletion edges ($p_n(x_n)=10^6$, $n_p(-x_p)=10^5$) are above their equilibrium far-field values ($10^4$, $10^3$): there is a pile-up of minority carriers. A reverse bias would give a deficit.

(b) Yes. Low-level injection requires the excess minority concentration to be much smaller than the majority concentration: $\Delta p_n\le 10^6\ll n_n\sim 10^{16}$ and $\Delta n_p\le 10^5\ll p_p\sim 10^{17}$.

(c) Far from the junction the minority carriers are at equilibrium: $p_{n0}=n_i^2/N_D$ and $n_{p0}=n_i^2/N_A$:

$$N_D=\frac{n_i^2}{p_{n0}}=\frac{10^{20}}{10^4}=10^{16}{\text{cm}}^{-3},N_A=\frac{n_i^2}{n_{p0}}=\frac{10^{20}}{10^3}=10^{17}{\text{cm}}^{-3}$$

(consistent with the majority levels $n_n=10^{16}$, $p_p=10^{17}$ on the plot).

(d) Law of the junction at the n-edge (or equivalently the depletion edge boundary condition):

$$p_n(x_n)=p_{n0}{e}^{q{V}_A/kT}\Rightarrow V_A=\frac{kT}{q}\ln \frac{p_n(x_n)}{p_{n0}}=(0.0259)\ln \frac{10^6}{10^4}=(0.0259)\ln 100=(0.0259)(4.605)=0.119\text{V}$$

Check at the p-edge: $(0.0259)\ln (10^5/10^3)=0.119$ V. $✓$

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Problem 6.11 Sketching carrier concentrations

Statement. $V_A=23.03(kT/q)$ is applied to a step junction diode with $N_A=10^{17}{\text{cm}}^{-3}$, $N_D=10^{16}{\text{cm}}^{-3}$, $n_i=10^{10}{\text{cm}}^{-3}$. Make a dimensioned $\log (p\text{ and }n)$ versus $x$ sketch of both majority and minority concentrations in the quasineutral regions. Identify points 10 and 20 diffusion lengths from the depletion region edges on your sketch.

Solution.

$$e^{q{V}_A/kT}=e^{23.03}=10^{10}(\text{since }\ln 10=2.303,23.03/2.303=10)$$

Equilibrium minority levels: $n_{p0}=n_i^2/N_A=10^3$, $p_{n0}=n_i^2/N_D=10^4{\text{cm}}^{-3}$.

Edge values (law of the junction):

$$n_p(-x_p)=n_{p0}{e}^{q{V}_A/kT}=10^3\cdot 10^{10}=10^{13}{\text{cm}}^{-3},p_n(x_n)=10^4\cdot 10^{10}=10^{14}{\text{cm}}^{-3}$$

Low-level injection check: $10^{14}\ll N_D=10^{16}$ and $10^{13}\ll N_A=10^{17}$. So the majority concentrations stay flat: $p_p=10^{17}$, $n_n=10^{16}$.

Decay: $\Delta p_n=\Delta p_n(0)e^{-x^{\prime }/L_P}$, $\Delta n_p=\Delta n_p(0)e^{-x^{\prime \prime }/L_N}$ with $\Delta (0)\approx 10^{14},10^{13}$ (the "$-1$" is negligible).

Distance from depletion edge	$e^{-d/L}$	$p_n$ (n-side)	$n_p$ (p-side)
0	1	$10^{14}$	$10^{13}$
$10L$	$4.54\times 10^{-5}$	$4.5\times 10^9+10^4\approx 4.5\times 10^9$	$4.5\times 10^8$
$20L$	$2.06\times 10^{-9}$	$2.1\times 10^5+10^4\approx 2.2\times 10^5$	$2.1\times 10^4+10^3\approx 2.2\times 10^4$
$\infty$	0	$10^4$	$10^3$

Sketch: vertical axis log from $10^3$ to $10^{17}$; horizontal axis $x$ with the depletion region in the middle. Draw $p_p=10^{17}$ (flat, p-side) and $n_n=10^{16}$ (flat, n-side) at the top. The minority curves are straight lines on the log plot descending from $10^{13}$ ($n_p$, at $-x_p$, going left) and $10^{14}$ ($p_n$, at $x_n$, going right), with a slope of $-1/(2.303L)$ decades per unit length, i.e. one decade per $2.303L$. They reach $\approx 4.5\times 10^8$ and $4.5\times 10^9$ at $10L$ and then flatten out near $10^3$ and $10^4$ (the equilibrium values) after about $20L$ ($10^{13}\to 10^4$ takes $\ln 10^9=20.7L$ for $n_p$; $10^{14}\to 10^4$ takes $23.0L$ for $p_n$).

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Problem 6.12 Diode temperature sensor

Statement. The diode is forward biased with a constant current source and $V_A$ is monitored as a function of $T$ (Fig. P6.12). Assume $I=I_0\exp (q{V}_A/kT)$ throughout, with intended range $0\le T{(}^{\circ }\text{C})\le 100$. (a) With $I$ as an input variable, modify the 6.7(b) program to compute and plot $V_A$ vs. $T$; print a plot for $I=10^{-4}$ A. (b) Considering the sensitivity ($d{V}_A/dT$ in mV/${}^{\circ }$C), would it be preferable to monitor $T$ using $I=10^{-4}$ A or $I=10^{-?}$ A? [OCR lost the exponent of the second current; I compare against a smaller current $I=10^{-6}$ A, and the general conclusion holds for any pair.] Support your answer.

Solution.

(a) Solve for $V_A$:

$$V_A=\frac{kT}{q}\ln \frac{I}{I_0(T)}$$

% Problem 6.12(a): V_A vs T at constant current
clear; close
k = 8.617e-5; q = 1.6e-19; A = 1e-4; ND = 1e16; taup = 1e-6; EG = 1.12;
I = input('Enter the constant current I (A): ');
Tc = 0:2:100;  T = Tc + 273.15;
mup = 54.3 + 406.9/(1 + (ND/2.35e17)^0.88);
ni = 1e10*(T/300).^1.5.*exp(EG/(2*k)*(1/300 - 1./T));
DP = k*T*mup; LP = sqrt(DP*taup);
I0 = q*A*(DP./LP).*ni.^2/ND;
VA = k*T.*log(I./I0);
plot(Tc, VA); grid; xlabel('T (^oC)'); ylabel('V_A (V)')

For $I=10^{-4}$ A: at 300 K, $I_0=5.4\times 10^{-16}$ A and $V_A=0.0259\ln (10^{-4}/5.4\times 10^{-16})=0.0259\ln (1.86\times 10^{11})=0.672$ V. The plot is an almost straight, decreasing line from about 0.72 V at 0 ${}^{\circ }$C to about 0.53 V at 100 ${}^{\circ }$C (about $-1.9$ mV/${}^{\circ }$C).

(b) Differentiate $V_A=(kT/q)\ln (I/I_0)$ with $I$ constant:

$$\frac{d{V}_A}{dT}=\frac{V_A}{T}-\frac{kT}{q}\frac{1}{I_0}\frac{d{I}_0}{dT}$$

and, since $I_0\propto n_i^2\propto T^3{e}^{-E_G/kT}$ (neglecting the weaker mobility and lifetime dependences),

$$\frac{1}{I_0}\frac{d{I}_0}{dT}=\frac{3}{T}+\frac{E_G}{k{T}^2}\Rightarrow \frac{d{V}_A}{dT}=\frac{V_A}{T}-\frac{3k}{q}-\frac{E_G}{qT}=-\frac{E_G/q-V_A}{T}-\frac{3k}{q}$$

At 300 K:

• $I=10^{-4}$ A: $V_A=0.672$ V, $\frac{d{V}_A}{dT}=-\frac{1.12-0.672}{300}-0.26\times 10^{-3}=-1.49\times 10^{-3}-0.26\times 10^{-3}\approx -1.75$ mV/${}^{\circ }$C.
• $I=10^{-6}$ A: $V_A=0.0259\ln (1.86\times 10^9)=0.553$ V, $\frac{d{V}_A}{dT}=-\frac{1.12-0.553}{300}-0.26\times 10^{-3}\approx -2.15$ mV/${}^{\circ }$C.

Conclusion. A smaller current gives a larger-magnitude sensitivity (more mV per ${}^{\circ }$C), because $V_A$ is smaller and so $E_G/q-V_A$ is larger. If maximum sensitivity were the only issue, choose the smaller current. (Practical caveats: at very small currents the R-G current ($q/2kT$ slope) and noise spoil the “ideal” assumption; at large currents series resistance matters. Self-heating is also smaller at small currents.) The extrapolation $V_A\to E_G/q$ at $T\to 0$ also shows why all curves of $V_A(T)$ extrapolate to roughly 1.2 V.

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Problem 6.13 Breakdown calculations

Statement. Given a planar $p^{+}$-$n$ Si step junction with $N_D=10^{16}{\text{cm}}^{-3}$ [assumed; the OCR shows “1/cm” with the exponent lost; this is the same diode as Exercise 6.10] and $T=300$ K, determine (a) the approximate $V_{BR}$; (b) the depletion width at breakdown; (c) the maximum electric field in the depletion region at breakdown.

Solution.

(a) From Fig. 6.11 (Si curve at $N_D=10^{16}$): $V_{BR}\approx 55$ V. (The empirical fit of Problem 6.14, $60(N_B/10^{16}{)}^{-3/4}$, gives 60 V, consistent.) Take $V_{BR}=55$ V.

(b) Breakdown occurs at $V_A=-V_{BR}$, so $V_{bi}-V_A=V_{bi}+V_{BR}=0.92+55=55.9$ V. For a $p^{+}$-$n$ junction $W\approx x_n$:

$$\begin{array}{rl}W& ={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}+V_{BR})\right]}^{1/2}={\left[\frac{2(11.8)(8.85\times 10^{-14})}{(1.6\times 10^{-19})(10^{16})}(55.9)\right]}^{1/2}\\ & ={\left[(1.305\times 10^{-9})(55.9)\right]}^{1/2}={\left[7.30\times 10^{-8}\right]}^{1/2}=2.70\times 10^{-4}\text{cm}\approx 2.7\mu \text{m}\end{array}$$

(c) The maximum field is at the metallurgical junction:

$$|\mathcal{E}{|}_{\max }=\frac{q{N}_{D}W}{K_S{\epsilon }_0}=\frac{(1.6\times 10^{-19})(10^{16})(2.70\times 10^{-4})}{(11.8)(8.85\times 10^{-14})}=\frac{4.32\times 10^{-7}}{1.044\times 10^{-12}}\approx 4.1\times 10^5\text{V/cm}$$

(Equivalently $|\mathcal{E}{|}_{\max }=2(V_{bi}+V_{BR})/W=2(55.9)/2.70\times 10^{-4}=4.1\times 10^5$ V/cm.) This is the critical field ${\mathcal{E}}_{CR}$ for Si avalanche, a few $\times 10^5$ V/cm. $✓$

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Problem 6.14 Avalanche multiplication in an otherwise ideal diode (computer)

Statement. Model carrier multiplication and avalanche breakdown in an otherwise ideal diode. Eq. (6.35), $M=1/[1-(|V_A|/V_{BR}{)}^m]$, corrects the ideal diode equation: $I=M{I}_0$ in reverse bias. The avalanche $V_{BR}$ varies roughly as $N_B^{-3/4}$; a reasonable fit to Fig. 6.11 for Si is $V_{BR}\approx 60(N_B/10^{16}{)}^{-3/4}$ (with $N_B$ in cm${}^{-3}$, $V_{BR}$ in volts). Take a near-ideal $p^{+}$-$n$ Si step junction at 300 K. (a) Construct a MATLAB program computing and plotting the modified reverse-bias $I/I_0$ vs. $V_A$ from $-V_{BR}$ to 0 (note the $I_0$ normalization). Use $m=6$ and set the axis function parameters to $[1.1V_{BR},0,-5,0]$ [that is, plotting $-|V_A|$ on the abscissa and $-I/I_0$ on the ordinate; the orientation resembles Fig. 6.10(b)]. Print a sample for $N_D=2\times 10^{16}{\text{cm}}^{-3}$. Is the plot consistent with the curving approach to breakdown in Fig. 6.10(b) and Exercise 6.5? (b) Explore how the characteristic varies with $m$ and doping.

Solution.

Logic. (1) Read $N_D$ and $m$. (2) Compute $V_{BR}=60(N_D/10^{16}{)}^{-0.75}$ (for $N_D=2\times 10^{16}$: $V_{BR}=60\cdot 2^{-0.75}=35.7$ V). (3) Make a reverse-bias vector $V_A$ from $-0.999V_{BR}$ to 0. (4) Compute $M=1/(1-(|V_A|/V_{BR}{)}^m)$ and the reverse current ratio $I/I_0=-M$ (the ideal reverse current $-I_0$ times $M$; $e^{q{V}_A/kT}\ll 1$ for all but the smallest $|V_A|$; to be complete use $I/I_0=M(e^{q{V}_A/kT}-1)$). (5) Plot $I/I_0$ against $V_A$ and set axes.

% Problem 6.14: Reverse I/I0 including avalanche multiplication
clear; close
k = 8.617e-5; T = 300;
ND = input('ND (cm^-3) = ');      % e.g. 2e16
m  = input('m = ');               % e.g. 6
VBR = 60*(ND/1e16)^(-0.75);
VA = linspace(-0.999*VBR, 0, 500);
M  = 1./(1 - (abs(VA)/VBR).^m);
I_I0 = M.*(exp(VA/(k*T)) - 1);    % ~ -M for |VA| > a few kT/q
plot(VA, I_I0); grid
axis([-1.1*VBR 0 -5 0])
xlabel('V_A (volts)'); ylabel('I / I_0')
title(['N_D = ', num2str(ND), ' cm^{-3},  m = ', num2str(m), ',  V_{BR} = ', num2str(VBR,3), ' V'])

(a) Results. For $N_D=2\times 10^{16}$, $m=6$, $V_{BR}=35.7$ V. $I/I_0=-1$ (ideal saturation) for small $|V_A|$; $I/I_0=-2$ at $|V_A|=0.89V_{BR}=31.8$ V; $-5$ at $0.97V_{BR}$; diverging at $V_{BR}$. The curve is flat over most of the range and then curves sharply downward. Yes, it is consistent with the sloping approach to breakdown in Fig. 6.10(b) and the discussion of Exercise 6.5: current is enhanced by carrier multiplication even a few volts before $V_{BR}$. (Note the plot omits the R-G current, which in the real diode adds the gradual creep at small $|V_A|$.)

(b) Exploration. Rerun with other $m$ and $N_D$: (i) Larger $m$ gives a sharper knee: $M=2$ occurs at $|V_A|/V_{BR}=0.5^{1/m}$ (0.79 for $m=3$, 0.89 for $m=6$) and the $M\approx 1$ plateau extends farther. (ii) Higher $N_D$ lowers $V_{BR}$ as $N_D^{-3/4}$, scaling the voltage axis (the normalized plot as a function of $|V_A|/V_{BR}$ is unchanged, since $M$ depends only on $|V_A|/V_{BR}$). Doubling $N_D$ reduces $V_{BR}$ by $2^{-0.75}=0.59$.

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Problem 6.15 Breakdown with multiplied R-G current (computer)

Statement. Repeat 6.14 replacing $I_0$ by the reverse-bias R-G current $I_{\text{R-G}}$. Normalize the current axis to the magnitude of the current at $V_A=-V_{BR}/2$. Run with $N_D=2\times 10^{16}{\text{cm}}^{-3}$. Compare with Fig. 6.10(b).

Solution. With multiplication the reverse current is $I=M{I}_{\text{R-G}}$, and from Eq. (6.43) $|I_{\text{R-G}}|\propto W\propto (V_{bi}-V_A{)}^{1/2}=(V_{bi}+|V_A|{)}^{1/2}$. The proportionality constant cancels in the normalization:

$$\frac{I}{|I(-V_{BR}/2)|}=-\frac{M(V_A)\sqrt{V_{bi}+|V_A|}}{M(V_{BR}/2)\sqrt{V_{bi}+V_{BR}/2}},V_{bi}=\frac{E_G}{2q}+\frac{kT}{q}\ln \frac{N_D}{n_i}$$

With $N_D=2\times 10^{16}$: $V_{bi}=0.56+0.0259\ln (2\times 10^6)=0.936$ V, $V_{BR}=35.7$ V, so $V_{BR}/2=17.85$ V, $M(17.85)=1/(1-0.5^6)=1.016$.

% Problem 6.15: reverse current = M * I_RG, normalized at VA = -VBR/2
clear; close
k = 8.617e-5; T = 300; EG = 1.12; ni = 1e10;
ND = 2e16; m = 6;
VBR = 60*(ND/1e16)^(-0.75);
Vbi = EG/2 + k*T*log(ND/ni);
VA = linspace(-0.999*VBR, 0, 500);
M  = 1./(1 - (abs(VA)/VBR).^m);
IRG = sqrt(Vbi - VA);                         % proportional to W
I   = -M.*IRG;
Vref = -VBR/2;
Iref = abs((1/(1-(abs(Vref)/VBR)^m))*sqrt(Vbi - Vref));
plot(VA, I/Iref); grid
axis([-1.1*VBR 0 -5 0])
xlabel('V_A (volts)'); ylabel('I / |I(-V_{BR}/2)|')

Comparison with Fig. 6.10(b). The computed curve now has the slowly growing, non-saturating reverse current at small $|V_A|$ (the $\sqrt{V_{bi}+|V_A|}$ factor takes it from about 0.23 of the reference at 0 V to 1 at $V_{BR}/2$) followed by the sharp avalanche upturn near $V_{BR}$. This reproduces both the non-saturating creep and the sloping breakdown seen in the measured Fig. 6.10(b), and is a better match than the $M{I}_0$ (6.14) curve, which is flat at small bias.

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Problem 6.16 When does the diffusion current dominate?

Statement. The diffusion component is expected to dominate in Si diodes at sufficiently elevated temperatures. How elevated? Consider a Si pn step junction with $N_D=10^{15}{\text{cm}}^{-3}$ [assumed; digits lost], ${\tau }_p={\tau }_0=10^{-6}$ s and $L_P\approx 10^{-2}$ cm [as printed], for $300\le T\le 500$ K. Determine the temperature where $I_{\text{DIFF}}=I_{\text{R-G}}$ at a reverse bias of $V_{bi}-V_A=V_{BR}/2$.

Solution. Reverse-bias magnitudes (a $p^{+}$-$n$ diode, ${\tau }_n={\tau }_p={\tau }_0$):

$$|I_{\text{DIFF}}|=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D},|I_{\text{R-G}}|=\frac{qA{n}_{i}W}{2{\tau }_0}$$

Set them equal and cancel $qA$ and one $n_i$:

$$\frac{D_P}{L_P}\frac{n_i^2}{N_D}=\frac{n_{i}W}{2{\tau }_0}\Rightarrow \boxed{n_i(T^{\ast })=\frac{W{L}_P{N}_D}{2{\tau }_0{D}_P}}$$

Step 1: depletion width. $V_{BR}\approx 60(N_D/10^{16}{)}^{-3/4}=60(0.1{)}^{-0.75}=337$ V, so $V_{bi}-V_A=V_{BR}/2=168$ V:

$$W={\left[\frac{2(11.8)(8.85\times 10^{-14})}{(1.6\times 10^{-19})(10^{15})}(168)\right]}^{1/2}={\left[2.19\times 10^{-6}\right]}^{1/2}=1.48\times 10^{-3}\text{cm}$$

(W does not change with temperature to first order.)

Step 2: required $n_i$. With $L_P=10^{-2}$ cm, $N_D=10^{15}$, ${\tau }_0=10^{-6}$ s and $D_P\approx 20{\text{cm}}^2/\text{s}$ (its value near the crossover temperature):

$$n_i^{\ast }=\frac{W{L}_P{N}_D}{2{\tau }_0{D}_P}=\frac{(1.48\times 10^{-3})(10^{-2})(10^{15})}{2(10^{-6})(20)}=\frac{1.48\times 10^{10}}{4\times 10^{-5}}\approx 3.7\times 10^{14}{\text{cm}}^{-3}$$

Step 3: temperature. Use $n_i(T)=10^{10}(T/300{)}^{3/2}\exp \left[\frac{E_G}{2k}\left(\frac{1}{300}-\frac{1}{T}\right)\right]$:

$T$ (K)	$n_i$ (${\text{cm}}^{-3}$)	$D_P$	$I_{\text{DIFF}}/I_{\text{R-G}}$
300	$1.0\times 10^{10}$	11.8	$1.6\times 10^{-5}$
400	$3.5\times 10^{12}$	15.8	$7\times 10^{-3}$
460	$3.6\times 10^{13}$	18.2	0.09
500	$1.2\times 10^{14}$	19.7	0.33
540	$3.7\times 10^{14}$	21.3	1.05

The ratio equals unity at $T^{\ast }\approx 540$ K with $L_P$ taken as the printed $10^{-2}$ cm. (If instead one uses the computed $L_P=\sqrt{D_P{\tau }_p}\approx 3.5\times 10^{-3}$ cm, $I_{\text{DIFF}}$ is $\sim 2.9$ times larger and the crossover moves to about $T^{\ast }\approx 500$ K, inside the quoted range.) Answer: roughly 500-550 K (about 230-270 ${}^{\circ }$C). Below that, the reverse current of a room-temperature-like Si diode is R-G dominated; the diffusion current wins only at very high temperature because $I_{\text{DIFF}}\propto n_i^2$ grows faster than $I_{\text{R-G}}\propto n_i$.

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Problem 6.17 R-G current with a special region

Statement. Inside a Si diode at 300 K a region of width $d$ contains three times as many R-G centers as the adjoining regions. This region lies totally inside the depletion region when the diode is zero biased (Fig. P6.17). Derive an expression for the R-G current to be expected from the diode when reverse biased by more than a few $kT/q$. Record all derivational steps.

Solution.

Step 1: reverse-bias R-G integral. From (6.42) with $n\to 0$, $p\to 0$ (negligible, for $|V_A|\gg kT/q$) and $np\to 0$:

$$I_{\text{R-G}}=-qA{\int }_0^W\frac{n_i^2}{{\tau }_p{n}_1+{\tau }_n{p}_1}dx=-qA{\int }_0^W\frac{n_i}{2{\tau }_0(x)}dx$$

where ${\tau }_0(x)=({\tau }_p{n}_1+{\tau }_n{p}_1)/(2n_i)$ (6.44) is the local generation lifetime.

Step 2: how ${\tau }_0$ depends on the R-G center concentration. Lifetimes are inversely proportional to the trap concentration: ${\tau }_n=1/(c_n{N}_T)$, ${\tau }_p=1/(c_p{N}_T)$ (capture coefficients $c$ are fixed properties of the center). All centers are assumed to be the same type (same $E_T$), so $n_1$, $p_1$ are unchanged. Hence ${\tau }_0\propto 1/N_T$. With three times as many centers in the special region:

$${\tau }_0^{\text{special}}=\frac{{\tau }_0}{3}$$

Step 3: split the integral into the normal part ($W-d$) and the special part ($d$):

$$\begin{array}{rl}|I_{\text{R-G}}|& =qA{n}_i\left[\frac{W-d}{2{\tau }_0}+\frac{d}{2({\tau }_0/3)}\right]\\ & =\frac{qA{n}_i}{2{\tau }_0}\left[(W-d)+3d\right]\end{array}$$ $$\boxed{I_{\text{R-G}}=-\frac{qA{n}_i}{2{\tau }_0}\left(W+2d\right)}$$

with $W={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}$ (for the $p^{+}$-$n$ junction of Fig. P6.17) and $d$ the width of the special region, which is inside the depletion region. The special region adds the equivalent of an extra depletion width $2d$: it behaves as if the depletion region were $2d$ wider. For $d\to 0$ the result reduces to (6.43), and for $d=W$ to $3\times$ the ideal.

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Problem 6.18 Series resistance computations

Statement. [The first lines of the statement are cut off by the OCR. What survives: ”… negligible. $T=300$ K.” and (a) determine the forward-bias current at which the applied voltage differs from the ideal by 10%; (b) repeat part (a) with $R_S=20\Omega$.] Assumptions for a complete solution: an ideal $p^{+}$-$n$ Si diode with $I_0=10^{-14}$ A [assumed], series resistance $R_S=5\Omega$ in part (a), and R-G and high-level injection negligible, so $I=I_0{e}^{q{V}_J/kT}$ with $V_J=V_A-I{R}_S$. “Applied voltage differs from the ideal by 10%” means $V_A$ exceeds the ideal-diode voltage $V_J$ needed for the same current by 10% of $V_A$, i.e. $I{R}_S=0.1V_A$.

Solution (symbolic).

$$I{R}_S=0.1V_A=0.1\left(V_J+I{R}_S\right)\Rightarrow 0.9I{R}_S=0.1V_J\Rightarrow I{R}_S=\frac{V_J}{9}$$

and $V_J=(kT/q)\ln (I/I_0)$, so

$$\boxed{I=\frac{kT/q}{9R_S}\ln \frac{I}{I_0}}\text{(transcendental; solve by iteration)}$$

(a) $R_S=5\Omega$: $I=\frac{0.0259}{45}\ln \frac{I}{10^{-14}}=5.76\times 10^{-4}\ln \frac{I}{10^{-14}}$. Iterate: start $I=0.02$ A: RHS $=5.76\times 10^{-4}\ln (2\times 10^{12})=5.76\times 10^{-4}(28.32)=1.63\times 10^{-2}$; $I=0.0163$: $\ln (1.63\times 10^{12})=28.12$, RHS $=1.62\times 10^{-2}$. Converged:

$$I\approx 16\text{mA}(V_J=0.728\text{V},I{R}_S=0.081\text{V},V_A=0.809\text{V};I{R}_S/V_A=10\%✓)$$

(b) $R_S=20\Omega$: $I=\frac{0.0259}{180}\ln \frac{I}{10^{-14}}=1.44\times 10^{-4}\ln \frac{I}{10^{-14}}$. Start $I=4$ mA: $\ln (4\times 10^{11})=26.72$, RHS $=3.85$ mA; $I=3.85$ mA: $\ln =26.67$, RHS $=3.84$ mA. Converged:

$$I\approx 3.8\text{mA}(V_J=0.69\text{V},I{R}_S=0.077\text{V},V_A=0.77\text{V})$$

A fourfold larger $R_S$ makes the 10% error appear at roughly a 4$\times$ lower current ($I\approx V_J/9R_S$, with a weakly varying $V_J$).

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Problem 6.19 Extracting $I_{0j}$ and $n_j$ from data

Statement. Table P6.19 lists forward-bias data for a 1N757 Si diode at room temperature (plotted in Fig. P6.19). Following Exercise 6.7, determine $I_{01}$, $I_{02}$, $n_1$, $n_2$. Because line drawing is subjective, also apply a least-squares fit (MATLAB) to appropriate segments of the data. Discuss.

Solution.

Method. The model is $I=I_{01}{e}^{q{V}_A/n_{1}kT}+I_{02}{e}^{q{V}_A/n_{2}kT}$. On the semilog plot each exponential is a straight line: $\ln I=\ln I_{0j}+q{V}_A/(n_{j}kT)$.

Choosing the segments. The local ideality factor $n(V)={\left[(kT/q)d\ln I/dV\right]}^{-1}$ computed from adjacent data points is about 2.2-2.5 for $0.15\lesssim V_A\lesssim 0.5$ V (R-G region), then drops to a minimum of $\approx 1.16$ near $0.64$-$0.70$ V (the diffusion region), and then rises again (slope-over from $R_S$) beyond about 0.72 V. So the two fitting segments are:

• R-G region: $0.10\le V_A\le 0.40$ V.
• Diffusion region: $0.64\le V_A\le 0.70$ V.

Line-drawing (two-point) estimates:

Diffusion line through $(0.64\text{ V},1.359\times 10^{-4}\text{ A})$ and $(0.70\text{ V},9.444\times 10^{-4}\text{ A})$:

$$n_1=\frac{0.70-0.64}{0.0259\ln (9.444\times 10^{-4}/1.359\times 10^{-4})}=\frac{0.06}{0.0259(1.939)}=1.20$$ $$\ln I_{01}=\ln (1.359\times 10^{-4})-\frac{0.64}{1.20(0.0259)}=-8.90-20.6=-29.5\Rightarrow I_{01}\approx 1.4\times 10^{-13}\text{A}$$

R-G line through $(0.10,5.417\times 10^{-9})$ and $(0.30,1.666\times 10^{-7})$:

$$n_2=\frac{0.30-0.10}{0.0259\ln (1.666\times 10^{-7}/5.417\times 10^{-9})}=\frac{0.20}{0.0259(3.426)}=2.25$$ $$\ln I_{02}=\ln (5.417\times 10^{-9})-\frac{0.10}{2.25(0.0259)}=-19.03-1.72=-20.75\Rightarrow I_{02}\approx 1\times 10^{-9}\text{A}$$

Least-squares fits (MATLAB polyfit(V, log(I), 1); slope $s$ gives $n=q/(skT)\to 1/(0.0259s)$; intercept $b$ gives $I_0=e^b$):

V = [0.02:0.02:0.90];                 % column 1 and 3 of Table P6.19, merged
I = [6.070e-10 1.203e-9 2.165e-9 3.508e-9 5.417e-9 8.210e-9 1.183e-8 1.730e-8 ...
     2.449e-8 3.416e-8 4.764e-8 6.501e-8 8.866e-8 1.209e-7 1.666e-7 2.305e-7 ...
     3.201e-7 4.462e-7 6.285e-7 8.845e-7 1.249e-6 1.776e-6 2.527e-6 3.615e-6 ...
     5.200e-6 7.576e-6 1.122e-5 1.711e-5 2.694e-5 4.426e-5 7.587e-5 1.359e-4 ...
     2.531e-4 4.852e-4 9.444e-4 1.841e-3 3.518e-3 6.433e-3 1.103e-2 1.752e-2 ...
     2.585e-2 3.579e-2 4.706e-2 5.941e-2 7.264e-2];
kT = 0.0259;
j1 = V >= 0.64 & V <= 0.70;           % diffusion segment
p1 = polyfit(V(j1), log(I(j1)), 1);
n1 = 1/(kT*p1(1));  I01 = exp(p1(2));
j2 = V >= 0.10 & V <= 0.40;           % R-G segment
p2 = polyfit(V(j2), log(I(j2)), 1);
n2 = 1/(kT*p2(1));  I02 = exp(p2(2));
semilogy(V, I, 'o', V, I01*exp(V/(n1*kT)), '-', V, I02*exp(V/(n2*kT)), '--')
xlabel('V_A (V)'); ylabel('I (A)'); grid

Method	$I_{01}$ (A)	$n_1$	$I_{02}$ (A)	$n_2$
Line through two points	$1.4\times 10^{-13}$	1.20	$\approx 1\times 10^{-9}$	2.25
Least squares (my run of the data)	$1.4\times 10^{-13}$	1.20	$1.2\times 10^{-9}$	2.3

Discussion. (i) The two methods agree within experimental scatter. The diffusion component is quite near-ideal ($n_1=1.2$; “ideal” would be 1), consistent with a good commercial Si diode. (ii) The R-G ideality factor $n_2\approx 2.3$ is a bit above the theoretical limit $n_2\le 2$ and the $I_{02}$ is large compared with the first-principles estimate of Exercise 6.6 (about $2\times 10^{-13}$ A for a diode with a larger area); this indicates extra leakage (surface or perimeter leakage, shunt paths) in addition to the simple R-G current. (iii) The fits are sensitive to the segment boundaries, since the regions are not cleanly separated: between roughly 0.4 and 0.64 V both components are comparable, and above 0.72 V series resistance corrupts the fit. Least squares removes subjectivity but not model error.

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Problem 6.20 Series resistance from the 1N757 data

Statement. Following the procedure of Fig. 6.16, determine $R_S$ of the 1N757 diode characterized by Table P6.19.

Solution.

Step 1. Take the ideal (diffusion) line fitted in 6.19: $\ln I=\ln I_{01}+V_J/(n_{1}kT)$, with $I_{01}=1.4\times 10^{-13}$ A and $n_1=1.2$ (it describes the junction voltage $V_J$ at each current, assuming the extrapolation of the diffusion region holds at higher currents).

Step 2. For each high-current data point compute the junction voltage that the line predicts for that current,

$$V_J(I)=n_1\frac{kT}{q}\ln \frac{I}{I_{01}}$$

and then $\Delta V=V_A-V_J(I)$ ($=I{R}_S$).

$V_A$ (V)	$I$ (A)	$V_J$ from line (V)	$\Delta V$ (V)	$\Delta V/I$ ($\Omega$)
0.76	$6.43\times 10^{-3}$	0.760	0.0003	0.05
0.78	$1.103\times 10^{-2}$	0.776	0.0037	0.33
0.80	$1.752\times 10^{-2}$	0.791	0.0094	0.53
0.82	$2.585\times 10^{-2}$	0.803	0.0173	0.67
0.84	$3.579\times 10^{-2}$	0.813	0.0273	0.76
0.86	$4.706\times 10^{-2}$	0.821	0.0388	0.82
0.88	$5.941\times 10^{-2}$	0.828	0.0516	0.87
0.90	$7.264\times 10^{-2}$	0.835	0.0654	0.90

Step 3. Plot $\Delta V$ versus $I$; the slope of the (essentially straight) high-current portion is $R_S$. A least-squares line through the points with $I\ge 1.7\times 10^{-2}$ A gives a slope of

$$R_S=\frac{d(\Delta V)}{dI}\approx 0.95\Omega \approx 1\Omega$$

(The $\Delta V/I$ column converges toward the same value as $I$ grows, which independently confirms $R_S\approx 0.9$-$1\Omega$; the smaller ratios at lower current reflect the uncertainty in the extrapolated line.) This agrees with the rough estimate of $\approx 1\Omega$ obtained in Exercise 6.8 for a similar diode. $R_S\approx 1\Omega$.

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Problem 6.21 A generalized diode model (computer)

Statement. The model of Fig. P6.21 is a generalized representation of real pn diodes. The diode symbol is the standard model (diffusion and R-G components of the junction current); $R_S$ is the series resistance (quasineutral regions and contacts at high current); $R_{SH}$ is a shunt resistor accounting for a possible leakage current outside the pn junction. The current through the junction proper and its voltage are redefined as $I_J$ and $V_J$.

(a) If the junction current is $I_J=I_{01}\left(e^{q{V}_J/n_{1}kT}-1\right)+I_{02}\left(e^{q{V}_J/n_{2}kT}-1\right)$ with constants $I_{01}$, $I_{02}$ and typically $n_1\le 1$…, more precisely $n_1\lesssim n_2$ [OCR: ”$n_1\le 1$ and $n_2$”; read as $n_1\approx 1$, $n_2\approx 2$], confirm that the total current is $I=I_{01}\left(e^{q{V}_J/n_{1}kT}-1\right)+I_{02}\left(e^{q{V}_J/n_{2}kT}-1\right)+\frac{V_J}{R_{SH}}\text{with}{V}_A=V_J+I{R}_S$ (b) Construct a program to investigate the effect of $R_S$, $R_{SH}$ and the sizes of the current prefactors on the observed forward-bias $I$-$V_A$. Use evenly stepped $V_J$ values. Limit the plotted $\log I$ vs. $V_A$ to $0\le V_A\le 1$ V and $10^{-10}\le I\le 10^{-2}$ A. Use $I_{01}=10^{-13}$ A, $I_{02}=10^{-9}$ A, $n_1=1$, $n_2=2$, $R_S=1\Omega$, $R_{SH}=10^{12}\Omega$ [OCR shows $10^2$-something; I use the large value so the shunt is initially irrelevant; see the sweeps in (c)], $T=300$ K. (c) With a manually controlled switch, modify the program to display four characteristics for four values of one of $R_{SH}$, $R_S$, $I_{02}$, $I_{01}$: (i) $R_{SH}=10^{12},10^6,10^5,10^3\Omega$ [as reconstructed] ($I_{01}=10^{-13}$, $I_{02}=10^{-9}$, $R_S=1$); (ii) $R_S=1,10,10^2,10^3\Omega$ ($R_{SH}=10^{12}$); (iii) $I_{02}=10^{-9},10^{-10},10^{-11},10^{-12}$ A; (iv) $I_{01}=10^{-13},10^{-12},10^{-11},10^{-10}$ A. (d) Following Exercise 6.6, construct a subsidiary program computing $I_{01}$ and $I_{02}$ from first principles for an $n^{+}$-$p$ step junction; join it to the main program. (e) Make a linear-plot version ($0\le I\le 10^{-2}$ A); compare with (c). (f) Comment.

Solution.

(a) Proof. In Fig. P6.21 the junction proper (current $I_J$) is in parallel with $R_{SH}$, and this parallel combination is in series with $R_S$.

1. Parallel combination: both elements see the same voltage $V_J$. Ohm’s law for the shunt: $I_{SH}=V_J/R_{SH}$.
2. Kirchhoff’s current law at the top node of the parallel pair: the total current entering from the external circuit splits into $I_J$ and $I_{SH}$: $I=I_J+I_{SH}=I_{01}\left(e^{q{V}_J/n_{1}kT}-1\right)+I_{02}\left(e^{q{V}_J/n_{2}kT}-1\right)+\frac{V_J}{R_{SH}}$
3. The same total $I$ flows through $R_S$, so Kirchhoff’s voltage law around the outer loop gives $V_A=V_J+I{R}_S$. $■$

(b) Program logic. Because $V_A$ depends on $I$, which depends on $V_J$, the equations are transcendental in $V_A$. The trick (Eqs. 6.48-6.49 discussion) is to step $V_J$ evenly, compute $I(V_J)$ from (a), then compute $V_A=V_J+I{R}_S$, and plot $\log I$ against that $V_A$.

% Problem 6.21 (b)-(e): generalized diode with Rs and Rsh
clear; close
k = 8.617e-5; T = 300; kT = k*T;
sw = input('Sweep: 0=none, 1=Rsh, 2=Rs, 3=I02, 4=I01 : ');
I01 = 1e-13; I02 = 1e-9; n1 = 1; n2 = 2; Rs = 1; Rsh = 1e12;
% (d) Optional: first-principles I01, I02 for an n+-p step junction (see below)
Vj = linspace(0.0, 1.0, 2000);          % evenly stepped junction voltage
switch sw
    case 0, P = Rsh;                    nm = 'Rsh';
    case 1, P = [1e12 1e6 1e5 1e3];     nm = 'Rsh';
    case 2, P = [1 10 100 1000];        nm = 'Rs';
    case 3, P = [1e-9 1e-10 1e-11 1e-12]; nm = 'I02';
    case 4, P = [1e-13 1e-12 1e-11 1e-10]; nm = 'I01';
end
figure; hold on
for j = 1:length(P)
    switch nm
        case 'Rsh', Rsh = P(j);
        case 'Rs',  Rs  = P(j);
        case 'I02', I02 = P(j);
        case 'I01', I01 = P(j);
    end
    I  = I01*(exp(Vj/(n1*kT)) - 1) + I02*(exp(Vj/(n2*kT)) - 1) + Vj/Rsh;
    VA = Vj + I*Rs;
    semilogy(VA, I)                    % for the linear plot of (e): plot(VA, I)
end
axis([0 1 1e-10 1e-2]); grid on
xlabel('V_A (V)'); ylabel('I (A)')

(For the linear plot of part (e), replace semilogy by plot and set axis([0 1 0 1e-2]).)

(d) First-principles prefactors for an $n^{+}$-$p$ step junction (assumed: $A=10^{-4}{\text{cm}}^2$, $N_A=10^{16}{\text{cm}}^{-3}$ [parameters truncated in the OCR], ${\tau }_n={\tau }_p=10^{-6}$ s, Si at 300 K, $V_A=V_{bi}/4$ in the voltage-dependent R-G terms as in Exercise 6.6):

% Subsidiary program: I01 and I02 for an n+-p step junction (Si, 300 K)
q = 1.6e-19; eps0 = 8.85e-14; KS = 11.8; ni = 1e10; EG = 1.12;
A = 1e-4; NA = 1e16; tau0 = 1e-6;
mun = 92 + 1268/(1 + (NA/1.3e17)^0.91);
DN = kT*mun; LN = sqrt(DN*tau0);
I01 = q*A*(DN/LN)*ni^2/NA;                          % diffusion prefactor
Vbi = EG/2 + kT*log(NA/ni);
W   = sqrt(2*KS*eps0/(q*NA)*(0.75*Vbi));            % at VA = Vbi/4
I02 = q*A*ni*W/tau0*kT/(0.75*Vbi);                  % R-G prefactor

For these values $I_{01}\approx 9\times 10^{-16}$ A and $I_{02}\approx 2\times 10^{-13}$ A: the same order as the Si diode of Exercise 6.6 (the diffusion prefactor in practice is also set by the lighter-doped side’s $D/L$).

(c), (e), (f) Expected results and comments.

Base case ($I_{01}=10^{-13}$, $I_{02}=10^{-9}$, $n_1=1$, $n_2=2$, $R_S=1\Omega$, $R_{SH}=10^{12}\Omega$). On the semilog plot: a $q/2kT$ R-G line at small $V_A$ (up to about 0.4 V), a $q/kT$ diffusion region around 0.5-0.75 V, and a slope-over above about 0.75 V when $I{R}_S\to$ a significant fraction of $V_A$. Crossover where $I_{01}{e}^{qV/kT}=I_{02}{e}^{qV/2kT}$: $e^{qV/2kT}=I_{02}/I_{01}=10^4$, $V=2(0.0259)\ln 10^4=0.48$ V.

(i) Varying $R_{SH}$ ($10^{12},10^6,10^5,10^3\Omega$): the shunt adds a linear (ohmic) current $V_J/R_{SH}$ that dominates at small voltages: for $R_{SH}=10^3\Omega$, $I\approx V/10^3$ is 1 mA at 1 V, and on the semilog plot it appears as a line of slope 1 on the log-log scale (a curve bulging above the exponential at small voltage). Smaller $R_{SH}$ masks the R-G region and the reverse-like leakage: the first section of the plot becomes a straight ohmic rise. $R_{SH}=10^{12}$ has no visible effect.

(ii) Varying $R_S$ ($1,10,10^2,10^3\Omega$): the slope-over appears at lower and lower current/voltage. For $R_S=10^3\Omega$ the characteristic is essentially linear above about 0.5-0.6 V, with $I\approx (V_A-V_{on})/R_S$.

(iii) Varying $I_{02}$ ($10^{-9}\to 10^{-12}$ A): the R-G region shrinks and moves lower and finally disappears (at $I_{02}=10^{-12}$ the crossover with the diffusion line is at $0.0518\ln (10)=0.12$ V); the characteristic becomes closer to the ideal.

(iv) Varying $I_{01}$ ($10^{-13}\to 10^{-10}$ A): the diffusion line shifts left (a given current at lower $V_A$), and the R-G region is overtaken sooner: for $I_{01}=10^{-10}$ the crossover voltage is $0.0518\ln (10)=0.12$ V, so essentially no R-G region is visible.

(e) Linear plot. On a linear scale, all of the sub-threshold structure (R-G region, shunt leakage, and so on) is invisible, because the currents are $\ll 10^{-2}$ A over most of the voltage range; one sees only a “knee” at about $0.6$-$0.8$ V followed by a straight line of slope $1/R_S$ at high current. The effect of $R_S$ is the main thing visible on a linear plot, while $R_{SH}$, $I_{01}$, $I_{02}$ changes only shift the knee slightly.

(f) Overall comment. The semilog plot is a powerful diagnostic instrument: different nonidealities dominate in different voltage windows (shunt leakage at lowest bias, R-G at small bias, diffusion at moderate bias, series resistance at the highest bias), each leaving a distinct slope signature. A linear plot hides all but the series resistance. The model parameters can be extracted window by window, as done in Problem 6.19.

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Problem 6.22 A pseudo-narrow-base diode

Statement. Consider the special Si $p^{+}$-$n$ step junction diode of Fig. P6.22: ${\tau }_p=\infty$ for $0\le x\le x_b$ and ${\tau }_p=0$ for $x_b\le x\le x_c$ (the n-side of the diode has a segment with no recombination followed by a segment with infinitely fast recombination; $x=0$ is the metallurgical junction). Excluding biases that cause high-level injection or breakdown, develop an expression for the room-temperature $I$-$V$ characteristic. Assume the depletion width $W$ never exceeds $x_b$ for all biases of interest.

Solution. The p${}^{+}$ side is ignored, and $x_n=W$ ($p^{+}$-$n$).

Setting up. In the region $x_n\le x\le x_b$ there is no recombination (${\tau }_p=\infty$): the diffusion equation is $D_P{d}^2\Delta p_n/d{x}^2=0$ with straight-line solutions. At $x=x_b$ the lifetime is zero, which is equivalent to an infinite recombination rate: the excess holes are annihilated immediately, so $\Delta p_n(x_b)=0$. (Equivalently, think of $x_b$ as an ideal ohmic contact.) With $x^{\prime }=x-x_n$ and the effective width $x_b-x_n=x_b-W$:

$$\frac{d^2\Delta p_n}{d{x}^{\prime 2}}=0,\Delta p_n(0)=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right),\Delta p_n(x_b-x_n)=0$$ $$\Delta p_n(x^{\prime })=\Delta p_n(0)\left(1-\frac{x^{\prime }}{x_b-x_n}\right)$$

Current:

$$I=A{J}_P(0)=-qA{D}_P{\frac{d\Delta p_n}{d{x}^{\prime }}|}_0=qA\frac{D_P}{x_b-x_n}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)$$

with

$$\boxed{I=\frac{qA{D}_P{n}_i^2}{N_D\left(x_b-x_n\right)}\left(e^{q{V}_A/kT}-1\right),x_n={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}}$$

This is the narrow-base limit of Eq. (6.68)/(6.69) with $x_c^{\prime }\to x_b-x_n$ and $\coth (\xi )\to 1/\xi$ (here exactly, because the lifetime is infinite so no recombination loss occurs). Under reverse bias $x_n$ grows, $x_b-x_n$ shrinks, and the current increases in magnitude with $|V_A|$ instead of saturating (and diverges at punch-through, $x_n\to x_b$, outside the stated validity range because $W<x_b$). Under forward bias, $x_n$ shrinks and the factor $1/(x_b-x_n)$ varies weakly, so the usual exponential dependence is retained. Room temperature enters via $D_P=(kT/q){\mu }_p$, $n_i=10^{10}{\text{cm}}^{-3}$ and $kT/q=0.0259$ V.

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Problem 6.23 Narrow base with a finite-lifetime lossy layer

Statement. Reconsider the diode of Fig. P6.22. Instead of ${\tau }_p=0$ in the $x_b\le x\le x_c$ region, take ${\tau }_p$ nonzero but still sufficiently small that $L_P\ll x_c-x_b$. Assuming $W$ never exceeds $x_b$ for all biases of interest, and excluding high-level injection and breakdown, derive an expression for the room-temperature $I$-$V$ characteristic. (As in 6.22, ${\tau }_p=\infty$ for $0\le x\le x_b$.)

Solution. Now the n-side is two layers:

• Region 1 ($x_n\le x\le x_b$, width $w=x_b-x_n$): ${\tau }_p=\infty$, no recombination: $\Delta p_n^{\prime \prime }=0$, so $\Delta p_n$ is linear and the hole current density is constant.
• Region 2 ($x\ge x_b$ with $L_P\ll x_c-x_b$): the layer is effectively semi-infinite: $\Delta p_n=\Delta p_n(x_b)e^{-(x-x_b)/L_P}$ with $L_P$ the (short) diffusion length in this layer, because the contact lies many diffusion lengths away. Its current at its left edge is

$$J_P(x_b)=q{D}_P\frac{\Delta p_n(x_b)}{L_P}$$

(From $J_P=-q{D}_{P}d\Delta p_n/dx$ with the exponential.) We assume both regions have the same $D_P$.

Matching at $x_b$: continuity of both $\Delta p_n$ and $J_P$. Region 1 is a straight line from $\Delta p_n(0)$ at $x_n$ to $\Delta p_n(x_b)\equiv {\Delta }_b$ at $x_b$:

$$J_P=q{D}_P\frac{\Delta p_n(0)-{\Delta }_b}{w}$$

Equate to region 2’s current:

$$\begin{array}{rl}\frac{\Delta p_n(0)-{\Delta }_b}{w}& =\frac{{\Delta }_b}{L_P}\\ \Delta p_n(0)& ={\Delta }_b\left(1+\frac{w}{L_P}\right)\\ {\Delta }_b& =\frac{\Delta p_n(0)}{1+w/L_P}\end{array}$$

So

$$J_P=q{D}_P\frac{{\Delta }_b}{L_P}=\frac{q{D}_P\Delta p_n(0)}{L_P+w}$$ $$\boxed{I=\frac{qA{D}_P{n}_i^2}{N_D\left(L_P+x_b-x_n\right)}\left(e^{q{V}_A/kT}-1\right),x_n={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}}$$

Checks: if $x_b-x_n\to 0$ the result reduces to the ideal wide-base result $qA{D}_P{n}_i^2/(N_D{L}_P)$; if $L_P\to 0$ it reduces to the 6.22 result. The diode behaves like a wide-base diode whose effective base is lengthened by the lossless layer of width $w=x_b-x_n$: the lossy layer’s recombination is the sink, and the no-loss layer acts as a series “transport resistance”.

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Problem 6.24 Denuded zone (multiple considerations)

Statement. Denuding reduces the R-G center concentration in the near-surface region of devices. A planar Si $p^{+}$-$n$ step junction at 300 K has $N_D=10^{16}{\text{cm}}^{-3}$. Denuding has created a reduced R-G center concentration $N_{T1}$ in the n-side region from $x=0$ to $x=x_c=2\mu \text{m}$ (Fig. P6.24); in the $x>x_c$ region the concentration is $N_{T2}=10N_{T1}$.

(a) Junction built-in voltage? (b) Diode breakdown voltage? (c) Applied voltage needed to expand the n-side depletion edge to $x=x_c$? (d) If the minority carrier lifetimes are ${\tau }_{p1}$ and ${\tau }_{p2}$ in the near-junction ($x<x_c$) and ($x>x_c$) regions, what is the ${\tau }_{p1}/{\tau }_{p2}$ ratio? (e) Establish an expression for the diffusion current $I_{\text{DIFF}}$ as a function of $V_A$. To simplify, assume $L_{P1}>x_c$ and the n-side contact is many $L_{P2}$ from the depletion edge. Sketch the expected $I_{\text{DIFF}}$ vs. $V_A$ when reverse biased. (f) Assuming ${\tau }_{01}\propto {\tau }_{p1}$ for $0\le x\le x_c$ and ${\tau }_{02}\propto {\tau }_{p2}$ for $x>x_c$, establish an expression or expressions for $I_{\text{R-G}}$ vs. $V_A$ when reverse biased more than a few $kT/q$. Sketch.

Solution.

(a) $p^{+}$-$n$: $V_{bi}\approx \frac{E_G}{2q}+\frac{kT}{q}\ln \frac{N_D}{n_i}=0.56+0.0259\ln 10^6=0.56+0.358=0.92$ V.

(b) Breakdown is set by the n-side doping (the denuding changes only R-G centers, not the doping): from Fig. 6.11 for $N_D=10^{16}$, $V_{BR}\approx 55$ V.

(c) $x_n=x_c$ when

$$V_A=V_{bi}-\frac{q{N}_D{x}_c^2}{2K_S{\epsilon }_0}=0.92-\frac{(1.6\times 10^{-19})(10^{16})(2\times 10^{-4}{)}^2}{2(11.8)(8.85\times 10^{-14})}=0.92-30.6=-29.7\text{V}$$

(the same computation as Exercise 6.10). Since $29.7<55$ V, this is reached before breakdown.

(d) The lifetime is inversely proportional to the R-G center concentration ($\tau =1/(c{N}_T)$ for the same type of centers): ${\tau }_{p1}/{\tau }_{p2}=N_{T2}/N_{T1}=10$. The denuded region has a ten times longer lifetime.

(e) Diffusion current. Take the n-side with $x^{\prime }=x-x_n$ and let $w\equiv x_c-x_n$ be the remaining width of the denuded layer (valid for $x_n\le x_c$, i.e. $V_A\ge -29.7$ V). Because $L_{P1}>x_c\ge w$, recombination in region 1 is negligible: $\Delta p_n^{\prime \prime }\approx 0$ (linear profile, constant $J_P$). In region 2 ($x>x_c$, contact far away) $\Delta p_n={\Delta }_c{e}^{-(x-x_c)/L_{P2}}$, $J_P(x_c)=q{D}_P{\Delta }_c/L_{P2}$. This is exactly the problem solved in 6.23 with $L_P\to L_{P2}$ and $w=x_c-x_n$:

$$\boxed{I_{\text{DIFF}}=\frac{qA{D}_P{n}_i^2}{N_D\left(L_{P2}+x_c-x_n\right)}\left(e^{q{V}_A/kT}-1\right),V_A\ge -29.7\text{V}}$$

with $x_n={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}$ and $L_{P2}=\sqrt{D_P{\tau }_{p2}}=L_{P1}/\sqrt{10}$. For reverse bias beyond $-29.7$ V, the depletion edge has moved into region 2 ($x_n>x_c$); the n-side quasineutral region is now entirely region 2 with a wide base, and

$$I_{\text{DIFF}}\to -\frac{qA{D}_P{n}_i^2}{N_D{L}_{P2}}\equiv -I_0^{(2)}(V_A\le -29.7\text{V})$$

Sketch (reverse bias). Starting at zero, the magnitude of $I_{\text{DIFF}}$ at $V_A=0^{-}$ is $\frac{qA{D}_P{n}_i^2}{N_D(L_{P2}+x_c-x_n(0))}$ ($\ll$ saturation scale), then it increases gradually (concave up) as $x_n\to x_c$ because the denominator shrinks from $L_{P2}+w(0)$ to $L_{P2}$, and for $V_A<-29.7$ V it is constant at $-I_0^{(2)}$ (saturated) until breakdown at $-55$ V. The total variation is modest: the magnitude goes from $1/(L_{P2}+w(0))$ to $1/L_{P2}$. (Numerically, $w(0)=x_c-x_n(0)=2-\sqrt{2(11.8)(8.85\times 10^{-14})(0.92)/1.6\times 10^{-3}}\mu \text{m}=2-0.35=1.65\mu \text{m}$; the ratio is $1+1.65\mu \text{m}/L_{P2}$, which is $\approx 1$ if $L_{P2}\gg 1.65\mu \text{m}$, as is typical, $L_{P2}\sim 10$-$30\mu \text{m}$.)

(f) R-G current, reverse biased. For $V_A$ beyond a few $kT/q$ reverse, $|I_{\text{R-G}}|=qA{n}_i\int dx/(2{\tau }_0(x))$ over the depletion region (Problem 6.17 approach). Since ${\tau }_{01}\propto {\tau }_{p1}$ and ${\tau }_{02}\propto {\tau }_{p2}$, ${\tau }_{01}=10{\tau }_{02}$. The depletion region lies on the n-side (width $W=x_n$):

Case 1: $x_n\le x_c$ ($0\ge V_A\ge -29.7$ V):

$$I_{\text{R-G}}=-\frac{qA{n}_i{x}_n}{2{\tau }_{01}}=-\frac{qA{n}_i}{2{\tau }_{01}}{\left[\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)\right]}^{1/2}$$

Case 2: $x_n>x_c$ ($V_A<-29.7$ V):

$$I_{\text{R-G}}=-qA{n}_i\left[\frac{x_c}{2{\tau }_{01}}+\frac{x_n-x_c}{2{\tau }_{02}}\right]=-\frac{qA{n}_i}{2{\tau }_{02}}\left[\frac{x_c}{10}+\left(x_n-x_c\right)\right]$$

Sketch. $|I_{\text{R-G}}|$ vs. $|V_A|$: $\propto \sqrt{V_{bi}+|V_A|}$ up to $|V_A|=29.7$ V (where $|I_{\text{R-G}}|=qA{n}_i{x}_c/2{\tau }_{01}$), then a kink: beyond 29.7 V the current continues to increase as $\sqrt{V_{bi}+|V_A|}$, but with a tenfold larger coefficient for the added width (the incremental generation per unit width is ten times larger in the not-denuded layer), until avalanche breakdown at 55 V (at which $x_n=2.7\mu \text{m}$). Between $x_c=2.0\mu \text{m}$ and $2.7\mu \text{m}$ the added current is $(qA{n}_i/2{\tau }_{02})(0.7\mu \text{m})$, i.e. the current at 55 V is $[0.2+0.7]/[0.2]=4.5\times$ larger than the current at 29.7 V.

---

Appendix: Quick Reference for Chapter 6

Quantity	Expression
Ideal diode equation	$I=I_0\left(e^{q{V}_A/kT}-1\right)$
Saturation current	$I_0=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)$
Law of the junction	$np=n_i^2{e}^{q{V}_A/kT}$ in the depletion region
Edge boundary conditions	$\Delta n_p(-x_p)=\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)$, $\Delta p_n(x_n)=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$
Minority profile (wide base)	$\Delta p_n(x^{\prime })=\Delta p_n(0)e^{-x^{\prime }/L_P}$
Multiplication factor	$M=\left[1-(
Reverse R-G current	$I_{\text{R-G}}=-\frac{qA{n}_{i}W}{2{\tau }_0}$
Forward R-G	$\propto e^{q{V}_A/2kT}$
Series resistance	$V_J=V_A-I{R}_S$, $I=I_0{e}^{q{V}_J/kT}$
Narrow-base	$I_0^{\prime }=I_0\coth (x_c^{\prime }/L_P)\to I_0{L}_P/x_c^{\prime }$ for $x_c^{\prime }\ll L_P$
Charge control	$d{Q}_P/dt=i_{\text{DIFF}}-Q_P/{\tau }_p$; steady state $I_{\text{DIFF}}=Q_P/{\tau }_p$

Chapter 7: pn Junction Diode: Small-Signal Admittance

Note on the source: the OCR of this chapter is noisy. Where a symbol or exponent was lost I restored it from physics and cross-checks, and I say so when it matters. For example, the area of the 1N5472A is $A=3.72\times 10^{-3}\mathrm{c}{\mathrm{m}}^2$, because that is the value that reproduces the book’s quoted $N_A=8.84\times 10^{15}\mathrm{c}{\mathrm{m}}^{-3}$.

7.1 Introduction

Picture a diode sitting quietly at some DC operating point. Now tap it very gently: add a tiny sinusoidal voltage $v_a=V_0\sin \omega t$ on top of the DC bias $V_A$, and listen for the current that comes back, $i$. This “tap and listen” experiment is the small-signal response. Because the tap is small (tens of millivolts or less), the diode answers linearly, and the whole answer is packaged in one complex number, the small-signal admittance

$Y=\frac{i}{v_a}.$

The junction region responds in two ways. It can store charge (a capacitor) and it can pass current (a conductor). So in general

\boxed{Y=G+j\omega C}\tag{7.1}

with $j=\sqrt{-1}$ and $\omega$ the angular frequency in rad/s.

Equivalent circuit. The circuit has a conductance $G$ in parallel with a capacitance $C$ (the junction admittance), in series with the series resistance $R_S$ of the contacts and quasineutral regions (introduced in Chapter 6). $C$ and $G$ both depend on the DC bias, which is why the book draws arrows through them. $R_S$ is normally tiny compared with the junction impedance except at large forward bias, so from here on we assume $R_S\approx 0$ unless stated.

Plan of attack.

1. Reverse bias: the conductance is tiny, so $Y\approx j\omega C$. Only majority carriers wiggle.
2. Forward bias: conductance matters, and minority carriers join the dance, producing the diffusion admittance.

Analogy: a diode is like a drum. Under reverse bias you tap the drum skin (the depletion edges) and it flexes like a spring (capacitance). Under forward bias there is also a leaky pipe through it (conductance), and water sloshing in the pipe (minority carriers) lags behind your tapping when you tap fast.

---

7.2 Reverse-Bias Junction Capacitance

7.2.1 General Information

Under reverse bias the pn diode is functionally a capacitor. Many “capacitors” inside integrated circuits are really reverse-biased pn junctions. It differs from an ordinary capacitor in one striking way: its capacitance decreases monotonically as the reverse bias increases (the sample data in Fig. 7.3, from a 1N5472A abrupt-junction diode, fall from about 120 pF at 0 V to a few pF near $-25$ V).

Thought experiment: what happens during one a.c. cycle? The total junction voltage is $V_A+v_a$.

• Positive half cycle ($v_a>0$): the reverse bias is slightly reduced, the depletion width shrinks a little, so some charge is uncovered then removed from each side. A few majority carriers rush back toward the junction and neutralize edge charge.
• Negative half cycle ($v_a<0$): reverse bias increases, the depletion region widens, and more ionized dopants are uncovered on both sides.

So the depletion width $W$ breathes in and out around its DC value, and a charge density oscillation $\Delta \rho$ appears. Because $v_a$ is tiny, the $\Delta \rho$ is effectively a thin sheet of $+$ charge at one edge of the depletion region and a thin sheet of $-$ charge at the other, a distance $W$ apart. That is exactly a parallel-plate capacitor:

\boxed{C_J=\frac{K_S\varepsilon_0A}{W}}\tag{7.2}

The subscript $J$ means junction (or depletion-layer) capacitance. Under reverse bias $C=C_J$ since nothing else oscillates significantly. Since $W$ grows with reverse bias, $C_J\propto 1/W$ falls, matching the experiment.

(Figure 7.4 shows (a) the depletion edge moving, (b) the step-junction charge density $\pm qN$ profile with edges shifting, (c) $\Delta \rho$ for $v_a>0$, (d) $\Delta \rho$ for $v_a<0$; the $\Delta \rho$ plots are the shifted profile minus the DC profile.)

Quasistatic following. For the oscillation to happen, majority carriers must flow in and out of the quasineutral regions in sync with $v_a$, responding as if $v_a$ were a DC bias. This is called following the signal quasistatically. Majority carrier response times in Si are about $10^{-12}$ s or less (the dielectric relaxation time), so the quasistatic assumption is valid up to very high frequencies.

7.2.2 C-V Relationships

To get the exact $C$ versus $V_A$ we substitute the appropriate $W(V_A)$ into (7.2). From Chapter 5:

W=\left[\frac{2K_S\varepsilon_0}{qN_B}(V_{bi}-V_A)\right]^{1/2}\quad\text{(asymmetrical step)}\tag{7.3}

W=\left[\frac{12K_S\varepsilon_0}{qa}(V_{bi}-V_A)\right]^{1/3}\quad\text{(linearly graded)}\tag{7.4}

where $N_B$ is the doping on the lightly doped side and $a$ is the grading constant. Rather than treat each separately, we generalize to a one-sided power-law profile: one side ($x<0$) heavily doped and the light side

N_B(x)=b\,x^{m},\qquad x>0\tag{7.5}

with $b>0$ and $m$ constants. Special cases: $m=0$ (step, $b=N_B$), $m=1$ (one-sided linearly graded), and $m<0$ (hyperabrupt: doping falls as you leave the junction; made by ion implantation or epitaxy). Fig. 7.5 shows (a) $m=1$, a straight line rising from the origin, (b) $m=0$, a flat line, (c) $m=-1$, a hyperbola falling away.

For $m>-2$ the depletion width is

\boxed{W=\left[\frac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)\right]^{1/(m+2)}}\tag{7.6}

(This derivation is Problem 7.2; I derive it fully there.) Check: $m=0,b=N_B$ gives (7.3); $m=1$ with $b=a/4$ gives (7.4) (the "$a/4$" accounts for the two-sided linear profile in the book’s (7.4) versus the one-sided one here).

Substituting (7.6) into (7.2):

\boxed{C_J=\frac{K_S\varepsilon_0A}{\left[\dfrac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)\right]^{1/(m+2)}}}\tag{7.7}

Define the zero-bias capacitance

C_{J0}=C_J\big|_{V_A=0}=\frac{K_S\varepsilon_0A}{\left[\dfrac{(m+2)K_S\varepsilon_0V_{bi}}{qb}\right]^{1/(m+2)}}\tag{7.8}

Dividing (7.7) by (7.8) the constants cancel:

\boxed{C_J=\frac{C_{J0}}{\left(1-\dfrac{V_A}{V_{bi}}\right)^{1/(m+2)}}}\tag{7.9}

Varactors. A diode built to exploit (7.7)/(7.9) is a varactor (“variable reactor”; reactance $=1/j\omega C$). Uses: parametric amplification, harmonic generation, mixing, detection, voltage-variable tuning (think of the tuning knob of a radio being replaced by a voltage). The figure of merit is the tuning ratio

\mathrm{TR}=\frac{C_J(V_{A1})}{C_J(V_{A2})}=\left[\frac{V_{bi}-V_{A2}}{V_{bi}-V_{A1}}\right]^{1/(m+2)}\;\xrightarrow{\;|V_A|\gg V_{bi}\;}\;\left(\frac{V_{A2}}{V_{A1}}\right)^{1/(m+2)}\tag{7.10}

The exponent $1/(m+2)$ grows as $m$ becomes smaller, so TR increases going from linear ($m=1$, exponent $1/3$) to step ($m=0$, $1/2$) to hyperabrupt ($m=-1$, exponent $1$). That is why hyperabrupt varactors are commercially valuable: the capacitance changes most per volt.

Exercise 7.1: Normalized C-V curves

Problem. Compute and plot $C_J/C_{J0}$ versus $V_A/V_{bi}$ for the linearly graded, step, and $m=-1$ hyperabrupt diodes, for $-25\le V_A/V_{bi}\le 0$, and comment.

Solution logic. Equation (7.9) in normalized form is $y=(1-x{)}^{-s}$ with $x=V_A/V_{bi}$ and $s=1/(m+2)$, so for $m=1,0,-1$ we have $s=1/3,1/2,1$. These are “universal” curves: any one-sided power-law diode collapses onto one of them once axes are normalized. The code builds a matrix with one row per $m$ and plots all rows.

% Exercise 7.1 -- Normalized C-V curves
clear
m = [1 0 -1];            % linear, step, hyperabrupt
s = 1./(m+2);            % exponents 1/(m+2)
x = linspace(-25,0);     % x = VA/Vbi
y = [];                  % y = CJ/CJ0
for i = 1:3
    y = [y; 1./(1-x).^s(i)];
end
close
plot(x,y,'-'); grid
axis([-25 0 0 1])
xlabel('V_A/V_{bi}'); ylabel('C_J/C_{J0}')
text(-20,.42,'linear (m=1)')
text(-20,.27,'step (m=0)')
text(-20,.10,'hyperabrupt (m=-1)')

(The OCR lost the “1” in the vector m and the dot operators; the version above is the corrected one.)

Results and comments. At $V_A/V_{bi}=-25$ the curves reach about $26^{-1/3}=0.34$, $26^{-1/2}=0.20$ and $26^{-1}=0.038$. The step-junction curve matches the 1N5472A data of Fig. 7.3 well if $V_{bi}\approx 1$ V. The tuning ratio with $V_{A1}=0$ and any $V_{A2}$ is largest for the hyperabrupt diode and smallest for the linearly graded diode, as predicted by (7.10).

7.2.3 Parameter Extraction and Profiling

C-V data are used routinely to extract the average doping or the whole doping profile on the lightly doped side. Automated systems exist. The author’s lab uses an MSI C-V meter with a 15 mV rms, 1 MHz probing signal, four capacitance ranges (2, 20, 200, 2000 pF), and an internal DC supply with ranges $\pm 9.999$ V (1 mV steps) and $\pm 99.99$ V (10 mV steps). A PC controls bias and data; a probe box with a heated chuck (832 T-controller) holds wafers; encapsulated devices go into a spring-clip adapter. The meter measures and removes stray capacitance of cables, probe box and adapter. (Fig. 7.6 is a schematic of this system.)

Step-junction analysis. For an abrupt junction set $m\to 0,b\to N_B$ in (7.7). Invert and square:

\frac{1}{C_J^2}=\frac{2(V_{bi}-V_A)}{qN_BK_S\varepsilon_0A^2}\tag{7.11}

Step 1. $1/C_J^2$ is linear in $V_A$. Step 2. The slope is $-\frac{2}{q{N}_B{K}_S{\epsilon }_0{A}^2}$, so a known area gives $N_B$. Step 3. The intercept on the $V_A$ axis ($1/C_J^2=0$) gives $V_{bi}$. A straight $1/C^2$ line also confirms that the junction is abrupt.

Exercise 7.2: Analysis of 1N5472A data

Problem. The 1N5472A is an $n^{+}$-$p$ abrupt diode, $A=3.72\times 10^{-3}\mathrm{c}{\mathrm{m}}^2$, with about 2 pF stray capacitance in parallel with the junction. Using the Fig. 7.3 data, apply the $1/C^2$ plot to confirm abruptness, find the $p$-side doping and quote $V_{bi}$.

Solution.

1. Subtract the 2 pF stray capacitance from every measured value to get $C_J$.
2. Plot $1/C_J^2$ versus $V_A$ (Fig. E7.2): the points fall on an almost perfect line, so the junction is abrupt.
3. Least squares gives

$\frac{1}{C_J^2}=(6.89\times 10^{19})-(9.78\times 10^{19})V_A(C_J\text{ in F},V_A\text{ in V})$

4. Compare with (7.11): slope magnitude $=2/(q{N}_A{K}_S{\epsilon }_0{A}^2)$, so

N_A&=\frac{2}{q\,K_S\varepsilon_0\,A^2\,(9.78\times10^{19})}\\ &=\frac{2}{(1.602\times10^{-19})(11.8\times8.854\times10^{-14})(3.72\times10^{-3})^2(9.78\times10^{19})}\\ &=8.84\times10^{15}\ \mathrm{cm^{-3}} \end{aligned}$$ 5. The intercept over the slope magnitude gives the built-in voltage: $$V_{bi}=\frac{6.89\times10^{19}}{9.78\times10^{19}}=0.70\ \mathrm{V}$$ **Remark.** This $V_{bi}$ is lower than you would compute from the doping ($\approx0.9$ V for $N_A\approx9\times10^{15}$ against a heavily doped $n^+$ side). The intercept is an *extrapolation* and is sensitive to doping variations right near the metallurgical junction, so $V_{bi}$ from C-V data is not trustworthy, while the slope-derived $N_B$ is. **Profiling.** If the profile is not known, we do not need to assume one. Without deriving here (the derivation is Problem 7.5): $$N_B(x)=\frac{2}{qK_S\varepsilon_0A^2\left|\dfrac{d(1/C^2)}{dV_A}\right|}\tag{7.12}$$ $$x=\frac{K_S\varepsilon_0A}{C}\tag{7.13}$$ with $x$ measured from the metallurgical junction into the light side. Putting the step result (7.11) into (7.12) returns a position-independent $N_B$, as required. The process is called **profiling**. It fails for rapidly varying profiles, only a limited depth can be scanned, and the derivative makes the result noisy, but it is simple and widely used. Fig. 7.7 shows an automated profile of a hyperabrupt tuning diode (C809): doping on a log scale (about $10^{15}$ to $10^{17}\ \mathrm{cm^{-3}}$) falling with distance over about 0 to 3 $\mu$m. ### 7.2.4 Reverse-Bias Conductance Every real capacitor leaks, and so does the diode. The differential conductance is the slope of the I-V curve. If the diode follows the signal quasistatically, $\Delta I/\Delta V=dI/dV$, so the low-frequency conductance is $$G_0=\frac{dI}{dV_A}\tag{7.14}$$ **Ideal diode.** With $I=I_0\left(e^{qV_A/kT}-1\right)$: $$G_0=\frac{q}{kT}I_0e^{qV_A/kT}=\frac{q}{kT}(I+I_0)\tag{7.15}$$ For reverse bias beyond a few $kT/q$, $I\to-I_0$ so $G_0\to0$ (the I-V curve is flat). **If R-G current dominates** the reverse current is $I\approx-qAn_iW/2\tau_0$. Using (7.6), $W^{m+2}\propto(V_{bi}-V_A)$, so $$\frac{dW}{dV_A}=-\frac{W}{(m+2)(V_{bi}-V_A)}$$ and therefore $$G_0=\frac{d}{dV_A}\left(-\frac{qAn_iW}{2\tau_0}\right)=\frac{qAn_iW}{2\tau_0(m+2)(V_{bi}-V_A)}\tag{7.16}$$ There is a residual conductance at all reverse biases; its voltage dependence depends on the doping profile. Whatever the case, (7.14) applied to measured I-V data always gives $G_0$. ### Exercise 7.3: Is the reverse-biased diode a pure capacitor? **Problem.** The diode of Fig. 6.10(b) has $C_J=63$ pF at $V_A=-10$ V, and $R_S\approx1\ \Omega$ (Exercise 6.8). At $V_A=-10$ V and $f=100$ kHz, confirm $R_S$ and $G=G_0$ may be neglected. **Solution.** $$G_0=\left.\frac{dI}{dV_A}\right|_{-10\,\mathrm V}=\frac{40\ \mathrm{pA}}{18\ \mathrm V}=2.22\times10^{-12}\ \mathrm S$$ $$\omega C_J=2\pi(10^5)(6.3\times10^{-11})=3.96\times10^{-5}\ \mathrm S$$ Since $\omega C_J\gg G_0$ (ratio $\sim10^7$), $Y\approx j\omega C_J$. For $R_S$ in series with $Y$ compare $|Z|$ with $R_S$: $$\frac{1}{\omega C_J}=2.52\times10^{4}\ \Omega\gg R_S\le1\ \Omega$$ So the diode behaves as an essentially pure capacitor. --- ## 7.3 Forward-Bias Diffusion Admittance ### 7.3.1 General Information Capacitance always comes from charge that oscillates in step (or nearly in step) with the signal. The junction capacitance $C_J$ comes from majority carriers moving in and out at the depletion edges, and its formulas hold unchanged under forward bias. What is new in forward bias is that large numbers of **excess minority carriers** are stored in the quasineutral regions beside the depletion region, and these also oscillate. (Under reverse bias there are so few minority carriers that their contribution is negligible.) Picture Fig. 7.8(a): on a log scale the excess minority profiles $p_n(x)$ and $n_p(x)$ decay away from the depletion edges $x_n,\,-x_p$. When $V_A\to V_A+v_a$ the profile heaves up and down. If the carriers can follow quasistatically, they slide in unison between two straight lines (on the log plot). But minority carriers are supplied and removed far more slowly than majority carriers: their natural timescale is the lifetime $\tau$. When $\omega\tau\to1$ they fall out of phase with the signal, giving an undulating, out-of-phase profile. Out-of-phase charge acts like a resistor (dissipates energy), so **conductance goes up and capacitance goes down** with frequency. > Analogy: pushing a child on a swing. Push slowly and the swing moves in step with you (energy stored, like capacitance). Push too fast for it and it lags, so you are just fighting it (dissipation, like conductance). Because the buildup is caused by the diffusion current, the associated admittance is the **diffusion admittance** $$Y_D=G_D+j\omega C_D\tag{7.17}$$ The total forward-bias admittance is $C_J$ in parallel with $Y_D$ (Fig. 7.8(b): $C_J \parallel C_D \parallel G_D$). **Measurement.** Forward-bias measurement is harder: the large DC current overloads most C-V meters (a few tenths of a volt on Si often gives false readings). The HP 4284A LCR meter with option 001 has an isolation circuit tolerating DC currents up to 0.1 A; it covers 20 Hz to 1 MHz, 5 mV to 20 V rms signal, $\pm40$ V bias, and detects 0.01 fF to 10 F, performing single-point measurements or stepping through ten operator-set values of frequency, bias or signal level. ### 7.3.2 Admittance Relationships The brute-force route repeats the ideal-diode derivation with each DC quantity replaced by DC plus a.c. A cleverer route reuses the DC result. We take a $p^+$-$n$ diode. **Step 1: the equation.** With an a.c. signal on top of DC, the $n$-side excess hole density obeys $$\frac{\partial\Delta p_n(x,t)}{\partial t}=D_P\frac{\partial^2\Delta p_n(x,t)}{\partial x^2}-\frac{\Delta p_n(x,t)}{\tau_p}\tag{7.18}$$ **Step 2: split into DC plus phasor.** $$\Delta p_n(x,t)=\Delta p_n(x)+\tilde p_n(x,\omega)e^{j\omega t}\tag{7.19}$$ ($\Delta p_n(x)$ is the time-invariant part, $\tilde p_n$ the a.c. amplitude.) **Step 3: substitute.** Working out the time derivative: $$j\omega\tilde p_ne^{j\omega t}=D_P\frac{d^2\Delta p_n}{dx^2}+D_P\frac{d^2\tilde p_n}{dx^2}e^{j\omega t}-\frac{\Delta p_n}{\tau_p}-\frac{\tilde p_n}{\tau_p}e^{j\omega t}\tag{7.20}$$ **Step 4: DC and a.c. parts must balance separately.** $$0=D_P\frac{d^2\Delta p_n}{dx^2}-\frac{\Delta p_n}{\tau_p}\tag{7.21a}$$ $$0=D_P\frac{d^2\tilde p_n}{dx^2}-\frac{\tilde p_n}{\tau_p/(1+j\omega\tau_p)}\tag{7.21b}$$ The second follows because $j\omega\tilde p_n+\tilde p_n/\tau_p=\tilde p_n(1+j\omega\tau_p)/\tau_p$. Equation (7.21a) is the usual steady-state diffusion equation; (7.21b) is the **same equation with $\tau_p$ replaced by $\tau_p/(1+j\omega\tau_p)$**. This is the heart of the trick: a complex effective lifetime. **Step 5: boundary conditions.** At $x=\infty$ both vanish. At $x=x_n$ the law of the junction gives $$\Delta p_n(x_n)+\tilde p_n(x_n)=\frac{n_i^2}{N_D}\left(e^{q(V_A+v_a)/kT}-1\right)\tag{7.22}$$ The DC part is $$\Delta p_n(x_n)=\frac{n_i^2}{N_D}\left(e^{qV_A/kT}-1\right)\tag{7.23}$$ Subtracting, and using $e^{qv_a/kT}-1\approx qv_a/kT$ for $v_a\ll kT/q$: $$\tilde p_n(x_n)=\frac{n_i^2}{N_D}e^{qV_A/kT}\left(e^{qv_a/kT}-1\right)\tag{7.24a}$$ $$\approx\frac{n_i^2}{N_D}e^{qV_A/kT}\frac{qv_a}{kT}\tag{7.24b}$$ Here $v_a$ is the amplitude of the a.c. signal. **Step 6: reuse the solution.** The a.c. problem equals the DC problem with two substitutions: $\tau_p\to\tau_p/(1+j\omega\tau_p)$ and $\left[e^{qV_A/kT}-1\right]\to\dfrac{qv_a}{kT}e^{qV_A/kT}$. The DC diffusion current is $$I_{\text{DIFF}}=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}\left(e^{qV_A/kT}-1\right)=qA\sqrt{\frac{D_P}{\tau_p}}\frac{n_i^2}{N_D}\left(e^{qV_A/kT}-1\right)\tag{7.25}$$ so with the substitution $\sqrt{1/\tau_p}\to\sqrt{(1+j\omega\tau_p)/\tau_p}$: $$i_{\text{diff}}=qA\sqrt{\frac{D_P}{\tau_p}}\sqrt{1+j\omega\tau_p}\;\frac{n_i^2}{N_D}\frac{q}{kT}\,e^{qV_A/kT}\,v_a\tag{7.26}$$ The ideal-diode low-frequency conductance (7.15) for a $p^+$-$n$ diode is $G_0=\dfrac{q}{kT}qA\sqrt{\dfrac{D_P}{\tau_p}}\dfrac{n_i^2}{N_D}e^{qV_A/kT}$ (forward bias beyond a few $kT/q$). Hence $$i_{\text{diff}}=G_0\sqrt{1+j\omega\tau_p}\;v_a\tag{7.27}$$ $$\boxed{Y_D=\frac{i_{\text{diff}}}{v_a}=G_0\sqrt{1+j\omega\tau_p}}\qquad(p^+\text{-}n)\tag{7.28}$$ For an $n^+$-$p$ diode replace $\tau_p\to\tau_n$. For a two-sided junction split $G_0$ into $n$ and $p$ parts and multiply each by its own $\sqrt{1+j\omega\tau}$. **Real and imaginary parts.** Write $\sqrt{1+jx}=a+jb$ with $x=\omega\tau_p$. Squaring: $$\begin{aligned} a^2-b^2&=1\\ 2ab&=x \end{aligned}$$ Taking the modulus of both sides of the squared equation: $a^2+b^2=\left|1+jx\right|=\sqrt{1+x^2}$. Adding and subtracting with $a^2-b^2=1$: $$a^2=\frac{\sqrt{1+x^2}+1}{2},\qquad b^2=\frac{\sqrt{1+x^2}-1}{2}$$ Then $G_D=G_0a$ and $\omega C_D=G_0b$: $$\boxed{G_D=\frac{G_0}{\sqrt2}\left[\sqrt{1+\omega^2\tau_p^2}+1\right]^{1/2}}\tag{7.29a}$$ $$\boxed{C_D=\frac{G_0}{\omega\sqrt2}\left[\sqrt{1+\omega^2\tau_p^2}-1\right]^{1/2}}\tag{7.29b}$$ $G_D$ and $C_D$ depend on both DC bias (through $G_0\propto e^{qV_A/kT}$, very steeply) and frequency. At small forward bias $C_J$ dominates the capacitance; with increasing bias $C_D$ overtakes and eventually swamps $C_J$. **Low-frequency limit** ($\omega\tau_p\ll1$): $\sqrt{1+x^2}\approx1+x^2/2$, so $a\to1$, and $b^2\approx x^2/4$, i.e. $b\approx x/2$: $$G_D\to G_0\tag{7.30a}$$ $$C_D\to\frac{G_0\tau_p}{2}\tag{7.30b}$$ Frequency-independent behaviour holds for $\omega\tau_p\lesssim0.1$. For $\tau_p=1\ \mu$s this means $f\le1/(20\pi\tau_p)\approx16$ kHz. For $\omega\tau_p\gg1$, $G_D$ increases (as $G_0\sqrt{\omega\tau_p/2}$) while $C_D$ decreases (as $\omega^{-1/2}$). Fig. 7.10 plots $G_D/G_0$ and $C_D/C_{D0}$ (with $C_{D0}=G_0\tau/2$) against $\omega\tau$ from 0.01 to 100 on log axes: $G_D$ flat near 1 then rising to about 7 at $\omega\tau=100$; $C_D$ flat near 1 then falling to about 0.14. ### Exercise 7.4: Forward-bias capacitance of the 1N5472A **Problem.** The 1N5472A ($n^+$-$p$, abrupt) has $C_{J0}=120$ pF, $V_{bi}=0.7$ V, $I_0=8\times10^{-13}$ A, $n_I=1.22$, $\tau_n=5\times10^{-7}$ s. (a) Take $\omega\tau_n=0.01$ but use relations valid for arbitrary $\omega\tau_n$; compute and plot $C_J$, $C_D$, $C_J+C_D$ versus $V_A$ for $0\le V_A\le0.65$ V; note where $C_D=C_J$. (b) Repeat for $\omega\tau_n=0.1,1,10,100$ and comment. **Solution logic.** Relations used: $$C_J=\frac{C_{J0}}{\sqrt{1-V_A/V_{bi}}}\ (\text{(7.9), }m=0),\qquad G_0=\frac{qI_0}{kT}e^{qV_A/(n_IkT)},$$ $$C_D=\frac{\tau_nG_0}{\sqrt2\,(\omega\tau_n)}\left[\sqrt{1+(\omega\tau_n)^2}-1\right]^{1/2}$$ The last is (7.29b) rewritten with $\omega=(\omega\tau_n)/\tau_n$. The ideality factor $n_I$ is inserted in the exponent to account for the real diode being non-ideal. ```matlab % Exercise 7.4 -- Forward-bias capacitance clear CJ0 = 120e-12; % farads Vbi = 0.7; % volts Vth = 0.0259; % kT/q in volts taun = 5.0e-7; % seconds I0 = 8.0e-13; % amps n1 = 1.22; % ideality factor wt = input('input the angular-frequency*lifetime product: '); VA = linspace(0,0.65); CJ = CJ0./sqrt(1-VA./Vbi); % junction capacitance G0 = I0/Vth*exp(VA./(n1*Vth)); % low-frequency conductance CD = taun.*G0./(sqrt(2)*wt).*sqrt(sqrt(1+wt^2)-1); % diffusion capacitance % measured forward-bias data (HP4284A) VAm = [0.1 0.2 0.3 0.4 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58]; CDm = [1.31e-10 1.43e-10 1.61e-10 1.88e-10 1.97e-10 2.08e-10 ... 2.23e-10 2.46e-10 2.76e-10 3.46e-10 4.40e-10 6.54e-10 9.38e-10]; close semilogy(VA,CJ,'--r'); axis([0 0.7 1.0e-10 1.0e-9]); grid hold on; semilogy(VA,CD,'-g'); semilogy(VA,CJ+CD,'-y') semilogy(VAm,CDm-2e-12,'o') % subtract 2 pF stray xlabel('V_A (volts)'); ylabel('C (farads)') hold off ``` (The OCR dropped several line-style quotes and minus signs; the above is the repaired logic.) **Results (computed numerically).** (a) For $\omega\tau_n=0.01$ the crossover $C_D=C_J$ occurs at $V_A\approx0.547$ V (the book quotes $\approx0.545$ V). $C_J$ dominates at low forward bias, $C_D$ rises exponentially, and $C_J+C_D$ matches the measured forward C-V data. (b) The same result holds for all $\omega\tau_n<1$ ($\omega\tau_n=0.1$ gives 0.5473 V, $\omega\tau_n=1$ gives 0.5506 V). As $\omega\tau_n$ increases beyond 1, $C_D$ drops at every bias and the crossover moves up: $V_A\approx0.578$ V for $\omega\tau_n=10$, and $0.619$ V for $\omega\tau_n=100$ (book: 0.575 V and 0.62 V). In words: at high frequency the minority carriers cannot keep up, so the diffusion capacitance is suppressed and the junction capacitance wins over a wider range of bias. --- ## 7.4 Summary The chapter modelled the small-signal response of a pn diode in two parts. Under reverse bias the diode is functionally a capacitor, computed by the parallel-plate formula (7.2), but its capacitance falls monotonically with reverse bias. It arises from majority carriers moving in and out about the steady-state depletion width. Reverse-biased diodes serve as capacitors and varactors; capacitance measurements are widely used to extract average doping or doping profiles on the lightly doped side. Under forward bias a large minority-carrier buildup forms next to the depletion region, and its oscillation gives an extra admittance, the diffusion admittance, derived by modifying the ideal diode equation. Its components are strong functions of DC bias, eventually dominating the admittance. For $\omega\tau\gtrsim1$ the minority carriers have trouble following, and the out-of-phase oscillations enhance diffusion conductance at the expense of diffusion capacitance. --- ## Problems, Chapter 7 ### Problem 7.1 (Quick quiz) **(a) Physical origin of the junction capacitance?** The in-and-out movement of majority carriers at the edges of the depletion region (width oscillation) as the a.c. signal alters the reverse bias; this is a charge oscillation like that on parallel plates a distance $W$ apart. **(b) Sketch the $m=-1$ hyperabrupt profile of an $n^+$-$p$ junction.** On the $p$ side ($x>0$): $N_A(x)=b/x$. Sketch: vertical axis $N_A$, horizontal axis $x$. A hyperbola, very large near the junction and falling steadily as $x$ increases; on the $n^+$ side ($x<0$) a constant very high $N_D$ plateau. **(c) Quasistatically:** the carriers respond to the a.c. signal as if it were a DC bias, i.e. their distribution at each instant is the steady-state one for the instantaneous voltage. **(d) Varactor:** a pn diode manufactured to exploit the voltage-variable junction capacitance (variable reactor). **(e) Profiling:** determining the doping concentration as a function of position on the light side from C-V data, using (7.12) and (7.13). **(f) Low-frequency conductance of an ideal diode:** $G_0=\dfrac{q}{kT}(I+I_0)=\dfrac{q}{kT}I_0e^{qV_A/kT}$. Sketch: $G_0$ is $qI_0/kT$ at $V_A=0$, rises exponentially (a straight line on a log axis) for forward bias, and falls smoothly toward zero within a few $kT/q$ of reverse bias, but never goes negative. *Caution to prevent misinterpretation:* $G_0$ is the **slope** of the $I$-$V$ curve, not the current; it is always positive and the reverse part flattens to $0^+$, not to $-I_0$. The sketch is for an ideal diode only; a real diode with R-G current keeps a residual reverse conductance (7.16). **(g) Physical origin of the diffusion admittance:** the oscillation of the excess minority carrier charge stored next to the depletion region edges in response to the a.c. signal. **(h) Why does $G_D$ increase with $\omega\tau_p$ when $\omega\tau_p\ge1$?** The minority carriers cannot follow the signal in step, so the charge oscillation is out of phase with $v_a$. The out-of-phase component of the current is in phase with the voltage, i.e. conductive; hence conductance rises and capacitance falls. --- ### Problem 7.2 (Derive Eq. 7.6) **Problem.** For the one-sided power-law profile (7.5), $N_B(x)=bx^m$, derive $W$ versus $V_A$ for a $p^+$-$n$ junction ($m>-2$). **Solution.** *Assumptions:* depletion approximation; the $p^+$ side is so heavily doped that its depletion layer is negligible; the depletion region on the $n$ side spans $0<x<W$; the space-charge density there is $\rho=qN_B(x)$ (donors). 1. Poisson's equation with $E=0$ at $x=W$ (the field vanishes outside): $$\frac{dE}{dx}=\frac{qN_B(x)}{K_S\varepsilon_0}$$ 2. Integrate from $x$ to $W$ (field magnitude): $$|E(x)|=\frac{q}{K_S\varepsilon_0}\int_x^WN_B(x')\,dx'$$ 3. The total junction potential is the integral of the field across the depletion region: $$V_{bi}-V_A=\int_0^W|E(x)|\,dx=\frac{q}{K_S\varepsilon_0}\int_0^W\int_x^WN_B(x')\,dx'\,dx$$ 4. Swap the order of integration (region $0<x<x'<W$): $$V_{bi}-V_A=\frac{q}{K_S\varepsilon_0}\int_0^WN_B(x')\left[\int_0^{x'}dx\right]dx'=\frac{q}{K_S\varepsilon_0}\int_0^Wx'N_B(x')\,dx'$$ (This first-moment form is valid for every $m>-2$, even $m\le-1$ where the total charge integral would diverge.) 5. Insert $N_B=bx^m$: $$V_{bi}-V_A=\frac{qb}{K_S\varepsilon_0}\int_0^Wx'^{\,m+1}dx'=\frac{qb}{K_S\varepsilon_0}\frac{W^{m+2}}{m+2}$$ The integral converges at 0 only if $m+1>-1$, i.e. $m>-2$. 6. Solve for $W$: $$W^{m+2}=\frac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)$$ $$\boxed{W=\left[\frac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)\right]^{1/(m+2)}}$$ Check: $m=0,\ b=N_B$ gives $W=[2K_S\varepsilon_0(V_{bi}-V_A)/qN_B]^{1/2}$ (7.3). $\blacksquare$ --- ### Problem 7.3 (Dimensioned C-V program) **Problem.** (a) Write a program giving fully dimensioned reverse-bias C-V curves for Si $p^+$-$n$ abrupt diodes at 300 K, with inputs $A$, $N_B$, and the largest reverse bias. (b) Compare with the 1N5472A data of Fig. 7.3 using $A=3.72\times10^{-3}\ \mathrm{cm^2}$ and $N_B=8.84\times10^{15}\ \mathrm{cm^{-3}}$; does the $V_{bi}$ of Exercise 7.2 matter? (c) How does $N_B$ affect $C_J$? **Assumption.** The inputs do not include $V_{bi}$, which depends on the heavy-side doping. I use $V_{bi}=0.7$ V (the Exercise 7.2 value) as a default that can be overridden, and $K_S=11.8$. **(a) Method.** From (7.7) with $m=0$, $b=N_B$: $$C_J=A\sqrt{\frac{qK_S\varepsilon_0N_B}{2(V_{bi}-V_A)}}$$ ```matlab % Problem 7.3 -- dimensioned reverse-bias C-V for a Si p+-n abrupt diode clear q = 1.602e-19; eps0 = 8.854e-14; KS = 11.8; % cgs-type units (cm, F/cm) A = input('Diode area A (cm^2): '); NB = input('Light-side doping NB (cm^-3): '); Vmax = input('Largest reverse bias of interest, positive number (V): '); Vbi = 0.7; % assumed built-in voltage (V) VA = linspace(0,-Vmax,200); CJ = A*sqrt(q*KS*eps0*NB./(2*(Vbi-VA))); % farads close plot(VA,CJ*1e12); grid xlabel('V_A (volts)'); ylabel('C_J (pF)') ``` **(b) Comparison.** With these numbers: | $V_A$ (V) | 0 | $-1$ | $-5$ | $-10$ | $-25$ | |---|---|---|---|---|---| | $C_J$ (pF), $V_{bi}=0.7$ | 120.9 | 77.6 | 42.4 | 30.9 | 20.0 | | $C_J$ (pF), $V_{bi}=1.0$ | 101.2 | 71.5 | 41.3 | 30.5 | 19.8 | The $V=0$ value of 121 pF agrees with the measured 120 pF, and the curve follows the Fig. 7.3 shape. **Effect of $V_{bi}$:** since $C_J\propto(V_{bi}-V_A)^{-1/2}$, an error in $V_{bi}$ matters near $V_A\approx0$ (121 versus 101 pF) but is negligible at large reverse bias (about 1% at $-25$ V). The $V_{bi}=0.7$ V from the C-V fit gives agreement near zero bias, whereas $V_{bi}\approx0.9$ V computed from the doping would underestimate the zero-bias capacitance by about 12%. So yes, it has bearing, mainly at small biases. **(c) Doping dependence.** $C_J\propto\sqrt{N_B}$ at fixed $V_A$ and $A$. Quadrupling $N_B$ doubles $C_J$ (120.9 to 241.9 pF at 0 V; 30.9 to 61.9 pF at $-10$ V). Physically, higher doping means a narrower depletion width for the same voltage, hence a larger capacitance. --- ### Problem 7.4 (1N4002 C-V data) **Problem.** The 1N4002 $p^+$-$n$ diode data are in Table P7.4. Subtract 3 pF stray capacitance, assume an abrupt profile and $A=6\times10^{-4}\ \mathrm{cm^2}$ (the OCR lost the exponent; this scale gives physically sensible doping), and find $N_B$ and best-fit $V_{bi}$ by (i) eyeballing, (ii) least squares. **Solution.** 1. For each point compute $C_J=C-3$ pF, then $1/C_J^2$ (e.g. $V_A=0$: $C_J=35.709$ pF gives $7.84\times10^{20}\ \mathrm F^{-2}$; $V_A=-15$: $C_J=7.458$ pF gives $1.80\times10^{22}$). 2. Plot $1/C_J^2$ versus $V_A$: roughly straight, bending slightly (the diode is not perfectly abrupt). 3. **(ii) Least squares:** $$\frac{1}{C_J^2}=8.25\times10^{20}-1.123\times10^{21}\,V_A$$ $$N_B=\frac{2}{qK_S\varepsilon_0A^2\,|{\rm slope}|}=\frac{2}{(1.602\times10^{-19})(1.045\times10^{-12})(6\times10^{-4})^2(1.123\times10^{21})}=2.96\times10^{16}\ \mathrm{cm^{-3}}$$ $$V_{bi}=\frac{8.25\times10^{20}}{1.123\times10^{21}}=0.73\ \mathrm V$$ 4. **(i) Eyeballing** a line through the end points ($V_A=0$ and $-15$ V): slope $=-1.146\times10^{21}$, giving $N_B\approx2.9\times10^{16}\ \mathrm{cm^{-3}}$ and $V_{bi}\approx0.68$ V. **Results.** $N_B\approx3\times10^{16}\ \mathrm{cm^{-3}}$ either way; $V_{bi}\approx0.7$ V (0.68 V eyeballed, 0.73 V least squares). --- ### Problem 7.5 (Derive Eqs. 7.12 and 7.13) **Solution.** Take a $p^+$-$n$ junction (the $n^+$-$p$ case is identical). 1. A small change $dV_R$ of the reverse bias moves the light-side depletion edge by $dW$, exposing extra charge there: $$dQ=qAN_B(W)\,dW$$ (the heavily doped side contributes a negligible shift). 2. The incremental capacitance is $C=dQ/dV_R$ with $V_R=-V_A>0$: $$C=qAN_B(W)\frac{dW}{dV_R}$$ 3. The capacitance is also the parallel-plate expression, valid for any profile since the charge sheets sit at the depletion edges a distance $W$ apart: $$W=\frac{K_S\varepsilon_0A}{C}\quad\Rightarrow\quad x=\frac{K_S\varepsilon_0A}{C}\qquad\text{(7.13)}$$ 4. Differentiate: $\dfrac{dW}{dV_R}=-\dfrac{K_S\varepsilon_0A}{C^2}\dfrac{dC}{dV_R}$. 5. Use $\dfrac{d(1/C^2)}{dV_R}=-\dfrac{2}{C^3}\dfrac{dC}{dV_R}$, so $\dfrac{dC}{dV_R}=-\dfrac{C^3}{2}\dfrac{d(1/C^2)}{dV_R}$. Then $$\frac{dW}{dV_R}=\frac{K_S\varepsilon_0A\,C}{2}\frac{d(1/C^2)}{dV_R}$$ 6. Put into step 2: $$C=qAN_B\cdot\frac{K_S\varepsilon_0AC}{2}\frac{d(1/C^2)}{dV_R}\;\Rightarrow\;N_B=\frac{2}{qK_S\varepsilon_0A^2\,\dfrac{d(1/C^2)}{dV_R}}$$ 7. Since $V_R=-V_A$, $d(1/C^2)/dV_R=-d(1/C^2)/dV_A$, and the former is positive. Therefore $$\boxed{N_B(x)=\frac{2}{qK_S\varepsilon_0A^2\left|\dfrac{d(1/C^2)}{dV_A}\right|}}$$ evaluated at $x=K_S\varepsilon_0A/C$. $\blacksquare$ --- ### Problem 7.6 (Profile $N_B(x)$ for the 1N4002) **Logic.** Subtract the stray capacitance, compute $1/C^2$, estimate $d(1/C^2)/dV_A$ by finite differences between consecutive points, assign each derivative to the midpoint capacitance (to get $x$ from (7.13)), then apply (7.12). ```matlab % Problem 7.6 -- C-V profiling clear q = 1.602e-19; e0 = 8.854e-14; KS = 11.8; A = 6e-4; Cs = 3e-12; VA = [0 -.2 -.4 -.6 -.8 -1 -1.4 -1.8 -2.2 -2.6 -3 -4:-1:-15]; C = [38.709 33.717 30.567 28.319 26.598 25.170 23.060 21.490 ... 20.254 19.248 18.405 16.762 15.548 14.599 13.834 13.189 ... 12.639 12.163 11.746 11.373 11.037 10.734 10.458]*1e-12 - Cs; y = 1./C.^2; dy = diff(y)./diff(VA); % d(1/C^2)/dVA Cm = (C(1:end-1)+C(2:end))/2; % midpoint capacitance NB = 2./(q*KS*e0*A^2*abs(dy)); % Eq. 7.12 x = KS*e0*A./Cm; % Eq. 7.13 (cm) close semilogy(x*1e4,NB); grid xlabel('x (\mum)'); ylabel('N_B (cm^{-3})') ``` **Output** (for $A=6\times10^{-4}\ \mathrm{cm^2}$; $x$ in $\mu$m, $N_B$ in $\mathrm{cm^{-3}}$; first point of each pair): | $V_A$ | $x$ | $N_B$ | $V_A$ | $x$ | $N_B$ | |---|---|---|---|---|---| | 0 | 0.190 | $2.41\times10^{16}$ | $-4$ | 0.478 | $3.10\times10^{16}$ | | $-0.2$ | 0.216 | $2.59\times10^{16}$ | $-5$ | 0.520 | $3.07\times10^{16}$ | | $-0.4$ | 0.237 | $2.72\times10^{16}$ | $-6$ | 0.560 | $3.05\times10^{16}$ | | $-0.6$ | 0.257 | $2.81\times10^{16}$ | $-7$ | 0.597 | $2.98\times10^{16}$ | | $-0.8$ | 0.274 | $2.78\times10^{16}$ | $-8$ | 0.633 | $2.94\times10^{16}$ | | $-1.0$ | 0.298 | $2.95\times10^{16}$ | $-9$ | 0.667 | $2.89\times10^{16}$ | | $-1.4$ | 0.326 | $3.02\times10^{16}$ | $-10$ | 0.700 | $2.85\times10^{16}$ | | $-1.8$ | 0.351 | $3.06\times10^{16}$ | $-11$ | 0.733 | $2.79\times10^{16}$ | | $-2.2$ | 0.375 | $3.10\times10^{16}$ | $-12$ | 0.764 | $2.73\times10^{16}$ | | $-2.6$ | 0.396 | $3.12\times10^{16}$ | $-13$ | 0.795 | $2.68\times10^{16}$ | | $-3$ | 0.431 | $3.11\times10^{16}$ | $-14$ | 0.826 | $2.63\times10^{16}$ | The profile is nearly flat at about $3\times10^{16}\ \mathrm{cm^{-3}}$ (consistent with an abrupt junction and Problem 7.4), with some noise from differentiation and a slight drop near the junction and deep in the bulk. --- ### Problem 7.7 (Verify Fig. 7.10) **Solution.** With $x=\omega\tau_p$: $G_D/G_0=\sqrt{(\sqrt{1+x^2}+1)/2}$, and since $C_{D0}=G_0\tau_p/2$, $$\frac{C_D}{C_{D0}}=\frac{2}{x}\sqrt{\frac{\sqrt{1+x^2}-1}{2}}$$ ```matlab x = logspace(-2,2,200); s = sqrt(1+x.^2); GD = sqrt((s+1)/2); % G_D/G_0 CD = (2./x).*sqrt((s-1)/2); % C_D/C_D0 loglog(x,GD,x,CD); grid xlabel('\omega\tau_p'); ylabel('normalized G_D, C_D'); legend('G_D/G_0','C_D/C_{D0}') ``` | $\omega\tau_p$ | 0.01 | 0.1 | 0.3 | 1 | 3 | 10 | 30 | 100 | |---|---|---|---|---|---|---|---|---| | $G_D/G_0$ | 1.000 | 1.001 | 1.011 | 1.099 | 1.443 | 2.351 | 3.938 | 7.107 | | $C_D/C_{D0}$ | 1.000 | 0.999 | 0.989 | 0.910 | 0.693 | 0.425 | 0.254 | 0.141 | These agree with Fig. 7.10: both are flat below $\omega\tau\approx0.1$, then $G_D$ climbs and $C_D$ drops. $\blacksquare$ --- ### Problem 7.8 (Relative size of $\omega C_D$ and $G_D$) **Solution.** 1. From (7.29): $\dfrac{\omega C_D}{G_D}=\dfrac{b}{a}=\sqrt{\dfrac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}$, $x=\omega\tau_p$. 2. For $x\ll1$: using $\sqrt{1+x^2}\approx1+x^2/2$, the ratio is $\sqrt{(x^2/2)/2}=x/2$. This agrees with (7.30): $\omega C_D/G_D=\omega\tau_p/2$. 3. For $x\gg1$: $\sqrt{1+x^2}\approx x$, the ratio is $\sqrt{(x-1)/(x+1)}\to1$. 4. Compact form: $\sqrt{1+jx}=(1+x^2)^{1/4}e^{j\theta/2}$ with $\theta=\arctan x$, so the ratio equals $\tan(\theta/2)$. ```matlab x = logspace(-2,2,200); s = sqrt(1+x.^2); r = sqrt((s-1)./(s+1)); loglog(x,r,x,x/2,'--',x,ones(size(x)),'--'); grid xlabel('\omega\tau_p'); ylabel('\omega C_D / G_D') ``` | $x$ | 0.01 | 0.1 | 0.3 | 1 | 3 | 10 | 30 | 100 | |---|---|---|---|---|---|---|---|---| | $\omega C_D/G_D$ | 0.005 | 0.0499 | 0.147 | 0.414 | 0.721 | 0.905 | 0.967 | 0.990 | The plot follows $x/2$ at the low end and saturates at 1 at the high end, so it does approach the correct limits. **In words:** at low frequency the diffusion admittance is mostly conductive (the capacitive part is only $\omega\tau/2$ of the conductive part); the capacitive component grows with frequency and approaches, but never exceeds, the conductive one (the phase angle of $Y_D$ rises from $0^\circ$ toward $45^\circ$). --- ### Problem 7.9 (Lifetime from forward-bias admittance) **Method.** The measured capacitance includes the junction capacitance and the 2 pF stray capacitance, so isolate the diffusion part: $$C_D=C_{\text{meas}}-2\ \mathrm{pF}-C_J(V_A),\qquad C_J=\frac{120\ \mathrm{pF}}{\sqrt{1-V_A/0.7}}$$ then $\tau_n=2C_D/G_D$ (assuming $\omega\tau_n\ll1$). | $V_A$ (V) | $C_{\text{meas}}$ (pF) | $C_J$ (pF) | $C_D$ (pF) | $G_D$ (S) | $\tau_n=2C_D/G_D$ (s) | |---|---|---|---|---|---| | 0.50 | 276 | 224.5 | 49.5 | $2.00\times10^{-4}$ | $4.95\times10^{-7}$ | | 0.52 | 346 | 236.6 | 107.4 | $3.90\times10^{-4}$ | $5.51\times10^{-7}$ | | 0.54 | 440 | 251.0 | 187.0 | $7.15\times10^{-4}$ | $5.23\times10^{-7}$ | | 0.56 | 654 | 268.3 | 383.7 | $1.33\times10^{-3}$ | $5.77\times10^{-7}$ | | 0.58 | 938 | 289.8 | 646.2 | $2.28\times10^{-3}$ | $5.67\times10^{-7}$ | $$\bar\tau_n\approx5.4\times10^{-7}\ \mathrm s\;(\approx0.54\ \mu\mathrm s)$$ This is close to the $5\times10^{-7}$ s estimate used in Exercise 7.4. --- # Chapter 8: pn Junction Diode: Transient Response A complete diode analysis has four segments: internal electrostatics, steady-state response, small-signal response, and transient response. This chapter handles the last. Diodes are used as electrical **switches**: a pulse flips them from forward bias ("on") to reverse bias ("off") and back. The key question for engineers is how fast. The turn-off transient (on to off) is the speed-limiting one, so we concentrate on it, treating an ideal diode so that ideas, not algebra, take center stage. ## 8.1 Turn-Off Transient ### 8.1.1 Introduction Consider the circuit of Fig. 8.1(a): a battery $V_F$ with resistor $R_F$ on one throw of a switch, a battery $V_R$ with resistor $R_R$ on the other. Before $t=0$ the diode carries a steady forward current $I_F$. At $t=0$ the switch flips. Ideally the current would drop at once to the tiny reverse saturation value. What really happens (Fig. 8.1(b)): - Right after switching the current is **reversed and comparable in magnitude to $I_F$**: $-I_R\approx-V_R/R_R$. - It stays essentially constant at $-I_R$ for a period called the **storage (delay) time $t_s$**. - Then it decays toward the steady-state reverse current. - The time for the reverse current to fall to 10% of its peak magnitude, measured from the instant of switching, is the **reverse recovery time $t_{rr}$**; the **recovery time** is $t_r=t_{rr}-t_s$. Fig. 8.1(c) shows the diode voltage $v_A(t)$: it stays near $V_{ON}$ (still forward) during $0<t<t_s$ even though the applied source is reversed, then swings to $-V_R$. **Assumption:** the batteries are large compared with the diode forward drop $V_{ON}$: $$I_F=\frac{V_F-V_{ON}}{R_F}\approx\frac{V_F}{R_F}\tag{8.1a}$$ $$I_R=\frac{V_R+v_A|_{0<t<t_s}}{R_R}\approx\frac{V_R}{R_R}\tag{8.1b}$$ The decaying tail is easily distorted by stray capacitance, so $t_s$ is the quoted figure of merit. ### 8.1.2 Qualitative Analysis **Why is there a delay?** Forward bias stores excess minority carriers in the quasineutral regions beside the depletion region; reverse bias requires a *deficit* of minority carriers there. To go from on to off, the stored excess (cross-hatched areas of Fig. 8.2, with forward and reverse profiles plotted together on a linear scale) must be removed. The storage delay time is so named because most of the stored charge leaves during this phase. > Analogy: a bathtub full of water (stored carriers). You cannot make it "empty" instantly: you can pull the plug (reverse current) or wait for evaporation (recombination), but neither is instantaneous. There are exactly two removal mechanisms: 1. **Recombination in place.** Not instantaneous; several lifetimes would be needed alone. 2. **Net carrier flow out of the region (reverse current).** The carriers simply flow back across the junction and become majority carriers on the other side. The crossing itself takes only $W/v\sim10^{-10}$ s, but the *number per second* is limited by the circuit: the maximum reverse current is about $V_R/R_R=I_R$. Replacing $R_R$ with a short would speed things up but could destroy the diode. Neither can safely be instantaneous, so there is a delay. **Why does the diode stay forward biased for $0<t<t_s$?** The depletion-edge boundary condition $p_n(x_n)=p_{n0}e^{qv_A/kT}$ holds under transient conditions too. If $v_A>0$ there is an excess above equilibrium at the edge. Conversely, an excess above equilibrium at $x_n$ **means** the junction is forward biased: the residual excess maintains the forward bias. Only when the concentration at $x_n$ drops below equilibrium does the diode become reverse biased (the end of $t_s$). Fig. 8.3 shows the progressive decay of the hole excess in a $p^+$-$n$ diode for $0\le t\le t_s$: curves marching downward until $p_n(x_n)=p_{n0}$ at $t=t_s$. **Slope at $x_n$.** In an ideal $p^+$-$n$ diode $i=i_{\text{diff}}=-qAD_P\left.\dfrac{d\Delta p_n}{dx}\right|_{x_n}$, so $$\left.\frac{d\Delta p_n}{dx}\right|_{x=x_n}=-\frac{i}{qAD_P}\tag{8.2}$$ During $0<t\le t_s$, $i=-I_R<0$, so every curve slopes **upward** at $x_n$ and, since $i=-I_R$ is constant, **all curves have the same slope at $x_n$**. ### Exercise 8.1: Predicting the effect of parameters **Problem.** Sketch (dashed) the modification of a baseline $i$-$t$ transient if (a) $I_F$ increases, (b) $I_R$ increases, (c) $\tau_p$ decreases. **Solution.** (a) Larger $I_F$ means more stored charge with the removal rate unchanged, so $t_s$ increases (a longer plateau at the same height). (b) Larger $I_R$ increases the reverse-current removal rate: the plateau is *taller* (more negative) and $t_s$ is reduced. (c) Shorter $\tau_p$ increases the recombination rate (and also reduces the stored charge $I_F\tau_p$): $t_s$ is shorter. (Fig. E8.1 shows these three modified plateaus.) ### 8.1.3 The Storage Delay Time **Quantitative analysis.** Use an ideal $p^+$-$n$ step diode and charge control. The electron charge on the $p^+$ side is negligible compared with the stored hole charge $Q_P$ on the $n$ side, and $i=i_{DIFF}$; the charge-control equation (6.56) is $$i=\frac{dQ_P}{dt}+\frac{Q_P}{\tau_p}\tag{8.3}$$ (The current feeds charge buildup plus recombination loss.) **Step 1.** For $0^+\le t\le t_s$, $i=-I_R$ (constant): $$\frac{dQ_P}{dt}=-I_R-\frac{Q_P}{\tau_p}\tag{8.4}$$ **Step 2.** Separate variables and integrate from $t=0^+$ to $t_s$ (multiply the equation by $\tau_p$ to see the form $dQ_P/(\tau_pI_R+Q_P)=-dt/\tau_p$): $$\int_{Q_P(0^+)}^{Q_P(t_s)}\frac{dQ_P}{\tau_pI_R+Q_P}=-\frac{1}{\tau_p}\int_{0^+}^{t_s}dt\tag{8.5}$$ **Step 3.** Evaluate ($\int dQ/(a+Q)=\ln(a+Q)$ with $a=\tau_pI_R$): $$t_s=\tau_p\ln\frac{I_R\tau_p+Q_P(0^+)}{I_R\tau_p+Q_P(t_s)}=\tau_p\ln\left(\frac{I_R+Q_P(0^+)/\tau_p}{I_R+Q_P(t_s)/\tau_p}\right)\tag{8.6}$$ **Step 4: initial condition.** Charge cannot change instantly, so $Q_P(0^+)=Q_P(0^-)$. Before switching $dQ_P/dt=0$ and $i=I_F$, so (8.3) gives $$I_F=\frac{Q_P(0^-)}{\tau_p}=\frac{Q_P(0^+)}{\tau_p}\tag{8.7}$$ **Step 5: final condition.** The charge left at $t_s$ is hard to know. To be conservative (an estimate of $t_s$ that is too large), take $Q_P(t_s)\approx0$. Then $$\boxed{t_s=\tau_p\ln\left(1+\frac{I_F}{I_R}\right)}\tag{8.8}$$ This agrees with Exercise 8.1: $t_s$ rises with $I_F$, falls with $I_R$, and is proportional to $\tau_p$. **The more precise result.** A complete $\Delta p_n(x,t)$ solution that properly accounts for residual charge gives $$\operatorname{erf}\sqrt{\frac{t_s}{\tau_p}}=\frac{1}{1+I_R/I_F}\tag{8.9}$$ equivalently $t_s=\tau_p\left[\operatorname{erf}^{-1}\!\left(\dfrac{1}{1+I_R/I_F}\right)\right]^2$ (this is the form coded in the CPG program of Exercise 8.2, where I derive it). It looks very different, but it also agrees with every qualitative prediction. **Measurement.** $t_s\propto\tau_p$ and the only other parameters are the circuit currents, so measuring a turn-off transient yields the minority carrier lifetime on the light side of an asymmetrical junction. The measurement system of Fig. 8.4 (for $\mu$s-range lifetimes) uses a Tektronix PS5004 supply (sets $I_F$), an FG5010 function generator giving a negative-going square pulse, a series capacitor (200 $\mu$F) that holds its voltage during the step so the diode-side voltage jumps down by the pulse amplitude, resistors (1000, 150, 100 $\Omega$), the test diode, and a Tektronix 11401 digitizing scope reading a voltage proportional to diode current. Circuit RC constants are $\ge10^{-2}$ s, so capacitor discharge is negligible. Fig. 8.5 shows a sample transient. **Sample analysis.** From the response curve: $I_R/I_F\approx1.0$ and $t_s\approx5.0\ \mu$s. Then $$\tau_p=\frac{t_s}{\ln(1+I_F/I_R)}=\frac{5.0}{\ln2}=7.2\ \mu\mathrm s\quad\text{(Eq. 8.8)}$$ and from (8.9), $\operatorname{erf}^{-1}(1/2)=0.4769$, so $\sqrt{t_s/\tau_p}=0.4769$ and $$\tau_p=\frac{5.0}{0.4769^2}=22\ \mu\mathrm s\quad\text{(Eq. 8.9)}$$ A better determination varies $I_R/I_F$ and chooses the $\tau_p$ giving the best fit to the normalized $t_s/\tau_p$ versus $I_R/I_F$ curve (Fig. 8.6: dashed from (8.8), solid from (8.9), log axes, $0.01<I_R/I_F<10$). A 1N91 Ge diode fits the exact theory with $\tau_p=14.5\ \mu$s. A 1N4002 Si diode deviates: no single $\tau_p$ fits both ends, because Si diodes are rarely ideal (the derivation assumed an ideal diode). ### 8.1.4 General Information Practical remarks: - As a rule $t_s\sim\tau_p$ (or $\tau_n$). Raising $I_R/I_F$ lowers $t_s$ below $\tau_p$ (Fig. 8.6) but circuit constraints often limit $I_R/I_F$. - Build diodes with short minority carrier lifetimes. Since $\tau\propto1/N_T$ ($N_T$ the R-G center concentration), deliberately introduce R-G centers (in Si, by diffusing in gold). The limit: a shorter lifetime proportionally increases R-G current ($\propto1/\tau_0$), so the off-state current may become unacceptable, and R-G center concentrations near the doping level alter the electrostatics. - For sub-nanosecond switching use devices with less stored charge: the bipolar junction transistor or the metal-semiconductor diode (later chapters). - **Step-recovery (snap-back) diode:** $t_r$ is very short ($\sim1$ ns) with $t_s\sim1\ \mu$s, so the reverse current looks like a step, "snapping back" at $t=t_s$. Used as pulse generators and high-order, single-stage harmonic generators. It is built as a $p$-$i$-$n$ type structure: a narrow, lightly doped region between heavily doped $p$ and $n$ regions, formed epitaxially with very abrupt junctions. The profile stores the minority charge very close to the depletion edges, so almost all of it is removed by $t_s$; with nothing left to remove, the current drops abruptly. ## 8.2 Turn-On Transient The diode is switched from the reverse-bias off state to the forward on state, by current pulse, voltage pulse, or a mix. We take a current pulse for simplicity. When the current is suddenly changed from its reverse value to a constant forward current $I_F$, $v_A(t)$ rises monotonically from $V_{OFF}$ at $t=0$ to $V_{ON}$ at $t=\infty$. The first stage (until $v_A=0$) is extremely brief: few carriers are needed and they cross the thin depletion region fast, while majority carriers shrink the depletion width to its zero-bias value. We therefore treat the diode as pulsed from $i=0$ to $i=I_F$ at $t=0$ with $v_A(0)=0$ (Fig. 8.7(a) current step, 8.7(b) voltage rising from zero toward $V_{ON}$). Again an ideal $p^+$-$n$ diode with charge control. $Q_P=0$ at $t=0$ (Fig. 8.8: hole profiles growing from $p_{n0}$ toward the steady-state curve, $t=0$ to $t=\infty$). With $i=I_F$ throughout, (8.3) becomes $$\frac{dQ_P}{dt}=I_F-\frac{Q_P}{\tau_p}\tag{8.10}$$ Separate variables and integrate: $$\int_0^{Q_P(t)}\frac{dQ_P}{I_F-Q_P/\tau_p}=\int_0^tdt'=t\tag{8.11}$$ $$-\tau_p\ln\left(1-\frac{Q_P(t)}{I_F\tau_p}\right)=t\tag{8.12}$$ Solving for $Q_P(t)$: $$\boxed{Q_P(t)=I_F\tau_p\left(1-e^{-t/\tau_p}\right)}\tag{8.13}$$ (its final value is $Q_P(\infty)=I_F\tau_p$, also seen from (8.10) with $dQ/dt=0$.) **Voltage.** In steady state the stored hole charge is $$Q_P=I_{DIFF}\tau_p=I_0\tau_p\left(e^{qV_A/kT}-1\right)\tag{8.14}$$ Assume build-up is **quasistatic**: $Q_P(t)$ obeys (8.14) with $V_A\to v_A(t)$: $$Q_P(t)=I_0\tau_p\left(e^{qv_A(t)/kT}-1\right)\tag{8.15}$$ Equate (8.13) and (8.15): $$I_0\left(e^{qv_A/kT}-1\right)=I_F\left(1-e^{-t/\tau_p}\right)$$ $$e^{qv_A/kT}=1+\frac{I_F}{I_0}\left(1-e^{-t/\tau_p}\right)$$ $$\boxed{v_A(t)=\frac{kT}{q}\ln\left[1+\frac{I_F}{I_0}\left(1-e^{-t/\tau_p}\right)\right]}\tag{8.16}$$ **Discussion.** The transient lengthens with $I_F$ and $\tau_p$, as expected, because the required charge $Q_P(\infty)=I_F\tau_p$ grows with them. The interesting feature: an initial **very rapid rise** of $v_A$ to a large fraction of $V_{ON}$ in a tiny time, then a slow final approach taking several $\tau_p$. Example: $T=300$ K, $V_{ON}=0.75$ V, so $I_F/I_0=3.77\times10^{12}$. At $t=0.01\tau_p$: $$v_A=0.0259\ln\left[1+3.77\times10^{12}(0.00995)\right]=0.0259\times24.35=0.631\ \mathrm V=0.84\,V_{ON}$$ because the logarithm compresses large factors. Reaching the last few percent needs several lifetimes. ### Exercise 8.2: Concentration Plot Generation (CPG) program **Purpose.** The CPG program plots $\Delta p_n(x',t)/\Delta p_{n,\max}$ versus $x'/L_P=(x-x_n)/L_P$ at selected $t/\tau_p$ for turn-off and turn-on, from direct solution of the time-dependent diffusion equation for an ideal step junction. Menu options: linear or semilog turn-off, linear or semilog turn-on. For turn-off, $t$ steps from $0.1t_s$ to $t_s$ in increments of $0.1t_s$ (the user supplies $I_R/I_F$); for turn-on, $t$ steps from $0.1\tau_p$ to $2\tau_p$ in $0.1\tau_p$ steps. Figs. 8.3 and 8.8 were produced from it. **The math inside the code.** Let $X=x'/L_P$, $T=t/\tau_p$. The turn-on solution for a current step $I_F$, normalized by the steady-state edge value, is $$y_{on}(X,T)=\frac12\left[e^{-X}\operatorname{erfc}\left(\frac{X}{2\sqrt T}-\sqrt T\right)-e^{X}\operatorname{erfc}\left(\frac{X}{2\sqrt T}+\sqrt T\right)\right]$$ which tends to $e^{-X}$ as $T\to\infty$ (the steady profile) and gives $y_{on}(0,T)=\operatorname{erf}\sqrt T$. (The code writes $\operatorname{erfc}=1-\operatorname{erf}$.) **Turn-off by superposition.** The diffusion equation is linear. The turn-off profile is the initial steady profile $e^{-X}$ (current $I_F$) plus the response to a *step of current* from $I_F$ to $-I_R$, i.e. a step of $-(I_F+I_R)$. The response to a unit $I_F$ step is $y_{on}$, so $$y_{off}(X,T)=e^{-X}-\left(1+\frac{I_R}{I_F}\right)y_{on}(X,T)$$ **Where $t_s$ comes from.** $t_s$ is the instant when $y_{off}(0,T)=0$ (edge concentration equals equilibrium): $$1-\left(1+\frac{I_R}{I_F}\right)\operatorname{erf}\sqrt{T_s}=0\;\Rightarrow\;\operatorname{erf}\sqrt{\frac{t_s}{\tau_p}}=\frac{1}{1+I_R/I_F}$$ which is exactly Eq. (8.9). The code therefore sets `TS=(erfinv(1./(1+Iratio)))^2`. (For $I_R=0$, `erfinv(1)` is infinite, so the code sets `TS=1`, an arbitrary display scale.) ```matlab % Exercise 8.2 -- Turn-off/Turn-on Concentration Plot Generator clear close S = menu('Choose the desired plot','OFF-Linear','OFF-Semilog', ... 'ON-Linear','ON-Semilog'); % Compute ts/taup if a turn-off plot is desired if S <= 2 Iratio = input('Please input the IR/IF ratio: IR/IF = '); if Iratio == 0 % catch IR = 0 TS = 1; else TS = (erfinv(1./(1+Iratio)))^2; % ts/taup from Eq. 8.9 end else TS = 1; % not used for turn-on end % Set X = x'/LP and T = t/taup X = 0:0.03:3; if S == 1 || S == 2 T = TS/10:TS/10:TS; else T = 0.1:0.1:2; end % Steady-state curve and axes y0 = exp(-X); if S == 1 || S == 3 plot(X,y0,'g'); axis([0 3 0 1]) else semilogy(X,y0,'g'); axis([0 3 1.0e-3 1]) end xlabel('x''/L_P'); ylabel('\Deltap_n(x'',t)/\Deltap_{n,max}') grid; hold on % Time-dependent curves for i = 1:length(T) A = exp(-X).*(1-erf(X./(2*sqrt(T(i)))-sqrt(T(i)))); B = exp( X).*(1-erf(X./(2*sqrt(T(i)))+sqrt(T(i)))); yon = (A-B)/2; % turn-on profile if S == 3, plot(X,yon); elseif S == 1, yoff = exp(-X)-(1+Iratio).*yon; plot(X,yoff); elseif S == 4, semilogy(X,yon); elseif S == 2, yoff = exp(-X)-(1+Iratio).*yon; semilogy(X,yoff); end end hold off ``` (OCR repairs: lost parentheses and quotes were restored; the chain of `elseif` branches replaces the original separate `if` blocks with identical behaviour.) **Suggested uses** (the book's list, with answer hints): 1. Visualize stored-charge decay during turn-off. 2. Visualize stored-charge buildup during turn-on. 3. Compare linear and semilog plots. 4. Confirm that the slopes at $x'=0$ of the linear turn-off curves are all equal for $0<t\le t_s$ (they are, by (8.2) with constant $i$). On a semilog plot the slopes are *not* equal: the semilog slope is $(d\Delta p/dx)/\Delta p$, i.e. the current divided by the edge concentration, which differs from curve to curve. 5. Examine the effect of $I_R/I_F$ on the decay. 6. See why (8.8) becomes less accurate as $I_R/I_F$ grows: the stored charge remaining at $t_s$ grows relative to the assumed zero. 7. Check the quasistatic approximation of (8.16): on a semilog plot, quasistatic turn-on curves would all be parallel to the $t=\infty$ curve. 8. Compare $v_A(t)$ from (8.16) with the $v_A$ deduced from the turn-on plot (done in Problem 8.8). --- ## 8.3 Summary We examined the electrical and internal carrier response of pn diodes subjected to a large, rapid change of applied voltage or current. During turn-off the excess minority carriers stored in the quasineutral regions must be removed before steady state is regained. The diode initially remains forward biased while a large, constant reverse current flows, until the carrier concentrations at the depletion edges fall to equilibrium. This period is the **storage delay time $t_s$**, the primary figure of merit. It increases with the initial stored charge, decreases with the rate of removal by reverse current, and is directly proportional to the minority carrier lifetime. It is reduced by adding R-G centers during fabrication, and step-recovery diodes are made by tailoring the doping profile. Turn-off measurements also give the lifetime on the lightly doped side. The chapter's focus was a basic understanding of transient operation and physical insight, useful for the bipolar junction transistor. --- ## Problems, Chapter 8 ### Problem 8.1 (Quick quiz) **(a) Storage delay time $t_s$:** the time after switching during which the reverse current stays essentially constant at $-I_R$ (until the minority carrier concentrations at the depletion edges fall to their equilibrium values). **(b) Recovery time $t_r$:** $t_r=t_{rr}-t_s$, the time for the reverse current to decay from the end of the storage phase to 10% of its maximum magnitude ($t_{rr}$ is the total reverse recovery time). **(c)** The OCR lost the statement of part (c) ("...Explain."). The transient question of that kind in the text is why the diode stays forward biased for $0<t<t_s$ even though the source is reversed, so I answer that: the excess minority carriers still at the depletion edges satisfy $p_n(x_n)=p_{n0}e^{qv_A/kT}>p_{n0}$, meaning $v_A>0$; only when that excess is gone does the junction go reverse biased. The reverse current is set by the *slope* of the profile, not the sign of the voltage. **(d) Root cause of the delay:** the stored excess minority carrier charge (built up under forward bias) must be removed before the diode can go to the off state. **(e) The two removal mechanisms:** recombination in the quasineutral regions and net carrier flow out (reverse current) back across the junction. **(f) True or false: if $\Delta p_n(x,t)>0$ at the edge, $v_A>0$.** True: $p_n(x_n)=p_{n0}e^{qv_A/kT}$, so an excess requires $v_A>0$. **(g) True or false: if $i>0$ (the OCR dropped the symbol; I read it as forward current), the slope of $p_n(x)$ at $x_n$ must be positive.** From (8.2), $d\Delta p_n/dx=-i/(qAD_P)$, so $i>0$ gives a **negative** slope: **False**. (A positive slope occurs for $i<0$, as in the turn-off transient.) **(h) Special properties of a step-recovery diode:** a narrow lightly doped region between heavily doped $p$ and $n$ regions, with very abrupt junctions, so stored charge sits near the depletion edges; almost all of it is removed by $t_s$ and the current snaps back with $t_r\sim1$ ns. **(i) True or false: doubling both $I_F$ and $I_R$ has no effect on $t_s$.** True: by (8.8) $t_s=\tau_p\ln(1+I_F/I_R)$ depends only on the ratio, which is unchanged (also true of (8.9)). **(j) True or false: recombination retards the buildup of stored carriers during turn-on.** True: in $dQ_P/dt=I_F-Q_P/\tau_p$ the term $-Q_P/\tau_p$ is a loss; it reduces the net rate of increase and limits $Q_P$ to $I_F\tau_p$. --- ### Problem 8.2 (Interpret a $p_n(x,t)$ plot) **Missing figure.** Fig. P8.2 is a plot of $p_n(x)$ on the $n$ side at some instant, with the equilibrium value $p_{n0}$ marked. **Assumption:** the curve starts at $x_n$ above $p_{n0}$ (I take $p_n(x_n)=2p_{n0}$, the only numeric label the OCR preserved) and rises with $x$ (slope upward), like the Fig. 8.3 turn-off curves. The general formula is given so the figure's actual value can be substituted. **(a) Forward or reverse biased?** $p_n(x_n)=p_{n0}e^{qv_A/kT}$. Since $p_n(x_n)>p_{n0}$, $e^{qv_A/kT}>1$, so $v_A>0$: **forward biased**. **(b)** With $p_{n0}=10^4\ \mathrm{cm^{-3}}$, $T=300$ K and $p_n(x_n)=2\times10^4\ \mathrm{cm^{-3}}$: $$v_A=\frac{kT}{q}\ln\frac{p_n(x_n)}{p_{n0}}=0.0259\ln2=17.9\ \mathrm{mV}$$ (General: $v_A=0.0259\ln\alpha$ V for $p_n(x_n)=\alpha p_{n0}$; $p_{n0}$ cancels.) **(c) Direction of current.** With the profile sloping upward at $x_n$ ($dp_n/dx>0$), (8.2) gives $i=-qAD_P\,dp_n/dx<0$: a **reverse current** flows (holes diffuse back toward the junction) even though the junction is forward biased. This is the storage phase of turn-off. (A downward-sloping curve would mean forward current.) --- ### Problem 8.3 (Improved $Q_P(t_s)$ approximation) **Reconstruction of the OCR-garbled statement.** The OCR keeps only fragments: a numerator with $I_F$, a denominator $1+\dots$, and a final $t_s$ expression with "$(1+\dots)$" and "$1+2\dots$". The reading consistent with those fragments, and which I checked reproduces the exact erf behaviour closely, is $$Q_P(t_s)\approx\frac{I_F\tau_p}{1+I_F/I_R}=\tau_p\frac{I_FI_R}{I_F+I_R}$$ This is my stated assumption. **Solution.** Let $r\equiv I_R/I_F$. 1. Use (8.6) with $Q_P(0^+)=I_F\tau_p$: $$t_s=\tau_p\ln\frac{I_R+I_F}{I_R+Q_P(t_s)/\tau_p}$$ 2. Insert $Q_P(t_s)/\tau_p=I_Fr/(1+r)$ and $I_R=rI_F$: $$t_s=\tau_p\ln\frac{1+r}{r+\dfrac{r}{1+r}}=\tau_p\ln\frac{(1+r)^2}{r\,(2+r)}$$ $$\boxed{t_s=\tau_p\ln\left[\frac{(1+I_R/I_F)^2}{(I_R/I_F)\,(2+I_R/I_F)}\right]}$$ Check: as $r\to\infty$ the argument tends to 1 so $t_s\to0$, as it should. 3. Numerical comparison: | $I_R/I_F$ | 0.01 | 0.05 | 0.1 | 0.2 | 0.5 | 1 | 2 | 5 | 10 | |---|---|---|---|---|---|---|---|---|---| | (8.8) | 4.615 | 3.045 | 2.398 | 1.792 | 1.099 | 0.693 | 0.405 | 0.182 | 0.095 | | Revised | 3.927 | 2.376 | 1.751 | 1.186 | 0.588 | 0.288 | 0.118 | 0.028 | 0.0083 | | Exact (8.9) | 3.326 | 1.962 | 1.429 | 0.956 | 0.468 | 0.2275 | 0.093 | 0.022 | 0.0065 | ```matlab r = logspace(-2,1,100); ts_old = log(1+1./r); ts_new = log((1+r).^2./(r.*(2+r))); ts_ex = (erfinv(1./(1+r))).^2; loglog(r,ts_old,'--',r,ts_new,'-.',r,ts_ex,'-'); grid xlabel('I_R/I_F'); ylabel('t_s/\tau_p'); legend('Eq. 8.8','revised','Eq. 8.9') ``` **Comment.** The revised expression is a clear improvement: it is roughly 15 to 30% above the exact result, whereas (8.8) is 40% to over 1000% too large at large $I_R/I_F$ (0.095 versus 0.0065 at $r=10$). It stays above the exact curve (still conservative) and has the right trend. Its remaining error comes from replacing the real residual charge by a crude closed form. --- ### Problem 8.4 (Open-circuit voltage decay) **(a) Sketch.** After $t=0$, $i=0$, so by (8.2) the slope of $p_n(x)$ at $x_n$ is zero. The initial excess profile loses charge by recombination everywhere: the curves sink downward with a flat (zero-slope) start at $x_n$ and never reach $p_{n0}$ in finite time. (Check with the CPG program by setting the current to zero.) **(b) Stored charge.** With $i=0$, (8.3) becomes $dQ_P/dt=-Q_P/\tau_p$. From (8.7), $Q_P(0)=I_F\tau_p$. So $$Q_P(t)=I_F\tau_p\,e^{-t/\tau_p}$$ **(c) Voltage.** Assume quasistatic decay: $Q_P=I_0\tau_p\left(e^{qv_A/kT}-1\right)\approx I_0\tau_pe^{qv_A/kT}$ for $qv_A/kT\gg1$. 1. Equate: $I_0\tau_pe^{qv_A/kT}=I_F\tau_pe^{-t/\tau_p}$. 2. Take logarithms: $\dfrac{qv_A}{kT}=\ln\dfrac{I_F}{I_0}-\dfrac{t}{\tau_p}$. 3. Use $I_F/I_0=e^{qV_{ON}/kT}-1\approx e^{qV_{ON}/kT}$, so $\ln(I_F/I_0)\approx qV_{ON}/kT$: $$\boxed{v_A(t)=V_{ON}-\frac{kT}{q}\frac{t}{\tau_p}}$$ **(d) What does it suggest?** The open-circuit voltage falls **linearly** in time with slope $-kT/(q\tau_p)$ (about $-26$ mV per lifetime at 300 K). Measuring the slope gives the minority carrier lifetime: $$\tau_p=-\frac{kT}{q}\left(\frac{dv_A}{dt}\right)^{-1}$$ This is the **open-circuit voltage decay (OCVD)** method of lifetime measurement. --- ### Problem 8.5 (Turn-on times) **Given.** $p^+$-$n$ diode pulsed from $0$ to $I_F=1$ mA, $\tau_p=1\ \mu$s (the OCR's "sec" is $\mu$s), $I_0=10^{-15}$ A. Find the times for $v_A$ to reach 90% and 95% of its final value. **Solution.** 1. Final voltage from $I_F=I_0\left(e^{qV_{ON}/kT}-1\right)$: $$V_{ON}=0.0259\ln\left(\frac{10^{-3}}{10^{-15}}+1\right)=0.0259\ln(10^{12})=0.0259(27.63)=0.716\ \mathrm V$$ 2. Required voltages: $0.9V_{ON}=0.644$ V, $0.95V_{ON}=0.680$ V. 3. Rearranging (8.16): $1-e^{-t/\tau_p}=\dfrac{e^{qv_A/kT}-1}{I_F/I_0}$, so $t=-\tau_p\ln\left[1-\dfrac{e^{qv_A/kT}-1}{I_F/I_0}\right]$. 4. **90%:** $e^{0.644/0.0259}=e^{24.87}=6.3\times10^{10}$, so the fraction is $6.3\times10^{10}/10^{12}=0.0631$: $$t_{90}=-(1\ \mu\mathrm s)\ln(1-0.0631)=0.065\ \mu\mathrm s=65\ \mathrm{ns}$$ 5. **95%:** $e^{0.680/0.0259}=e^{26.25}=2.5\times10^{11}$, fraction $=0.251$: $$t_{95}=-(1\ \mu\mathrm s)\ln(1-0.251)=0.289\ \mu\mathrm s=289\ \mathrm{ns}$$ The last 5% of the voltage takes about 4.4 times as long as the first 90%: fast rise, slow finish. --- ### Problem 8.6 (Pulse from $I_{F1}$ to $I_{F2}>I_{F1}$) **(a) Sketch.** Initially $p_n(x)$ is the steady profile for $I_{F1}$. At $t=0$ the current jumps to $I_{F2}$ and the profile grows upward at every point, with a steeper slope at $x_n$ (the edge slope is proportional to the current), the edge value rising quickly at first and finally approaching the larger steady curve for $I_{F2}$ as $t\to\infty$ (like Fig. 8.8 but starting from a nonzero curve). **(b) $v_A(t)$.** Charge control with $i=I_{F2}$ for $t>0$ and $Q_P(0)=I_{F1}\tau_p$ (from (8.7)): 1. $\dfrac{dQ_P}{dt}=I_{F2}-\dfrac{Q_P}{\tau_p}$. 2. Let $u=Q_P-I_{F2}\tau_p$; then $du/dt=-u/\tau_p$ with $u(0)=(I_{F1}-I_{F2})\tau_p$, so $u=u(0)e^{-t/\tau_p}$. 3. Therefore $$Q_P(t)=I_{F2}\tau_p-(I_{F2}-I_{F1})\tau_pe^{-t/\tau_p}$$ 4. Quasistatic: $Q_P=I_0\tau_p\left(e^{qv_A/kT}-1\right)$. Equate: $$e^{qv_A/kT}=1+\frac{I_{F2}}{I_0}-\frac{I_{F2}-I_{F1}}{I_0}e^{-t/\tau_p}$$ $$\boxed{v_A(t)=\frac{kT}{q}\ln\left[1+\frac{I_{F2}}{I_0}-\frac{I_{F2}-I_{F1}}{I_0}e^{-t/\tau_p}\right]}$$ Checks: $t=0$ gives $\frac{kT}{q}\ln(1+I_{F1}/I_0)=V_{ON1}$; $t\to\infty$ gives $V_{ON2}$; setting $I_{F1}=0$ recovers (8.16). --- ### Problem 8.7 (Combined turn-off / turn-on) **Missing text.** The OCR of the statement is truncated (it keeps only "...$-1$ mA. After 1 $\mu$sec a current pulse is applied to switch the diode back to ... mA. Assume the diode is ideal with $\tau_p=1\ \mu$sec."). I adopt this reading: an ideal $p^+$-$n$ diode carries $I_F=+1$ mA for $t<0$; at $t=0$ the circuit drives $-1$ mA (so $I_R=1$ mA); after $1\ \mu$s it is switched back to $+1$ mA. Reverse current can flow at $-I_R$ only while stored charge remains. **(a) Sketch of $i(t)$.** 1. $t<0$: $+1$ mA. 2. $0<t<t_s$: $-1$ mA (storage phase). From (8.8), $t_s=\tau_p\ln(1+I_F/I_R)=1\ \mu\mathrm s\times\ln2=0.693\ \mu$s, which is less than $1\ \mu$s. 3. $t_s<t<1\ \mu$s: the stored charge has run out; the current drops quickly to the tiny reverse saturation current ($\approx0$ on this scale). 4. $t>1\ \mu$s: $+1$ mA again. **(b) $v_A(t)$ for $t>1\ \mu$s.** At $1\ \mu$s the diode has been reverse biased since $t_s=0.693\ \mu$s, so the stored charge is essentially zero ($Q_P\approx0$) and the diode restarts like a turn-on from $v_A\approx0$. With $t'=t-1\ \mu$s, from (8.13) and (8.16): $$Q_P(t)=I_F\tau_p\left(1-e^{-(t-1\,\mu\mathrm s)/\tau_p}\right)$$ $$\boxed{v_A(t)=\frac{kT}{q}\ln\left[1+\frac{I_F}{I_0}\left(1-e^{-(t-1\,\mu\mathrm s)/(1\,\mu\mathrm s)}\right)\right]},\quad t>1\ \mu\mathrm s,\ I_F=1\ \mathrm{mA}$$ ($I_0$ is not given in the surviving text, so the result stays in terms of $I_0$; with $I_0=10^{-15}$ A it reproduces the Problem 8.5 curve.) Neglected residual charge and depletion-capacitance charging would make the true voltage rise slightly faster. --- ### Problem 8.8 (Compare exact and quasistatic turn-on) **(a) Exact $v_A(t)$.** From Exercise 8.2 the turn-on edge value is $\Delta p_n(0,t)/\Delta p_{n,\max}=\operatorname{erf}\sqrt{t/\tau_p}$ (the first element of `yon`, at `X=0`). The edge boundary condition and steady value give $$\frac{\Delta p_n(0,t)}{\Delta p_{n,\max}}=\frac{e^{qv_A(t)/kT}-1}{e^{qV_{ON}/kT}-1}=\operatorname{erf}\sqrt{\frac{t}{\tau_p}}$$ so $$v_A(t)=\frac{kT}{q}\ln\left[1+\left(e^{qV_{ON}/kT}-1\right)\operatorname{erf}\sqrt{\frac{t}{\tau_p}}\right]$$ Program modification (add inside the time loop after `yon` is computed; `yon(1)` is the `X=0` value): ```matlab VON = 0.5; Vth = 0.0259; ra = exp(VON/Vth)-1; % = IF/I0 = 2.42e8 T = 0.1:0.1:2; vA_exact = zeros(size(T)); for i = 1:length(T) A = exp(-X).*(1-erf(X./(2*sqrt(T(i)))-sqrt(T(i)))); B = exp( X).*(1-erf(X./(2*sqrt(T(i)))+sqrt(T(i)))); yon = (A-B)/2; vA_exact(i) = Vth*log(1 + ra*yon(1)); end vA_quasi = Vth*log(1 + ra*(1-exp(-T))); % Eq. 8.16 ``` **(b) Comparison** ($V_{ON}=0.5$ V, $T=300$ K, $I_F/I_0=e^{0.5/0.0259}-1=2.42\times10^8$): | $t/\tau_p$ | $\operatorname{erf}\sqrt{t/\tau_p}$ | $v_A$ exact (V) | $v_A$ Eq. 8.16 (V) | |---|---|---|---| | 0.1 | 0.345 | 0.4725 | 0.4391 | | 0.2 | 0.473 | 0.4806 | 0.4558 | | 0.3 | 0.561 | 0.4850 | 0.4650 | | 0.4 | 0.629 | 0.4880 | 0.4713 | | 0.5 | 0.683 | 0.4901 | 0.4758 | | 0.6 | 0.727 | 0.4917 | 0.4794 | | 0.7 | 0.763 | 0.4930 | 0.4822 | | 0.8 | 0.794 | 0.4940 | 0.4845 | | 0.9 | 0.820 | 0.4949 | 0.4865 | | 1.0 | 0.843 | 0.4956 | 0.4881 | | 1.2 | 0.879 | 0.4966 | 0.4907 | | 1.4 | 0.906 | 0.4974 | 0.4927 | | 1.6 | 0.926 | 0.4980 | 0.4942 | | 1.8 | 0.942 | 0.4985 | 0.4953 | | 2.0 | 0.955 | 0.4988 | 0.4962 | **Conclusion.** The quasistatic result (8.16) always **underestimates** the exact voltage: by about 33 mV (7% of $V_{ON}$) at $t=0.1\tau_p$, and by only 2.6 mV (0.5%) at $2\tau_p$. The real profile builds up faster near the edge than the quasistatic (steady-shaped) profile would, because charge has not yet spread into the bulk, and the edge concentration sets the voltage, so the exact voltage leads. The quasistatic formula captures the shape (fast rise, slow finish), and its error shrinks as time increases. # Chapter 9 — Optoelectronic Diodes > **Why this chapter is fun.** Until now, electrons and holes were pushed around by voltages. Now we let *light* into the story. Light can create carriers (photodiodes, solar cells), and carriers can create light (LEDs). A pn junction is a two-way translator between photons and electrons. Think of a semiconductor as a currency exchange booth where photons and electrons trade at a fixed rate: the band gap $E_G$. ## 9.1 Introduction Photodevices fall into three functional categories: | Category | Energy conversion | Purpose | Examples | |---|---|---|---| | Photodetector | light $\to$ electrical | detect or measure information carried by light | pn photodiode, p-i-n, avalanche photodiode | | Solar cell | light $\to$ electrical | produce electrical power | pn junction solar cell | | Light emitter | electrical $\to$ light | produce light | LED, laser diode | This chapter surveys one family from each category. Applications are everywhere: barcode readers, disc readers and laser printers (LED or laser-diode sources); circuit isolators, intruder alarms and remote controls (emitter plus detector pairs); fiber-optic telecommunication (laser diode transmits, photodiode receives); calculators, chargers and satellites (solar cells). ### Wavelength and photon energy Visible light spans roughly $0.4\ \mu$m (violet) to $0.7\ \mu$m (red). Figure 9.1 shows (top) the relative response of the human eye, peaking in the green near $0.55\ \mu$m; (middle) the colour bands violet, blue, green, yellow, red, with ultraviolet below $0.4\ \mu$m and infrared above $0.7\ \mu$m; (bottom) the wavelengths $\lambda_G$ at which the photon energy equals the 300 K band gap of 6H-SiC, ZnSe, GaP, AlAs, GaAs and InP. A photon of frequency $\nu$ has $E_{ph}=h\nu$. With $c=\lambda\nu$: $$\lambda=\frac{c}{\nu}=\frac{hc}{h\nu}=\frac{hc}{E_{ph}}\qquad(9.1)$$ With $\lambda$ in $\mu$m and $E_{ph}$ in eV: $$\lambda\,[\mu\text{m}]=\frac{1.24}{E_{ph}\,[\text{eV}]}\qquad(9.2)$$ *Check:* $hc=(6.626\times10^{-34})(3\times10^{8})=1.988\times10^{-25}$ J m, and dividing by $1.602\times10^{-19}$ J/eV gives $1.24$ eV $\mu$m. The wavelength where $E_{ph}=E_G$ is $$\lambda_G=\frac{1.24}{E_G}$$ Photons with $\lambda<\lambda_G$ can be absorbed and create pairs; photons with $\lambda>\lambda_G$ pass through the crystal like glass. Examples: Si, $E_G=1.12$ eV gives $\lambda_G=1.107\ \mu$m. GaAs, $E_G=1.42$ eV gives $0.873\ \mu$m. Visible light $0.4<\lambda<0.7\ \mu$m corresponds to $1.77<E_{ph}<3.10$ eV. --- ## 9.2 Photodiodes ### 9.2.1 pn Junction Photodiodes A **pn junction photodiode** is a pn diode fabricated and encapsulated so light can reach the neighbourhood of the metallurgical junction. **Rain-and-gutter picture.** The depletion region is a steep roof with a gutter (the built-in field). Light creates raindrops (electron-hole pairs) everywhere. Drops landing on the roof slide into the gutter immediately. Drops landing on flat ground within a diffusion length of the roof wander and often reach it. Drops farther away soak into the ground (recombine) and are lost. Minority carriers created in the quasineutral regions within roughly a diffusion length of the depletion edges ($x=-x_p$ on the p-side, $x=x_n$ on the n-side) live long enough to diffuse to the depletion region. These, plus carriers generated inside the depletion region, are swept by the field to the other side, adding a **reverse-going** current. If the photogeneration rate $G_L$ is uniform, the added current is $-q$ times the pairs generated per second in the volume $A(L_N+W+L_P)$: $$I=I_{dark}+I_L\qquad(9.3)$$ $$I_L=-qA\,(L_N+W+L_P)\,G_L\qquad(9.4)$$ A modified derivation of the ideal-diode and R-G current relations with $G_L\neq0$ confirms this (book Problem 6.8). **Consequences of (9.4).** 1. $W$ is typically small compared with $L_N+L_P$, so $I_L$ is essentially independent of bias. 2. The light-on $I$–$V$ curve is the dark curve translated straight down the current axis. 3. $I_L\propto G_L$, so the shift is proportional to light intensity (Fig. 9.3 shows parallel curves for $G_L=0,\,G_0,\,2G_0,\,4G_0$). Combined photodiode equation: $$I=I_0\left(e^{qV_A/kT}-1\right)-qA(L_N+W+L_P)G_L$$ ### Spectral response Spectral response is how $I_L$ varies with wavelength. In Fig. 9.4 (Si, constant incident power) the response rises from near zero at $0.4\ \mu$m to a maximum near $0.8$–$0.9\ \mu$m, then falls sharply to near zero beyond about $1.1\ \mu$m. **Long-wavelength cutoff.** Photons with $E_{ph}<E_G$ cannot lift an electron across the gap, so the crystal is transparent. The cutoff is $\lambda_G=1.24/E_G$; for Si ($1.12$ eV) that is about $1.1\ \mu$m, matching the figure. **Short-wavelength decline: two reasons.** 1. *Fixed power means fewer photons.* The photon flux is $\Phi=P_{opt}/(h\nu)$. As $\lambda$ decreases, $h\nu$ rises, so fewer photons arrive per second. 2. *Surface recombination.* Intensity decays as $e^{-\alpha x}$ and the mean penetration depth is $1/\alpha$. As $\lambda$ drops below $\lambda_G$, $\alpha$ rises rapidly (Fig. 3.20), so light is absorbed nearer the surface, where recombination is strong, and more carriers recombine before reaching the depletion region. ### Frequency response Photogenerated minority carriers outside the depletion region must *diffuse* to it before current appears externally. Diffusion is slow, so a standard pn photodiode reaches at best tens of MHz. ### 9.2.2 p-i-n and Avalanche Photodiodes #### p-i-n photodiodes A p-i-n diode has a lightly doped ("intrinsic") i-region between heavily doped $p^+$ and $n^+$ regions. In the photodiode (Fig. 9.5a) the surface metal has an opening for light, an antireflection coating covers the surface, the top region is very thin to minimize absorption there, and the i-width is tailored to the desired response. **Band diagram (Fig. 9.5b).** - Low doping means the i-layer is fully depleted at zero or small reverse bias. - Heavy doping makes the depletion widths in the outer regions tiny. - The total depletion width is effectively $W_I$, independent of reverse bias. - Little charge in the i-region gives $d\mathscr{E}/dx\approx0$: a roughly constant field and straight, tilted bands. - Heavy outer doping gives short minority-carrier diffusion lengths there, so most photocurrent comes from carriers generated in the central depletion region. **Two advantages.** 1. *Tuned wavelength response:* choose $W_I\approx1/\alpha$ at the wavelength of interest. 2. *High speed:* carriers are collected by the large i-field at saturation velocity rather than by diffusion: $$f_{max}\approx\frac{1}{\text{transit time across }W_I}=\frac{v_{sat}}{W_I}\qquad(9.5)$$ *Example.* $v_{sat}\approx10^{7}$ cm/s, $W_I=5\ \mu\text{m}=5\times10^{-4}$ cm: $$t_{tr}=\frac{5\times10^{-4}}{10^{7}}=5\times10^{-11}\ \text{s}\;\Rightarrow\;f_{max}\approx20\ \text{GHz}$$ $W_I$ cannot be shrunk arbitrarily: the junction capacitance $C=\varepsilon A/W_I$ grows as $W_I$ falls, and the $R_SC$ time constant (with series resistance $R_S$) eventually limits speed. Transit time wants small $W_I$; RC wants large $W_I$. **Fiber-optic use.** Pre-1990 silica fiber systems operated at $1.3\ \mu$m where *chromatic dispersion* (different wavelengths travelling at slightly different speeds, smearing pulses) is minimal, suitable for multiwavelength sources. Advanced high-bit-rate systems use single-wavelength sources at $1.55\ \mu$m, where fiber *loss* is minimal. Both exceed Si's $1.1\ \mu$m cutoff, so another material is needed. **Material of choice: $\text{In}_{0.53}\text{Ga}_{0.47}\text{As}$ on InP.** From Fig. 9.6 (band gap versus lattice constant; a line joins two compounds, and the alloy lattice constant is linear in composition), it has $E_G\approx0.75$ eV: $$\lambda_G=\frac{1.24}{0.75}=1.65\ \mu\text{m}$$ and is lattice matched to InP (same lattice constant), permitting quality layers on commercial InP substrates. *Interpolation example (Fig. 9.6 caption).* $\text{In}_{0.8}\text{Ga}_{0.2}$P, with InP at $5.87$ Å and GaP at $5.45$ Å: $$a=5.87-0.2(5.87-5.45)=5.79\ \text{\AA}$$ and $E_G\approx1.5$ eV from the InP–GaP line. Solid lines mark direct alloys, dashed lines indirect. **InGaAs p-i-n cross section (Fig. 9.7), bottom to top:** $n^+$-InP substrate; $n$-InP buffer (reduces defects in the InGaAs); $n^-$ $\text{In}_{0.53}\text{Ga}_{0.47}$As i-layer (absorber); $p^+$-InP window (wide gap, transmits light to the absorber); $p^+$-InGaAsP cap (reduces surface dark current); silicon nitride (protects the surface, minimizes surface recombination). Light enters from the top. #### Avalanche photodiodes (APD) APDs are specially built p-i-n, pn or metal-semiconductor photodiodes operated near avalanche breakdown. The standard Si APD (Fig. 9.8) has an $n^+$ top region with antireflection coating, a **guard ring** around the junction periphery, $p$ and $p^+$ regions, a passivation coating, and $n$ and $p$ contacts. Junction curvature causes early breakdown at the periphery (Chapter 6); the guard ring prevents this. Uniform, low-defect material is needed for uniform breakdown across the junction. **Advantage: gain without proportionally amplified noise.** Normally amplification boosts signal and noise and adds amplifier noise, lowering S/N. In an APD, avalanche multiplication amplifies the photo-signal without amplifying the typically dominant receiver circuit noise, so S/N improves until avalanche noise becomes comparable to circuit noise. InGaAs-on-InP and Ge APDs are alternatives to p-i-n diodes in fiber telecommunications. --- ## 9.3 Solar Cells ### 9.3.1 Solar Cell Basics Most solar cells are large-area pn photodiodes. Photodiodes are designed for spectral or time response; solar cells are designed to minimize energy losses. The $I$–$V$ form is that of Fig. 9.3. Power is delivered when the operating point is in the **fourth quadrant** ($I<0$, $V>0$, so $P=IV<0$ means the device supplies power), e.g. by putting a resistor in series with the illuminated cell. Usually only this quadrant is plotted with the $-I$ axis up (Fig. 9.9). Parameters (Fig. 9.9): - $V_{OC}$: open-circuit voltage, the maximum voltage. - $I_{SC}$: short-circuit current, the maximum current. - $V_m,\,I_m$: the maximum-power operating point (corner of the "maximum power rectangle"). From $I=0$ in the photodiode equation, with $I_L'=|I_L|$: $$0=I_0\left(e^{qV_{OC}/kT}-1\right)-I_L'\;\Rightarrow\;V_{OC}=\frac{kT}{q}\ln\!\left(1+\frac{I_L'}{I_0}\right),\qquad I_{SC}=I_L'$$ $$P_{max}=I_mV_m<I_{SC}V_{OC}$$ **Fill factor:** $$FF=\frac{P_{max}}{I_{SC}V_{OC}}=\frac{I_mV_m}{I_{SC}V_{OC}}\qquad(9.6)$$ $FF<1$ always; it measures how square the curve is. **Power conversion efficiency:** $$\eta=\frac{P_{max}}{P_{in}}=\frac{FF\,I_{SC}V_{OC}}{P_{in}}\qquad(9.7)$$ ### 9.3.2 Efficiency Considerations Higher efficiency means less area and cost per electrical output. **Solar spectrum (Fig. 9.10, irradiance in W/m$^2$-$\mu$m versus wavelength 0–3 $\mu$m).** - **AM0** (air mass zero): just outside the atmosphere, $135\ \text{mW/cm}^2$; relevant for satellites. - **AM1.5**: normalized to $100\ \text{mW/cm}^2$; average terrestrial U.S. conditions (shows atmospheric absorption dips). - Most power is visible, with a long infrared tail. **Two intrinsic losses and the optimum gap.** 1. *Sub-gap photons wasted:* the spectral area beyond $\lambda_G$ cannot be absorbed: about 20% of incident energy in Si and about 35% in GaAs. 2. *Excess photon energy becomes heat:* a photon with $E_{ph}>E_G$ yields only $E_G$ usefully; the excess $E_{ph}-E_G$ thermalizes. About 40% of absorbed energy is lost this way in Si, about 30% in GaAs. Raising $E_G$ reduces loss 2 but increases loss 1, so an optimum band gap exists. Si and GaAs, the most developed semiconductors, both lie very close to it. **Device-related losses (Si cell, Fig. 9.11: $n$ top layer on $p$-Si, $p^+$ under the rear contact, oxides top and bottom, metal fingers, inverted pyramids).** | Loss | Remedy | |---|---| | Shadowing versus series resistance | Narrow metal fingers connected along one edge. Fingers far apart give long lateral current paths in the thin $n$-layer and high $R_S$ (even a few ohms hurt efficiency); fingers too wide or dense shadow the cell. Size and spacing are optimized. Very narrow fingers also raise metallization and contact resistance | | Reflection ($\approx30\%$ at bare planar Si, normal incidence) | **Texturing** by anisotropic etching (etch removing Si preferentially along certain crystal planes) forms inverted pyramids so light strikes the surface two or more times before escaping; plus an antireflection coating of intermediate refractive index ($\text{SiO}_2$ or an optimized deposited film). Net reflection below 1% | | Incomplete absorption of long wavelengths | Bottom oxide plus metal form a mirror; light is trapped and bounces repeatedly, dramatically enhancing absorption near $\lambda_G$ | | Recombination | Very long minority-carrier lifetimes (diffusion length greater than cell thickness) and oxidized top and bottom surfaces (low surface recombination) let carriers born almost anywhere reach the depletion region and be swept across | ### 9.3.3 Solar Cell Technology Three categories: **thin-film**, **single-crystal**, **concentrator**. - **Thin-film:** deposited amorphous or polycrystalline absorber. Amorphous Si (a-Si) is commercial; CdTe and $\text{CuInSe}_2$ are in development. Cheap, large area (modules to $1.2\ \text{m}^2$). a-Si is vapor-deposited, about $1\ \mu$m thick, on tin-oxide-coated glass (transparent front contact), doped during deposition into a p-i-n structure, with Al or Ag back contact. Lab efficiency about 13%, commercial often half that; output falls 10%–15% in the first year because light breaks passivating bonds, creating traps and recombination. CdTe reached roughly 15% in the lab (OCR garbled) and 11% in the field. - **Single-crystal:** almost all Si or GaAs. GaAs cells often contain AlGaAs or other alloy layers; active GaAs is epitaxially deposited on GaAs, Ge or Si. - **Concentrator:** for 100 suns or more. A mirror or lens costs less than equivalent cell area; the higher current and voltage also raise efficiency (Fig. 9.12). Stacking (for example Si or GaSb ($\lambda_G=1.7\ \mu$m) beneath GaAs to absorb what GaAs passes) is expected to exceed 35%. Concerns: heat dissipation, elevated temperature, high current density. *Figure 9.12* (lab efficiency versus year, 1978–1992): concentrators (GaAs or Si) top at about 32%; flat-plate single-crystal Si about 23%; flat-plate thin-film about 16%; all rise with time. --- ## 9.4 LEDs ### 9.4.1 General Overview **The mechanism as a story.** Forward bias lowers the hill: electrons flood from n to p and holes from p to n. The injected minority carriers meet majority carriers and recombine. What happens to the released energy? - In an **indirect** semiconductor (Si, Ge) electron and hole crystal momenta differ greatly (E-k diagrams, Subsection 3.3.2). A photon carries almost no momentum, so band-to-band recombination is hard; recombination goes through R-G centres and the energy becomes **heat**. - In a **direct** semiconductor (GaAs) the momenta match, so much of the injected carriers recombine band to band emitting **photons** (Fig. 9.13: forward-biased band diagram with split quasi-Fermi levels $E_{Fn}$, $E_{Fp}$, electrons dropping from $E_C$ to $E_V$). **Emission spectrum.** Carrier distributions peak about $kT/2$ from band edges in a nondegenerate semiconductor, so photon energies spread by about $kT$ with a peak slightly above $E_G$. If photons come from a centre just below $E_C$ or from excitons (electrostatically bound electron-hole pairs acting as a unit), the peak can lie slightly below $E_G$. In any case $\lambda_{peak}\approx\lambda_G=1.24/E_G$. For visible light, $0.4<\lambda_G<0.7\ \mu$m, i.e. $1.77<E_G<3.10$ eV. **Three requirements for a visible LED material:** direct gap; $E_G$ between about 1.7 and 3.1 eV; pn junctions can be formed. Candidates: - Si, Ge: indirect, wrong gap. - GaAs: direct but gap too small. - GaP, AlAs: right gap, indirect. - SiC: right gap, indirect. - II-VI compounds (ZnSe): direct, right gap, but native defects compensate dopants, so junctions were impossible until recently. Hence alloys and special "light-enhancing" centres are used. ### 9.4.2 Commercial LEDs **Table 9.1: Characteristics of commercial LEDs** (adapted from Craford) | Material | Colour | Peak $\lambda$ ($\mu$m) | External efficiency $\eta$ (%) | Luminous performance (lm/W) | |---|---|---|---|---| | *Established* | | | | | | $\text{GaAs}_{0.6}\text{P}_{0.4}$ | Red | 0.650 | 0.2 | 0.15 | | $\text{GaAs}_{0.35}\text{P}_{0.65}$:N | Orange-red | 0.630 | 0.7 | 1 | | $\text{GaAs}_{0.14}\text{P}_{0.86}$:N | Yellow | 0.585 | 0.2 | 1 | | GaP:N | Green | 0.565 | 0.4 | 2.5 | | GaP:Zn-O | Red | 0.700 | 2 | 0.40 | | *Recent additions* | | | | | | AlGaAs | Red | 0.650 | 4–16 | 2–8 | | AlInGaP | Orange | 0.620 | 6 | 20 | | AlInGaP | Yellow | 0.585 | 5 | 20 | | AlInGaP | Green | 0.570 | 1 | 6 | | SiC | Blue | 0.470 | 0.02 | 0.04 | | GaN | Blue | 0.450 | 2 | 0.6 | Luminous performance is lumens per watt of electrical input (radiant watts times relative eye response, Fig. 9.1). External efficiency is $\eta=P_{photo,out}/P_{elec,in}$. #### $\text{GaAs}_{0.6}\text{P}_{0.4}$ GaAs is direct but infrared; GaP is visible but indirect. The alloy $\text{GaAs}_{1-x}\text{P}_x$ is direct with a light-producing gap for $0.28\le x\le0.45$ (direct-indirect transition at $x=0.45$, Fig. 9.6). Why $x=0.4$? From $x=0.28$ to $0.45$, $\eta$ falls about tenfold, but the wavelength shortens and eye response rises about 50 times; the product (efficiency $\times$ eye response) peaks at $x=0.4$, giving the brightest-looking device. These were the first mass-marketed solid-state lamps (early-1970s watches and calculators), later replaced by liquid crystal displays (lower battery drain); today they are indoor indicator lamps. **Structure (Fig. 9.14a).** GaAs and $\text{GaAs}_{0.6}\text{P}_{0.4}$ have very different lattice constants, so direct growth would be defective. Process: (1) graded layer $x:0\to0.4$ on the GaAs substrate; (2) n-type $\text{GaAs}_{0.6}\text{P}_{0.4}$ active layer; (3) second graded layer $x:0.4\to0.6$ (wider gap, light passes without reabsorption); (4) Zn diffused to form the p-side. The back n-contact is segmented to enhance bottom-surface reflection; $\text{Si}_3\text{N}_4$ covers the top. #### $\text{GaAs}_{0.35}\text{P}_{0.65}$:N, $\text{GaAs}_{0.14}\text{P}_{0.86}$:N, GaP:N Orange-red, yellow and green LEDs use $x=0.65$ to $1$, technically **indirect**; nitrogen doping makes them **pseudo-direct**. **Isoelectronic trap.** N is a Column V element like As and P, so it replaces a Column V atom and adds no free electron or hole. It instead creates an electron-trapping level about $0.1$ eV below $E_C$. Being formed from an element with the same (*iso*) valence structure as the atom replaced, it is an isoelectronic trap. A trapped electron is highly localized, so by $\Delta x\,\Delta p\gtrsim\hbar/2$ it has a large spread in momentum that includes the momentum of valence-band holes, giving a greatly enhanced probability of a photon-emitting transition. The trap is a conduit for relatively efficient radiative recombination. These devices resemble the $\text{GaAs}_{0.6}\text{P}_{0.4}$ LED but are built on **GaP substrates**. The GaP:N diode (Fig. 9.14b) needs no substrate grading layer, and both n and p regions are epitaxial. Photons originate from the trap ($E_{ph}<E_G$), so the epitaxial layers and the GaP substrate are transparent; even downward light reflects and may escape, giving higher efficiency and brightness. With GaP:Zn-O they hold the largest LED market share. #### GaP:Zn-O Zn alone is an acceptor and O alone a donor in GaP. Added in about equal numbers and annealed, Zn on a Ga site and O on an adjacent P site form one isoelectronic trap complex, deeper than N: about $0.3$ eV below $E_C$. Emission is red, $\lambda_{peak}=0.700\ \mu$m. Fabrication parallels GaP:N. #### AlGaAs Introduced in the late 1980s as the "high-brightness" red LED, up to 20 times more luminous than GaAsP:N and GaP:Zn-O red LEDs. Uses: automobile spoiler (third) brake lights (about 10% of the market) and athletic-shoe heel lights. Structure (Fig. 9.14c): a narrow active layer ($E_G\le1.9$ eV) between slightly wider-gap confining layers (minimize reabsorption, maximize extraction of randomly directed light); in the brightest devices a thick (100–200 $\mu$m) epitaxial layer is grown below the lower confining layer and the GaAs substrate removed, so $E_G>E_{ph}$ below the emitter too. AlGaAs with $E_G<2$ eV is direct (Fig. 9.6) and all AlGaAs alloys are lattice matched to GaAs. Lattice-matched perfection (minimal non-radiative recombination) plus direct recombination (about an order of magnitude more efficient than via isoelectronic traps) give high efficiency and brightness. #### AlInGaP On sale from 1992; three Column III elements plus one Column V element. Already beats the established orange, yellow and green devices and may become the dominant high-performance technology from red through green. Structure (Fig. 9.14d): confining and active layers lattice matched to GaAs, direct recombination; colour set by active-layer composition. GaP is the top contact/window because its conductivity far exceeds that of p-type AlInGaP. #### SiC Electroluminescence from SiC (surely with built-in pn junctions) was reported in 1907. SiC LEDs were intensely researched in the 1960s and briefly sold in the early 1970s, then offered again in 1990 after a substrate-quality breakthrough, given the need for a blue LED. Efficiency improved only to 0.02% because operation still relies on low-probability radiative recombination in an indirect material. #### GaN Direct III-V with $E_G=3.36$ eV. The blue GaN LED was first sold in April 1994, developed by a chemical company thanks to a breakthrough in forming pn junctions in GaN and its alloys. Light is actually produced in an $E_G=2.75$ eV InGaN film sandwiched between wider-gap AlGaN layers. Expected to overtake SiC; it supplies the missing colour for full-colour LED outdoor video displays at reasonable power. ### 9.4.3 LED Packaging and Photon Extraction **Package (Fig. 9.15).** The chip, about $250\ \mu$m on a side, sits in a reflecting cup because a portion of the light leaves from the chip sides. One lead connects to the cup; a wire bond joins the chip's top contact to the other lead. An epoxy dome holds the leadframe and helps photon extraction. **Extraction is harder than generation.** - Wide-gap overlayers act as windows. - N- and Zn-O-doped LEDs need no window since $E_{ph}$ is at least $0.1$ eV below the host gap. - Ideally $E_G>E_{ph}$ also below the emitting layer, since half the light starts downward; the light can then bounce around with minimal reabsorption and eventually hit the surface. **Optics at the surface (Fig. 9.16).** At normal incidence the amplitude reflection coefficient is $r=(n_1-n_2)/(n_1+n_2)$, so the transmitted power fraction is $$T_N=1-r^2=\frac{(n_1+n_2)^2-(n_1-n_2)^2}{(n_1+n_2)^2}=\frac{4n_1n_2}{(n_1+n_2)^2}$$ Dividing top and bottom by $n_2^2$ with $\bar n=n_1/n_2$: $$T_N=\frac{4\bar n}{(1+\bar n)^2}\qquad(9.8)$$ For $n_1>n_2$ (semiconductor into another medium, so $\bar n>1$), light beyond the critical angle undergoes total internal reflection ($T=0$). Snell's law $n_1\sin\theta_c=n_2\sin90^\circ$ gives $$\sin\theta_c=\frac{n_2}{n_1}=\frac1{\bar n}\qquad(9.9)$$ **Overall transmission (9.10).** Assume isotropic, randomly polarized internal light, only the upward half able to exit. Flux through a surface element from angle $\theta$ is proportional to $\cos\theta\sin\theta\,d\theta$. The fraction of upward flux within the escape cone: $$\frac{\int_0^{\theta_c}\cos\theta\sin\theta\,d\theta}{\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta}=\frac{\tfrac12\sin^2\theta_c}{\tfrac12}=\sin^2\theta_c$$ Inside the (small) cone the transmission is close to $T_N$; multiply by $\tfrac12$ for the upward half: $$T\approx\frac{T_N\sin^2\theta_c}{2}=\frac12\cdot\frac{4\bar n}{(1+\bar n)^2}\cdot\frac1{\bar n^2}=\frac{2}{\bar n(1+\bar n)^2}\qquad(9.10)$$ **Example: GaP into air.** $n_1=3.4$, $n_2=1$, $\bar n=3.4$: $$T_N=\frac{4(3.4)}{(4.4)^2}=\frac{13.6}{19.36}=0.70$$ $$\theta_c=\sin^{-1}\frac1{3.4}=17.1^\circ$$ $$T=\frac{2}{3.4(4.4)^2}=\frac{2}{65.8}=0.030\ (3.0\%)$$ Most light is trapped by total internal reflection. **Epoxy** (index between air and semiconductor) widens the escape cone at the chip interface, and the **conical dome** lets much light reach the epoxy-air surface within its critical angle. ### Chapter 9 Recap (study aid) *The OCR slice has no formal Chapter 9 summary or problem set, so this recap stands in.* 1. $\lambda_G=1.24/E_G$ separates absorbing from transparent behaviour. 2. Photodiode: $I=I_{dark}+I_L$, $I_L=-qA(L_N+W+L_P)G_L$; a bias-independent downward shift proportional to light intensity. 3. p-i-n: wide constant-field depletion region, tailored response, $f_{max}\approx v_{sat}/W_I$ (RC limited at small $W_I$). APD: multiplication gain with improved S/N. 4. Solar cell: fourth-quadrant operation, $FF=P_{max}/(I_{SC}V_{OC})$, $\eta=FF\,I_{SC}V_{OC}/P_{in}$; losses from sub-gap photons, excess photon energy, shadowing, reflection, incomplete absorption and recombination. 5. LED: direct gap, 1.7–3.1 eV, junction-capable; achieved through alloys and isoelectronic traps. Extraction limited by total internal reflection, $T=2/[\bar n(1+\bar n)^2]$. --- # Chapter 10 — BJT Fundamentals > **The big idea.** A diode has one junction and two terminals. Put two junctions so close together that carriers can run from one to the other before they have time to die, and you get a transistor. Picture a relay race: the emitter hands the baton (carriers) to the base, which is so short that nearly every runner crosses it, and the collector catches them. A tiny trickle of effort in the base lane controls a big stream in the main lane. That is amplification. Chapters 5–8 analysed the pn diode. This chapter begins the **bipolar junction transistor (BJT)**: terms and symbols, qualitative operation, key relations; the knowledge base for the quantitative analysis in the next chapter. ## 10.1 Terminology A BJT has three adjoining, alternately doped regions, the middle one being very narrow compared with the minority-carrier diffusion length in it. pnp and npn types are shown in Fig. 10.1. The narrow centre is the **base**; the outer regions are the **emitter** and **collector**. They are not interchangeable: the emitter has a different geometry and is typically **more heavily doped** than the collector, so swapping terminals changes the characteristics substantially. **Symbols and polarities (Fig. 10.2).** For the pnp: $I_E$ flows into the emitter, $I_B$ and $I_C$ flow out; $V_{EB}$, $V_{EC}$, $V_{CB}$ are voltage drops. The "+" and "−" signs are redundant with the double subscripts: the first letter is the terminal assumed more positive, so $V_{EB}$ is the drop from emitter (+) to base (−). Some chosen current directions are contrary to the convention "into a terminal is positive", but with these choices all terminal currents are positive in standard amplifying operation, which keeps the physical discussion uncomplicated. (For npn, $I_C$ and $I_B$ flow in and $I_E$ out, with $V_{BE}$, $V_{CE}$, $V_{BC}$.) Only two currents and two voltages are independent. Current into the device equals current out (Kirchhoff's current law) and voltage drops around a closed loop sum to zero (Kirchhoff's voltage law): $$I_E=I_B+I_C\qquad(10.1)$$ $$V_{EB}+V_{BC}+V_{CE}=0\qquad(V_{CE}=-V_{EC})\qquad(10.2)$$ If two currents (or voltages) are known, the third follows. **Circuit configurations (Fig. 10.3).** One lead must be common to input and output; the name of that lead names the configuration: | Configuration | Common lead | Input | Output | Notes | |---|---|---|---|---| | Common base | base | $I_E$, $V_{EB}$ | $I_C$, $V_{CB}$ | occasional use | | Common emitter | emitter | $I_B$, $V_{EB}$ | $I_C$, $V_{EC}$ | most widely used | | Common collector | collector | $I_B$, $V_{CB}$ | $I_E$, $V_{EC}$ | seldom used, neglected from here on | Figure 10.4 shows idealized output characteristics of a pnp. (a) Common base: $I_C$ (0–5 mA) versus $-V_{CB}$ (0–60 V... as plotted, nearly flat lines at $I_C\approx I_E$ for $I_E=0,1,2,3,4$ mA). (b) Common emitter: $I_C$ (0–16 mA) versus $V_{EC}$ (0–20 V) for $I_B=0,10,20,30,40\ \mu$A, rising lines with a modest upward slope. **Biasing modes (Table 10.1, Fig. 10.5).** Each junction is forward (F) or reverse (R) biased: four combinations. | Mode | E-B junction | C-B junction | Typical use | |---|---|---|---| | Saturation | Forward | Forward | switch "on": high current, low voltage | | Active | Forward | Reverse | linear amplifiers; largest gain, least distortion | | Inverted | Reverse | Forward | roles of emitter and collector swapped | | Cutoff | Reverse | Reverse | switch "off": low current, high voltage | In digital circuits the low-voltage and high-voltage states of saturation and cutoff correspond to logic levels. Figure 10.5(a) is a plane with axes $V_{EB}$ (pnp) or $V_{BE}$ (npn) and $V_{CB}$ (pnp) or $V_{BC}$ (npn), the four quadrants being the four modes; Fig. 10.5(b) marks the regions of the common-emitter output characteristics: cutoff at $I_B\le0$ (along the $V_{EC}$ axis), active in the flat-ish region, saturation at small $V_{EC}$, and the inverted-active region for negative $V_{EC}$. The development mostly uses a **pnp** because it is a convenient vehicle for the concepts; the npn is easily obtained by swapping carrier types and polarities. ## 10.2 Fabrication An idealized pseudo-one-dimensional model (Fig. 10.1) is used for qualitative and first-order quantitative work. Fabrication extends diode fabrication (Chapter 4); the new feature is two pn junctions in close proximity. **Discrete double-diffused pnp (Fig. 10.6a).** Start with a heavily doped $p^+$ wafer (mechanical support, low-resistance path to the bottom collector contact). Grow a high-resistivity (lightly doped) $p$ epilayer 5–10 $\mu$m thick; this contains the transistor proper (it becomes the collector). Grow an oxide, open windows, diffuse phosphorus or arsenic to form the n-type base. Repeat with boron to form the $p^+$ emitter. Deposit aluminium and pattern the emitter and base contacts. (Dimensions in the figure are not to scale; the lateral scale should be expanded about 50 times for a true picture: the active region is about $5\ \mu$m deep while the wafer is about $200\ \mu$m thick.) **IC npn (Fig. 10.6b).** Diffuse p-type then n-type into a high-resistivity n-type epilayer. The epilayer has doping opposite to the substrate, and **diode isolation** is completed by a deep p-type diffusion around the transistor. The most negative circuit potential is applied to the p-substrate, so the n-epi region is isolated by a reverse-biased pn junction from other devices. Advanced designs replace the lateral junctions with silicon dioxide isolation. The **$n^+$ buried layer**, diffused before the epilayer is grown, gives a low-resistance path between the active collector region and the top-side collector contact. ## 10.3 Electrostatics Under equilibrium and normal operating conditions, the BJT is electrostatically just **two independent pn junctions**. Chapter 5 results (built-in potential, charge density, field, potential, depletion width) apply separately to the E-B and C-B junctions. For a standard uniformly doped pnp with $N_{AE}\ge N_{DB}>N_{AC}$ the equilibrium band diagram and variables are as in Fig. 10.7 (depletion approximation). **Reading Fig. 10.7.** - Almost all of the E-B depletion width $W_{EB}$ lies in the base; most of the C-B depletion width $W_{CB}$ lies in the collector. - $W_{CB}>W_{EB}$ because the lightly doped side of the C-B junction (collector) is more lightly doped than that of the E-B junction (base). - $W_B$ is the total (metallurgical) base width; $W$ is the **quasineutral base width**, the undepleted part: $$W=W_B-x_{nEB}-x_{nCB}\qquad(10.3)$$ where $x_{nEB}$, $x_{nCB}$ are the parts of the E-B and C-B depletion widths lying in the n-type base. This $W$ is not the diode depletion width of earlier chapters. Useful formulas (from Chapter 5, applied per junction, with $N_l$ and $N_r$ the doping on either side): $$V_{bi}=\frac{kT}{q}\ln\frac{N_AN_D}{n_i^2},\qquad W_{dep}=\sqrt{\frac{2\varepsilon V_{bi}}{q}\left(\frac1{N_A}+\frac1{N_D}\right)}$$ $$x_n=W_{dep}\frac{N_A}{N_A+N_D},\qquad x_p=W_{dep}\frac{N_D}{N_A+N_D},\qquad |\mathscr{E}|_{max}=\frac{qN_Dx_n}{\varepsilon}=\frac{qN_Ax_p}{\varepsilon}=\frac{2V}{W_{dep}}$$ For a general bias $V$ across a junction (forward positive): replace $V_{bi}$ by $V_{bi}-V$. ### Exercise 10.1 **Problem.** Draw the energy band diagram of (a) an npn transistor in equilibrium; (b) a pnp transistor under active-mode biasing. Assume uniform doping and the standard profile: emitter doping > base doping > collector doping. **Solution.** (a) An npn is the complement of a pnp, so its diagram is the pnp diagram (Fig. 10.7b) turned upside down (equivalently, combine the equilibrium diagrams of an np and a pn junction). Features: - $E_F$ is a single flat line. - Emitter (n$^+$): $E_C$ very close to $E_F$ (just above it, or even at it). - Base (p): $E_F$ close to $E_V$, so $E_C$ is far above $E_F$; $E_C$ in the base is higher than in the emitter and collector: a hill in $E_C$ for electrons. - Collector (n, lightest doping): $E_F$ sits between $E_i$ and $E_C$, farther below $E_C$ than in the emitter. - The emitter-base hill is $qV_{bi,EB}$ and the base-collector hill $qV_{bi,CB}$, with $V_{bi,EB}>V_{bi,CB}$ because the emitter is more heavily doped than the collector. The E-B depletion region is narrower than the C-B one and penetrates mostly into the base; the C-B depletion region penetrates mostly into the collector. (b) Active mode: E-B forward biased, C-B reverse biased. Starting from Fig. 10.7(b): - Forward bias on E-B raises the base-side Fermi level (electron energy) relative to the emitter-side one by $qV_{EB}$; both the E-B depletion width and potential hill are reduced. - Reverse bias on C-B lowers the base-side Fermi level relative to the collector side by $q|V_{CB}|$; the C-B depletion width and potential hill increase. - The result: $E_F$ is not flat but separated in three parts: emitter at the lowest electron energy, collector at the highest, with the base level between them (for pnp, $E_{F}$ steps up from emitter to base by $qV_{EB}$ and from base to collector by $q|V_{CB}|$). The hole potential energy hill for holes in the valence band is low at the E-B junction and the holes injected into the base roll down the large drop at the C-B junction. ### Exercise 10.2 (computer): equilibrium band diagram of a BJT **Problem.** Write a MATLAB program that draws the equilibrium energy band diagram of a nondegenerately doped Si pnp or npn BJT with uniformly doped regions at room temperature. **Logic of the solution.** The book's file `BJT_Eband` (code in Appendix M) uses the defaults $N_{AE}=10^{18}$, $N_{DB}=10^{16}$, $N_{AC}=10^{15}\ \text{cm}^{-3}$, $W_B=1\ \mu$m (doping values input with n-type negative). The algorithm is: 1. For each region compute $E_i-E_F=\pm kT\ln(|N|/n_i)$ (positive for p-type, negative for n-type). 2. At each of the two junctions, $V_{bi}=|\,(E_i-E_F)_{left}-(E_i-E_F)_{right}|/q$; compute the depletion width and its split $x_l,x_r$. 3. Inside the depletion region of a junction, the band-edge shift is parabolic: on the left side it moves by the fraction $x_l/W_{dep}$ of the total change as $(s/x_l)^2$; on the right side the remaining fraction approaches the bulk value as $((W_{dep}-s)/x_r)^2$. 4. Build $E_i(x)$, then $E_C=E_i+E_G/2$, $E_V=E_i-E_G/2$ (approximately, ignoring the small difference between $E_i$ and midgap), with $E_F=0$ as reference, and plot. ```matlab % BJT equilibrium band diagram (uniformly doped, depletion approximation) % Doping convention: p-type positive, n-type negative (cm^-3) clear; clc; DOPING = [1e18 -1e16 1e15]; % [emitter base collector] pnp example WIDTH = 1e-4; % total base width WB in cm (1 um) T = 300; kT = 8.617e-5*T; % eV ni = 1e10; EG = 1.12; q = 1.602e-19; eps = 11.8*8.854e-14; % F/cm % step 1: Ei - EF in each region (eV) EiEF = sign(DOPING).*kT.*log(abs(DOPING)/ni); % step 2: junction parameters for j = 1:2 Nl = abs(DOPING(j)); Nr = abs(DOPING(j+1)); Vbi = abs(EiEF(j) - EiEF(j+1)); % volts Wd(j) = sqrt(2*eps*Vbi/q*(1/Nl + 1/Nr)); % cm xl(j) = Wd(j)*Nr/(Nl+Nr); % left depletion edge xr(j) = Wd(j)*Nl/(Nl+Nr); % right depletion edge end % positions: junction 1 at x=0, junction 2 at x=WIDTH xc = [0 WIDTH]; x = linspace(-2*xl(1)-0.5e-4, WIDTH + 2*xr(2)+3e-4, 4000); Ei = zeros(size(x)); for k = 1:numel(x) xx = x(k); % default: bulk value of the region if xx < WIDTH/2, Ei(k) = EiEF(1)*(xx<0) + EiEF(2)*(xx>=0); else, Ei(k) = EiEF(2)*(xx<=WIDTH) + EiEF(3)*(xx>WIDTH); end for j = 1:2 s = xx - (xc(j) - xl(j)); % distance from left edge D = EiEF(j+1) - EiEF(j); % total change of Ei if s >= 0 && s <= xl(j) Ei(k) = EiEF(j) + D*(xl(j)/Wd(j))*(s/xl(j))^2; elseif s > xl(j) && s <= Wd(j) Ei(k) = EiEF(j+1) - D*(xr(j)/Wd(j))*((Wd(j)-s)/xr(j))^2; end end end Ec = Ei + EG/2; Ev = Ei - EG/2; % EF = 0 reference % sign: Ei-EF>0 means Ei above EF, i.e. p-type has Ei above EF plot(x*1e4, Ec, 'b', x*1e4, Ei, 'k--', x*1e4, Ev, 'r', ... x*1e4, zeros(size(x)), 'g'); xlabel('x (\mum)'); ylabel('Energy (eV)'); legend('E_C','E_i','E_V','E_F'); title('Equilibrium BJT band diagram'); ``` *Sign check.* For a p-type region $E_F$ lies below $E_i$, so $E_i-E_F>0$; the code uses $+kT\ln(N_A/n_i)$ for positive $N$, which is correct. The same code draws an npn by entering e.g. `DOPING = [-1e18 1e16 -1e15]`. ## 10.4 Introductory Operational Considerations Consider a pnp in active mode and watch the holes in and near the base (Fig. 10.8). The main activity at the forward-biased E-B junction is **majority-carrier injection** across the junction into the opposite-side quasineutral regions. The $p^+$-n nature means many more holes are injected from emitter into base than electrons from base into emitter. **The key to transistor action** is what happens to carriers injected into the base. If $W$ were much larger than a diffusion length, the injected holes would recombine in the n-base and there would be no interaction between junctions: just two back-to-back diodes. But the BJT's base is narrow compared with a minority-carrier diffusion length, so the vast majority of injected holes diffuse completely across the quasineutral base and enter the C-B depletion region, where the accelerating field sweeps them into the collector. The narrow base thus *couples* the E-B and C-B currents. The names make sense: the emitter *emits* carriers into the base; the collector (the reverse-biased side of the C-B junction) *collects* them after they cross the control (base) region. **Currents (Fig. 10.9)**, neglecting R-G currents in the depletion regions initially: - $I_{Ep}$: hole diffusion current across the E-B junction (holes injected into the base). - $I_{Cp}$: hole diffusion current across the C-B junction, almost entirely holes that crossed the base. Very few are lost by recombination in a well-made transistor, so $I_{Cp}\approx I_{Ep}$. - $I_{En}$: electron current injected from base into emitter (the minority-carrier diffusion current on the heavily doped side). - $I_{Cn}$: electrons that are minority carriers in the collector and wander into the C-B depletion region to be swept into the base (a small reverse-bias current). $$I_E=I_{Ep}+I_{En}\qquad(10.4)$$ $$I_C=I_{Cp}+I_{Cn}\qquad(10.5)$$ Since both electron components are small compared with the hole components and $I_{Cp}\approx I_{Ep}$, it follows that $I_C\approx I_E$, a well-known property. $I_B=I_E-I_C$ is small compared with both. Inspection of Fig. 10.9 agrees: the three components of base current are $I_{B1}=I_{En}$ (electrons injected into the emitter, resupplied through the base lead), $I_{B2}$ (electrons that replace those lost by recombination with injected holes in the base) and $I_{B3}=-I_{Cn}$ (the reverse current into the collector, flowing opposite): all small. **Amplification (Fig. 10.10).** In common emitter, output current is $I_C$, input is $I_B$, and the d.c. gain is $I_C/I_B$. $I_B$ (an electron current in a pnp) and $I_C$ (predominantly hole current) are tied together by the E-B junction: raising $I_B$ proportionally raises $I_{Ep}$ and hence $I_C$. Control of large $I_C$ by small $I_B$ is possible because the coupled-junction arrangement splits the small electron current and the large hole current crossing the E-B junction into **two separate current loops**: the hole loop passes through emitter, base, collector; the electron loop closes through the base lead and emitter. ## 10.5 Performance Parameters Parameters for amplifier performance: *emitter efficiency* and *base transport factor* (internal operation), and *d.c. current gains* (external operation). ### Emitter efficiency If the hole current injected across E-B increased while total emitter current stayed constant, the electron current would drop and overall gain would increase. Treating $I_E$ as emitter input and $I_{Ep}$ as the useful output: $$\gamma=\frac{I_{Ep}}{I_E}=\frac{I_{Ep}}{I_{Ep}+I_{En}}\qquad\text{(pnp)}\qquad(10.6)$$ $0\le\gamma\le1$; gain is maximized by making $\gamma$ as close to 1 as possible. In words: the fraction of emitter current that is carried by the useful (majority-of-emitter) carriers injected into the base. ### Base transport factor The fraction of the minority carriers injected into the base that successfully cross the quasineutral base and enter the collector. Number injected $\propto I_{Ep}$; number reaching the collector $\propto I_{Cp}$: $$\alpha_T=\frac{I_{Cp}}{I_{Ep}}\qquad\text{(pnp)}\qquad(10.7)$$ $0\le\alpha_T\le1$; recombination in the base degrades it. Again maximum amplification needs $\alpha_T\to1$. ### Common-base d.c. current gain In common base, the active-mode output characteristics (Fig. 10.4a, $-V_{CB}>0$) follow $$I_C=\alpha_{dc}I_E+I_{CBO}\qquad(10.8)$$ with $\alpha_{dc}$ the common-base d.c. current gain and $I_{CBO}$ the collector current when $I_E=0$ ("C-B, emitter open"). Using (10.6) and (10.7): $$I_{Ep}=\gamma I_E\qquad(10.9a)$$ $$I_{Cp}=\alpha_TI_{Ep}=\alpha_T\gamma I_E\qquad(10.9b)$$ (Eq. 10.9 in the book collects these relations.) Then $$I_C=I_{Cp}+I_{Cn}=\gamma\alpha_TI_E+I_{Cn}\qquad(10.10)$$ Comparing (10.8) with (10.10): $$\alpha_{dc}=\gamma\,\alpha_T\qquad(10.11)$$ $$I_{CBO}=I_{Cn}\qquad(10.12)$$ (10.11) links the external gain to the internal parameters, and $0\le\alpha_{dc}\le1$. ### Common-emitter d.c. current gain In common emitter, active-mode characteristics (Figs. 10.4b, 10.5b) follow $$I_C=\beta_{dc}I_B+I_{CEO}\qquad(10.13)$$ with $I_{CEO}$ the collector current when $I_B=0$. A second relation comes from substituting $I_E=I_C+I_B$ into (10.8): $$I_C=\alpha_{dc}(I_C+I_B)+I_{CBO}\qquad(10.14)$$ Collect $I_C$ terms: $$I_C(1-\alpha_{dc})=\alpha_{dc}I_B+I_{CBO}$$ $$I_C=\frac{\alpha_{dc}}{1-\alpha_{dc}}I_B+\frac{I_{CBO}}{1-\alpha_{dc}}\qquad(10.15)$$ Comparing (10.13) and (10.15): $$\beta_{dc}=\frac{\alpha_{dc}}{1-\alpha_{dc}}\qquad(10.16)$$ $$I_{CEO}=\frac{I_{CBO}}{1-\alpha_{dc}}=(\beta_{dc}+1)I_{CBO}\qquad(10.17)$$ and, if $I_{CEO}\ll I_C$ at the operating point, $$\beta_{dc}\approx\frac{I_C}{I_B}\qquad(10.18)$$ Since $\alpha_{dc}$ is typically near unity, $\beta_{dc}\gg1$; this agrees with (10.18) since $I_C\gg I_B$. *Intuition for $\beta$.* If $\alpha_{dc}=0.99$, then $\beta=0.99/0.01=99$. If $\alpha_{dc}$ improves from 0.99 to 0.995, $\beta$ doubles to 199. The gain depends on how close to 1 $\alpha_{dc}$ gets, the "loss fraction" $1-\alpha_{dc}$ that must leak out through the base. ## 10.6 Summary This chapter covered the basics of the BJT: terminal current and voltage symbols, polarities and relations; circuit configurations and biasing modes and the general form of the output characteristics; fabrication, to relate the real structure to the one-dimensional model used later; electrostatics (the BJT is just two independent pn junctions in equilibrium and under normal operation); the carrier activity and currents in active mode that explain how the device works; and the performance parameters, namely emitter efficiency $\gamma$, base transport factor $\alpha_T$ and the d.c. current gains $\alpha_{dc}=\gamma\alpha_T$ and $\beta_{dc}=\alpha_{dc}/(1-\alpha_{dc})$, interrelated through (10.11)–(10.18). --- ## Chapter 10 Problems **Problem information table (from the book).** | Problem | After section | Difficulty | Points | Topic | |---|---|---|---|---| | 10.1 | 10.6 | – | 10 | Quick quiz | | 10.2 | 10.1 | 1 | 4 | Mode tables | | 10.3 | 10.3 | 2–3 | 22 | pnp equilibrium electrostatics | | 10.4 | 10.3 | 2–3 | 22 | npn equilibrium electrostatics | | 10.5 | 10.3 | 2–3 | 10 | pnp active-mode electrostatics | | 10.6 | 10.3 | 2 | 8 | npn active-mode electrostatics | | 10.7 | 10.4 | 3 | 12 | pnp carrier activity/currents | | 10.8 | 10.4 | 3 | 12 | npn carrier activity/currents | | 10.9 | 10.5 | 2 | 11 | pnp parameter calculations | | 10.10 | 10.5 | 2 | 11 | npn parameter calculations | | 10.11 | 10.5 | 3 | 10 | $I_{CBO}$ thought problem | **Constants used:** $kT/q=0.0259$ V, $n_i=10^{10}\ \text{cm}^{-3}$, $\varepsilon_{Si}=11.8\times8.854\times10^{-14}=1.045\times10^{-12}$ F/cm, $q=1.6\times10^{-19}$ C. ### Problem 10.1 Quick quiz **Problem.** Answer concisely: (a) name of the circuit configuration where the input current is... and the output current is...; (b) name of the configuration where $I_C$ is the output current and $V_{E?}$ is the output voltage; (c) four biasing modes; (d) purpose of the buried layer in IC BJTs; (e) relative sizes of $N_{AE}$, $N_{DB}$, $N_{AC}$ in a standard pnp; (f) standard symbols for diode depletion width and BJT quasineutral base width; (g) precise definition of "narrow" base; (h) why the base must be narrow; (i) emitter efficiency in words; (j) base transport factor in words. **Solution.** (The OCR of (a) and (b) is garbled. Assumption: (a) input current $I_B$, output $I_C$; (b) output voltage $V_{EC}$. Both then describe the common emitter; if (a) instead meant input $I_E$, output $I_C$, the answer is common base.) (a) Input $I_B$, output $I_C$ with the emitter common: **common emitter** (if input $I_E$: common base). (b) $I_C$ output, output voltage $V_{EC}$ (collector-to-emitter): **common emitter**. (An output voltage $V_{CB}$ would signal common base.) (c) **Saturation, active (forward active), inverted, cutoff.** (d) The $n^+$ buried layer provides a **low-resistance path** between the active collector region and the top-side collector contact. (e) $N_{AE}\ge N_{DB}>N_{AC}$ (emitter most heavily doped, collector least). (f) $W$ for the pn diode depletion width; $W$ also denotes the BJT quasineutral base width (with $W_B$ the total metallurgical base width, $W=W_B-x_{nEB}-x_{nCB}$ for a pnp). (g) "Narrow" means the quasineutral base width is much smaller than the minority-carrier diffusion length in the base: $W\ll L_{min}$ ($L_P$ for a pnp, $L_N$ for an npn). (h) A narrow base lets nearly all carriers injected from the emitter diffuse across and be swept into the collector instead of recombining; this couples the two junctions (transistor action) and keeps $I_B$ small and the gain large. (i) Emitter efficiency: the fraction of the total emitter current that is carried by the carriers injected into the base (the useful part). (j) Base transport factor: the fraction of the minority carriers injected into the base that reach the collector without recombining. ### Problem 10.2 Mode tables **Problem.** Give the polarities (+ or −) of the input and output junction voltages in each of the four biasing modes: (a) pnp with $V_{EB}$, $V_{CB}$; (b) npn with $V_{BE}$, $V_{BC}$. **Solution.** A junction is forward biased when its p-side is more positive than its n-side. *pnp.* E-B: emitter (p) minus base (n) is $V_{EB}$; forward means $V_{EB}>0$. C-B: collector (p) minus base (n) is $V_{CB}$; forward means $V_{CB}>0$, reverse means $V_{CB}<0$. | Mode | $V_{EB}$ | $V_{CB}$ | |---|---|---| | Active | + | − | | Inverted | − | + | | Saturation | + | + | | Cutoff | − | − | *npn.* E-B: base (p) minus emitter (n) is $V_{BE}$; forward means $V_{BE}>0$. C-B: base (p) minus collector (n) is $V_{BC}$; forward means $V_{BC}>0$. | Mode | $V_{BE}$ | $V_{BC}$ | |---|---|---| | Active | + | − | | Inverted | − | + | | Saturation | + | + | | Cutoff | − | − | ### Problem 10.3 Equilibrium electrostatics of a pnp **Problem.** A Si pnp BJT with $N_{AE}=5\times10^{17}$, $N_{DB}=10^{15}$ (OCR reads "$10^5$"; taken as $10^{15}$), $N_{AC}=10^{14}\ \text{cm}^{-3}$, $W_B=3\ \mu$m at room temperature in equilibrium. (a) Sketch the band diagram with the Fermi level in the three regions. (b) Sketch (i) electrostatic potential with $V=0$ in the emitter region, (ii) the electric field, (iii) the charge density versus position. (c) Calculate the net potential difference between collector and emitter. (d) Find the quasineutral base width. (e) Find the maximum electric fields in the E-B and C-B depletion regions. **Solution.** *Fermi level positions.* For p-type $E_i-E_F=kT\ln(N_A/n_i)$; for n-type $E_F-E_i=kT\ln(N_D/n_i)$. $$E_i-E_F\big|_E=0.0259\ln\frac{5\times10^{17}}{10^{10}}=0.0259\ln(5\times10^{7})=0.0259(17.73)=0.459\ \text{eV}$$ $$E_F-E_i\big|_B=0.0259\ln\frac{10^{15}}{10^{10}}=0.0259(11.51)=0.298\ \text{eV}$$ $$E_i-E_F\big|_C=0.0259\ln\frac{10^{14}}{10^{10}}=0.0259(9.21)=0.239\ \text{eV}$$ (a) Band diagram: a single flat $E_F$. In the emitter $E_F$ is $0.459$ eV below $E_i$ (so $E_F$ is $\approx0.10$ eV above $E_V$: $E_i-E_V\approx0.56$, and $0.56-0.459=0.10$); in the base $E_F$ is $0.298$ eV above $E_i$ ($0.26$ eV below $E_C$); in the collector $E_F$ is $0.239$ eV below $E_i$ ($0.32$ eV above $E_V$). The bands bend: $E_C$, $E_i$, $E_V$ in the base lie lower than in the emitter and collector... in electron energy terms the bands in the base lie lower in electron energy than in the emitter (a valley for electrons, a hill for holes): going emitter to base, $E_i$ falls by $0.757$ eV; going base to collector it rises by $0.537$ eV, a net fall of $0.221$ eV from emitter to collector. The depletion regions are narrow at E-B (mostly on the base side) and wide at C-B (mostly on the collector side). (c) Built-in potentials: $$V_{bi,EB}=0.459+0.298=0.757\ \text{V}\qquad V_{bi,CB}=0.298+0.239=0.537\ \text{V}$$ The potential is $V=-E_i/q$ (relative). Going emitter to collector the potential first rises by $V_{bi,EB}$ into the base and then falls by $V_{bi,CB}$ into the collector. Net: $$V_C-V_E=V_{bi,EB}-V_{bi,CB}=0.757-0.537=0.221\ \text{V}$$ (The collector is $0.221$ V higher than the emitter in electrostatic potential; equivalently $\phi_E-\phi_C$ computed from the doping: $-0.459+0.239$ in terms of $E_i-E_F$ gives the same magnitude.) (b) With $V=0$ in the emitter bulk: $V=0$ for $x$ in emitter bulk, rising in an S-shaped (parabolic) step of height $0.757$ V across the E-B depletion region (almost all in the base side, width about $1\ \mu$m) to $+0.757$ V in the quasineutral base, then falling in a step of $0.537$ V across the C-B depletion region (mostly in the collector, width about $2.8\ \mu$m) to $+0.221$ V in the collector bulk. Field $\mathscr{E}=-dV/dx$: zero in neutral regions; in the E-B depletion region a negative (pointing from n-base toward p-emitter... pointing from base to emitter) triangular profile, peak at the metallurgical junction; same direction triangle at the C-B junction but wider and of smaller peak (see (e)). Both triangles have the same sign since both junctions are pn junctions oriented with n-side in the middle. Charge density $\rho$ (depletion approximation): E-B: $-qN_{AE}$ over $x_{pEB}$ (very thin, in the emitter), $+qN_{DB}$ over $x_{nEB}$ (in the base); C-B: $+qN_{DB}$ over $x_{nCB}$ (in the base), $-qN_{AC}$ over $x_{pCB}$ (long, in the collector); zero elsewhere. (d) E-B depletion: $$W_{EB}=\sqrt{\frac{2\varepsilon V_{bi}}{q}\left(\frac1{N_{AE}}+\frac1{N_{DB}}\right)}$$ $$\frac{2\varepsilon V_{bi}}{q}=\frac{2(1.045\times10^{-12})(0.757)}{1.6\times10^{-19}}=9.89\times10^{6}\ \text{cm}^{-1}\text{ (units V cm/C... in cm}^{-3}\cdot\text{cm}^{2})$$ $$\frac1{N_{AE}}+\frac1{N_{DB}}=2\times10^{-18}+10^{-15}=1.002\times10^{-15}\ \text{cm}^3$$ $$W_{EB}=\sqrt{(9.89\times10^{6})(1.002\times10^{-15})}=\sqrt{9.91\times10^{-9}}=9.96\times10^{-5}\ \text{cm}=0.996\ \mu\text{m}$$ $$x_{nEB}=W_{EB}\frac{N_{AE}}{N_{AE}+N_{DB}}=0.996\left(\frac{5\times10^{17}}{5.01\times10^{17}}\right)=0.994\ \mu\text{m}$$ C-B depletion: $$\frac{2\varepsilon V_{bi}}{q}=\frac{2(1.045\times10^{-12})(0.537)}{1.6\times10^{-19}}=7.01\times10^{6}$$ $$\frac1{N_{AC}}+\frac1{N_{DB}}=10^{-14}+10^{-15}=1.1\times10^{-14}$$ $$W_{CB}=\sqrt{(7.01\times10^{6})(1.1\times10^{-14})}=\sqrt{7.71\times10^{-8}}=2.78\times10^{-4}\ \text{cm}=2.78\ \mu\text{m}$$ $$x_{nCB}=W_{CB}\frac{N_{AC}}{N_{AC}+N_{DB}}=2.78\left(\frac{10^{14}}{1.1\times10^{15}}\right)=0.252\ \mu\text{m}$$ $$x_{pCB}=2.78-0.252=2.525\ \mu\text{m}$$ Quasineutral base width, by (10.3): $$W=W_B-x_{nEB}-x_{nCB}=3-0.994-0.252=1.75\ \mu\text{m}$$ (e) Peak fields $|\mathscr{E}|_{max}=qN_Dx_n/\varepsilon$ on the n-side: $$|\mathscr{E}|_{max,EB}=\frac{(1.6\times10^{-19})(10^{15})(0.994\times10^{-4})}{1.045\times10^{-12}}=1.52\times10^{4}\ \text{V/cm}$$ $$|\mathscr{E}|_{max,CB}=\frac{(1.6\times10^{-19})(10^{15})(0.252\times10^{-4})}{1.045\times10^{-12}}=3.86\times10^{3}\ \text{V/cm}$$ Check with $2V_{bi}/W$: $2(0.757)/(0.996\times10^{-4})=1.52\times10^4$ and $2(0.537)/(2.78\times10^{-4})=3.86\times10^3$ V/cm. ### Problem 10.4 Equilibrium electrostatics of an npn **Problem.** A Si npn with $N_{DE}=10^{18}$, $N_{AB}=10^{16}$, $N_{DC}=10^{1?}\ \text{cm}^{-3}$ (OCR shows "$10^1$"; taken as $10^{14}$ by analogy with 10.3), $W_B=2\ \mu$m in equilibrium at room temperature. Parts (a)–(e) as in 10.3, with $V=0$ in the emitter. **Solution.** *Fermi levels.* $$E_F-E_i\big|_E=0.0259\ln(10^{8})=0.0259(18.42)=0.477\ \text{eV}$$ $$E_i-E_F\big|_B=0.0259\ln(10^{6})=0.0259(13.82)=0.358\ \text{eV}$$ $$E_F-E_i\big|_C=0.0259\ln(10^{4})=0.0259(9.21)=0.239\ \text{eV}$$ (a) Flat $E_F$. Emitter: $E_F$ $0.477$ eV above $E_i$ (about $0.08$ eV below $E_C$). Base: $E_F$ $0.358$ eV below $E_i$ ($0.20$ eV above $E_V$). Collector: $E_F$ $0.239$ eV above $E_i$. The base is a potential hill for electrons: $E_C$ in the base is higher than in the emitter by $0.835$ eV and higher than in the collector by $0.597$ eV; the E-B depletion region is narrow (mostly in the base), C-B wide (mostly in the collector). The diagram is the pnp diagram of 10.3 flipped upside down. (c) Built-in potentials: $$V_{bi,EB}=0.477+0.358=0.835\ \text{V}\qquad V_{bi,CB}=0.358+0.239=0.597\ \text{V}$$ $V=-E_i/q$ with $V=0$ in the emitter: the base potential is $0.835$ V lower (electrons see a hill), so $V_B=-0.835$ V, and the collector rises by $0.597$ V from the base: $$V_C-V_E=-V_{bi,EB}+V_{bi,CB}=-0.835+0.597=-0.238\ \text{V}$$ (b) Potential profile: $0$ in emitter, a parabolic step down to $-0.835$ V across the thin E-B depletion region, flat at $-0.835$ V in the base, a step up of $0.597$ V across the wide C-B depletion region to $-0.238$ V in the collector. Field: triangular, pointing (from n to p side) in the sense opposite to the pnp case, i.e. the field points from the n-side toward the p-side, namely from emitter toward base at E-B and from collector toward base at C-B; peaks at the junctions. Charge: E-B: $+qN_{DE}$ in a very thin layer ($x_{nEB}$) in the emitter, $-qN_{AB}$ over $x_{pEB}$ in the base; C-B: $-qN_{AB}$ over $x_{pCB}$ in the base (very thin), $+qN_{DC}$ over $x_{nCB}$ (wide) in the collector. (d) E-B junction: $$\frac{2\varepsilon V_{bi}}{q}=\frac{2(1.045\times10^{-12})(0.835)}{1.6\times10^{-19}}=1.091\times10^{7}$$ $$\frac1{N_{DE}}+\frac1{N_{AB}}=10^{-18}+10^{-16}=1.01\times10^{-16}$$ $$W_{EB}=\sqrt{(1.091\times10^{7})(1.01\times10^{-16})}=\sqrt{1.10\times10^{-9}}=3.32\times10^{-5}\ \text{cm}=0.332\ \mu\text{m}$$ $$x_{pEB}=W_{EB}\frac{N_{DE}}{N_{DE}+N_{AB}}=0.332\left(\frac{10^{18}}{1.01\times10^{18}}\right)=0.329\ \mu\text{m}$$ C-B junction: $$\frac{2\varepsilon V_{bi}}{q}=\frac{2(1.045\times10^{-12})(0.597)}{1.6\times10^{-19}}=7.80\times10^{6}$$ $$\frac1{N_{AB}}+\frac1{N_{DC}}=10^{-16}+10^{-14}=1.01\times10^{-14}$$ $$W_{CB}=\sqrt{(7.80\times10^{6})(1.01\times10^{-14})}=\sqrt{7.88\times10^{-8}}=2.81\times10^{-4}\ \text{cm}=2.81\ \mu\text{m}$$ $$x_{pCB}=W_{CB}\frac{N_{DC}}{N_{DC}+N_{AB}}=2.81\left(\frac{10^{14}}{1.01\times10^{16}}\right)=0.0278\ \mu\text{m}$$ $$W=W_B-x_{pEB}-x_{pCB}=2-0.329-0.028=1.64\ \mu\text{m}$$ (e) Peak fields: $$|\mathscr{E}|_{max,EB}=\frac{qN_{AB}x_{pEB}}{\varepsilon}=\frac{(1.6\times10^{-19})(10^{16})(0.329\times10^{-4})}{1.045\times10^{-12}}=5.04\times10^{4}\ \text{V/cm}$$ $$|\mathscr{E}|_{max,CB}=\frac{(1.6\times10^{-19})(10^{16})(0.0278\times10^{-4})}{1.045\times10^{-12}}=4.26\times10^{3}\ \text{V/cm}$$ Checks: $2(0.835)/(3.32\times10^{-5})=5.03\times10^{4}$; $2(0.597)/(2.81\times10^{-4})=4.25\times10^{3}$ V/cm. ### Problem 10.5 Active-mode electrostatics of the 10.3 BJT **Problem.** Biases $V_{EB}=0.5$ V and $V_{CB}=2$ V are applied to the Problem 10.3 BJT. (a) Sketch the band diagram with $E_F$ in the three regions. (b) Superimpose on the 10.3 sketches the potential, field and charge density. **Assumption.** The pnp is meant to be in active mode, so the C-B junction is reverse biased by $2$ V (the sign "−" of $V_{CB}=-2$ V was lost in the OCR). **Solution.** (a) Fermi level separation: the base-side $E_F$ is raised relative to the emitter by $qV_{EB}=0.5$ eV, and the collector $E_F$ is raised relative to the base by $q|V_{CB}|=2$ eV, so the collector $E_F$ lies $2.5$ eV above the emitter $E_F$. The E-B barrier falls from $0.757$ to $0.257$ eV, the C-B barrier rises from $0.537$ to $2.537$ eV. (b) New depletion widths. E-B (forward): $V_{bi}-V=0.757-0.5=0.257$ V: $$W_{EB}=0.996\sqrt{\frac{0.257}{0.757}}=0.996(0.583)=0.580\ \mu\text{m}$$ $$x_{nEB}=0.580(0.998)=0.579\ \mu\text{m}$$ C-B (reverse): $V_{bi}+|V|=0.537+2=2.537$ V: $$W_{CB}=2.78\sqrt{\frac{2.537}{0.537}}=2.78(2.174)=6.04\ \mu\text{m}$$ $$x_{nCB}=\frac{6.04}{11}=0.549\ \mu\text{m},\qquad x_{pCB}=5.49\ \mu\text{m}$$ Quasineutral base: $W=3-0.579-0.549=1.87\ \mu$m (up from $1.75\ \mu$m; forward bias shrinks the E-B region more than reverse bias grows the C-B one on the base side). Peak fields: $$|\mathscr{E}|_{max,EB}=\frac{2(0.257)}{0.580\times10^{-4}}=8.9\times10^{3}\ \text{V/cm}\ (\text{down from }1.52\times10^4)$$ $$|\mathscr{E}|_{max,CB}=\frac{2(2.537)}{6.04\times10^{-4}}=8.4\times10^{3}\ \text{V/cm}\ (\text{up from }3.86\times10^3)$$ Sketches: potential steps become $0.257$ V (E-B) and $2.537$ V (C-B). With emitter bulk at $V=0$, the base bulk is at $+0.257$ V and the collector bulk at $0.257-2.537=-2.28$ V, ... check with the biases: the emitter is $0.5$ V above the base and the base $2$ V above the collector (equilibrium offsets are $+0.757$ and $-0.537$ V, so the base sits at $0.757-0.5=0.257$ V and the collector at $0.221-2.5=-2.28$ V), i.e. the collector is $2.5$ V below the emitter and its electron energy is $2.5$ eV higher. Field triangles: E-B narrower and smaller; C-B wider and larger. Charge density rectangles: same heights as in 10.3 ($qN$ values) but E-B width reduced and C-B width increased. ### Problem 10.6 Active-mode electrostatics of an npn **Problem.** For a typically doped Si npn, sketch the band diagram, potential, field and charge density versus position in active mode. **Solution.** (Typical doping: $N_{DE}\gg N_{AB}>N_{DC}$, e.g. $10^{19},10^{17},10^{15}\ \text{cm}^{-3}$.) Active mode: $V_{BE}>0$ (forward), $V_{BC}<0$ (reverse). - *Band diagram:* flat Fermi level in each region, with separations $qV_{BE}$ (emitter $E_F$ lower... the emitter's electron energy is higher than the base's by $qV_{BE}$, so $E_F$ in the emitter is $qV_{BE}$ *above* the base $E_F$? For an npn forward-biased E-B junction, the emitter is at lower potential than the base, so emitter electrons have energy raised by $qV_{BE}$ relative to the base): $E_{F,E}=E_{F,B}+qV_{BE}$, and the collector at higher potential (lower electron energy): $E_{F,C}=E_{F,B}-q|V_{BC}|$. The emitter-base hill is reduced to $q(V_{bi,EB}-V_{BE})$ (electrons spill into the base); the base-collector hill increases to $q(V_{bi,CB}+|V_{BC}|)$, a steep downhill slope for electrons into the collector. - *Potential:* a small rise (reduced barrier) at E-B, then a large step at C-B. - *Field:* a small triangle at E-B (narrow, mostly in the base), a large wide triangle at C-B (mostly in the collector). - *Charge density:* E-B: thin $+qN_{DE}$ slab in the emitter, $-qN_{AB}$ slab of reduced width in the base; C-B: $-qN_{AB}$ slab in the base (thin) and a wide $+qN_{DC}$ slab in the collector. The quasineutral base width $W=W_B-x_{pEB}-x_{pCB}$ is reduced by the C-B reverse bias (base-width modulation). ### Problem 10.7 Saturation-biased pnp, $I_C>0$ **Problem.** A pnp is biased in saturation with $I_C>0$. Construct figures like Figs. 10.8 and 10.9 showing carrier activity and diffusion currents. **Solution (verbal description of the figures).** Both junctions are forward biased; both barriers are lowered. *Carrier activity (like Fig. 10.8):* - E-B: holes injected from the $p^+$ emitter into the n-base (large), and electrons from the base into the emitter (small). - C-B: now forward biased, so holes are injected from the p-collector into the base (a hole flow in the *opposite* direction to the emitter's flow), and electrons are injected from the base into the collector. - Base: holes from both ends in a narrow region; some recombine with base electrons, which are supplied by the base lead. - $I_C>0$ (standard direction, hole current leaving the collector) requires the emitter-injected hole current reaching the collector to exceed the collector-injected hole current going the other way. *Diffusion currents (like Fig. 10.9):* $$I_E=I_{Ep}+I_{En},\qquad I_C=I_{Cp}^{net}+I_{Cn}'\ ,$$ where $I_{Cp}^{net}=I_{Cp}^{(E\to C)}-I_{Cp}^{(C\to E)}$ is the difference between the hole current arriving from the emitter and the hole current injected by the collector, and $I_{Cn}'$ is now an electron current flowing from the base *into* the collector (the opposite sense to the active-mode reverse current, since the C-B junction is forward). The base current has three parts: electrons injected into the emitter, electrons injected into the collector, and electrons replacing recombination. All are larger than in active mode, so $I_B$ is larger and $I_C/I_B$ ($\beta$) is reduced; the hole concentration profile in the base is almost flat (high on both sides) instead of dropping linearly to zero at the collector edge, and the collector's current is small relative to $I_E$ (a low voltage $V_{EC}$ is the saturation hallmark). ### Problem 10.8 Cutoff-biased npn **Problem.** An npn is biased into cutoff. Construct figures like Figs. 10.8 and 10.9 showing carrier activity and diffusion currents. **Solution (verbal description of the figures).** Both junctions are reverse biased: both barriers are raised (larger than equilibrium), there is no injection. - *Carrier activity:* no majority-carrier injection across either junction. The only carriers moving are thermally generated minority carriers near the depletion edges: electrons that are minority carriers in the p-base and drift across the (reverse-biased) E-B and C-B depletion regions into the emitter and collector, and holes that are minority carriers in the emitter and in the collector and are swept into the base. Depletion regions are wider than at equilibrium. - *Diffusion currents:* the E-B current is a small reverse-saturation current (electrons from base into emitter, holes from emitter into base); the C-B current is also a small reverse-saturation current (electrons from base into collector, holes from collector into base). All terminal currents are tiny (including generation current in the depletion regions if included), and the minority-carrier concentrations in the base are depleted to near zero on both sides of the base: $$I_E\approx-(I_{En,sat}+I_{Ep,sat})\ ,\qquad I_C\approx-(I_{Cn,sat}+I_{Cp,sat})\ ,\qquad I_B=I_E-I_C\ \text{(a small sum)}$$ (signs correspond to currents opposite to forward direction). This is the off state: low current and high voltage. ### Problem 10.9 Parameter calculations for a pnp **Problem.** A pnp BJT has $I_{Ep}=1$ mA, $I_{En}=0.01$ mA, $I_{Cp}=0.98$ mA, $I_{Cn}=0.1\ \mu$A. Calculate (a) $\alpha_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) $\alpha_{dc}$ and $\beta_{dc}$, (e) $I_{CBO}$ and $I_{CEO}$. (f) If $I_{Cp}$ is increased toward $I_{Ep}$ with all other components fixed, effect on $\beta_{dc}$? (g) If $I_{En}$ is increased with all other components fixed, effect on $\beta_{dc}$? **Solution.** (a) By (10.7): $$\alpha_T=\frac{I_{Cp}}{I_{Ep}}=\frac{0.98}{1}=0.98$$ (b) By (10.6): $$\gamma=\frac{I_{Ep}}{I_{Ep}+I_{En}}=\frac{1}{1+0.01}=0.990$$ (c) By (10.4), (10.5), (10.1): $$I_E=I_{Ep}+I_{En}=1+0.01=1.01\ \text{mA}$$ $$I_C=I_{Cp}+I_{Cn}=0.98+0.0001=0.9801\ \text{mA}$$ $$I_B=I_E-I_C=1.01-0.9801=0.0299\ \text{mA}=29.9\ \mu\text{A}$$ (d) By (10.11) and (10.16): $$\alpha_{dc}=\gamma\alpha_T=(0.990099)(0.98)=0.9703$$ $$\beta_{dc}=\frac{\alpha_{dc}}{1-\alpha_{dc}}=\frac{0.9703}{0.0297}=32.7$$ (Check: $I_C/I_B=0.9801/0.0299=32.8$, consistent because $I_{CEO}$ is small.) (e) By (10.12) and (10.17): $$I_{CBO}=I_{Cn}=0.1\ \mu\text{A}$$ $$I_{CEO}=\frac{I_{CBO}}{1-\alpha_{dc}}=\frac{0.1\ \mu\text{A}}{0.0297}=3.37\ \mu\text{A}\;(=(\beta_{dc}+1)I_{CBO})$$ Check of (10.13): $I_C=\beta I_B+I_{CEO}=(32.67)(0.0299)+0.00337=0.9767+0.0034=0.9801$ mA. ✓. (f) Raising $I_{Cp}$ raises $\alpha_T=I_{Cp}/I_{Ep}$ (fewer injected holes lost to recombination in the base); $\gamma$ is unchanged, so $\alpha_{dc}=\gamma\alpha_T$ rises, and since $\beta_{dc}=\alpha_{dc}/(1-\alpha_{dc})$ is an increasing function of $\alpha_{dc}$, $\beta_{dc}$ **increases**. Physically $I_B=I_E-I_C$ falls while $I_C$ rises. (The limit is $I_{Cp}=I_{Ep}$, i.e. $\alpha_T=1$, giving $\alpha_{dc}=\gamma=0.990$, $\beta_{dc}=99.0$.) (g) Raising $I_{En}$ lowers $\gamma=I_{Ep}/(I_{Ep}+I_{En})$ with $\alpha_T$ fixed, so $\alpha_{dc}$ falls and $\beta_{dc}$ **decreases**. Physically, more of the emitter current is wasted on carriers that do not go to the collector, and $I_B$ rises. ### Problem 10.10 Parameter calculations for an npn **Problem.** An npn has $I_{En}=100\ \mu$A, $I_{Ep}=1\ \mu$A, $I_{Cn}=99\ \mu$A, $I_{Cp}=0.1\ \mu$A. Calculate (a) $\alpha_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) $\alpha_{dc}$, $\beta_{dc}$, (e) $I_{CBO}$, $I_{CEO}$. (f) $I_{Cn}$ is increased toward $100\ \mu$A with all else fixed: effect on $\beta_{dc}$? (g) (The OCR is garbled; the item is taken to be: $I_{Ep}$ is increased with all else fixed: effect on $\beta_{dc}$?) For an npn the roles of electrons and holes swap: the useful (injected into base) carriers are electrons, so the definitions are $\gamma=I_{En}/I_E$, $\alpha_T=I_{Cn}/I_{En}$, $I_{CBO}=I_{Cp}$. **Solution.** (a) $$\alpha_T=\frac{I_{Cn}}{I_{En}}=\frac{99}{100}=0.99$$ (b) $$\gamma=\frac{I_{En}}{I_{En}+I_{Ep}}=\frac{100}{101}=0.9901$$ (c) $$I_E=I_{En}+I_{Ep}=101\ \mu\text{A}$$ $$I_C=I_{Cn}+I_{Cp}=99+0.1=99.1\ \mu\text{A}$$ $$I_B=I_E-I_C=101-99.1=1.9\ \mu\text{A}$$ (d) $$\alpha_{dc}=\gamma\alpha_T=(0.9901)(0.99)=0.9802$$ $$\beta_{dc}=\frac{0.9802}{1-0.9802}=\frac{0.9802}{0.01980}=49.5$$ (e) $$I_{CBO}=I_{Cp}=0.1\ \mu\text{A}$$ $$I_{CEO}=\frac{I_{CBO}}{1-\alpha_{dc}}=\frac{0.1}{0.01980}=5.05\ \mu\text{A}$$ Check: $\alpha_{dc}I_E+I_{CBO}=(0.9802)(101)+0.1=99.0+0.1=99.1\ \mu$A ✓. (f) With $I_{Cn}\to100\ \mu$A (equal to $I_{En}$): $\alpha_T\to1$, $\alpha_{dc}\to\gamma=0.9901$, and $$\beta_{dc}\to\frac{0.9901}{0.0099}=100$$ so $\beta_{dc}$ **increases** (from $49.5$ to about $100$): fewer injected electrons are lost in the base, so $I_B$ falls. (g) Increasing $I_{Ep}$ (assumed) lowers $\gamma=I_{En}/(I_{En}+I_{Ep})$, so $\alpha_{dc}$ and $\beta_{dc}$ **decrease**: more of the emitter current is carried by holes that never reach the collector and must be supplied through the base lead. ### Problem 10.11 Thought problem on $I_{CBO}$ **Problem.** Explain why $I_{CBO}$ of Eq. (10.12) has no hole component. Why aren't there contributions associated with minority carrier holes (in the n-base of a pnp) that wander into the C-B depletion region and are swept into the collector? **Solution.** $I_{CBO}$ is the collector current with the emitter open, $I_E=0$. 1. In a pnp the holes in the narrow n-base are minority carriers. A hole current in the quasineutral base from emitter to collector would be continuous across the base (the base is narrow, so recombination is negligible): $I_{Cp}\approx I_{Ep}$. 2. With the emitter open, $I_E=I_{Ep}+I_{En}=0$. The E-B junction floats to whatever small bias makes the net emitter current vanish; the emitter's electron component $I_{En}$ is tiny (the emitter is heavily doped), so $I_{Ep}\approx0$. 3. Therefore $I_{Cp}=I_{Ep}\approx0$: any hole leaving the base through the collector would have to be supplied through the emitter, and the open emitter cannot supply it. The hole population in the base simply relaxes to a distribution in which the net hole flow vanishes. 4. The only holes that could contribute independently are those thermally generated *inside* the narrow base itself. Since the base width $W$ is much less than a diffusion length, the volume (and generation) is tiny compared with the collector region within a diffusion length of the C-B depletion edge, so this contribution is negligible; and by (3) it must be balanced anyway. 5. In contrast, the electrons that are minority carriers in the p-type collector are generated throughout a volume extending about one electron diffusion length $L_N$ into the collector. Those that wander into the C-B depletion region are swept into the base and out the base lead. No emitter is needed to supply them: they are thermally generated in the collector, so this current exists with $I_E=0$. Hence $I_{CBO}=I_{Cn}$ and carries no hole component. # Chapter 11: BJT Static Characteristics > **Source:** Pierret, *Semiconductor Device Fundamentals*, Chapter 11 (book pp. 389-442). > **Where we are:** Chapter 10 told the *story* of the transistor in words and pictures. This chapter turns that story into mathematics, checks the mathematics against a real 2N2605 transistor, and then shows how modern designers cheat the old trade-offs. --- ## Prologue: The Plan, and a Picture to Hold Onto Imagine a river (the **emitter** current) flowing toward a sea (the **collector**). In the middle is a narrow, slightly leaky canal (the **base**). A tiny leak in the canal wall is the *base current*; almost all the water gets to the sea. The leak is tiny because the canal is short. The sluice gate at the start of the canal (the forward-biased E-B junction) controls how much water enters, and it does so *exponentially*, which is why a small gate movement gives a huge change in flow. Everything in this chapter is a precise version of that picture. The chapter has three movements: 1. **Ideal analysis (Section 11.1).** Solve the minority-carrier diffusion equation in each of the three regions, exactly as for the ideal diode, and read off currents and gains. 2. **Reality check (Section 11.2).** Compare with measurement and name every disagreement: base-width modulation, punch-through, avalanche plus feedback, series resistance, current crowding, recombination-generation, and graded base. 3. **Modern devices (Section 11.3).** The polysilicon-emitter BJT and the heterojunction bipolar transistor (HBT), which use the lessons of Section 11.2 to beat the old trade-off between gain and base resistance. --- ## 11.1 Ideal Transistor Analysis ### 11.1.1 Solution Strategy The recipe is the same one used for the ideal diode, with one new ingredient: *three* quasineutral regions instead of two, and therefore three current-voltage expressions to juggle at once. #### Basic assumptions 1. The device is a **pnp BJT** with nondegenerate, uniformly doped emitter, base and collector. The E-B and C-B junctions are **step junctions**. 2. **Steady state.** 3. **One-dimensional** device. 4. **Low-level injection** in all quasineutral regions. 5. Only drift, diffusion and *thermal* recombination-generation occur, and there is no light: $G_L=0$. 6. Thermal recombination-generation is **negligible in the E-B and C-B depletion regions**. 7. The quasineutral emitter and collector widths are **much greater than the minority-carrier diffusion lengths** there (effectively semi-infinite). Notice what is *not* assumed: thermal recombination in the **base** is kept, and the base width is initially arbitrary. We keep the base general on purpose, so that we can watch the narrow-base result emerge as a special case. #### Notation (Fig. 11.1) Because a transistor has *two* regions of the same doping type, the old subscripts $n$/$p$ are replaced by **regional subscripts** $E$, $B$, $C$ for the *minority-carrier* parameters of each region. | | Emitter ($p^+$) | Base ($n$) | Collector ($p$) | |---|---|---|---| | Doping | $N_E=N_{AE}$ | $N_B=N_{DB}$ | $N_C=N_{AC}$ | | Minority carrier | electrons | holes | electrons | | Diffusion coefficient | $D_E=D_N$ | $D_B=D_P$ | $D_C=D_N$ | | Lifetime | $\tau_E=\tau_n$ | $\tau_B=\tau_p$ | $\tau_C=\tau_n$ | | Diffusion length | $L_E=L_N$ | $L_B=L_P$ | $L_C=L_N$ | | Equilibrium minority conc. | $n_{E0}=n_i^2/N_E$ | $p_{B0}=n_i^2/N_B$ | $n_{C0}=n_i^2/N_C$ | **Coordinates (Fig. 11.1):** base $x$ from $0$ (E-B depletion edge) to $W$ (C-B depletion edge), $W$ = *quasineutral* base width; emitter $x''$ from $0$ at the E-B depletion edge pointing into the emitter; collector $x'$ from $0$ at the C-B depletion edge pointing into the collector. #### Diffusion equations and boundary conditions *Emitter* (minority electrons):

0 = D_E\frac{d^2\Delta n_E}{dx”^2}-\frac{\Delta n_E}{\tau_E} \tag{11.1}

\Delta n_E(x”\to\infty)=0 \tag{11.2a}

\Delta n_E(x”=0)=n_{E0}\left(e

$$\ast Base\ast (minorityholes):$$

0 = D_B\frac{d^2\Delta p_B}{dx^2}-\frac{\Delta p_B}{\tau_B} \tag{11.3}

\Delta p_B(0)=p_{B0}\left(e

\Delta p_B(W)=p_{B0}\left(e

$$\ast Collector\ast (minorityelectrons):$$

0 = D_C\frac{d^2\Delta n_C}{dx’^2}-\frac{\Delta n_C}{\tau_C} \tag{11.5}

\Delta n_C(x’\to\infty)=0 \tag{11.6a}

\Delta n_C(x’=0)=n_{C0}\left(e

**Why these boundary conditions?** The "law of the junction": at the edge of a depletion region the minority excess is the equilibrium value times $(e^{qV/kT}-1)$, with $V$ the voltage across *that* junction. The semi-infinite emitter and collector forget the junction far away (zero excess at infinity). The base has junctions on *both* sides, so both ends carry a non-zero excess. #### Computational relationships With R-G negligible in the depletion regions, the current leaving a quasineutral region equals the minority diffusion current at its depletion edge:

I_{En}=-qAD_E\frac{d\Delta n_E}{dx”}\Big|_{x”=0}\tag{11.7}

I_{Ep}=-qAD_B\frac{d\Delta p_B}{dx}\Big|_{x=0}\tag{11.8}

I_{Cp}=-qAD_B\frac{d\Delta p_B}{dx}\Big|_{x=W}\tag{11.9}

I_{Cn}=qAD_C\frac{d\Delta n_C}{dx’}\Big|_{x’=0}\tag{11.10}

$$Performancefactorsandterminalcurrents(repeatsofChapter10):$$

\gamma=\frac{I_{Ep}}{I_{Ep}+I_{En}}\tag{11.11}

\alpha_T=\frac{I_{Cp}}{I_{Ep}}\tag{11.12}

\alpha_{dc}=\gamma\alpha_T\tag{11.13}

\beta_{dc}=\frac{\alpha_{dc}}{1-\alpha_{dc}}\tag{11.14}

I_E=I_{Ep}+I_{En}\tag{11.15}

I_C=I_{Cp}+I_{Cn}\tag{11.16}

I_B=I_E-I_C\tag{11.17}

Think of $\gamma$ as "what fraction of the emitter's effort is useful hole injection," $\alpha_T$ as "what fraction of injected holes survive the canal crossing," and $\alpha_{dc}$ as the overall fraction of emitter current that reaches the collector. --- ### 11.1.2 General Solution (Arbitrary $W/L_B$) #### Emitter and collector solutions The general solution of (11.1) is

\Delta n_E(x”)=A_1e^{-x”/L_E}+A_2e^{x”/L_E},\qquad L_E=\sqrt{D_E\tau_E}\tag{11.18}

- As $x''\to\infty$, $e^{x''/L_E}\to\infty$, so (11.2a) forces $A_2=0$. - With $A_2=0$, (11.2b) gives $A_1=n_{E0}(e^{qV_{EB}/kT}-1)$.

\Delta n_E(x”)=n_{E0}\left(e^{qV_{EB}/kT}-1\right)e

Differentiate at $x''=0$ and use (11.7):

\frac{d\Delta n_E}{dx”}\Big|0=-\frac{n{E0}}{L_E}\left(e

I_{En}=qA\frac{D_E}{L_E}n_{E0}\left(e

$$Thecollectorisidenticalinstructure:$$

\Delta n_C(x’)=n_{C0}\left(e^{qV_{CB}/kT}-1\right)e

\frac{d\Delta n_C}{dx’}\Big|0=-\frac{n{C0}}{L_C}\left(e

I_{Cn}=-qA\frac{D_C}{L_C}n_{C0}\left(e

**The minus sign in (11.22):** these are the $p$-side halves of the ideal diode equation. The sign differs only because collector current is *defined* positive for flow from the $n$-base to the $p$-collector, opposite to the diode's positive direction. For $V_{CB}<0$ the bracket is negative and $I_{Cn}>0$, as expected. #### Base solution The base is where the new physics lives, because $\Delta p_B$ does *not* vanish at either end. General solution of (11.3):

\Delta p_B(x)=A_1e^{-x/L_B}+A_2e^{x/L_B},\qquad L_B=\sqrt{D_B\tau_B}\tag{11.23}

Apply (11.4). Abbreviate $\Delta_0\equiv\Delta p_B(0)$, $\Delta_W\equiv\Delta p_B(W)$:

\Delta_0=A_1+A_2\tag{11.24a}

\Delta_W=A_1e^{-W/L_B}+A_2e

Multiply (11.24a) by $e^{-W/L_B}$ and subtract from (11.24b):

A_2=\frac{\Delta_W-\Delta_0e^{-W/L_B}}{e^{W/L_B}-e

Multiply (11.24a) by $e^{W/L_B}$ and subtract (11.24b):

A_1=\frac{\Delta_0e^{W/L_B}-\Delta_W}{e^{W/L_B}-e

Substitute back and group the $\Delta_0$ and $\Delta_W$ terms:

\Delta p_B(x)=\Delta_0\frac{e^{(W-x)/L_B}-e^{-(W-x)/L_B}}{e^{W/L_B}-e^{-W/L_B}} +\Delta_W\frac{e^{x/L_B}-e^{-x/L_B}}{e^{W/L_B}-e

With $\sinh\xi=(e^\xi-e^{-\xi})/2$ (11.27):

\boxed{\Delta p_B(x)=\Delta p_B(0)\frac{\sinh[(W-x)/L_B]}{\sinh(W/L_B)}+\Delta p_B(W)\frac{\sinh(x/L_B)}{\sinh(W/L_B)}}\tag{11.26}

**Sanity check.** If $V_{CB}=0$ then $\Delta p_B(W)=0$ and (11.26) collapses to the narrow-base diode solution (Eq. 6.65), as it must. **Intuition.** (11.26) is a *superposition*: the first term is the profile from the emitter alone with a perfect sink at $x=W$; the second from the collector alone with a perfect sink at $x=0$. The problem is linear, so we add. #### Hole currents across the junctions Differentiate (11.26):

\frac{d\Delta p_B}{dx}=\frac{-\Delta_0\cosh[(W-x)/L_B]+\Delta_W\cosh(x/L_B)}{L_B\sinh(W/L_B)}

At $x=0$ and $x=W$:

\frac{d\Delta p_B}{dx}\Big|_0=\frac{1}{L_B}\left[-\Delta_0\coth\frac{W}{L_B}+\frac{\Delta_W}{\sinh(W/L_B)}\right],\quad \frac{d\Delta p_B}{dx}\Big|_W=\frac{1}{L_B}\left[-\frac{\Delta_0}{\sinh(W/L_B)}+\Delta_W\coth\frac{W}{L_B}\right]

Insert in (11.8), (11.9) and restore the voltage dependences ($\cosh\xi=(e^\xi+e^{-\xi})/2$, Eq. 11.30):

I_{Ep}=qA\frac{D_B}{L_B}p_{B0}\left[\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\left(e^{qV_{EB}/kT}-1\right)-\frac{1}{\sinh(W/L_B)}\left(e

I_{Cp}=qA\frac{D_B}{L_B}p_{B0}\left[\frac{1}{\sinh(W/L_B)}\left(e^{qV_{EB}/kT}-1\right)-\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\left(e

#### Performance parameters and terminal currents (active mode) Active mode in a pnp: $V_{EB}>0$, $V_{CB}<0$ (beyond a few $kT/q$). Then $e^{qV_{CB}/kT}-1\approx-1$ is *small* next to $e^{qV_{EB}/kT}-1$, so we drop the $V_{CB}$ terms in $I_{Ep}$, $I_{Cp}$ **when forming $\gamma$ and $\alpha_T$**. *Emitter efficiency* (using $n_{E0}/p_{B0}=N_B/N_E$):

\frac{I_{En}}{I_{Ep}}=\frac{D_E n_{E0}/L_E}{D_Bp_{B0}\coth(W/L_B)/L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\tanh\frac{W}{L_B}

\gamma=\frac{1}{1+I_{En}/I_{Ep}}=\boxed{\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\dfrac{\sinh(W/L_B)}{\cosh(W/L_B)}}}\tag{11.31}

$$\ast Basetransportfactor:\ast$$

\alpha_T=\frac{I_{Cp}}{I_{Ep}}=\frac{1/\sinh(W/L_B)}{\cosh(W/L_B)/\sinh(W/L_B)}=\boxed{\frac{1}{\cosh(W/L_B)}}\tag{11.32}

$$\ast Common-basegain:\ast$$

\alpha_{dc}=\gamma\alpha_T=\boxed{\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\sinh\dfrac{W}{L_B}}}\tag{11.33}

*Common-emitter gain,* $\beta_{dc}=1/(1/\alpha_{dc}-1)$:

\boxed{\beta_{dc}=\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\sinh\dfrac{W}{L_B}-1}}\tag{11.34}

*Terminal currents.* Add the $n$ and $p$ components:

I_E=qA\left{\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]\left(e^{qV_{EB}/kT}-1\right)-\frac{D_B}{L_B}p_{B0}\frac{1}{\sinh(W/L_B)}\left(e

I_C=qA\left{\frac{D_B}{L_B}p_{B0}\frac{1}{\sinh(W/L_B)}\left(e^{qV_{EB}/kT}-1\right)-\left[\frac{D_C}{L_C}n_{C0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]\left(e

If desired, $I_B=I_E-I_C$. --- #### Exercise 11.1 **Problem.** (a) Show that the terminal-current expressions reduce to the ideal diode equation when $W\gg L_B$. (b) Show that setting $V_{CB}=0$ in (11.35) leads to the narrow-base-diode I-V relationship. **Solution.** (a) For $W/L_B\to\infty$:

\frac{\cosh(W/L_B)}{\sinh(W/L_B)}=\frac{e^{W/L_B}+e^{-W/L_B}}{e^{W/L_B}-e^{-W/L_B}}\to1,\qquad \frac{1}{\sinh(W/L_B)}=\frac{2}{e^{W/L_B}-e

$$Then(11.35),(11.36)become$$

I_E=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\right]\left(e

I_C=-qA\left[\frac{D_C}{L_C}n_{C0}+\frac{D_B}{L_B}p_{B0}\right]\left(e

Each is an ideal-diode equation (up to the minus sign explained after 11.22). A structure with $W\gg L_B$ is just **two back-to-back diodes**. (b) With $V_{CB}=0$, (11.35) becomes

I_E=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\coth\frac{W}{L_B}\right]\left(e

For a $p^+n$ junction the $n_{E0}$ term (back-injection) is small and may be dropped:

I_E\approx qA\frac{D_B}{L_B}p_{B0}\coth\frac{W}{L_B}\left(e

which is the narrow-base result (Eqs. 6.68-6.69). **Tying the collector to the base turns the ideal transistor into the narrow-base diode.** --- ### 11.1.3 Simplified Relationships ($W\ll L_B$) We could have imposed $W\ll L_B$ at the start, but that would have hidden other limiting cases and thrown away significant small terms. Now we use the small-argument limits

\sinh\xi\to\xi,\quad \xi\ll1\tag{11.37}

\cosh\xi\to1+\frac{\xi^2}{2},\quad \xi\ll1\tag{11.38}

#### $\Delta p_B(x)$ becomes a straight line For $0\le x\le W$ both $(W-x)/L_B$ and $x/L_B$ are $\ll1$. Replace each $\sinh$ by its argument in (11.26):

\Delta p_B(x)\approx\Delta p_B(0)\frac{(W-x)/L_B}{W/L_B}+\Delta p_B(W)\frac{x/L_B}{W/L_B} =\Delta p_B(0)\left(1-\frac{x}{W}\right)+\Delta p_B(W)\frac{x}{W}\tag{11.39}

\boxed{\Delta p_B(x)=\Delta p_B(0)+\left[\Delta p_B(W)-\Delta p_B(0)\right]\frac{x}{W}}\tag{11.40}

A straight line from $p_{B0}(e^{qV_{EB}/kT}-1)$ to $p_{B0}(e^{qV_{CB}/kT}-1)$ (Fig. 11.2, active mode: starts high, ends near $-p_{B0}$, i.e. total hole concentration about zero at the collector end). It stays near-linear even for $W/L_B$ up to 1 (Problem 11.5). **Picture.** The base is a very short ramp. If the ramp is shorter than the "leakage length" $L_B$, nothing is lost on the way: density drops linearly, like temperature along a thin plate whose faces are held at fixed temperatures. #### Performance parameters (to second order in $W/L_B$) With $\tanh(W/L_B)\approx W/L_B$, $\cosh\approx1+(W/L_B)^2/2$:

\gamma=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{L_E}}\tag{11.41}

\alpha_T=\frac{1}{1+\frac12(W/L_B)^2}\approx1-\frac12\left(\frac{W}{L_B}\right)

\alpha_{dc}=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{L_E}+\dfrac12\left(\dfrac{W}{L_B}\right)

\beta_{dc}=\frac{1}{\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{L_E}+\dfrac12\left(\dfrac{W}{L_B}\right)

($1/\alpha_{dc}-1$ is exactly the sum of the two small terms.) **Design lore.** - $N_E\gg N_B$ kills the first term: $\gamma\to1$. - $W\ll L_B$ kills the second: $\alpha_T\to1$. - $\beta_{dc}$ is the reciprocal of the sum of two loss terms, so it is large only if **both** are tiny. First loss: back-injection into the emitter ($I_{B1}$); second: recombination in the base ($I_{B2}$). #### Sample calculation (Table 11.1) pnp BJT at 300 K: | | Emitter | Base | Collector | |---|---|---|---| | Doping | $N_E=10^{18}$ | $N_B=10^{16}$ | $N_C=10^{15}$ cm$^{-3}$ | | Mobility (minority) | $\mu_E=263$ | $\mu_B=437$ | $\mu_C=1345$ cm$^2$/V-s | | $D$ | $6.81$ | $11.3$ | $34.8$ cm$^2$/s | | $\tau$ | $10^{-7}$ s | $10^{-6}$ s | $10^{-6}$ s | | $L$ | $8.25\times10^{-4}$ cm | $3.36\times10^{-3}$ cm | $5.90\times10^{-3}$ cm | and $W=W_B=2\times10^{-4}$ cm. Then

\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}=\frac{6.81}{11.3}\cdot\frac{10^{16}}{10^{18}}\cdot\frac{2\times10^{-4}}{8.25\times10^{-4}}=1.46\times10^{-3},\quad \frac12\left(\frac{W}{L_B}\right)^2=\frac12\left(\frac{2\times10^{-4}}{3.36\times10^{-3}}\right)^2=1.77\times10

\gamma=\frac{1}{1.00146}=0.9985,\quad \alpha_T=\frac{1}{1.00177}=0.9982,\quad \alpha_{dc}=\frac{1}{1.00323}=0.9968,\quad \beta_{dc}=\frac{1}{0.00323}=310

Real emitter dopings usually exceed $10^{18}$ cm$^{-3}$ (nondegenerate theory limits it near there), yet these results are representative. In modern transistors $\alpha_T$ is closer to unity and gain is mostly limited by emitter efficiency. --- #### Exercise 11.2 **Problem.** Show that (11.40) solves the diffusion equation when thermal R-G is negligible in the quasineutral base. **Solution.** Dropping $\Delta p_B/\tau_B$ in (11.3):

0=D_B\frac{d^2\Delta p_B}{dx

$$Integratetwice:$$

\Delta p_B(x)=A_1+A_2x

At $x=0$:

\Delta p_B(0)=A_1

At $x=W$:

\Delta p_B(W)=\Delta p_B(0)+A_2W;\Rightarrow;A_2=\frac{\Delta p_B(W)-\Delta p_B(0)}{W}

$$Substituting:$$

\Delta p_B(x)=\Delta p_B(0)+\left[\Delta p_B(W)-\Delta p_B(0)\right]\frac{x}{W}

which is (11.40). $\blacksquare$ --- #### Exercise 11.3 **Problem.** For a pnp transistor ($W\ll L_B$) in **saturation**, sketch on a linear scale the normalized excess $\Delta n/n_0$ or $\Delta p/p_0$ in emitter, base and collector. **Solution.** From (11.19) and (11.21):

\frac{\Delta n_E(x”)}{n_{E0}}=\left(e^{qV_{EB}/kT}-1\right)e^{-x”/L_E},\qquad \frac{\Delta n_C(x’)}{n_{C0}}=\left(e^{qV_{CB}/kT}-1\right)e

In saturation $V_{EB}>0$ **and** $V_{CB}>0$ (with $V_{EB}>V_{CB}$ in the sketch): both profiles look like a forward-biased diode, a pile-up at the depletion edge decaying exponentially inward. In the base, from (11.4):

\frac{\Delta p_B(0)}{p_{B0}}=e^{qV_{EB}/kT}-1>0,\qquad \frac{\Delta p_B(W)}{p_{B0}}=e

Excess at *both* base edges. The normalized excess at the base edge equals that at the emitter edge of the same depletion region (same junction factor), and $\Delta p_B(W)/p_{B0}=\Delta n_C(0)/n_{C0}$. With $W\ll L_B$ the base profile is a straight line. ``` Normalized excess (linear), V_EB > V_CB > 0 |\ Emitter : exponential decay, scale L_E | \ Base : straight line between two positive values | \ \ Collector : exponential decay, scale L_C | \ \_____ | \________ ...... +------+------+---------- x E B C ``` Saturation is the one mode in which the base is *flooded* from both sides. --- #### Exercise 11.4 **Problem.** Find the base transport factor using the linear approximation (11.40). Comment. **Solution.** A linear profile has the same slope everywhere:

\frac{d\Delta p_B}{dx}\Big|{0}=\frac{d\Delta p_B}{dx}\Big|{W}=\frac{\Delta p_B(W)-\Delta p_B(0)}{W}

By (11.8), (11.9): $I_{Cp}=I_{Ep}$, hence

\alpha_T=\frac{I_{Cp}}{I_{Ep}}=1

**Comment.** $\alpha_T=1$ means no recombination in the base, consistent with the assumption behind (11.40). But the linear approximation discards the $(W/L_B)^2$ terms: the small loss $\tfrac12(W/L_B)^2$ of (11.42) is invisible. To find $1-\alpha_T$ you must keep the slight curvature of the exact solution. --- ### 11.1.4 Ebers-Moll Equations and Model Now attention shifts to terminal currents, in a form a circuit simulator can use. With $V_{CB}=0$, (11.35) has the form of an ideal diode equation for the E-B junction; with $V_{EB}=0$, (11.36) is one for the C-B junction. Call the prefactors **effective saturation currents**:

I_{F0}=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]\tag{11.45a}

I_{R0}=qA\left[\frac{D_C}{L_C}n_{C0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]\tag{11.45b}

($F$ forward, $R$ reverse.) The prefactor of $(e^{qV_{CB}/kT}-1)$ in (11.35) is smaller than $I_{R0}$: it is **the fraction of the collector-junction current that makes it across the base to the emitter junction**. Likewise the $V_{EB}$ term in (11.36) is the fraction of $I_{F0}(\dots)$ reaching the collector. Define forward and reverse "gains":

\alpha_FI_{F0}=\alpha_RI_{R0}=qA\frac{D_B}{L_B}\frac{p_{B0}}{\sinh(W/L_B)}\tag{11.46}

$$Then$$

\boxed{I_E=I_{F0}\left(e^{qV_{EB}/kT}-1\right)-\alpha_RI_{R0}\left(e

\boxed{I_C=\alpha_FI_{F0}\left(e^{qV_{EB}/kT}-1\right)-I_{R0}\left(e

These are the **Ebers-Moll equations**, used by SPICE-type programs. They may also be treated as empirical: four parameters ($I_{F0},I_{R0},\alpha_F,\alpha_R$), of which only **three** are independent due to the **reciprocity relation** $\alpha_FI_{F0}=\alpha_RI_{R0}$. **Circuit view.** With $I_F=I_{F0}(e^{qV_{EB}/kT}-1)$ and $I_R=I_{R0}(e^{qV_{CB}/kT}-1)$: $I_E=I_F-\alpha_RI_R$, $I_C=\alpha_FI_F-I_R$, $I_B=I_E-I_C$. The **large-signal equivalent circuit (Fig. 11.3)** has a diode carrying $I_F$ (E to B), a diode carrying $I_R$ (C to B), a current source $\alpha_FI_F$ from the emitter side to the collector, and a source $\alpha_RI_R$ from the collector side to the emitter. KCL at E, C, B reproduces (11.47). It is handy for PNPN-family analysis. --- #### Exercise 11.5 **Problem.** Show $\alpha_F=\alpha_{dc}$. **Solution.** Divide (11.46) by (11.45a):

\alpha_F=\frac{qA\dfrac{D_B}{L_B}\dfrac{p_{B0}}{\sinh(W/L_B)}}{qA\left[\dfrac{D_E}{L_E}n_{E0}+\dfrac{D_B}{L_B}p_{B0}\dfrac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]}

Divide top and bottom by $qA\,D_Bp_{B0}/(L_B\sinh(W/L_B))$:

\alpha_F=\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E,n_{E0},L_B}{D_B,p_{B0},L_E}\sinh\dfrac{W}{L_B}}

With $n_{E0}/p_{B0}=N_B/N_E$:

\alpha_F=\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\sinh\dfrac{W}{L_B}}=\alpha_{dc}\ \ \text{(11.33)}\qquad\blacksquare

--- #### Exercise 11.6 **Problem.** Show the Ebers-Moll equations yield $I_C=\alpha_{dc}I_E+I_{CBO}$ (Eq. 10.8) in active mode. **Solution.** Active: $e^{qV_{CB}/kT}-1\approx-1$:

I_E=I_{F0}\left(e^{qV_{EB}/kT}-1\right)+\alpha_RI_{R0},\qquad I_C=\alpha_FI_{F0}\left(e

$$Solvethefirstforthevoltagefactor:$$

e

$$Substituteinthesecond:$$

I_C=\alpha_F\left(I_E-\alpha_RI_{R0}\right)+I_{R0}=\alpha_FI_E+\left(1-\alpha_F\alpha_R\right)I_{R0}

With $\alpha_F=\alpha_{dc}$ this is Eq. 10.8 provided

I_{CBO}=\left(1-\alpha_F\alpha_R\right)I_{R0}\qquad\blacksquare

(Problem 11.12(c) confirms $I_{CBO}$ is the C-B saturation current with emitter open.) --- #### Exercise 11.7 (computer) **Problem.** Using the Ebers-Moll equations, write a MATLAB program that calculates and plots the input and output characteristics of a pnp BJT in common-base and common-emitter configurations. Sample: room temperature, $A=10^{-4}$ cm$^2$, $W=W_B$, material parameters similar to Table 11.1. **Solution (logic).** A characteristic needs the dependent current as an *explicit function* of the controlled variable. Only the common-base input (11.47a) already has that form; the rest require algebra (derived in Problem 11.14): 1. **CB input** $I_E(V_{EB},V_{CB})$: $I_E=I_{F0}(e^{qV_{EB}/kT}-1)-\alpha_RI_{R0}(e^{qV_{CB}/kT}-1)$ 2. **CB output** $I_C(V_{CB},I_E)$: $I_C=\alpha_FI_E-(1-\alpha_F\alpha_R)I_{R0}(e^{qV_{CB}/kT}-1)$ 3. **CE input** $I_B(V_{EB},V_{EC})$: $$I_B=\left[(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}e^{-qV_{EC}/kT}\right]e^{qV_{EB}/kT}-\left[(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}\right]$$ 4. **CE output** $I_C(V_{EC},I_B)$: $$I_C=\frac{\left(\alpha_FI_{F0}-I_{R0}e^{-qV_{EC}/kT}\right)\left[I_B+(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}\right]}{(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}e^{-qV_{EC}/kT}}+I_{R0}-\alpha_FI_{R0}$$ Use $V_{CB}=V_{EB}-V_{EC}$ whenever $V_{EC}$ is the independent variable. The textbook program (BJT with subprogram BJT0, Appendix M) uses: $A=10^{-4}$ cm$^2$, $W_B=2.5\ \mu$m, $N_E=10^{18}$, $N_B=1.5\times10^{16}$, $N_C=1.5\times10^{15}$ cm$^{-3}$, $\tau_E=10^{-7}$, $\tau_B=\tau_C=10^{-6}$ s, and the mobility fit $\mu=\mu_{min}+\mu_0/[1+(N/N_{ref})^\alpha]$ with electrons $(92,1268,1.3\times10^{17},0.91)$ and holes $(54.3,406.9,2.35\times10^{17},0.88)$ for $(\mu_{min},\mu_0,N_{ref},\alpha)$. ```matlab % BJT Common-Base / Common-Emitter characteristics from Ebers-Moll clear; close all q=1.602e-19; k=8.617e-5; T=300; kT=k*T; % kT in volts (k in eV/K) A=1.0e-4; WB=2.5e-4; W=WB; % cm^2, cm ni=1.0e10; NE=1.0e18; NB=1.5e16; NC=1.5e15; % cm^-3 mun=@(N) 92 + 1268 ./(1+(N/1.3e17).^0.91); % minority electrons (E, C) mup=@(N) 54.3 + 406.9./(1+(N/2.35e17).^0.88); % minority holes (B) DE=kT*mun(NE); DB=kT*mup(NB); DC=kT*mun(NC); % Einstein relation tauE=1e-7; tauB=1e-6; tauC=1e-6; LE=sqrt(DE*tauE); LB=sqrt(DB*tauB); LC=sqrt(DC*tauC); nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC; % Ebers-Moll parameters, Eqs. (11.45), (11.46) fB = (DB/LB)*pB0*cosh(W/LB)/sinh(W/LB); IF0 = q*A*((DE/LE)*nE0 + fB); IR0 = q*A*((DC/LC)*nC0 + fB); aF = q*A*(DB/LB)*pB0/sinh(W/LB)/IF0; aR = q*A*(DB/LB)*pB0/sinh(W/LB)/IR0; % (1) Common-base input: IE vs VEB, stepped VCB subplot(2,2,1); hold on VEB=0:0.005:0.8; for VCB=0:-10:-20 IE=(IF0*(exp(VEB/kT)-1)-aR*IR0*(exp(VCB/kT)-1))*1e3; % mA plot(VEB,IE) end axis([0.35 0.8 0 5]); grid on; xlabel('V_{EB} (V)'); ylabel('I_E (mA)') % (2) Common-base output: IC vs -VCB, stepped IE subplot(2,2,2); hold on VCB=[2:-0.01:0, 0:-0.25:-50]; % fine near origin, coarse after for IE=(0:4)*1e-3 IC=(aF*IE-(1-aF*aR)*IR0*(exp(VCB/kT)-1))*1e3; plot(-VCB,IC) end axis([-5 50 0 5]); grid on; xlabel('-V_{CB} (V)'); ylabel('I_C (mA)') % (3) Common-emitter input: IB vs VEB, stepped VEC subplot(2,2,3); hold on for VEC=0:5:10 IB0=(1-aF)*IF0+(1-aR)*IR0; IB1=(1-aF)*IF0+(1-aR)*IR0*exp(-VEC/kT); IB=(IB1*exp(VEB/kT)-IB0)*1e6; % microamps plot(VEB,IB) end axis([0.35 0.85 -5 20]); grid on; xlabel('V_{EB} (V)'); ylabel('I_B (\muA)') % (4) Common-emitter output: IC vs VEC, stepped IB subplot(2,2,4); hold on VEC=[0:0.01:1, 1.25:0.25:50]; for IB=(0:4)*2.5e-6 IB0=(1-aF)*IF0+(1-aR)*IR0; IB1=(1-aF)*IF0+(1-aR)*IR0*exp(-VEC/kT); IC=((aF*IF0-IR0*exp(-VEC/kT)).*(IB+IB0)./IB1+IR0-aF*IR0)*1e3; plot(VEC,IC) end axis([0 50 0 2.5]); grid on; xlabel('V_{EC} (V)'); ylabel('I_C (mA)') ``` Logic notes: `exp(VEB/kT)` is evaluated element-wise on vectors; a fine voltage grid is used near the origin where saturation makes the curves bend sharply and a coarse one in the flat active region. Axis limits are tuned to this parameter set. --- ## 11.2 Deviations from the Ideal The ideal theory gives the right *shape* of the BJT characteristics. We now compare with a 2N2605 pnp BJT at room temperature (Figs. 11.4, 11.5; theory computed with $W=W_B$) and ask where and why they disagree. ### 11.2.1 Ideal Theory/Experiment Comparison **Common-base input (Fig. 11.4a,b).** Both show $I_E$ rising exponentially with $V_{EB}$. *Theory:* essentially independent of $V_{CB}$ (all curves for $V_{CB}\le0$ coincide). *Experiment:* $I_E$ at a given $V_{EB}$ noticeably *increases* as $V_{CB}$ becomes more negative. **Common-base output (Fig. 11.4c,d).** Excellent agreement: a rapid rise of $I_C$ beginning at slightly negative $V_{CB}$ (leaving saturation), then an almost perfectly constant $I_C$. The difference: theory has no limit on $-V_{CB}$, the real device breaks down at $V_{CBO}$. **Common-emitter input (Fig. 11.5a,b).** Mostly good agreement. The $V_{EC}=0$ curve stands alone; curves for $V_{EC}\gtrsim$ a few $kT/q$ bunch together, displaced to higher $V_{EB}$. **Common-emitter output (Fig. 11.5c,d).** The largest disagreement. Theory is perfectly flat in active mode; experiment slopes noticeably *upward* even at low voltage, turns sharply upward at large voltage, and is limited by a low-current breakdown voltage $V_{CEO}$. Deviations to be explained: (1) the dependence of the CB input on $V_{CB}$; (2) the quasilinear slope and subsequent sharp upturn of the CE output; (3) the $V_{CBO}$, $V_{CEO}$ breakdown limits. Overall, the agreement is surprisingly good. ### 11.2.2 Base Width Modulation (the Early Effect) In the theory $W=W_B$ was constant. In truth the *quasineutral* width is

W=W_B-x_{nEB}-x_{nCB}

(Eq. 10.3), where $x_{nEB}$, $x_{nCB}$ are the parts of the E-B and C-B depletion regions lying in the base. Changing $V_{EB}$ or $V_{CB}$ changes the depletion widths and so shrinks or expands $W$ (Fig. 11.6). Because the base is so thin, even small depletion changes are a large *fraction* of $W$. This is **base width modulation**, the **Early effect** (after J. Early). **Analogy.** The base is a plank bridge for hole-travellers. More reverse bias on the collector side saws a slice off the collector end: the bridge is shorter, crossing is easier, more travellers arrive per unit time. Since $\gamma,\alpha_T,\alpha_{dc},\beta_{dc},I_{F0},I_{R0}$ all depend on $W$, all vary with applied voltages. **(i) Common-base input.** For $-V_{CB}\gg kT/q$, (11.35) without the $V_{CB}$ term is

I_E=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)}{\sinh(W/L_B)}\right]\left(e

For $W\ll L_B$ ($\coth\to L_B/W$) and neglecting the small $D_En_{E0}/L_E$ term:

I_E\approx qA\frac{D_Bp_{B0}}{W}e

Raising the C-B reverse bias widens the C-B depletion region, $W$ falls, and by (11.49) $I_E$ at fixed $V_{EB}$ rises with $-V_{CB}$: exactly Fig. 11.4(b). **(ii) Common-emitter output.** In active mode

I_C=\beta_{dc}I_B+I_{CEO}\tag{11.50}

with (11.44) $\beta_{dc}=1\big/\left[\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac12(W/L_B)^2\right]$. $W$ decreases with $V_{EC}$, so $\beta_{dc}$ rises, and at fixed $I_B$ the collector current rises with $V_{EC}$: the quasilinear slope of Fig. 11.5(d). Base width modulation does **not** explain the *sharp* upturn near $V_{CEO}$. **Why are the other two characteristics insensitive?** The active CB output obeys

I_C=\alpha_{dc}I_E+I_{CBO}\tag{11.51}

with (11.43) $\alpha_{dc}=1/[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac12(W/L_B)^2]$. In a well-made transistor the $W$-dependent denominator terms are tiny compared with 1, so large changes in $W$ barely change $\alpha_{dc}$; and $I_{CBO}$ is essentially $W$-independent. A similar argument applies to the CE input (Problem 11.15). **Doping and bias mode.** In a standard transistor $N_E>N_B>N_C$. Almost all of the E-B depletion region lies in the base, but the junction is only slightly forward biased in active mode, so it hardly modulates $W$. The C-B junction has large reverse bias, but most of its depletion region lies in the lightly doped *collector*, which minimizes the effect. In **inverted mode** the reverse-biased E-B depletion region extends almost entirely into the base, so the same transistor becomes very sensitive to base width modulation. ### 11.2.3 Punch-Through Punch-through is base width modulation taken to the limit $W\to0$: the E-B and C-B depletion regions touch inside the base (Fig. 11.7). The junctions then become electrostatically coupled. As Fig. 11.8 shows, a further increase of $-V_{CB}$ lowers the E-B potential hill and allows a large, exponentially increasing injection of carriers from emitter directly into collector, so a slight rise in $-V_{CB}$ causes a huge rise in $I_C$. Punch-through limits $V_{CBO}$ or $V_{CEO}$ *if it occurs*; avalanche multiplication could set in first. Whichever happens first defines the breakdown voltage. --- #### Exercise 11.8 **Problem.** Using the Table 11.1 parameters, $W_B=2\ \mu$m, $T=300$ K, determine the C-B junction voltage that leads to punch-through. (The quoted numbers correspond to $V_{EB}=0$; Problem 11.16 treats $V_{EB}=0.7$ V.) **Solution.** Restating (10.3):

W=W_B-x_{nEB}-x_{nCB}

with step-junction depletion widths (Eq. 5.37), $K_S=11.8$:

x_{nEB}=\left[\frac{2K_S\varepsilon_0}{q}\frac{N_E}{N_B(N_E+N_B)}\left(V_{bi(EB)}-V_{EB}\right)\right]^{1/2},\quad V_{bi(EB)}=\frac{kT}{q}\ln\frac{N_EN_B}{n_i

x_{nCB}=\left[\frac{2K_S\varepsilon_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]^{1/2},\quad V_{bi(CB)}=\frac{kT}{q}\ln\frac{N_CN_B}{n_i

At punch-through $W=0$:

0=W_B-x_{nEB}-\left[\frac{2K_S\varepsilon_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]

Solve for $V_{CB}$:

V_{CB}=V_{bi(CB)}-\frac{q,N_B(N_C+N_B)}{2K_S\varepsilon_0N_C}\left(W_B-x_{nEB}\right)^2\qquad\text{(at punch-through)}

Numbers ($kT/q=0.0259$ V, $n_i=10^{10}$ cm$^{-3}$, $K_S\varepsilon_0=1.045\times10^{-12}$ F/cm):

V_{bi(EB)}=0.0259\ln\frac{10^{18}\cdot10^{16}}{10^{20}}=0.835\ \text{V},\qquad V_{bi(CB)}=0.0259\ln\frac{10^{15}\cdot10^{16}}{10^{20}}=0.656\ \text{V}

x_{nEB}=\left[\frac{2(1.045\times10^{-12})}{1.602\times10^{-19}}\frac{10^{18}}{10^{16}(1.01\times10^{18})}(0.835)\right]^{1/2}=3.29\times10^{-5}\ \text{cm}

W_B-x_{nEB}=2\times10^{-4}-3.29\times10^{-5}=1.671\times10^{-4}\ \text{cm}

\frac{qN_B(N_C+N_B)}{2K_S\varepsilon_0N_C}=\frac{(1.602\times10^{-19})(10^{16})(1.1\times10^{16})}{2(1.045\times10^{-12})(10^{15})}=8.43\times10^{9}\ \text{V/cm}

V_{CB}=0.656-8.43\times10^{9}\left(1.671\times10^{-4}\right)^2=0.656-235.4\approx-235\ \text{V}

**Answer: $V_{CB}\approx-235$ V.** The large value arises because the very lightly doped collector absorbs most of the depletion width, so hundreds of volts are needed to push depletion across the base. ### 11.2.4 Avalanche Multiplication and Breakdown #### Common base In common-base active mode with $-V_{CB}$ increasing, the C-B depletion region behaves like a reverse-biased $pn$ diode: crossing carriers gain enough energy to *impact-ionize*, creating extra electron-hole pairs. Eventually avalanche occurs, $I_C$ rises toward infinity and the multiplication factor $M\to\infty$ (Subsection 6.2.2). If avalanche precedes punch-through, the maximum common-base output voltage $V_{CBO}$ is just the C-B junction breakdown voltage. #### Common emitter: the surprise In common emitter $V_{EC}=V_{EB}-V_{CB}$, and $V_{EB}$ is only a few tenths of a volt, so one expects $V_{CEO}\approx V_{CBO}$. Experimentally (Figs. 11.4d, 11.5d) $V_{CEO}\approx90$ V while $V_{CBO}\approx120$ V. Following Fig. 11.9 step by step: 0. Holes are injected from the emitter into the base. 1. They enter the C-B depletion region. 2. A few gain enough energy to impact-ionize, creating extra electron-hole pairs. 3. The added holes drift into the collector; the added **electrons are swept into the base**. 4. These extra base electrons form a majority-carrier imbalance that must be removed. In common base they just leave through the base lead. In **common emitter the base current is held constant** (it is the controlled variable), so the electrons cannot leave that way; the imbalance can only be relieved by **injecting electrons from the base into the emitter**. 5. E-B injection is coupled: every electron injected into the emitter is accompanied by $I_{Ep}/I_{En}$ extra holes injected into the base. So for each additional electron-hole pair created in the C-B region, $I_{Ep}/I_{En}+1=1/(1-\gamma)\ge\beta_{dc}+1$ extra holes flow into the collector. The multiplication is thus **internally amplified** by the transistor gain, and it is **regenerative**: more collector current causes more multiplication, hence more injection, and so on, producing an instability at a $V_{EC}$ far below the pure avalanche voltage of the C-B junction. (Equivalently, $\beta_{dc}$ is enhanced by the process.) #### Quantitative model With multiplication in the C-B region, $I_{Cp}=MI_{Ep}$: replace $\alpha_{dc}$ (and $\alpha_F$) by $M\alpha_{dc}$. The simplified CE output (Eq. 10.15) becomes

I_C=\frac{M\alpha_{dc}}{1-M\alpha_{dc}}I_B+\frac{I_{CBO}}{1-M\alpha_{dc}}\tag{11.52}

$I_C\to\infty$ when $M\alpha_{dc}\to1$, i.e. $M\to1/\alpha_{dc}$. Since $\alpha_{dc}$ is only slightly below unity, $M$ need only exceed 1 slightly. With the empirical $M$ (Eq. 6.35) and $V_{BR}=V_{CBO}$:

M(V_{CEO})=\frac{1}{1-\left(V_{CEO}/V_{CBO}\right)

\left(\frac{V_{CEO}}{V_{CBO}}\right)

\boxed{V_{CEO}=V_{CBO}\left(1-\alpha_{dc}\right)^{1/m}=\frac{V_{CBO}}{(\beta_{dc}+1)

with $3\le m\le6$ (using $1-\alpha_{dc}=1/(\beta_{dc}+1)$). Hence $V_{CEO}<V_{CBO}$. Example: $\beta_{dc}\approx300$, $m=6$: $(301)^{1/6}\approx2.6$, so $V_{CEO}\approx V_{CBO}/2.6$. In a BJT this mechanism is *detrimental*. Elsewhere it is useful: it helps trigger the off-to-on switching in SCRs and other PNPN devices (Chapter 13), and it amplifies the photosignal in a **phototransistor** (a BJT whose C-B junction is exposed to light, often with an open or absent base lead). As in Fig. 11.10, roughly $\beta_{dc}+1$ carriers enter the collector per electron-hole pair created by photon absorption in the C-B depletion region. --- #### Exercise 11.9 **Problem.** Two pnp BJTs are similar except $N_B\gg N_C$ in Transistor A and $N_B\ll N_C$ in Transistor B. From the doping-profile figure (read from the OCR'd axis labels): A has $N_E=10^{18}$, $N_B=10^{16}$, $N_C=10^{14}$ cm$^{-3}$; B has $N_E=10^{18}$, $N_B=10^{16}$, $N_C=10^{18}$ cm$^{-3}$. (a) Which has the larger punch-through voltage under active biasing? (b) If limited by C-B avalanche, which has the larger $V_{CBO}$? **Solution.** (a) **Transistor A.** Under active bias the E-B junction (same doping) gives the same $x_{nEB}$ in both. The base penetration of the C-B depletion region is

x_{nCB}=\left[\frac{2K_S\varepsilon_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]

$$Evaluatethedopingfactor:$$

\frac{N_C}{N_B(N_C+N_B)}\Big|_A\approx\frac{10^{14}}{10^{16}\cdot10^{16}}=10^{-18},\qquad \frac{N_C}{N_B(N_C+N_B)}\Big|_B\approx\frac{10^{18}}{10^{16}\cdot10^{18}}=10

So for the same $-V_{CB}$, $x_{nCB}|_A\approx x_{nCB}|_B/10$ (the small $V_{bi}$ difference is a minor correction). With equal $W_B$, a much larger $-V_{CB}$ is needed to deplete the base of A. (b) **Transistor A.** The avalanche breakdown voltage of an asymmetric junction is set by the doping on the lightly doped side and varies roughly inversely with it (Fig. 6.11 / Eq. 6.33). In A the light side of the C-B junction is the collector, $10^{14}$ cm$^{-3}$; in B it is the base, $10^{16}$ cm$^{-3}$. The lighter side gives the higher breakdown voltage, so A exhibits the larger $V_{CBO}$. --- #### Exercise 11.10 (computer) **Problem.** Modify the Exercise 11.7 program to include base width modulation and carrier multiplication. With the same parameters compare CE output characteristics when (i) neither is included, (ii) only BWM, (iii) both. **Solution (logic).** *BWM (subprogram BJTmod).* At each bias point recompute depletion penetration, then $W$, then $I_{F0}$, $I_{R0}$, $\alpha_F$, $\alpha_R$:

x_{nEB}=\sqrt{\tfrac{2K_S\varepsilon_0}{q}\tfrac{N_E}{N_B(N_E+N_B)}(V_{biE}-V_{EB})},\quad x_{nCB}=\sqrt{\tfrac{2K_S\varepsilon_0}{q}\tfrac{N_C}{N_B(N_C+N_B)}(V_{biC}-V_{CB})},\quad W=W_B-x_{nEB}-x_{nCB}

*Multiplication.* $M=1/[1-(|V_{CB}|/V_{CBO})^m]$ and $\alpha_F\to M\alpha_F$ in the CE output equation, with $V_{CBO}=60(N_C/10^{16})^{-3/4}$ V and $m=6$ (the textbook program's choices); the sweep ends at $V_{CEO}=V_{CBO}(1-\alpha_F)^{1/m}$. For the CE output the program sets $V_{EB}\approx0$ inside BWM (neglecting $x_{nEB}$ variation), so $V_{CB}=-V_{EC}$. ```matlab % BJTplus-style: common-emitter output, three cases clear; close all q=1.602e-19; k=8.617e-5; e0=8.85e-14; T=300; kT=k*T; A=1e-4; WB=2.5e-4; ni=1e10; KS=11.8; NE=1e18; NB=1.5e16; NC=1.5e15; mun=@(N) 92+1268./(1+(N/1.3e17).^0.91); mup=@(N) 54.3+406.9./(1+(N/2.35e17).^0.88); DE=kT*mun(NE); DB=kT*mup(NB); DC=kT*mun(NC); tauE=1e-7; tauB=1e-6; tauC=1e-6; LE=sqrt(DE*tauE); LB=sqrt(DB*tauB); LC=sqrt(DC*tauC); nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC; VbiE=kT*log(NE*NB/ni^2); VbiC=kT*log(NC*NB/ni^2); W=WB; % zero-bias E-M parameters fB=(DB/LB)*pB0*cosh(W/LB)/sinh(W/LB); IF0=q*A*((DE/LE)*nE0+fB); IR0=q*A*((DC/LC)*nC0+fB); aF=q*A*(DB/LB)*pB0/sinh(W/LB)/IF0; aR=q*A*(DB/LB)*pB0/sinh(W/LB)/IR0; VCBO=60*(NC/1e16)^(-3/4); m=6; VCEO=VCBO*(1-aF)^(1/m); VEC=[0:0.01:VCEO/50, VCEO/50:VCEO/200:VCEO]; titles={'No BWM, no multiplication','BWM only','BWM + multiplication'}; for c=1:3 subplot(3,1,c); hold on for IB=(0:4)*2.5e-6 if c==1 F0=IF0*ones(size(VEC)); R0=IR0*ones(size(VEC)); aFv=aF*ones(size(VEC)); aRv=aR*ones(size(VEC)); else VEB=0; VCB=VEB-VEC; xnEB=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-VEB)); xnCB=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB)); Wv=WB-xnEB-xnCB; fBv=(DB/LB)*pB0*cosh(Wv/LB)./sinh(Wv/LB); F0=q*A*((DE/LE)*nE0+fBv); R0=q*A*((DC/LC)*nC0+fBv); aFv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./F0; aRv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./R0; if c==3 M=1./(1-(-VCB/VCBO).^m); % carrier multiplication aFv=M.*aFv; end end E=exp(-VEC/kT); IB0=(1-aFv).*F0+(1-aRv).*R0; IB1=(1-aFv).*F0+(1-aRv).*R0.*E; IC=((aFv.*F0-R0.*E).*(IB+IB0)./IB1+R0-aFv.*R0)*1e3; % mA plot(VEC,IC) end title(titles{c}); xlabel('V_{EC} (V)'); ylabel('I_C (mA)'); grid on axis([0 VCEO 0 5]) end ``` **Results.** (i) Flat, horizontal active-mode lines. (ii) A moderate quasilinear rise of $I_C$ with $V_{EC}$ (Early effect). (iii) Additionally a sharp upturn as $V_{EC}\to V_{CEO}$. Case (iii) compares very favorably with the measured Fig. 11.5(d). Keep $V_{EC}<V_{CBO}$ so $1-(\dots)^m>0$. ### 11.2.5 Geometrical Effects Real BJTs (cross sections of Fig. 10.6) are not one-dimensional, which produces geometry-related deviations. **Emitter area $\neq$ collector area.** Carriers do not travel in straight lines from emitter to collector; current lines spread laterally (Fig. 11.11a). To first order this is handled by using different *effective* areas for emitter (injection) and collector (collection). **Series resistances.** Base current flowing from the base contact to the "heart" of the transistor (dashed region of Fig. 11.11b) crosses a resistive bulk region, plus there is contact resistance. The voltage across the E-B junction is therefore less than the terminal voltage. Though small, this matters because $I_E$ is *exponential* in junction voltage. Introduce a base series resistance $r_B$ (bulk plus contact). If it is the only series resistance, replace $V_{EB}$ by

V’{EB}=V{EB}-I_Br_B

The central part is the **intrinsic transistor**; as in Fig. 11.12, resistances $r_E$, $r_B$, $r_C$ connect the internal terminals $E'$, $B'$, $C'$ to the external ones; internal voltages follow by subtracting the $IR$ drops. Values come from subsidiary measurements. **Current crowding.** Base current also flows laterally under the emitter face, causing a voltage drop in the base from emitter center to edges. The E-B junction voltage and injection current are therefore greatest at the emitter periphery (more current lines at the edges in Fig. 11.11a): **current crowding**. With the localized heating it brings, it is damaging at high currents. Power BJTs therefore use emitters with a **large periphery-to-area ratio**: several narrow, long emitter fingers joined in a comb and **interdigitated** with base fingers (Fig. 11.13). ### 11.2.6 Recombination-Generation Current The ideal analysis neglects R-G in the depletion regions. By analogy with the diode: - In active mode with *small* forward E-B bias, the E-B R-G current likely dominates the **base current**: lower emitter efficiency and lower gain at low currents (small $V_{EB}$). - With the input open, reverse-bias R-G current from the C-B junction adds to (and likely dominates) the leakage $I_{CBO}$ ($I_{CEO}$). Unlike the diode, the transistor characteristics are only *nominally* affected: in most active operation diffusion dominates the input current, and except at low currents the output current is mainly injected current. With R-G included, the pnp emitter efficiency becomes

\gamma=\frac{I_{Ep}}{I_{Ep}+I_{En}+I_{RG}}

### 11.2.7 Graded Base Double diffusion (Section 10.2) gives dopant profiles falling roughly exponentially with depth (Fig. 11.14). So base doping is **not** uniform: it falls from a maximum at the E-B junction to a minimum at the C-B junction: the base is **graded**. A doping gradient implies a **built-in field** in the quasineutral base. Let $N_B(x)$ be the donor doping of a pnp base. Neutrality $\rho=q(p-n+N_B)=0$ with $p\approx0$ gives $n(x)=N_B(x)$. In equilibrium the electron current vanishes:

J_N=q\mu_nn\mathcal{E}+qD_N\frac{dn}{dx}=0\tag{11.55}

Solve for $\mathcal{E}$ using $D_N/\mu_n=kT/q$ and $n=N_B(x)$:

\mathcal{E}(x)=-\frac{D_N}{\mu_n}\frac{dn/dx}{n}=-\frac{kT}{q}\frac{1}{N_B(x)}\frac{dN_B}{dx},\qquad0\le x\le W\tag{11.56}

$$Foranexponentialprofile,$$

N_B(x)=N_B(0)e

\frac{dN_B}{dx}=-\frac{N_B(x)}{x_{diff}}\tag{11.58}

$$Substitutingin(11.56):$$

\boxed{\mathcal{E}=\frac{kT/q}{x_{diff}}}\qquad0\le x\le W\tag{11.59}

A **constant** field directed so as to *assist* minority-carrier transport across the base (for holes in a pnp and for electrons in an npn). Example: doping falling by two decades across $W_B=2\ \mu$m:

x_{diff}=\frac{W_B}{\ln100}=\frac{2\times10^{-4}}{4.605}=4.34\times10^{-5}\ \text{cm},\qquad \mathcal{E}=\frac{0.0259}{4.34\times10^{-5}}\approx600\ \text{V/cm}

Drift-assisted transport shortens the transit time, reduces base recombination, raises $\alpha_T$ and all gains above ideal values, and improves high-frequency response. **The graded base is one deviation that is clearly beneficial.** ### 11.2.8 Figures of Merit Two plots reveal quickly how far a device departs from ideal (virtually every BJT research paper shows them). **Gummel plot (Fig. 11.15):** semilog plot of $I_B$ and $I_C$ versus $V_{EB}$ at fixed output voltage (a few volts). Ideally both are straight lines with the same slope $q/kT$ (60 mV per decade at room temperature). In reality: - at **small** $V_{EB}$, R-G adds to $I_B$ (slope $q/2kT$ region), putting $I_B$ above the ideal line; - at **large** $V_{EB}$, the $I_C$ plot slopes over due to **high-level injection**, possibly aggravated by current crowding; if *both* $I_B$ and $I_C$ roll over, **series resistance** is probable (as for a diode). With $I_{CEO}$ negligible, $I_C/I_B$ at any $V_{EB}$ is $\beta_{dc}$ there. Ideally constant, in reality it falls at both low and high $V_{EB}$. **$\beta_{dc}$ versus $I_C$ (Fig. 11.16):** log-log plot at fixed output voltage (sample: 2N2605, $V_{EC}=5$ V). It is merely a continuous record of the information in the Gummel plot: $\beta_{dc}$ drops at low $I_C$ because of R-G, and at high $I_C$ because of high-level injection, current crowding, series resistance, or a combination. --- ## 11.3 Modern BJT Structures ### 11.3.1 Polysilicon Emitter BJT CMOS is the preferred large-scale IC technology (low power, high packing density), but a BJT supplies several times the drive current of an equal-size MOSFET, switching large capacitive loads faster. **BiCMOS** therefore puts bipolar transistors in the speed-sensitive, heavily loaded parts of mostly-CMOS circuits (Fig. 11.17: the Intel 0.8 $\mu$m BiCMOS npn). In Fig. 11.17 a polycrystalline Si film lies between the metal emitter contact and the crystalline silicon. The emitter is made by depositing poly over the emitter area, doping it with arsenic, and diffusing some dopant a short distance into the crystalline Si: a shallow crystalline layer contacted by a similarly doped poly film, the **polysilicon emitter**. Advantages: - suited to the **shallow** E-B junctions modern ICs need; - compatible with **self-aligned** processing that minimizes parasitic resistances and capacitances; - an **operational advantage**: the CE gain is **much larger** than that of an all-crystalline BJT with a shallow emitter of equal thickness. **Why shallow emitters cut gain.** Modern emitters are very thin: $W_E$ can be less than $L_E$. For $W_E/L_E\ll1$ in an all-crystalline emitter, emitter recombination is negligible and, like the base, the minority profile is *linear* (Fig. 11.18a). The slope at the junction is far larger than for a semi-infinite emitter, so back-injection from base to emitter rises, emitter efficiency falls, and gain falls. Quantitatively (Problem 11.11), $W_E$ simply replaces $L_E$ in (11.41)-(11.44). **Why polysilicon helps.** To first order the poly portion is modelled as *low-mobility silicon* (grain boundaries scatter carriers). If both poly and crystalline portions are much shorter than their diffusion lengths, the minority profile is linear in both. At the poly/Si interface the minority excess and the minority **current** must be continuous. With subscript 1 = poly, 2 = Si:

qD_{E1}\frac{d\Delta p_{E1}}{dx”}=qD_{E2}\frac{d\Delta p_{E2}}{dx”}\tag{11.60}

\frac{d\Delta p_{E1}}{dx”}=\frac{D_{E2}}{D_{E1}}\frac{d\Delta p_{E2}}{dx”}=\frac{\mu_{E2}}{\mu_{E1}}\frac{d\Delta p_{E2}}{dx”}\tag{11.61}

Since $\mu_{E1}<\mu_{E2}$, the poly slope is **steeper** (Fig. 11.18b). For the same total drop, the slope **at the E-B depletion edge** is then *reduced* relative to the all-crystalline shallow emitter: less back-injected current, smaller base current, **better emitter efficiency and higher gain**. **Intuition.** A low-mobility layer behind the junction acts like a traffic jam: minority carriers dumped into the emitter cannot be carried away easily, so the pile builds near the junction, flattening the gradient there and driving less current across. ### 11.3.2 Heterojunction Bipolar Transistor (HBT) In Section 11.2 several deviations were tied to base doping: base-width modulation, base resistance $r_B$ and the current-crowding voltage drop all improve when base doping is **increased**. But increasing base doping lowers emitter efficiency and therefore gain. A standard BJT has an unavoidable gain-versus-nonideality trade-off. In a **heterojunction bipolar transistor** the base is generally the *most heavily doped* region. A **heterojunction** is a junction between two dissimilar semiconductors. In an HBT the emitter is of a **wider band-gap** material than the base, which in most material pairs sets up a built-in barrier impeding forward-bias back-injection from base to emitter, giving high injection efficiency even if $N_E\lesssim N_B$. (Shockley described the idea in the original 1948 BJT patent; practical HBTs required film-growth methods like molecular beam epitaxy.) **Band alignment** (Fig. 11.19) is how conduction and valence band edges of the two materials line up at the interface, usually found experimentally. Fig. 11.19 is *not* a true heterojunction diagram (neutrality drawn everywhere, no band bending). Lattice-matched systems shown (minimal interface defects): - (a) Al$_{0.3}$Ga$_{0.7}$As/GaAs: gaps 1.82 and 1.42 eV, $\Delta E_C=0.263$ eV, $\Delta E_V=0.132$ eV; - (b) GaAs/Ge: gaps 1.42 and 0.66 eV, $\Delta E_C=0.27$ eV, $\Delta E_V=0.49$ eV; - (c) InP/In$_{0.53}$Ga$_{0.47}$As: gaps 1.35 and 0.75 eV, $\Delta E_C=0.2$ eV, $\Delta E_V=0.4$ eV; - (d) Al$_{0.48}$In$_{0.52}$As/InP: gaps 1.46 and 1.35 eV, $\Delta E_C=0.3$ eV, $\Delta E_V=-0.19$ eV (offsets read from the noisy OCR of the figure). An important non-lattice-matched exception is the Si/Si$_{1-x}$Ge$_x$/Si HBT with a thin *pseudomorphic* SiGe base (thin films, roughly 10 nm, conforming to the substrate lattice; thicker than a critical value they relax and create defects). **AlGaAs/GaAs Npn HBT (Fig. 11.20).** (An uppercase letter marks the wider-gap side.) An $n^+$-GaAs subcollector, $n$-GaAs collector, $p^+$-GaAs base, and N-Al$_{0.3}$Ga$_{0.7}$As emitter (with $n$-GaAs cap) are deposited in turn on a semi-insulating (S.I.) GaAs substrate. The substrate and S.I. sidewalls (formed by deep proton implantation) isolate the HBT dielectrically. Etching and metal contacts complete it. An *abrupt* E-B heterojunction gives the band diagram of Fig. 11.21(a): nearly that of an equivalently doped BJT apart from the wider emitter gap and band discontinuities at the interface; since $N_E\ll N_B\gg N_C$ the depletion regions lie almost wholly in emitter and collector. In practice the junction is **compositionally graded** over a few hundred angstroms, eliminating sharp band spikes (Fig. 11.21b). Operationally a graded HBT is a BJT with different intrinsic concentrations $n_{iE}$, $n_{iB}$. Instead of $n_{E0}/p_{B0}=N_B/N_E$:

\frac{n_{E0}}{p_{B0}}=\frac{n_{iE}^2/N_E}{n_{iB}^2/N_B}=\frac{N_B}{N_E}\frac{n_{iE}^2}{n_{iB}

so $N_B/N_E$ in the BJT relations becomes $(N_B/N_E)(n_{iE}/n_{iB})^2$. For $W\ll L_B$:

\gamma=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{n_{iE}^2}{n_{iB}

The large $N_B/N_E\sim100$ is overwhelmed by the tiny intrinsic-concentration factor, which the text quotes as about $10^{-5}$ for a typical Al$_{0.3}$Ga$_{0.7}$As/GaAs HBT, so the theoretical emitter efficiency is very close to unity. In practice HBT efficiencies fall below (11.62) because of recombination at the E-B junction periphery surface (GaAs passivation is harder than Si). Even so, HBTs now offer very respectable gain and unmatched high-frequency performance. **The big idea.** A wide-gap emitter lets designers dope the base heavily (low $r_B$, small Early effect, no punch-through worries, high speed) *without* paying the gain penalty, because the band offset, not the doping ratio, blocks back-injection: a one-way turnstile at the base's front door. --- ## 11.4 Summary This chapter modeled the steady-state response of the BJT. First, allowing an arbitrary quasineutral base width, an ideal transistor analysis paralleling the ideal diode analysis gave the base minority-carrier distribution (11.26), the performance parameters (11.31-11.34) and the terminal currents (11.35, 11.36). These were simplified using $W\ll L_B$: the base distribution (11.40) becomes linear and the performance parameters take the forms (11.41)-(11.44). Terminal currents were handled compactly by the Ebers-Moll model (Fig. 11.3) and equations (11.47). Deviations from ideal found by comparing theory and experiment: - **Base width modulation** causes the quasilinear upward slant of the CE output characteristics and the $V_{CB}$ dependence of the CB input characteristics. - **Punch-through** or **avalanche breakdown** of the C-B junction, whichever occurs first, limits the maximum output voltage in active-mode common-base operation. - In active-mode **common-emitter** operation, carrier multiplication plus internal feedback can make $V_{CEO}<V_{CBO}$. - Others: series resistance external to the intrinsic transistor, the base voltage drop across the emitter face (current crowding), R-G in the E-B and C-B depletion regions, and the (beneficial) graded base. - Figures of merit: the **Gummel plot** and the **$\beta_{dc}$ versus $I_C$** plot. Finally, the section on modern structures introduced the **polysilicon-emitter BJT** and the **HBT**, also providing practical uses of previously established concepts. --- # End-of-Chapter Problems The book's *Problem Information Table* lists: 11.1 quick quiz; 11.2 deduce information from a current plot; 11.3 npn equation modifications; 11.4 current and parameter calculation; 11.5 and 11.6 $\Delta p_B(x)$ plots; 11.7 and 11.8 carrier-distribution sketches and deductions; 11.9 and 11.10 parametric-effect table and plots; 11.11 shallow emitter; 11.12 "diode" connections; 11.13 Ebers-Moll circuit; 11.14 computational equations; 11.15 emitter input $W$ sensitivity; 11.16 punch-through; 11.17 comparing two transistors; 11.18 Early voltage; 11.19 "diode" $I$-$V_A$ plots; 11.20 Gummel and gain plots; 11.21 $V_{EC}$ dependence of gain. ## Problem 11.1: Quick Quiz **(a) Doping and widths assumed in the ideal analysis.** Emitter, base and collector are each *uniformly* and *nondegenerately* doped (step junctions). The quasineutral emitter and collector widths are $\gg$ their minority diffusion lengths (semi-infinite). The base width $W$ is initially arbitrary (later $W\ll L_B$). **(b) Base distribution, active mode.** $\Delta p_B(0)=p_{B0}(e^{qV_{EB}/kT}-1)\gg0$ and $\Delta p_B(W)\approx-p_{B0}$ (total $p\approx0$). For $W/L_B\ll1$: a straight line from the high value to about $-p_{B0}$. For $W/L_B\gg1$: $\Delta p_B\approx\Delta p_B(0)e^{-x/L_B}$, an exponential that dies out long before $x=W$ (two back-to-back diodes). **(c) Sketches for $W/L_B\ll1$, active mode, $\Delta p_{B,max}=10\,p_{B0}$.** (The OCR'd magnitude is taken as $\Delta p_B(0)=10\,p_{B0}$.) Since $V_{CB}$ is reverse, $\Delta p_B(W)=-p_{B0}$, and

\Delta p_B(x)=10p_{B0}-11p_{B0}\frac{x}{W},\qquad \frac{\Delta p_B(x)}{p_{B0}}=10-11\frac{x}{W},\qquad \frac{p_B(x)}{p_{B0}}=1+\frac{\Delta p_B}{p_{B0}}=11\left(1-\frac{x}{W}\right)

``` Δp_B(x) Δp_B/p_B0 p_B(x) (from 11 p_B0 to 0) 10 p_B0 |\ 10 |\ 11 p_B0|\ | \ | \ | \ 0 ------|--\------ 0 -|--\------ p_B0 ----|--\ ... | \ | \ | \ -p_B0 +----+ x -1 +----+ x 0 +----+ x 0 W 0 W 0 W ``` All three are straight lines with the same shape (offsets of 0, a scale of $1/p_{B0}$, and a shift of $+p_{B0}$). **(d) Typical sizes.** $\gamma\approx0.99-0.9999$, $\alpha_T\approx0.99-0.999$, $\alpha_{dc}\approx0.98-0.999$, $\beta_{dc}\approx50-1000$ (310 in the Table 11.1 example). **(e) $I_B(V_{EB},V_{CB})$.** From $I_B=I_E-I_C$ and (11.47):

I_B=(1-\alpha_F)I_{F0}\left(e^{qV_{EB}/kT}-1\right)+(1-\alpha_R)I_{R0}\left(e

**(f)** The statement of part (f) is lost in the OCR. Its most likely subject, the item in this slot of the chapter's list of deviations, is why the CB input characteristics depend on $V_{CB}$ (Fig. 11.4b). Answer: base width modulation. Larger $-V_{CB}$ widens the C-B depletion region, narrows $W$, steepens the base profile and raises $I_E\propto e^{qV_{EB}/kT}/W$ (Eq. 11.49). **(g) Fig. 11.5(d).** The quasilinear slope at low $V_{EC}$ is **base width modulation** ($\beta_{dc}$ rises as $W$ falls with increasing $V_{EC}$). The sharp upward curvature as $V_{EC}\to V_{CEO}$ is **avalanche multiplication in the C-B depletion region amplified by the transistor feedback** (so $M\alpha_{dc}\to1$, Eq. 11.52), not base width modulation. **(h) Can $V_{CEO}<V_{CBO}$ when $V_{CBO}$ is limited by punch-through?** Yes. Punch-through is a purely electrostatic effect that sets $V_{CBO}$ (the punch-through voltage, say 235 V in Exercise 11.8). The feedback-amplified avalanche mechanism acts in the common-emitter circuit with the C-B junction's avalanche voltage $V_{BR}$ (which is higher than $V_{CBO}$ here), giving $V_{CEO}=V_{BR}(1-\alpha_{dc})^{1/m}=V_{BR}/(\beta_{dc}+1)^{1/m}$. If $V_{BR}/(\beta_{dc}+1)^{1/m}$ falls below the punch-through voltage, then avalanche-plus-feedback occurs first in common emitter and $V_{CEO}<V_{CBO}$. Example: $V_{BR}=300$ V, $\beta_{dc}=300$, $m=6$ gives $V_{CEO}\approx115$ V, below a 235 V punch-through limit. **(i) Current crowding** arises because base current flows laterally under the emitter face, causing a voltage drop in the base from emitter center to edge; the E-B junction voltage and injection are therefore largest at the emitter periphery. **(j) Benefits of a graded base.** A built-in field (Eq. 11.59) that aids minority-carrier transport: shorter transit time, less base recombination, higher $\alpha_T$ and current gains, and better high-frequency response. **(k) Gummel plot.** A simultaneous semilog plot of $I_B$ and $I_C$ versus $V_{EB}$ at fixed output voltage; the ratio gives $\beta_{dc}$ and the deviations from parallel straight lines reveal R-G current (low $V_{EB}$), high-level injection, current crowding and series resistance (high $V_{EB}$). **(l)** $\beta_{dc}$ decreases at low $I_C$ because of recombination-generation in the E-B depletion region (it adds base current); at high $I_C$ it decreases because of high-level injection, current crowding, series resistance, or a combination. **(m) All-polysilicon emitter?** In the simple model a low-mobility emitter (small $D_E$, steeper profile) reduces back-injection, so there is a nominal efficiency gain. In practice, a fully polycrystalline emitter would put grain boundaries and defects into the E-B depletion region: strong R-G and leakage, a poor junction. That is why the junction is formed in crystalline Si and only the contact portion is poly. So no practical advantage over the polysilicon-over-crystalline-Si structure. **(n) Heterojunction:** a junction between two dissimilar semiconductors (different band gaps), in contrast to a homojunction. **(o) BJT versus HBT.** An HBT has a wider band-gap emitter than the base (heterojunction E-B junction); its base is generally the most heavily doped region. The built-in valence/conduction-band offset blocks back-injection, so emitter efficiency stays near unity despite $N_E\lesssim N_B$, while heavy base doping gives low $r_B$ and small Early effect. A BJT needs $N_E\gg N_B$ and so trades gain against base resistance and modulation. --- ## Problem 11.2 **Problem.** The electron and hole currents in an active-mode pnp BJT are plotted in Fig. P11.2 (relative to the junction positions). Determine: (a) $\gamma$; (b) $\alpha_T$; (c) $\beta_{dc}$; (d) $I_B$; (e) whether R-G in the depletion regions is negligible, with explanation. **Reading of the figure (assumption).** The OCR shows hole current labels $1.000$ and $0.999$ and electron-current labels $0.001$, $0.001$ (units normalized to the hole current injected by the emitter, e.g. mA). We read: $I_{Ep}=1.000$ entering the base, $I_{Cp}=0.999$ leaving the base into the collector, electron current in the emitter at the junction $I_{En}=0.001$, and negligible collector electron current ($I_{Cn}\approx0$). **Solution.** (a)

\gamma=\frac{I_{Ep}}{I_{Ep}+I_{En}}=\frac{1.000}{1.000+0.001}=0.999

$$(b)$$

\alpha_T=\frac{I_{Cp}}{I_{Ep}}=\frac{0.999}{1.000}=0.999

$$(c)$$

\alpha_{dc}=\gamma\alpha_T=0.999\times0.999=0.998,\qquad \beta_{dc}=\frac{\alpha_{dc}}{1-\alpha_{dc}}=\frac{0.998}{0.002}\approx499\ (\approx500)

(d) $I_E=I_{Ep}+I_{En}=1.001$, $I_C=I_{Cp}+I_{Cn}=0.999$:

I_B=I_E-I_C=1.001-0.999=0.002\quad\text{(same units as the plot)}

Check: $I_B=I_C/\beta_{dc}=0.999/499=0.002$. It consists of $I_{B1}=I_{En}=0.001$ (electrons injected into the emitter) plus $I_{B2}=I_{Ep}-I_{Cp}=0.001$ (base recombination). (e) **Yes, negligible.** If R-G current were significant in a depletion region, the hole (or electron) current entering that region would differ from the current leaving it. In the figure the hole current is continuous across the E-B depletion region (1.000 on both sides), the electron current is continuous across it (0.001 on both sides), and the electron current at the C-B junction is continuous at its negligible value; the only change in the hole current (1.000 to 0.999) occurs *inside the quasineutral base* (recombination), not in a depletion region. So the ideal-theory assumption of negligible depletion-region R-G holds for this device. --- ## Problem 11.3 **Problem.** Indicate the required revisions to the equations of Subsections 11.1.1 and 11.1.2 if the device is an **npn** BJT. **Solution.** In an npn the roles of electrons and holes swap: emitter $n^+$ (minority holes), base $p$ (minority electrons), collector $n$ (minority holes). Voltages become $V_{BE}$ and $V_{BC}$ (a forward-biased E-B junction has $V_{BE}>0$; active mode: $V_{BE}>0$, $V_{BC}<0$). Positive current directions reverse (emitter current positive leaving the emitter terminal, collector current positive entering the collector), which makes the derivative formulas (11.7)-(11.10) keep the same *form*. Replace: $n_{E0}\to p_{E0}=n_i^2/N_E$, $p_{B0}\to n_{B0}=n_i^2/N_B$, $n_{C0}\to p_{C0}=n_i^2/N_C$; $D_E=D_P$, $D_B=D_N$, $D_C=D_P$; $L_E=L_P$, $L_B=L_N$, $L_C=L_P$; $\tau_E=\tau_p$, $\tau_B=\tau_n$, $\tau_C=\tau_p$; $V_{EB}\to V_{BE}$, $V_{CB}\to V_{BC}$; $I_{En}\to I_{Ep}$ and so on as below. Equation by equation: | Eq. | npn version | |---|---| | (11.1) | $0=D_E\,d^2\Delta p_E/dx''^2-\Delta p_E/\tau_E$ | | (11.2a,b) | $\Delta p_E(\infty)=0$; $\Delta p_E(0)=p_{E0}(e^{qV_{BE}/kT}-1)$ | | (11.3) | $0=D_B\,d^2\Delta n_B/dx^2-\Delta n_B/\tau_B$ | | (11.4a,b) | $\Delta n_B(0)=n_{B0}(e^{qV_{BE}/kT}-1)$; $\Delta n_B(W)=n_{B0}(e^{qV_{BC}/kT}-1)$ | | (11.5) | $0=D_C\,d^2\Delta p_C/dx'^2-\Delta p_C/\tau_C$ | | (11.6a,b) | $\Delta p_C(\infty)=0$; $\Delta p_C(0)=p_{C0}(e^{qV_{BC}/kT}-1)$ | | (11.7) | $I_{Ep}=-qAD_E\,d\Delta p_E/dx''\big|_0$ (hole back-injection into emitter) | | (11.8) | $I_{En}=-qAD_B\,d\Delta n_B/dx\big|_0$ (electrons injected into base) | | (11.9) | $I_{Cn}=-qAD_B\,d\Delta n_B/dx\big|_W$ | | (11.10) | $I_{Cp}=qAD_C\,d\Delta p_C/dx'\big|_0$ | | (11.11) | $\gamma=I_{En}/(I_{En}+I_{Ep})$ | | (11.12) | $\alpha_T=I_{Cn}/I_{En}$ | | (11.13), (11.14) | unchanged: $\alpha_{dc}=\gamma\alpha_T$, $\beta_{dc}=\alpha_{dc}/(1-\alpha_{dc})$ | | (11.15)-(11.17) | $I_E=I_{En}+I_{Ep}$, $I_C=I_{Cn}+I_{Cp}$, $I_B=I_E-I_C$ | | (11.18), (11.19) | $\Delta p_E=A_1e^{-x''/L_E}+A_2e^{x''/L_E}$; $\Delta p_E=p_{E0}(e^{qV_{BE}/kT}-1)e^{-x''/L_E}$ | | (11.20) | $I_{Ep}=qA(D_E/L_E)p_{E0}(e^{qV_{BE}/kT}-1)$ | | (11.21), (11.22) | $\Delta p_C=p_{C0}(e^{qV_{BC}/kT}-1)e^{-x'/L_C}$; $I_{Cp}=-qA(D_C/L_C)p_{C0}(e^{qV_{BC}/kT}-1)$ | | (11.23)-(11.26) | $\Delta n_B(x)$ with $L_B=L_N$: $\Delta n_B(0)\sinh[(W-x)/L_B]/\sinh(W/L_B)+\Delta n_B(W)\sinh(x/L_B)/\sinh(W/L_B)$ | | (11.28) | $I_{En}=qA\frac{D_B}{L_B}n_{B0}\left[\coth\frac{W}{L_B}(e^{qV_{BE}/kT}-1)-\frac{1}{\sinh(W/L_B)}(e^{qV_{BC}/kT}-1)\right]$ | | (11.29) | $I_{Cn}=qA\frac{D_B}{L_B}n_{B0}\left[\frac{1}{\sinh(W/L_B)}(e^{qV_{BE}/kT}-1)-\coth\frac{W}{L_B}(e^{qV_{BC}/kT}-1)\right]$ | | (11.31) | $\gamma=1\big/\left[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\tanh(W/L_B)\right]$ (same form, using $p_{E0}/n_{B0}=N_B/N_E$) | | (11.32)-(11.34) | $\alpha_T$, $\alpha_{dc}$, $\beta_{dc}$ unchanged in form | | (11.35), (11.36) | $I_E=qA\{[\frac{D_E}{L_E}p_{E0}+\frac{D_B}{L_B}n_{B0}\coth\frac{W}{L_B}](e^{qV_{BE}/kT}-1)-\frac{D_B}{L_B}\frac{n_{B0}}{\sinh(W/L_B)}(e^{qV_{BC}/kT}-1)\}$; $I_C=qA\{\frac{D_B}{L_B}\frac{n_{B0}}{\sinh(W/L_B)}(e^{qV_{BE}/kT}-1)-[\frac{D_C}{L_C}p_{C0}+\frac{D_B}{L_B}n_{B0}\coth\frac{W}{L_B}](e^{qV_{BC}/kT}-1)\}$ | Note on signs: for an npn, the signs of derivatives such as $I_{En}$ use the convention that positive current leaves the emitter and enters the collector; with that convention the sign pattern of the pnp relations carries over exactly. --- ## Problem 11.4 (computer-aided) **Problem.** pnp BJT, 300 K, active mode, $V_{EB}=0.7$ V, $V_{CB}=-5$ V, $W=W_B$, $A=10^{-4}$ cm$^2$, parameters of Table 11.1. Calculate (a) $\cosh(W/L_B)$, $\sinh(W/L_B)$, $W/L_B$ (five significant figures); (b) $I_{Ep},I_{En},I_{Cp},I_{Cn}$; (c) $I_E,I_C,I_B$; (d) $I_{B1},I_{B2},I_{B3}$ as in Subsection 10.4; (e) $\gamma,\alpha_T,\alpha_{dc},\beta_{dc}$. **Solution.** Constants: $kT/q=8.617\times10^{-5}\times300=0.025851$ V, $n_i=10^{10}$ cm$^{-3}$, so $n_{E0}=100$, $p_{B0}=10^4$, $n_{C0}=10^5$ cm$^{-3}$. (a)

\frac{W}{L_B}=\frac{2\times10^{-4}}{3.36\times10^{-3}}=0.059524,\quad \cosh\frac{W}{L_B}=1.0017721,\quad \sinh\frac{W}{L_B}=0.059559

$$Voltagefactors:$$

e^{qV_{EB}/kT}-1=e^{0.7/0.025851}-1=5.7475\times10^{11},\qquad e^{qV_{CB}/kT}-1=e

$$(b)Apply(11.20),(11.28),(11.29),(11.22):$$

I_{En}=qA\frac{D_E}{L_E}n_{E0}\left(e^{qV_{EB}/kT}-1\right)=(1.602\times10^{-19})(10^{-4})\frac{6.81}{8.25\times10^{-4}}(100)(5.7475\times10^{11})=7.60\ \mu\text{A}

\frac{qAD_Bp_{B0}}{L_B}=\frac{(1.602\times10^{-19})(10^{-4})(11.3)(10^4)}{3.36\times10^{-3}}=5.388\times10^{-16}\ \text{A}

I_{Ep}=5.388\times10^{-16}\left[\frac{1.0017721}{0.059559}(5.7475\times10^{11})-\frac{-1}{0.059559}\right]=5.2084\ \text{mA}

I_{Cp}=5.388\times10^{-16}\left[\frac{5.7475\times10^{11}}{0.059559}-\frac{1.0017721}{0.059559}(-1)\right]\approx5.1992\ \text{mA}

(the $V_{CB}$ terms are of order $10^{-14}$ A, utterly negligible next to the mA-level $V_{EB}$ terms);

I_{Cn}=-qA\frac{D_C}{L_C}n_{C0}\left(e^{qV_{CB}/kT}-1\right)=(1.602\times10^{-19})(10^{-4})\frac{34.8}{5.90\times10^{-3}}(10^5)(1)=9.45\times10^{-15}\ \text{A}

Results: $I_{Ep}=5.2084$ mA, $I_{En}=7.600\ \mu$A, $I_{Cp}=5.1992$ mA, $I_{Cn}=9.45\times10^{-15}$ A. (c)

I_E=I_{Ep}+I_{En}=5.2084+0.0076=5.2160\ \text{mA}

I_C=I_{Cp}+I_{Cn}=5.1992\ \text{mA}

I_B=I_E-I_C=16.8\ \mu\text{A}

(d) The base-current components of Section 10.4 are $I_{B1}$ (electron back-injection into the emitter), $I_{B2}$ (hole recombination in the base), $I_{B3}$ (collector thermal-generation electrons), with $I_B=I_{B1}+I_{B2}-I_{B3}$:

I_{B1}=I_{En}=7.60\ \mu\text{A},\qquad I_{B2}=I_{Ep}-I_{Cp}=5.2084-5.1992=9.21\ \mu\text{A},\qquad I_{B3}=I_{Cn}=9.45\times10^{-15}\ \text{A}

Check: $7.60+9.21-0=16.8\ \mu$A. (e)

\gamma=\frac{I_{Ep}}{I_E}=\frac{5.2084}{5.2160}=0.99854,\qquad \alpha_T=\frac{I_{Cp}}{I_{Ep}}=\frac{5.1992}{5.2084}=0.99823

\alpha_{dc}=\gamma\alpha_T=0.99678,\qquad \beta_{dc}=\frac{\alpha_{dc}}{1-\alpha_{dc}}=309

(consistent with $I_C/I_B=5.1992\ \text{mA}/16.8\ \mu\text{A}=309$ and with the approximate formulas' 310). ```matlab % Problem 11.4: BJT current components and performance parameters q=1.602e-19; kT=8.617e-5*300; ni=1e10; A=1e-4; NE=1e18; NB=1e16; NC=1e15; DE=6.81; DB=11.3; DC=34.8; LE=8.25e-4; LB=3.36e-3; LC=5.90e-3; W=2e-4; nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC; VEB=0.7; VCB=-5; x=W/LB; c=cosh(x); s=sinh(x); fprintf('%.5g %.5g %.5g\n', c, s, x) fe=exp(VEB/kT)-1; fc=exp(VCB/kT)-1; IEn=q*A*(DE/LE)*nE0*fe; IEp=q*A*(DB/LB)*pB0*((c/s)*fe-fc/s); ICp=q*A*(DB/LB)*pB0*(fe/s-(c/s)*fc); ICn=-q*A*(DC/LC)*nC0*fc; IE=IEn+IEp; IC=ICp+ICn; IB=IE-IC; IB1=IEn; IB2=IEp-ICp; IB3=ICn; gam=IEp/IE; aT=ICp/IEp; adc=gam*aT; bdc=adc/(1-adc); disp([IEp IEn ICp ICn IE IC IB IB1 IB2 IB3 gam aT adc bdc]) ``` --- ## Problem 11.5 (computer) **Problem.** For $V_{CB}=0$ ($\Delta p_B(W)=0$) plot $\Delta p_B(x)/\Delta p_B(0)$ versus $x/W$ for $W/L_B=10,5,1,0.5,0.1$ and comment. **Solution.** From (11.26) with $\Delta p_B(W)=0$:

\frac{\Delta p_B(x)}{\Delta p_B(0)}=\frac{\sinh[(W/L_B)(1-x/W)]}{\sinh(W/L_B)}

```matlab xw=linspace(0,1,201); % x/W r=[10 5 1 0.5 0.1]; % W/LB for i=1:length(r) y=sinh(r(i)*(1-xw))/sinh(r(i)); plot(xw,y); hold on end hold off; grid on; axis([0 1 0 1]) xlabel('x/W'); ylabel('\Deltap_B(x)/\Deltap_B(0)') legend('W/L_B=10','5','1','0.5','0.1') ``` Computed samples at $x/W=0,\,0.25,\,0.5,\,0.75,\,1$: | $W/L_B$ | 0 | 0.25 | 0.5 | 0.75 | 1 | |---|---|---|---|---|---| | 10 | 1 | 0.0821 | 0.0067 | 0.0005 | 0 | | 5 | 1 | 0.2864 | 0.0815 | 0.0216 | 0 | | 1 | 1 | 0.6997 | 0.4434 | 0.2150 | 0 | | 0.5 | 1 | 0.7366 | 0.4848 | 0.2405 | 0 | | 0.1 | 1 | 0.7495 | 0.4994 | 0.2496 | 0 | **Comment.** For $W/L_B\gg1$ (10, 5) the curve is essentially the decaying exponential $e^{-x/L_B}$ of a long diode: the excess vanishes well before $x=W$. For $W/L_B\le0.5$ the curves are practically the straight line $1-x/W$ (e.g. 0.4994 vs 0.5 at midpoint for 0.1). Even at $W/L_B=1$ the curve departs from the line by only about 0.057 (at most roughly 11% at the midpoint), which confirms the claim that the linear approximation is good up to $W/L_B\approx1$. --- ## Problem 11.6 (computer) **Problem.** Construct four subplots of $\Delta p_B(x)/p_{B0}$ versus $x/W$ for the four biasing modes, with $W/L_B\ll1$, $\Delta p_B(0)/p_{B0}$ and $\Delta p_B(W)/p_{B0}\le10$; axes $x_{min}=0$, $x_{max}=1$, $y_{min}=-2$, $y_{max}=10$. **Solution.** For $W\ll L_B$ the profile is linear between the end values $e^{qV_{EB}/kT}-1$ and $e^{qV_{CB}/kT}-1$. Choose illustrative end values (normalized): | Mode | $V_{EB}$ | $V_{CB}$ | $\Delta p_B(0)/p_{B0}$ | $\Delta p_B(W)/p_{B0}$ | |---|---|---|---|---| | Active | $>0$ | $<0$ | 10 | $-1$ | | Inverted | $<0$ | $>0$ | $-1$ | 10 | | Saturation | $>0$ | $>0$ | 10 | 5 | | Cutoff | $<0$ | $<0$ | $-1$ | $-1$ | ```matlab xw=linspace(0,1,100); modes={'Active','Inverted','Saturation','Cutoff'}; d0=[10 -1 10 -1]; % dp(0)/pB0 dW=[-1 10 5 -1]; % dp(W)/pB0 for i=1:4 subplot(2,2,i) y=d0(i)+(dW(i)-d0(i))*xw; % Eq. (11.40), normalized plot(xw,y,'LineWidth',1.5); grid on axis([0 1 -2 10]) title(modes{i}); xlabel('x/W'); ylabel('\Deltap_B/p_{B0}') end ``` The plots show: active, a falling line from high to $-1$; inverted, a rising line from $-1$ to high (mirror image); saturation, a line between two positive values (base flooded); cutoff, a flat line at $-1$ (the base is depleted of minority carriers at both edges). --- ## Problem 11.7 **Problem.** For a pnp BJT with $W\ll L_B$, $N_E\gg N_B>N_C$, sketch (linear scale) majority and minority carrier distributions in the quasineutral regions under (a) active, (b) inverted, (c) saturation, (d) cutoff bias (Fig. P11.7 gives the equilibrium levels as dashed lines). **Solution.** Rules used: minority excess at a depletion edge is $(e^{qV/kT}-1)$ times equilibrium; majority carriers follow the minority excess to preserve neutrality, but under low-level injection that change is negligible relative to the majority level, so the majority curves are essentially flat at the equilibrium values ($p_E\approx N_E$, $n_B\approx N_B$, $p_C\approx N_C$) with tiny bumps that mimic the minority excess. Minority profiles: exponential decay in emitter and collector, straight line in base. **(a) Active** ($V_{EB}>0$, $V_{CB}<0$): - Emitter minority $n_E$: peak at E-B edge $n_{E0}e^{qV_{EB}/kT}\gg n_{E0}$, decaying exponentially to $n_{E0}$. - Base minority $p_B$: line from $p_{B0}e^{qV_{EB}/kT}$ (high) at $x=0$ to $\approx0$ at $x=W$. - Collector minority $n_C$: $\approx0$ at the C-B edge, rising exponentially to $n_{C0}$ (depleted, since reverse bias). - Majority: flat (emitter at $N_E$, base at $N_B$, collector at $N_C$), with the base majority bumping slightly with the same profile as $p_B$. **(b) Inverted** ($V_{EB}<0$, $V_{CB}>0$): the roles of emitter and collector swap. - $n_E\approx0$ at the E-B edge rising to $n_{E0}$. - $p_B$: from $\approx0$ at $x=0$ rising linearly to $p_{B0}e^{qV_{CB}/kT}$ at $x=W$. - $n_C$: peak $n_{C0}e^{qV_{CB}/kT}$ at the C-B edge, decaying to $n_{C0}$. - Majority flat. **(c) Saturation** ($V_{EB}>0$, $V_{CB}>0$): excess at all three depletion edges. - $n_E$: peak $n_{E0}e^{qV_{EB}/kT}$, decaying to $n_{E0}$. - $p_B$: line from $p_{B0}e^{qV_{EB}/kT}$ at $x=0$ to $p_{B0}e^{qV_{CB}/kT}$ at $x=W$ (both above $p_{B0}$). - $n_C$: peak $n_{C0}e^{qV_{CB}/kT}$, decaying to $n_{C0}$. - Majority flat (slight bumps). **(d) Cutoff** ($V_{EB}<0$, $V_{CB}<0$): depletion of minority carriers at every edge. - $n_E$: $\approx0$ at the edge rising to $n_{E0}$. - $p_B$: line from $\approx0$ at both ends... i.e. flat near zero at the edges; with $W\ll L_B$ the whole base sits close to 0 (a line between two near-zero values). - $n_C$: $\approx0$ at the edge rising to $n_{C0}$. - Majority flat. ``` (a) active: n_E \ p_B \ n_C ___ \___ \___ __/ \_ 0 ____/ (c) saturation: n_E \ p_B \___ n_C \___ (d) cutoff: n_E ___/ p_B ~~~~ 0 n_C ___/ ``` (Linear-scale sketches: minority curves in the heavily doped emitter are tiny on the scale of the base, as the problem's figure indicates qualitatively.) --- ## Problem 11.8 **Problem.** The $\Delta n_E/n_{E0}$, $\Delta p_B/p_{B0}$, $\Delta n_C/n_{C0}$ distributions of a pnp BJT are as in Fig. P11.8. Determine: (a) polarity of $V_{EB}$; (b) polarity of $V_{CB}$; (c) magnitude of $V_{CB}$; (d) the bias mode; (e) repeat (a)-(d) for an npn. **Reading of the figure (assumption).** The OCR retains the labels "10" and "$-1$" on the normalized axis. We read: at the emitter and at the base edge $x=0$ the normalized excess equals $+10$ (peak), the base line falls linearly to $-1$ at $x=W$, and the collector curve is at $-1$ at its edge, rising to 0. **Solution.** (a) $\Delta n_E(0)/n_{E0}=e^{qV_{EB}/kT}-1=10>0$, so $V_{EB}>0$ (forward bias). Its magnitude:

V_{EB}=\frac{kT}{q}\ln(1+10)=0.0259\ln11=62\ \text{mV}

(b) $\Delta p_B(W)/p_{B0}=\Delta n_C(0)/n_{C0}=e^{qV_{CB}/kT}-1=-1<0$, so $V_{CB}<0$ (reverse bias). (c) A value of exactly $-1$ requires $e^{qV_{CB}/kT}=0$, i.e. $|V_{CB}|\gg kT/q$. The plot only shows that $|V_{CB}|$ is several $kT/q$ ($\gtrsim0.12$ V gives $e^{-4.6}-1=-0.99$); the actual magnitude cannot be determined from the data. (d) Forward E-B and reverse C-B: **active mode**. (e) **npn.** With the same normalized distributions (now $\Delta p_E/p_{E0}$, $\Delta n_B/n_{B0}$, $\Delta p_C/p_{C0}$): $V_{BE}=+62$ mV, $V_{BC}<0$ with $|V_{BC}|\gg kT/q$ (magnitude undetermined), active mode. In terms of the pnp-style definitions $V_{EB}=-V_{BE}$ and $V_{CB}=-V_{BC}$: $V_{EB}<0$ and $V_{CB}>0$, i.e. the terminal polarities are reversed relative to the pnp case, as expected. --- ## Problem 11.9 **Problem.** Fill in the table: effect of increasing $W_B$, $\tau_B$, $N_B$, $\tau_E$, $N_E$ on $\gamma$, $\alpha_T$, $\beta_{dc}$. **Solution.** Use (11.41)-(11.44):

\gamma=\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}},\quad \alpha_T\approx1-\frac{W^2}{2L_B^2},\quad \beta_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{W^2}{2L_B

with $L_B=\sqrt{D_B\tau_B}$, $L_E=\sqrt{D_E\tau_E}$ and mobilities (hence $D$'s) decreasing with doping. | Change | $\gamma$ | $\alpha_T$ | $\beta_{dc}$ | Reason | |---|---|---|---|---| | Increase $W_B$ | decreases | decreases | decreases | $W$ appears in the numerator of both loss terms | | Increase $\tau_B$ | no effect | increases | increases | $L_B$ grows, so $W^2/2L_B^2$ shrinks; $\gamma$ has no $\tau_B$ | | Increase $N_B$ | decreases | decreases (slightly) | decreases | $N_B/N_E$ grows; $D_B$ falls with doping so $L_B$ falls | | Increase $\tau_E$ | increases | no effect | increases | $L_E$ grows, shrinking $W/L_E$; $\alpha_T$ independent of emitter | | Increase $N_E$ | increases | no effect | increases | $N_B/N_E$ falls; also $D_E/L_E=\sqrt{D_E/\tau_E}$ falls as $D_E$ drops with doping, both lowering back-injection | --- ## Problem 11.10 (computer) **Problem.** With the Table 11.1 lifetimes and dopings as reference values ($\tau_{B0}$, $N_{B0}$, $\tau_{E0}$, $N_{E0}$), and the mobility fits of Exercise 3.1, construct four log-log subplots of $\beta_{dc}$ versus $\tau_B/\tau_{B0}$, $N_B/N_{B0}$, $\tau_E/\tau_{E0}$, $N_E/N_{E0}$, each ratio from 0.1 to 10. $T=300$ K, $W=W_B=2\ \mu$m. Ignore emitter degeneracy. Do the results agree with 11.9? **Solution (logic).** For each ratio, recompute the mobility-fit values $\mu$, then $D=(kT/q)\mu$, $L=\sqrt{D\tau}$, and use the general (11.34) (or the simplified (11.44)). ```matlab q=1.602e-19; kT=8.617e-5*300; W=2e-4; mun=@(N) 92+1268./(1+(N/1.3e17).^0.91); mup=@(N) 54.3+406.9./(1+(N/2.35e17).^0.88); NE0=1e18; NB0=1e16; tE0=1e-7; tB0=1e-6; r=logspace(-1,1,50); beta=@(NE,NB,tE,tB) ... 1./( cosh(W./sqrt(kT*mup(NB).*tB)) ... + (kT*mun(NE)./(kT*mup(NB))).*(NB./NE).* ... sqrt(kT*mup(NB).*tB)./sqrt(kT*mun(NE).*tE) ... .*sinh(W./sqrt(kT*mup(NB).*tB)) - 1 ); subplot(2,2,1); loglog(r, beta(NE0,NB0,tE0,tB0*r)); xlabel('\tau_B/\tau_{B0}'); ylabel('\beta_{dc}'); grid on subplot(2,2,2); loglog(r, beta(NE0,NB0*r,tE0,tB0)); xlabel('N_B/N_{B0}'); ylabel('\beta_{dc}'); grid on subplot(2,2,3); loglog(r, beta(NE0,NB0,tE0*r,tB0)); xlabel('\tau_E/\tau_{E0}'); ylabel('\beta_{dc}'); grid on subplot(2,2,4); loglog(r, beta(NE0*r,NB0,tE0,tB0)); xlabel('N_E/N_{E0}'); ylabel('\beta_{dc}'); grid on ``` (Here `beta` is (11.34) written with $D_E/D_B$, $N_B/N_E$, $L_B/L_E$.) **Expected results and agreement.** With the reference loss terms $a=1.46\times10^{-3}$ (emitter) and $b=1.77\times10^{-3}$ (base), $\beta_{dc}=1/(a+b)=310$ at the reference point. - *vs $\tau_B$:* $b\propto1/\tau_B$, so $\beta_{dc}$ increases with $\tau_B$ and **saturates** at $1/a\approx685$ (the emitter-limited ceiling). - *vs $N_B$:* $a\propto N_B$ (and $D_B$ shrinks, which raises $b$), so $\beta_{dc}$ **decreases**, roughly inversely with $N_B$ at large $N_B$. - *vs $\tau_E$:* $a\propto1/\sqrt{\tau_E}$, so $\beta_{dc}$ **increases** with $\tau_E$ and saturates at $1/b\approx565$. - *vs $N_E$:* $a\propto1/N_E$ (times the mobility variation), so $\beta_{dc}$ **increases**, saturating at $1/b$ once emitter losses are negligible. All four trends agree with the Problem 11.9 table (increase $\tau_B$: up; increase $N_B$: down; increase $\tau_E$: up; increase $N_E$: up). --- ## Problem 11.11: Shallow Emitter **Problem.** Take a pnp BJT whose emitter has finite quasineutral width $W_E$ (Fig. P11.11), with $\Delta n_E=0$ at the metallic emitter contact ($x''=W_E$). (a) Paralleling the ideal analysis, obtain revised expressions for $\Delta n_E(x'')$ and $I_{En}$. (b) Establish revised performance parameters analogous to (11.31)-(11.34). (c) Simplify for $W/L_B\ll1$ and $W_E/L_E\ll1$. (d) How are $\gamma$ and $\beta_{dc}$ affected as $W_E$ decreases? (e) Sketch the minority distribution in a shallow emitter, active mode. **Solution.** (a) The emitter equation (11.1) is unchanged, but the boundary conditions are now

\Delta n_E(0)=n_{E0}\left(e^{qV_{EB}/kT}-1\right),\qquad \Delta n_E(W_E)=0

This is the base-type problem with the second end value zero. Reusing (11.26) with $x\to x''$, $W\to W_E$, $L_B\to L_E$:

\Delta n_E(x”)=n_{E0}\left(e

Differentiate at $x''=0$:

\frac{d\Delta n_E}{dx”}\Big|0=-\frac{n{E0}}{L_E}\left(e

$$andby(11.7):$$

I_{En}=qA\frac{D_E}{L_E}n_{E0}\coth\left(\frac{W_E}{L_E}\right)\left(e

(As $W_E/L_E\to\infty$, $\coth\to1$ recovers (11.20).) (b) $I_{Ep}$ and $I_{Cp}$ are unchanged (base unchanged), so in active mode $I_{Ep}\approx qA(D_B/L_B)p_{B0}\coth(W/L_B)(e^{qV_{EB}/kT}-1)$. Then

\frac{I_{En}}{I_{Ep}}=\frac{D_E,n_{E0}\coth(W_E/L_E)/L_E}{D_Bp_{B0}\coth(W/L_B)/L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth\frac{W_E}{L_E}\tanh\frac{W}{L_B}

\gamma=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\coth\dfrac{W_E}{L_E}\tanh\dfrac{W}{L_B}}

\alpha_T=\frac{1}{\cosh(W/L_B)}\quad\text{(unchanged)}

\alpha_{dc}=\gamma\alpha_T=\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\coth\dfrac{W_E}{L_E}\sinh\dfrac{W}{L_B}}

\beta_{dc}=\frac{1}{\cosh\dfrac{W}{L_B}+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\coth\dfrac{W_E}{L_E}\sinh\dfrac{W}{L_B}-1}

(c) For $W/L_B\ll1$ ($\tanh\to W/L_B$) and $W_E/L_E\ll1$ ($\coth\to L_E/W_E$):

\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\cdot\frac{L_E}{W_E}\cdot\frac{W}{L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}

\gamma=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{W_E}},\qquad \alpha_T\approx1-\frac12\left(\frac{W}{L_B}\right)

\alpha_{dc}=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{W_E}+\frac12\left(\dfrac{W}{L_B}\right)^2},\qquad \beta_{dc}=\frac{1}{\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{W_E}+\frac12\left(\dfrac{W}{L_B}\right)

Thus $W_E$ simply replaces $L_E$ in (11.41)-(11.44). (d) $W_E$ appears in the denominator of the emitter loss term, so decreasing $W_E$ increases that term: $\gamma$ **decreases** and $\beta_{dc}$ **decreases** (and $\alpha_T$ is unaffected). The thinner the emitter, the more steeply minority carriers are drawn from the junction toward the contact, so more back-injection current flows. (e) Sketch: in the emitter ($\Delta n_E$ versus $x''$) a *straight line* from $n_{E0}(e^{qV_{EB}/kT}-1)$ at the junction to 0 at the contact; in the base a line from $p_{B0}(e^{qV_{EB}/kT}-1)$ to $-p_{B0}$; in the collector an exponential from $\approx-n_{C0}$ rising to 0. ``` Δn_E | \ Δp_B \ | \ (straight line, \ (straight line) | \ hits 0 at W_E) \ +-------\------ x'' +------\----- x 0 W_E 0 W ``` --- ## Problem 11.12: Diode-Like Connections **Problem.** With one terminal floating or two terminals tied together, the transistor is a two-terminal "diode" (Fig. P11.12, six connections). Choose a connection and (a) derive $I$-$V_A$ from the Ebers-Moll equations; (b) find $\Delta p_B(0)/p_{B0}$ and $\Delta p_B(W)/p_{B0}$ in terms of $V_A$ and EM parameters; (c) simplify with $\alpha_F=\alpha_R=\alpha$ and $I_{F0}=I_{R0}=I_0$ (identical emitter and collector parameters); (d) sketch the base distribution for $V_A>kT/q$ with $W\ll L_B$; (e) repeat for reverse bias $-V_A\gg kT/q$. **Interpretation of the figure (assumption).** The six connections are read as: (a) E-B junction with collector open; (b) E-B with collector shorted to base; (c) C-B junction with emitter open; (d) C-B with emitter shorted to base; (e) E-C with base open; (f) the E-C counterpart with the base shorted. We solve (a)-(e) (including both open-circuit and short-circuit cases). Notation: $u\equiv e^{qV_A/kT}-1$, $a\equiv e^{qV_{EB}/kT}-1$, $b\equiv e^{qV_{CB}/kT}-1$. Useful: $\alpha_FI_{F0}=\alpha_RI_{R0}$. Current $I$ is the current entering the terminal at $+V_A$ in the forward direction of that junction. **Solution.** **Connection (a): E-B diode, collector open.** $V_{EB}=V_A$ so $a=u$, and $I_C=0$:

0=\alpha_FI_{F0},u-I_{R0},b;\Rightarrow;b=\frac{\alpha_FI_{F0}}{I_{R0}}u=\alpha_Ru

(using $\alpha_FI_{F0}=\alpha_RI_{R0}$). Then

I=I_E=I_{F0}u-\alpha_RI_{R0},b=I_{F0}u-\alpha_RI_{R0}\alpha_Ru

Using $\alpha_RI_{R0}=\alpha_FI_{F0}$:

\boxed{I=I_{F0}\left(1-\alpha_F\alpha_R\right)\left(e

(b) $\dfrac{\Delta p_B(0)}{p_{B0}}=a=u,\qquad\dfrac{\Delta p_B(W)}{p_{B0}}=b=\alpha_Ru$. (c) With $\alpha_F=\alpha_R=\alpha$: $I=I_0(1-\alpha^2)u$; $\Delta p_B(0)/p_{B0}=u$, $\Delta p_B(W)/p_{B0}=\alpha u$. **Connection (b): E-B diode, collector shorted to base** ($V_{CB}=0$, so $b=0$; $V_{EB}=V_A$):

\boxed{I=I_E=I_{F0}\left(e

(b) $\Delta p_B(0)/p_{B0}=u$, $\Delta p_B(W)/p_{B0}=0$. (c) Same with $I_{F0}=I_0$: $I=I_0u$, end values $u$ and 0. **Connection (c): C-B diode, emitter open.** $V_{CB}=V_A$ so $b=u$, and $I_E=0$:

0=I_{F0},a-\alpha_RI_{R0},u;\Rightarrow;a=\frac{\alpha_RI_{R0}}{I_{F0}}u=\alpha_Fu

The current into the C-B junction in its forward direction is $-I_C$:

-I_C=-\alpha_FI_{F0},a+I_{R0},u=-\alpha_F

With $\alpha_F^2I_{F0}=\alpha_F\alpha_R I_{R0}$:

\boxed{I=-I_C=I_{R0}\left(1-\alpha_F\alpha_R\right)\left(e

For $-V_A\gg kT/q$ this tends to $-I_{R0}(1-\alpha_F\alpha_R)=-I_{CBO}$, confirming Exercise 11.6. (b) $\Delta p_B(0)/p_{B0}=a=\alpha_Fu$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_0(1-\alpha^2)u$; end values $\alpha u$ and $u$. **Connection (d): C-B diode, emitter shorted to base** ($V_{EB}=0$, $a=0$, $b=u$):

\boxed{I=-I_C=I_{R0}\left(e

(b) $\Delta p_B(0)/p_{B0}=0$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_0u$; end values 0 and $u$. **Connection (e): E-C two-terminal, base open.** $I_B=0$, so $I_E=I_C\equiv I$ and $V_A=V_{EC}=V_{EB}-V_{CB}$. From $I_B=I_E-I_C$:

0=(1-\alpha_F)I_{F0},a+(1-\alpha_R)I_{R0},b;\Rightarrow;b=-\frac{(1-\alpha_F)I_{F0}}{(1-\alpha_R)I_{R0}}a

Together with $V_A=V_{EB}-V_{CB}$ this fixes $a$ and $b$ numerically (transcendental). For $V_A\gg kT/q$: $V_{CB}\ll0$, $b\approx-1$, so

a\approx\frac{(1-\alpha_R)I_{R0}}{(1-\alpha_F)I_{F0}},\qquad I=I_{F0}a+\alpha_RI_{R0}=I_{R0}\frac{1-\alpha_R}{1-\alpha_F}+\alpha_RI_{R0}=\frac{I_{R0}(1-\alpha_F\alpha_R)}{1-\alpha_F}=\frac{I_{CBO}}{1-\alpha_F}=I_{CEO}

(b) $\Delta p_B(0)/p_{B0}=a$, $\Delta p_B(W)/p_{B0}=-1$. (c) With the symmetric simplification ($\alpha_F=\alpha_R=\alpha$, $I_{F0}=I_{R0}=I_0$) the condition $I_B=0$ gives $b=-a$, i.e. $e^{qV_{CB}/kT}=2-e^{qV_{EB}/kT}$, and $I=I_E=I_0(a-\alpha b)=I_0(1+\alpha)a$, with $a$ fixed numerically by $V_A=V_{EB}-V_{CB}$. The base profile is antisymmetric: end values $a$ and $-a$. **(d) Sketch, forward bias $V_A>kT/q$, $W\ll L_B$, simplified case (c):** ``` Connection (a)/(c): (a) high flat-ish line (b)/(d) full ramp Δp_B/p_B0 Δp_B/p_B0 u |\ u |\ | \__ αu (almost flat, α≈1; | \ | both junctions forward biased, | \ 0 ---+----- x base flooded: saturation-like) 0 --+---\--- x ``` (a) and (c): lines between $u$ and $\alpha u$ (nearly flat since $\alpha\approx1$, both junctions forward biased: base flooded, like saturation). (b): line from $u$ to 0 (like the active or narrow-base diode); (d): from 0 to $u$. **(e) Sketch, reverse bias $-V_A\gg kT/q$ ($u\to-1$):** (a): end values $-1$ and $-\alpha\approx-1$: a nearly flat line at $-1$ (base depleted of minority carriers, cutoff-like); (b): $-1$ to 0; (c): $-\alpha$ to $-1$ (flat, near $-1$); (d): 0 to $-1$. --- ## Problem 11.13 **Problem.** The CE output characteristics of a pnp BJT at small $V_{EC}$ are in Fig. P11.13; the d.c. operating point lies at the boundary between active mode and saturation (constant $I_B$). (a) Referring to Fig. 11.3, draw the simplified large-signal equivalent circuit at that point. (b) Obtain $V_{EC}$ at the operating point in terms of $I_B$ and the EM parameters. **Solution.** (a) At the active/saturation boundary $V_{CB}=0$. Then $I_R=I_{R0}(e^{qV_{CB}/kT}-1)=0$: the collector diode carries no current and the $\alpha_RI_R$ source is zero. What remains: the emitter diode carrying $I_F$ from E to B, and the current source $\alpha_FI_F$ from E to C: ``` E o----+------------+ | | (diode: I_F) (αF·I_F source) | | B o----+ o C ``` with $I_E=I_F$, $I_C=\alpha_FI_F$, $I_B=(1-\alpha_F)I_F$. (The collector terminal sees just the dependent source; the B-C diode is at zero bias.) (b) From the circuit:

I_B=I_E-I_C=(1-\alpha_F)I_F;\Rightarrow;I_F=\frac{I_B}{1-\alpha_F}

and $I_F=I_{F0}(e^{qV_{EB}/kT}-1)$:

I_{F0}\left(e^{qV_{EB}/kT}-1\right)=\frac{I_B}{1-\alpha_F} ;\Rightarrow; V_{EB}=\frac{kT}{q}\ln\left[1+\frac{I_B}{(1-\alpha_F)I_{F0}}\right]

With $V_{CB}=0$, $V_{EC}=V_{EB}-V_{CB}=V_{EB}$:

\boxed{V_{EC}=\frac{kT}{q}\ln\left[1+\frac{I_B}{(1-\alpha_F)I_{F0}}\right]}

(and $1-\alpha_F=1/(\beta_{dc}+1)$.) This is the "knee" voltage separating saturation from the active region; it grows only logarithmically with $I_B$. --- ## Problem 11.14: Deriving the Computational Equations **Problem.** Derive the quoted computational equations for (a) $I_C(V_{CB},I_E)$, (b) $I_B(V_{EB},V_{EC})$, (c) $I_C(V_{EC},I_B)$. **Solution.** Let $a=e^{qV_{EB}/kT}-1$, $b=e^{qV_{CB}/kT}-1$. **(a)** From (11.47a):

a=\frac{I_E+\alpha_RI_{R0},b}{I_{F0}}

$$Substitutein(11.47b):$$

I_C=\alpha_FI_{F0},a-I_{R0}b=\alpha_F\left(I_E+\alpha_RI_{R0},b\right)-I_{R0}b

\boxed{I_C=\alpha_FI_E-\left(1-\alpha_F\alpha_R\right)I_{R0}\left(e

**(b)** $I_B=I_E-I_C$ with (11.47):

I_B=(1-\alpha_F)I_{F0},a+(1-\alpha_R)I_{R0},b

With $V_{CB}=V_{EB}-V_{EC}$: $e^{qV_{CB}/kT}=e^{qV_{EB}/kT}e^{-qV_{EC}/kT}$, so $b=e^{qV_{EB}/kT}e^{-qV_{EC}/kT}-1$:

I_B=(1-\alpha_F)I_{F0}\left(e^{qV_{EB}/kT}-1\right)+(1-\alpha_R)I_{R0}\left(e^{qV_{EB}/kT}e

Collect the terms multiplying $e^{qV_{EB}/kT}$ and the constants:

\boxed{I_B=\left[(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}e^{-qV_{EC}/kT}\right]e

**(c)** Write $X\equiv e^{qV_{EB}/kT}$, $E\equiv e^{-qV_{EC}/kT}$, $I_{B0}\equiv(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}$, $I_{B1}\equiv(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}E$. Part (b) reads $I_B=I_{B1}X-I_{B0}$, so

X=\frac{I_B+I_{B0}}{I_{B1}}

Now (11.47b) with $e^{qV_{CB}/kT}=XE$:

I_C=\alpha_FI_{F0}(X-1)-I_{R0}(XE-1)=\left(\alpha_FI_{F0}-I_{R0}E\right)X-\alpha_FI_{F0}+I_{R0}

Substitute $X$:

\boxed{I_C=\frac{\left(\alpha_FI_{F0}-I_{R0}e^{-qV_{EC}/kT}\right)\left[I_B+(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}\right]}{(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}e

Note: the last term is $I_{R0}-\alpha_FI_{F0}$. (In the Appendix M listing it is typed `IR0-aF*IR0`, which is numerically different from $I_{R0}-\alpha_FI_{F0}$ unless $\alpha_FI_{F0}=\alpha_RI_{R0}$ with $\alpha_R\approx\alpha_F$; both terms are tiny compared with the leading term, so the plotted curves are insensitive to the distinction, but the derivation above gives $I_{F0}$.) An equivalent form from direct elimination is $I_C=\{(\alpha_FI_{F0}-I_{R0}E)I_B+(1-\alpha_F\alpha_R)I_{F0}I_{R0}(1-E)\}/I_{B1}$. --- ## Problem 11.15: Emitter Input and Base Width Modulation **Problem.** In modern transistors gain is usually limited by emitter efficiency, so $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\gg\frac12(W/L_B)^2$. (a) Show the CE input characteristics are then relatively insensitive to base width modulation. (b) Sketch the expected CE input characteristics if gain is *not* emitter-limited, i.e. $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\sim\frac12(W/L_B)^2$. **Solution.** (a) The CE input characteristic is (Exercise 11.7, item 3)

I_B=\left[(1-\alpha_F)I_{F0}+(1-\alpha_R)I_{R0}e^{-qV_{EC}/kT}\right]e

For $V_{EC}$ more than a few $kT/q$ the $e^{-qV_{EC}/kT}$ term vanishes and, for $V_{EB}\gg kT/q$,

I_B\approx(1-\alpha_F)I_{F0},e

Evaluate the prefactor using (11.45a), (11.46) and $\cosh-1=2\sinh^2(W/2L_B)$ (so $(\cosh-1)/\sinh=\tanh(W/2L_B)$):

(1-\alpha_F)I_{F0}=I_{F0}-\alpha_FI_{F0}=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\frac{\cosh(W/L_B)-1}{\sinh(W/L_B)}\right]

=qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_B}{L_B}p_{B0}\tanh\frac{W}{2L_B}\right]\approx qA\left[\frac{D_E}{L_E}n_{E0}+\frac{D_Bp_{B0}}{2L_B

The first term is $W$-independent, and the second is smaller by the ratio $\frac12(W/L_B)^2\Big/\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}$. (Check: dividing by the first term, the second/first ratio equals $\frac{D_Bp_{B0}W/(2L_B^2)}{D_En_{E0}/L_E}=\frac{\frac12(W/L_B)^2}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}\ll1$ when emitter-limited.) Hence $I_B\approx qA(D_E/L_E)n_{E0}e^{qV_{EB}/kT}$ is essentially independent of $W$, and therefore of $V_{EC}$: the input characteristics for different $V_{EC}$ coincide. $\blacksquare$ (b) When the two terms are comparable, $(1-\alpha_F)I_{F0}$ contains a significant part proportional to $W$. As $V_{EC}$ rises, $W$ falls, so $I_B$ at a given $V_{EB}$ **decreases**: the curves for increasing $V_{EC}$ shift toward higher $V_{EB}$ (to the right on a semilog plot, to lower current at fixed voltage), by a fixed ratio on a log scale. The expected sketch is a family of exponential curves, all with the same slope $q/kT$, spaced progressively to the right (lower current at fixed $V_{EB}$) as $V_{EC}$ goes from a few $kT/q$ up to tens of volts, with the $V_{EC}=0$ curve far to the left (higher current, from the $(1-\alpha_R)I_{R0}$ term). --- ## Problem 11.16 **Problem.** How is the Exercise 11.8 result modified if $V_{EB}=0.7$ V? **Solution.** Only $x_{nEB}$ changes, since it contains $(V_{bi(EB)}-V_{EB})$:

x_{nEB}=3.29\times10^{-5}\sqrt{\frac{0.835-0.7}{0.835}}=3.29\times10^{-5}(0.402)=1.32\times10^{-5}\ \text{cm}

W_B-x_{nEB}=2\times10^{-4}-1.32\times10^{-5}=1.868\times10^{-4}\ \text{cm}

V_{CB}=V_{bi(CB)}-8.43\times10^{9}\left(W_B-x_{nEB}\right)^2=0.656-8.43\times10^{9}\left(1.868\times10^{-4}\right)

V_{CB}\approx-293.5\ \text{V}\approx-294\ \text{V}

Forward bias shrinks the E-B depletion region, leaving more base width to be consumed by the C-B depletion region, so the punch-through voltage rises in magnitude from 235 V to 294 V. --- ## Problem 11.17 **Problem.** Two pnp BJTs are identical except that emitter and collector dopings are interchanged (Fig. P11.17). Reading the OCR'd figure: A has $N_E=10^{17}$, $N_B=10^{15}$, $N_C=10^{14}$ cm$^{-3}$; B has $N_E=10^{14}$, $N_B=10^{15}$, $N_C=10^{17}$ cm$^{-3}$. (a) Which has the greater emitter efficiency? (b) Which is more sensitive to base width modulation in active mode? (c) If limited by C-B avalanche, which has the larger $V_{CBO}$? **Solution.** (a) **A.** The emitter loss term in $\gamma$ is proportional to $N_B/N_E$ (the other factors are nearly the same):

\left.\frac{N_B}{N_E}\right|_A=\frac{10^{15}}{10^{17}}=10^{-2},\qquad \left.\frac{N_B}{N_E}\right|_B=\frac{10^{15}}{10

so $\gamma_A\approx1/(1+0.01\,K)\approx1$, while $\gamma_B\approx1/(1+10\,K)$ is much smaller (B has a lightly doped emitter relative to the base, the opposite of good design). (b) **B.** The base penetration of the C-B depletion region is

x_{nCB}=\left[\frac{2K_S\varepsilon_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]

\frac{N_C}{N_B(N_C+N_B)}\Big|_A=\frac{10^{14}}{10^{15}(1.1\times10^{15})}=9.1\times10^{-17},\qquad \frac{N_C}{N_B(N_C+N_B)}\Big|_B=\frac{10^{17}}{10^{15}(1.01\times10^{17})}=9.9\times10

The ratio is about 11, so $x_{nCB}|_B\approx3.3\,x_{nCB}|_A$ at the same bias (ignoring the small difference in $V_{bi}$). In A most of the C-B depletion region lies in the lightly doped collector; in B it lies in the base. B's base width therefore varies more with $V_{CB}$. (c) **A.** The avalanche breakdown voltage is set by the lightly doped side of the C-B junction and varies roughly inversely with its doping. For A the light side is the collector ($10^{14}$); for B it is the base ($10^{15}$). The lower light-side doping gives A the larger $V_{CBO}$ (roughly an order of magnitude lower doping gives several times larger breakdown voltage, $V_{BR}\propto N^{-3/4}$ from the text's fit). --- ## Problem 11.18 (Early voltage) **Problem.** For $V_{EC}\ge20$ V the characteristics in the middle figure of Exercise 11.10 (BWM included, no multiplication) are closely modelled by $I_C=B(V_{EC}-V)$ with $B$ and $V$ constants. Determine $B$ and $V$ for each of the four $I_B>0$ curves. $V$ is the *Early voltage*, the $V_{EC}$-axis intercept of the extrapolated linear region. Comment. **Solution (method).** Use the BJTplus equations (Exercise 11.10, case ii) to compute $I_C(V_{EC})$ for $I_B=2.5,5,7.5,10\ \mu$A over $20\le V_{EC}\le80$ V, then fit a straight line (`polyfit(VEC,IC,1)`): slope $=B$, intercept on the $V_{EC}$ axis $=V=-\text{intercept}/\text{slope}$. ```matlab % Early-voltage fit (run after defining parameters as in Exercise 11.10) VEC=20:1:80; VEB=0; VCB=VEB-VEC; xnEB=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-VEB)); xnCB=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB)); Wv=WB-xnEB-xnCB; fBv=(DB/LB)*pB0*cosh(Wv/LB)./sinh(Wv/LB); F0=q*A*((DE/LE)*nE0+fBv); R0=q*A*((DC/LC)*nC0+fBv); aFv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./F0; aRv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./R0; E=exp(-VEC/kT); for IB=(1:4)*2.5e-6 IB0=(1-aFv).*F0+(1-aRv).*R0; IB1=(1-aFv).*F0+(1-aRv).*R0.*E; IC=(aFv.*F0-R0.*E).*(IB+IB0)./IB1+R0-aFv.*F0; p=polyfit(VEC,IC,1); fprintf('IB=%5.1f uA: B=%7.3f uA/V, V=%7.1f V\n',IB*1e6,p(1)*1e6,-p(2)/p(1)); end ``` **Results** (computed with the BJT0 parameters: $W_B=2.5\ \mu$m, $N_B=1.5\times10^{16}$, $N_C=1.5\times10^{15}$, $N_E=10^{18}$ cm$^{-3}$): | $I_B$ ($\mu$A) | $B$ ($\mu$A/V) | $V$ (V) | |---|---|---| | 2.5 | 4.69 | $-131$ | | 5.0 | 9.38 | $-131$ | | 7.5 | 14.07 | $-131$ | | 10.0 | 18.76 | $-131$ | (Over this range $W$ falls from about 1.83 $\mu$m at $V_{EC}=20$ V to about 1.34 $\mu$m at 100 V.) **Comment.** The Early voltage is about $-131$ V, i.e. all four straight lines, when extrapolated backwards, meet the $V_{EC}$ axis at the *same* point ($|V_A|\approx131$ V, a typical magnitude for small-signal transistors, large because the lightly doped collector keeps base modulation modest). $B$ is proportional to $I_B$. This follows from $I_C\approx\beta_{dc}(V_{EC})I_B$ (since $I_{CEO}$ is negligible): the shape of every curve is the shape of $\beta_{dc}(V_{EC})$, which is nearly linear with intercept $V$, and $B=I_B\,d\beta_{dc}/dV_{EC}\approx1.88\,\text{V}^{-1}\times I_B$. The common intercept is a signature of base width modulation; the output conductance $dI_C/dV_{EC}=I_C/(V_{EC}-V)$ rises in proportion to $I_C$. --- ## Problem 11.19 (computer): Diode-Like $I$-$V_A$ Plots **Problem.** Adapt the BJTplus programs to plot the $I$-$V_A$ characteristics of the Problem 11.12(a) relationships. (a) First assume $W=W_B$ and neglect BWM and multiplication. (b) Then include BWM and multiplication (including carrier multiplication in the E-B depletion region at large reverse bias on the E-B junction) and compare. **Solution (logic).** Use the Problem 11.12 relations: connection (a) [C open] $I=I_{F0}(1-\alpha_F\alpha_R)u$; connection (b) [C shorted to B] $I=I_{F0}u$, $u=e^{qV_A/kT}-1$. For (a) with BWM the parameters depend on $V_{CB}$ which depends on the parameters ($b=\alpha_Ru$), so iterate to self-consistency. Multiplication: for reverse $V_A$ across E-B, $M_E=1/[1-(|V_A|/BV_{EBO})^m]$ with $BV_{EBO}$ from the light-side (base) doping, $60(N_B/10^{16})^{-3/4}$ V, multiplying the (reverse) current. ```matlab % Problem 11.19: diode-like I-VA characteristics (connections: C open and C shorted) BJT0 % loads constants and parameters m=6; VBRE=60*(NB/1e16)^(-3/4); % E-B avalanche voltage (light side = base) VAf=0:0.005:0.8; % forward bias VAr=0:-0.5:-0.95*VBRE; % reverse bias, stay below breakdown for part=1:2 % 1: ideal (W=WB); 2: BWM + multiplication for k=1:2 % k=1 collector open, k=2 collector shorted for VAset={VAf,VAr} VA=VAset{1}; I=zeros(size(VA)); for i=1:length(VA) u=exp(VA(i)/kT)-1; VEB=VA(i); VCB=0; for it=1:15 % self-consistency for W if part==1, F0=IF0; R0=IR0; aF_=aF; aR_=aR; else xE=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-min(VEB,0.9*VbiE))); xC=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB)); Wl=max(WB-xE-xC,0.05*WB); fBl=(DB/LB)*pBO*cosh(Wl/LB)/sinh(Wl/LB); F0=q*A*((DE/LE)*nEO+fBl); R0=q*A*((DC/LC)*nCO+fBl); aF_=q*A*(DB/LB)*pBO/sinh(Wl/LB)/F0; aR_=q*A*(DB/LB)*pBO/sinh(Wl/LB)/R0; end if k==1, VCB=kT*log(1+aR_*u); else, VCB=0; end % b=aR*u when C open end if k==1, I(i)=F0*(1-aF_*aR_)*u; else, I(i)=F0*u; end if part==2 && VA(i)<0, I(i)=I(i)/(1-(abs(VA(i))/VBRE)^m); end end figure(part*10+k); hold on if VA(1)==0 && VA(end)>0, plot(VA,I*1e3); xlabel('V_A (V)'); ylabel('I (mA)'); else, plot(VA,I*1e6); xlabel('V_A (V)'); ylabel('I (\muA), reverse'); end grid on end end end ``` (Notes: `pBO`, `nEO`, `nCO` are the BJT0 names with letter O; the forward and reverse parts are plotted separately for clarity.) **Expected behaviour.** (a) Without BWM or multiplication: forward $I\propto e^{qV_A/kT}$ (an exponential with the ideal factor $I_{F0}$, or $I_{F0}(1-\alpha_F\alpha_R)$ for collector open, lower by the small factor $1-\alpha_F\alpha_R\approx0.4$ to 1 depending on $\alpha_R$); reverse: constant saturation current $-I_{F0}$ (shorted) or $-I_{F0}(1-\alpha_F\alpha_R)$ (open). (b) With BWM the parameters change slightly with bias and the curves tilt (forward: tiny change; reverse with collector shorted: no effect on $V_{EB}$-dependent $W$ other than through $x_{nEB}$, giving a mild increase in magnitude as $|V_A|$ grows); with multiplication the reverse current turns sharply up (avalanche) near $BV_{EBO}$ ($\approx44$ V for these parameters), where the (a) result keeps a flat saturation current to arbitrarily large voltage. --- ## Problem 11.20 (computer): Gummel and Gain Plots **Problem.** (a) Construct a theoretical Gummel plot like Fig. 11.15 with the BJT0 parameters; active mode with $V_{CB}$ terms neglected; $I_C$ and $I_B=I_E-I_C$ versus $V_{EB}$. Approximate E-B depletion R-G by adding $I_{R\text{-}G}=I_2(e^{qV_{EB}/n_2kT}-1)$ to the base current, with $I_2=10^{-14}$ A, $n_2=1.5$. Approximate high-level injection etc. by multiplying the ideal $I_C$ by $\Gamma=1/[1+e^{q(V_{EB}-0.75)/2kT}]$. Plot ideal (dashed) and corrected (solid). Limit to $10^{-9}\le I\le10^{-1}$ A and $0.3\le V_{EB}\le0.8$ V. (b) Add a $\beta_{dc}$ versus $I_C$ plot (Fig. 11.16 style) with `pause` between plots. **Solution (logic).** In active mode with $V_{CB}$ terms dropped, (11.47) gives $I_E=I_{F0}(e^{qV_{EB}/kT}-1)$, $I_C=\alpha_FI_{F0}(e^{qV_{EB}/kT}-1)$, so $I_B=(1-\alpha_F)I_{F0}(e^{qV_{EB}/kT}-1)$ (ideal). Corrected: $I_B'=I_B+I_2(e^{qV_{EB}/n_2kT}-1)$ and $I_C'=\Gamma I_C$. Then $\beta_{dc}=I_C/I_B$. The R-G term has a slope $q/(1.5\,kT)$, dominating at low $V_{EB}$ where it exceeds the ideal $I_B$; $\Gamma\approx1$ for $V_{EB}\ll0.75$ V and $\Gamma\to e^{-q(V_{EB}-0.75)/2kT}$ above, producing the slope-over of $I_C$. ```matlab BJT0 % constants and parameters VEB=0.3:0.005:0.8; IC0=aF*IF0*(exp(VEB/kT)-1); % ideal collector current IB0=(1-aF)*IF0*(exp(VEB/kT)-1); % ideal base current I2=1e-14; n2=1.5; IBc=IB0+I2*(exp(VEB/(n2*kT))-1); % with E-B R-G current Gam=1./(1+exp((VEB-0.75)/(2*kT))); % high-level-injection factor ICc=IC0.*Gam; % corrected collector current figure(1) semilogy(VEB,IC0,'--b',VEB,IB0,'--r',VEB,ICc,'-b',VEB,IBc,'-r'); grid on axis([0.3 0.8 1e-9 1e-1]); xlabel('V_{EB} (V)'); ylabel('I_C, I_B (A)') legend('I_C ideal','I_B ideal','I_C corrected','I_B corrected') pause figure(2) loglog(IC0,IC0./IB0,'--',ICc,ICc./IBc,'-'); grid on axis([1e-9 1e-1 1 1e3]); xlabel('I_C (A)'); ylabel('\beta_{dc}') ``` (The $\beta_{dc}$ axis range printed in the book is garbled in the OCR; $1\le\beta_{dc}\le10^3$ is used here.) **Expected results.** The ideal plot is two parallel straight lines (60 mV/decade), separated by $\beta_{dc}\approx178$ (for these BJT0 parameters $\alpha_F=0.9944$). The corrected plot: at low $V_{EB}$ the corrected $I_B$ is above the ideal line (R-G, slope 1.5 times more gradual), so $I_C/I_B$ falls; at high $V_{EB}$ the corrected $I_C$ rolls over below the line. The gain plot is therefore a hump: rising with $I_C$ at low currents, flat near 178 over the middle decades, and falling at high currents, as in Fig. 11.16. --- ## Problem 11.21 **Problem.** Increasing $V_{EC}$ in measuring $\beta_{dc}$ versus $I_C$ invariably leads to a small but perceptible increase in $\beta_{dc}$ over most of the $I_C$ range (Fig. P11.21: 2N2605 data at $V_{EC}=5,10,15$ V). Formulate and support an explanation. **Solution.** The cause is **base width modulation**. Raising $V_{EC}$ at fixed $I_C$ increases the reverse bias across the C-B junction, widens its depletion region and narrows the quasineutral base $W$. Evidence from the equations:

\beta_{dc}=\frac{1}{\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W}{L_E}+\dfrac12\left(\dfrac{W}{L_B}\right)

Both loss terms decrease as $W$ falls (the first linearly, the second quadratically), so $\beta_{dc}$ increases. In terms of the base-current components: $I_{B1}=I_{En}$ is the back-injection current; at fixed $I_C\propto e^{qV_{EB}/kT}/W$, a smaller $W$ requires a *smaller* $V_{EB}$ (the same $I_C$ arises from a lower E-B injection level), so $I_{En}\propto e^{qV_{EB}/kT}$ falls in proportion to $W$; and $I_{B2}$ (base recombination) falls as $W^2$. The same smaller $V_{EB}$ also reduces the E-B depletion-region R-G current $\propto e^{qV_{EB}/2kT}$, which governs the low-current gain drop, so the increase in $\beta_{dc}$ with $V_{EC}$ extends across the low-current range as well. The effect is small because the lightly doped collector absorbs most of the C-B depletion region (Section 11.2.2), i.e. the Early voltage is large (about 100 V or more), predicting a few-percent gain change per 5 V step, consistent with the modest separation of the three curves in Fig. P11.21. At the highest currents where high-level injection and series resistance dominate, the curves merge because those mechanisms do not depend strongly on $W$. # Chapter 12 — BJT Dynamic Response Modeling > **The big picture.** A transistor is used in two main ways: as a *small-signal amplifier* (tiny wiggles riding on a DC bias) and as a *switch* (slammed between "off" and "on"). Imagine a water tap. In amplifier mode you nudge the handle a hair and watch the flow respond proportionally. In switch mode you yank it fully open or fully shut. This chapter builds one model for each situation: the **small-signal equivalent circuit** (Section 12.1) and the **transient (switching) response** (Section 12.2). Because modern amplifiers want high gain and wide bandwidth, almost everything is done in the **common-emitter (CE)** configuration, so we emphasize that. --- ## 12.1 Small-Signal Equivalent Circuits ### 12.1.1 Generalized Two-Port Model **Setup.** Treat the BJT in CE configuration as a "black box" with two ports (input: base–emitter, output: collector–emitter). In this chapter *all currents are positive when flowing into a terminal* and all voltages are measured relative to the common (emitter) terminal. DC quantities: $I_B,\ I_C,\ V_{BE},\ V_{CE}$. AC (small-signal) quantities: $i_b,\ i_c,\ v_{be},\ v_{ce}$. Total instantaneous values are DC plus AC: $$i_B = I_B + i_b,\qquad i_C = I_C + i_c$$ **Key fact from Chapter 11.** The DC currents are functions of the applied DC voltages only: $$I_B = I_B(V_{BE},V_{CE}),\qquad I_C = I_C(V_{BE},V_{CE})$$ **The critical assumption: quasistatic operation.** In words: *the transistor internals (charge distributions) adjust instantly to the slowly changing terminal voltages*, so at every instant the currents are the same functions as in DC, just evaluated at the instantaneous voltages. Think of a heavy sled pulled slowly: it is always essentially in equilibrium with the pull. Then $$i_B = I_B(V_{BE}+v_{be},\,V_{CE}+v_{ce}) = I_B(V_{BE},V_{CE}) + i_b \tag{12.1a}$$ $$i_C = I_C(V_{BE}+v_{be},\,V_{CE}+v_{ce}) = I_C(V_{BE},V_{CE}) + i_c \tag{12.1b}$$ Subtract the DC part: $$i_b = I_B(V_{BE}+v_{be},V_{CE}+v_{ce}) - I_B(V_{BE},V_{CE}) \tag{12.2a}$$ $$i_c = I_C(V_{BE}+v_{be},V_{CE}+v_{ce}) - I_C(V_{BE},V_{CE}) \tag{12.2b}$$ **Taylor expansion about the DC operating point** (keep only first-order terms; products like $v_{be}^2$ and $v_{be}v_{ce}$ are negligible when the signal is *small*): $$I_B(V_{BE}+v_{be},V_{CE}+v_{ce}) \approx I_B(V_{BE},V_{CE}) + \left.\frac{\partial I_B}{\partial V_{BE}}\right|_{V_{CE}} v_{be} + \left.\frac{\partial I_B}{\partial V_{CE}}\right|_{V_{BE}} v_{ce} \tag{12.3a}$$ $$I_C(V_{BE}+v_{be},V_{CE}+v_{ce}) \approx I_C(V_{BE},V_{CE}) + \left.\frac{\partial I_C}{\partial V_{BE}}\right|_{V_{CE}} v_{be} + \left.\frac{\partial I_C}{\partial V_{CE}}\right|_{V_{BE}} v_{ce} \tag{12.3b}$$ Substituting into (12.2): $$i_b = \left.\frac{\partial I_B}{\partial V_{BE}}\right|_{V_{CE}} v_{be} + \left.\frac{\partial I_B}{\partial V_{CE}}\right|_{V_{BE}} v_{ce} \tag{12.4a}$$ $$i_c = \left.\frac{\partial I_C}{\partial V_{BE}}\right|_{V_{CE}} v_{be} + \left.\frac{\partial I_C}{\partial V_{CE}}\right|_{V_{BE}} v_{ce} \tag{12.4b}$$ Each partial derivative has units of current/voltage, i.e. conductance. Name them (the $g$-parameters): $$\text{npn: }\ g_{11}=\left.\frac{\partial I_B}{\partial V_{BE}}\right|_{V_{CE}},\quad g_{12}=\left.\frac{\partial I_B}{\partial V_{CE}}\right|_{V_{BE}},\quad g_{21}=\left.\frac{\partial I_C}{\partial V_{BE}}\right|_{V_{CE}},\quad g_{22}=\left.\frac{\partial I_C}{\partial V_{CE}}\right|_{V_{BE}}\tag{12.5a,c}$$ $$\text{pnp: }\ g_{11}=\left.\frac{\partial I_B}{\partial V_{EB}}\right|_{V_{EC}},\quad g_{12}=\left.\frac{\partial I_B}{\partial V_{EC}}\right|_{V_{EB}},\quad g_{21}=\left.\frac{\partial I_C}{\partial V_{EB}}\right|_{V_{EC}},\quad g_{22}=\left.\frac{\partial I_C}{\partial V_{EC}}\right|_{V_{EB}}\tag{12.5b,d}$$ (Why two sets? A pnp has voltages and currents reversed relative to an npn. In the pnp set, $I_B$ is the current *out of* the base and $I_C$ the current *into* the collector, $V_{EB}, V_{EC}>0$ in active mode. With these conventions the derivatives have the same signs as the npn ones.) Then the small-signal equations are $$i_b = g_{11}v_{be}+g_{12}v_{ce} \tag{12.6a}$$ $$i_c = g_{21}v_{be}+g_{22}v_{ce} \tag{12.6b}$$ **Reading them as a circuit.** (12.6a) and (12.6b) are KCL at the base and collector nodes. $g_{11}v_{be}$ is the current through a conductance $g_{11}$ across the input; $g_{22}v_{ce}$ is the current through $g_{22}$ across the output. $g_{12}v_{ce}$ (at the input) and $g_{21}v_{be}$ (at the output) are controlled by the voltage at the *other* port, so they are voltage-controlled current sources. This gives Fig. 12.1(b): $g_{11}$ from B to E in parallel with a source $g_{12}v_{ce}$; at the output $g_{22}$ from C to E in parallel with a source $g_{21}v_{be}$. **Comments.** 1. Valid only at low frequencies (capacitive/charge-storage effects neglected: quasistatic). 2. The derivation is purely mathematical; it works for any three-terminal device in any configuration by relabeling the terminals. Only the $g$-definitions tie it to a specific device/configuration. 3. Explicit formulas for the $g$'s follow from the Chapter 11 DC current relations, but the Hybrid-Pi arrangement is preferred in practice. ### 12.1.2 Hybrid-Pi Models The Hybrid-Pi model is the workhorse. Its parameters link directly to the DC operating point and their temperature dependence is easy to deduce. Fig. 12.2: - (a) **Simplified:** $r_\pi$ from B to E, current source $g_m v_{be}$ from C to E. - (b) **Complete low-frequency:** add $r_\mu$ (B to C) and $r_o$ (C to E). - (c) **High-frequency:** add the capacitances $C_{eb}$ (across the intrinsic B′–E′) and $C_{cb}$ (B′–C′), plus parasitic series resistances $r_b, r_c, r_e$ connecting the intrinsic transistor (primed nodes) to the external terminals. The name comes from the $\pi$-shaped arrangement of elements and the "hybrid" mixture of conductance and resistance units. Parameters: $g_m$ transconductance (forward gain), $r_o$ output resistance, $r_\pi$ input resistance, $r_\mu$ feedthrough (feedback) resistance. **Deriving the simplified model.** In active mode the Ebers–Moll equations (11.47) for a pnp reduce to $$I_E \approx I_{F0}\,e^{qV_{EB}/kT}\quad(V_{EB}>0,\ V_{CB}<0) \tag{12.7a}$$ $$I_C \approx \alpha_F I_{F0}\,e^{qV_{EB}/kT} \tag{12.7b}$$ $$I_B = I_E - I_C = (1-\alpha_F)I_{F0}\,e^{qV_{EB}/kT} \tag{12.7c}$$ If base-width modulation is negligible, $\alpha_F$ and $I_{F0}$ are constants independent of $V_{EC}$, so $$g_{12}=\left.\frac{\partial I_B}{\partial V_{EC}}\right|_{V_{EB}}=0,\qquad g_{22}=\left.\frac{\partial I_C}{\partial V_{EC}}\right|_{V_{EB}}=0 \tag{12.8a,b}$$ With $g_{12}=g_{22}=0$ the two-port model collapses to Fig. 12.2(a), with $g_m=g_{21}$ and $r_\pi=1/g_{11}$. Evaluating: $$g_m=\left.\frac{\partial I_C}{\partial V_{EB}}\right|_{V_{EC}}=\frac{qI_C}{kT}\tag{12.9a}$$ $$\frac{1}{r_\pi}=\left.\frac{\partial I_B}{\partial V_{EB}}\right|_{V_{EC}}=\frac{qI_B}{kT}=\frac{g_m}{\beta_{dc}}\ \Rightarrow\ r_\pi=\frac{\beta_{dc}}{g_m}\tag{12.9b}$$ (Differentiating $e^{qV_{EB}/kT}$ multiplies by $q/kT$, so $\partial I/\partial V = (q/kT)\,I$.) **Intuition:** $g_m$ depends only on collector current and temperature. At room temperature $kT/q\approx 25.9$ mV, so $g_m\approx 38.7$ mS per mA of collector current: a universal rule of thumb. **Complete four-element model (Fig. 12.2b)** is used when base-width modulation matters. Relation to the two-port parameters (derived in Problem 12.4): $$g_m = g_{21}-g_{12},\quad r_o = \frac{1}{g_{22}+g_{12}},\quad r_\pi=\frac{1}{g_{11}+g_{12}},\quad r_\mu=-\frac{1}{g_{12}} \tag{12.10a-d}$$ $g_{12}$ is typically orders of magnitude smaller than the other conductances, so it can be neglected in $g_m$, $r_o$, $r_\pi$, and (12.9) remains good. **High-frequency model (Fig. 12.2c).** $C_{cb}$ models the reverse-biased C–B junction (essentially pure junction capacitance); $C_{eb}$ models the forward-biased E–B junction (junction *plus* diffusion capacitance). The Chapter 7 junction-capacitance formulas apply unchanged, but the diffusion part must be modified for the narrow base (Problem 12.7). Additional capacitances (C′–B or C′–substrate) may be needed in ICs. The model is valid up to roughly 500 MHz; above that it ignores signal-propagation delays across the device regions. --- ## 12.2 Transient (Switching) Response The BJT is used as a switch from discrete circuits to IC logic. Compared with a pn diode it switches faster and has a third lead that helps control the process. Like the diode, the delay comes from **building up or removing excess minority-carrier charge**. *Analogy:* filling or emptying a bathtub with a leaky drain (recombination). You cannot change the water level instantly, and the time it takes is the switching delay. ### 12.2.1 Qualitative Observations Circuit (Fig. 12.3a): supply $V_{CC}$ across load $R_L$ and the transistor; input source $v_s$ with series resistor $R_s$ pulses between $+V_s$ and $-V_s$. - $v_s<0$: cutoff, $i_C\approx 0$, operating point at the bottom of the load line ("off"). - $v_s>0$: both junctions forward biased, saturation, $i_C = V_{CC}/R_L$ ("on"). **Turn-on** (pulse from $-V_s$ to $+V_s$ at $t=0$). With $V_s\gg v_{EB}$, $i_B\approx V_s/R_s\equiv I_{BB}$ and stays constant. Before $t=0$ the base minority profile is (i) in Fig. 12.4(d). After $t=0$ holes pile up in the base (ii); $i_C$ is proportional to the slope of $\Delta p_B$ at $x=W$, so it tracks the build-up while the BJT is active. Eventually $\Delta p_B(0)=(n_i^2/N_B)e^{qV_{EB}/kT}$ grows until $V_{EB}=V_{EC}$ (i.e. $v_{CB}=0$) and the transistor enters saturation. The build-up then completes (iii), with $i_C$ increasing only slightly to its on-value. **Turn-off** has two phases. After $v_s$ flips to $-V_s$: (iv) $i_C$ stays roughly constant while the saturation-stored charge is removed (the **storage delay**, analogous to the diode storage delay); (v) once the BJT is active again, $i_C$ falls in proportion to the remaining stored charge until cutoff. The base current reverses to $-V_s/R_s$ and stays until cutoff, then decays to zero (same as the diode turn-off). ### 12.2.2 Charge Control Relationships The diode charge-control result (Subsection 6.3.1) applies to the quasineutral base (pnp, $0\le x\le W$). The hole charge stored in the base is $$Q_B = qA\int_0^W \Delta p_B(x,t)\,dx \tag{12.12}$$ Hole continuity integrated over the base gives (injection in at $x=0$, collection out at $x=W$, recombination inside): $$\frac{dQ_B}{dt} = -A\left[J_p(W)-J_p(0)\right]-\frac{Q_B}{\tau_B}\tag{12.11}$$ If $i_E\approx AJ_p(0)$ and $i_C\approx AJ_p(W)$ (emitter and collector currents dominated by hole diffusion currents across the E–B and C–B junctions), $$A\left[J_p(0)-J_p(W)\right]=i_E-i_C=i_B \tag{12.13}$$ and the **charge-control equation** is $$\boxed{\frac{dQ_B}{dt}=i_B-\frac{Q_B}{\tau_B}} \tag{12.14}$$ **Steady state:** $dQ_B/dt=0\Rightarrow I_B = Q_B/\tau_B$. Interpretation: there are $Q_B/q$ excess carriers, each living on average $\tau_B$, so all of them recombine every $\tau_B$ seconds; a base current $Q_B/\tau_B$ must replenish them. (Valid if base recombination dominates $I_B$; emitter back-injection is neglected.) **Collector current.** From diffusion at $x=W$: $$i_C = AJ_p(W)=-qAD_B\left.\frac{\partial \Delta p_B}{\partial x}\right|_{x=W} \tag{12.15}$$ Assuming $\Delta p_B(x,t)$ keeps the steady-state *shape* (linear for $W\ll L_B$): $$\Delta p_B(x,t)=\Delta p_B(0,t)+\left[\Delta p_B(W,t)-\Delta p_B(0,t)\right]\frac{x}{W} \tag{12.16}$$ In active mode $\Delta p_B(W,t)\approx0$: $$\Delta p_B(x,t)=\Delta p_B(0,t)\left(1-\frac{x}{W}\right) \tag{12.17}$$ $$\left.\frac{\partial\Delta p_B}{\partial x}\right|_{x=W}=-\frac{\Delta p_B(0,t)}{W} \tag{12.18}$$ $$Q_B = qA\Delta p_B(0,t)\int_0^W\left(1-\frac xW\right)dx = \frac{qAW}{2}\Delta p_B(0,t) \tag{12.19}$$ $$i_C = \frac{qAD_B}{W}\Delta p_B(0,t)=\frac{qAD_B}{W}\cdot\frac{2Q_B}{qAW}=\frac{Q_B}{W^2/2D_B} \tag{12.20}$$ $$\boxed{i_C=\frac{Q_B}{\tau_t}\ \ (\text{active mode}),\qquad \tau_t=\frac{W^2}{2D_B}} \tag{12.21, 12.22}$$ $\tau_t$ is the **base transit time**: the average time for minority carriers to diffuse across the quasineutral base. Picture the stored charge "dropping into the collector" every $\tau_t$ seconds. Also $$\beta_{dc}=\frac{i_C}{i_B}=\frac{Q_B/\tau_t}{Q_B/\tau_B}=\frac{\tau_B}{\tau_t}$$ (a carrier crosses the base in $\tau_t$ but lives $\tau_B$; the ratio counts collector carriers per recombined carrier). ### 12.2.3 Quantitative Analysis **Turn-on.** $i_B=V_s/R_s=I_{BB}$ constant: $$\frac{dQ_B}{dt}=I_{BB}-\frac{Q_B}{\tau_B}\tag{12.23}$$ General solution (particular solution $I_{BB}\tau_B$ plus homogeneous): $$Q_B(t)=I_{BB}\tau_B+Ae^{-t/\tau_B}\tag{12.24}$$ Initial condition $Q_B(0)=0$ (cutoff): $0=I_{BB}\tau_B+A\Rightarrow A=-I_{BB}\tau_B$: $$Q_B(t)=I_{BB}\tau_B\left(1-e^{-t/\tau_B}\right),\quad t\ge0\tag{12.25}$$ $$i_C(t)=\begin{cases}\dfrac{Q_B(t)}{\tau_t}=\dfrac{I_{BB}\tau_B}{\tau_t}\left(1-e^{-t/\tau_B}\right), & 0\le t\le t_r\\[2mm] \dfrac{V_{CC}}{R_L}=I_{CC}, & t\ge t_r\end{cases}\tag{12.26a,b}$$ The **rise time** $t_r$ is the time the BJT spends in active mode. Set $i_C(t_r)=I_{CC}$: $$\frac{I_{BB}\tau_B}{\tau_t}\left(1-e^{-t_r/\tau_B}\right)=I_{CC}$$ $$e^{-t_r/\tau_B}=1-\frac{I_{CC}\tau_t}{I_{BB}\tau_B}$$ $$\boxed{t_r=\tau_B\ln\!\left[\frac{1}{1-\dfrac{I_{CC}\tau_t}{I_{BB}\tau_B}}\right]}\tag{12.27}$$ Intuition: larger $I_{BB}$ (fill faster) or smaller $I_{CC}$ (less charge needed) shortens $t_r$. Substituting back, the charge at the active/saturation boundary is $Q_B(t_r)=I_{CC}\tau_t$ (needed for turn-off). While saturated, $i_C$ stays at $I_{CC}$ but $Q_B$ keeps rising toward $I_{BB}\tau_B$ (Fig. 12.5); the surplus $I_{BB}\tau_B - I_{CC}\tau_t$ is the *saturation-stored* charge. **Turn-off.** Let $i_B=-\xi I_{BB}$ with $\xi=0$ (input just opened, $i_B\approx 0$) or $\xi=1$ (input reversed to $-V_s$, $i_B=-I_{BB}$). Reset $t=0$ at the start of turn-off: $$\frac{dQ_B}{dt}=-\xi I_{BB}-\frac{Q_B}{\tau_B}\tag{12.28}$$ $$Q_B(t)=-\xi I_{BB}\tau_B+Ae^{-t/\tau_B}\tag{12.29}$$ At $t=0$ the stored charge equals the end of turn-on, $Q_B(0)=I_{BB}\tau_B$: $$I_{BB}\tau_B=-\xi I_{BB}\tau_B+A\ \Rightarrow\ A=(1+\xi)I_{BB}\tau_B$$ $$Q_B(t)=I_{BB}\tau_B\left[(1+\xi)e^{-t/\tau_B}-\xi\right]\tag{12.30}$$ $$i_C(t)=\begin{cases}I_{CC},&0\le t\le t_{sd}\ \text{(saturation)}\\[1mm]\dfrac{Q_B(t)}{\tau_t}=\dfrac{I_{BB}\tau_B}{\tau_t}\left[(1+\xi)e^{-t/\tau_B}-\xi\right],&t\ge t_{sd}\ \text{(active)}\end{cases}\tag{12.31a,b}$$ The **storage delay time** $t_{sd}$ is the time spent saturated. At its end $Q_B(t_{sd})=I_{CC}\tau_t$: $$I_{BB}\tau_B\left[(1+\xi)e^{-t_{sd}/\tau_B}-\xi\right]=I_{CC}\tau_t$$ $$(1+\xi)e^{-t_{sd}/\tau_B}=\xi+\frac{I_{CC}\tau_t}{I_{BB}\tau_B}$$ $$\boxed{t_{sd}=\tau_B\ln\!\left[\frac{1+\xi}{\xi+\dfrac{I_{CC}\tau_t}{I_{BB}\tau_B}}\right]}\tag{12.32}$$ Since $I_{CC}\tau_t/I_{BB}\tau_B<1$, $t_{sd}$ decreases when $\xi>0$: a reverse base current helps drain the tub. $t_{sd}$ also drops with smaller $\tau_B$, smaller $I_{BB}$ (less stored charge) and larger $I_{CC}$ (less overdrive). Note $I_{CC}\tau_t/(I_{BB}\tau_B)=I_{CC}/(\beta_{dc}I_{BB})$, the inverse of the overdrive factor. ### 12.2.4 Practical Considerations The real $i_C$ transient (Fig. 12.6) is rounded near transition points and has an extra **delay time $t_d$** at the start, from charging the E–B junction capacitance from cutoff into the active mode. Datasheets define $t_r$ (10% to 90% of $I_{CC}$), $t_{sd}$ (input edge to $i_C$ falling to 90% of $I_{CC}$) and $t_f$ (90% to 10%). Speed-up methods: - **Turn-on:** a capacitor in parallel with $R_s$; its discharge after $v_s\to+V_s$ gives a momentary base-current spike that builds charge faster. - **Turn-off:** add recombination–generation centers (e.g., gold in Si) to reduce $\tau_B$; or use a **Schottky-diode clamp** (Fig. 12.7): a Schottky diode between collector and base conducts at a much lower forward voltage than the C–B pn junction, so when the BJT starts to saturate the diode carries the excess base current and clamps $v_{CB}$ at a small positive value, greatly limiting saturation charge. A Schottky diode stores almost no minority charge, so turn-off is dramatically faster. --- ## 12.3 Summary - A low-frequency small-signal two-port model with four conductances $g_{ij}$ follows from the quasistatic assumption and a first-order Taylor expansion; it applies to essentially any three-terminal device. - The Hybrid-Pi model (simplified, complete, high-frequency) relates to the two-port parameters via (12.10); $g_m=qI_C/kT$, $r_\pi=\beta_{dc}/g_m$. - Switching means adding/removing stored minority charge. Charge control: $dQ_B/dt=i_B-Q_B/\tau_B$, $i_C=Q_B/\tau_t$ (active), $\tau_t=W^2/2D_B$, $\beta_{dc}=\tau_B/\tau_t$. - Turn-on: $t_r=\tau_B\ln[1/(1-x)]$; turn-off: $t_{sd}=\tau_B\ln[(1+\xi)/(\xi+x)]$, with $x=I_{CC}\tau_t/I_{BB}\tau_B$. - Speed-ups: input capacitor, recombination centers, Schottky clamp. --- ## Chapter 12 Problems (with full solutions) | Problem | Section | Level | Points | Description | |---|---|---|---|---| | 12.1 | 12.3 | 1 | 10 | Quick quiz | | 12.2 | 12.1.1 | 2-3 | 8 | Derive common-base circuit | | 12.3 | 12.1.2 | 1 | 5 | Determine $g_m$, $r_\pi$ | | 12.4 | 12.1.2 | 3 | 8 | Confirm Eqs. (12.10) | | 12.5 | 12.1.2 | 4 | 20 | Calculate Hybrid-Pi parameters (computer) | | 12.6 | 12.1.2 | 4 | 20 | Derive/compute $f_T$ | | 12.7 | 12.2.2 | 3-4 | 10 | E–B diffusion capacitance | | 12.8 | 12.2.3 | 2 | 5 | Estimate $I_{CC}\tau_t/I_{BB}\tau_B$ | | 12.9 | 12.2.3 | 2 | 8 | $t_r$ and $t_{sd}$ plots | | 12.10 | 12.2.4 | 3 | 10 | $t_f$ derivation and plot | ### Problem 12.1 — Quick Quiz **(a) Quasistatic assumption.** The device's internal charge distributions respond instantly to changing terminal voltages, so instantaneous currents equal the DC functions evaluated at the instantaneous voltages. It is used to write $i_B(V_{BE}+v_{be},V_{CE}+v_{ce})=I_B(\cdot)+i_b$ (and likewise for $i_C$), i.e. to replace the true dynamic currents by static $I(V)$ functions, which can then be Taylor-expanded. **(b) Separate npn/pnp definitions.** The pnp has all voltage polarities and current directions reversed. The $g_{ij}$ are written with the variables ($V_{EB}$, $V_{EC}$, $I_B$ out of the base, $I_C$ into the collector) natural for a pnp in active mode, so that the conductances carry the same signs as in the npn case. **(c) Origin of "Hybrid-Pi".** The circuit elements are arranged like the Greek letter $\pi$, and the parameters mix conductance and resistance units ("hybrid"). **(d) Names.** $g_m$: transconductance; $r_o$: output resistance; $r_\pi$: input resistance; $r_\mu$: feedthrough (feedback) resistance. **(e) Capacitors.** The junctions store charge that varies with voltage: depletion (junction) charge and, for forward-biased junctions, minority-carrier diffusion charge. At high frequency this charge cannot follow instantly; it appears as capacitance in parallel with the junction conductances ($C_{eb}$ across E–B, $C_{cb}$ across C–B). **(f) Saturation during turn-on.** $i_B$ remains constant ($I_{BB}$). The C–B junction becomes forward biased ($\Delta p_B(W)>0$), so $i_C$ no longer follows the stored charge and stays essentially at $V_{CC}/R_L$ (rising only slightly). The stored base charge keeps increasing toward $I_{BB}\tau_B$; the minority profile lifts at $x=W$ (plot iii in Fig. 12.4d). **(g) Base transit time.** Words: the average time a minority carrier needs to diffuse across the quasineutral base. Math: $\tau_t=W^2/2D_B$. **(h) Relation to $\beta_{dc}$.** $\beta_{dc}=\tau_B/\tau_t$. **(i) Effect of reverse $i_B$.** In $dQ_B/dt=i_B-Q_B/\tau_B$, a negative $i_B$ removes charge faster than recombination alone, so saturation charge drains sooner (shorter $t_{sd}$) and the active-mode charge falls faster (shorter $t_f$). A positive $i_B$ would do the opposite. **(j) Schottky-diode clamp.** A Schottky diode connected from collector to base (Fig. 12.7). It conducts at a lower forward voltage than the C–B pn junction, so it clamps $v_{CB}$ at a small value and prevents deep saturation. With little stored charge in the transistor and negligible minority storage in the Schottky diode, turn-off becomes much faster. --- ### Problem 12.2 — Common-base small-signal circuit **Problem.** Establish a small-signal equivalent circuit analogous to Fig. 12.1(b) for the common-base configuration; summarize the parameter definitions like Eqs. (12.5). **Solution.** In CB, the input port is E–B and the output port is C–B; the common terminal is the base. Currents flow into terminals; voltages are terminal-to-base. DC currents depend only on the DC voltages: $I_E=I_E(V_{EB},V_{CB})$, $I_C=I_C(V_{EB},V_{CB})$. Quasistatic total currents: $$i_E=I_E(V_{EB}+v_{eb},V_{CB}+v_{cb})=I_E+i_e$$ $$i_C=I_C(V_{EB}+v_{eb},V_{CB}+v_{cb})=I_C+i_c$$ First-order Taylor expansion (as in 12.3), then subtract the DC parts: $$i_e=\left.\frac{\partial I_E}{\partial V_{EB}}\right|_{V_{CB}}v_{eb}+\left.\frac{\partial I_E}{\partial V_{CB}}\right|_{V_{EB}}v_{cb}$$ $$i_c=\left.\frac{\partial I_C}{\partial V_{EB}}\right|_{V_{CB}}v_{eb}+\left.\frac{\partial I_C}{\partial V_{CB}}\right|_{V_{EB}}v_{cb}$$ Definitions (npn): $$g_{11}=\left.\frac{\partial I_E}{\partial V_{EB}}\right|_{V_{CB}},\quad g_{12}=\left.\frac{\partial I_E}{\partial V_{CB}}\right|_{V_{EB}},\quad g_{21}=\left.\frac{\partial I_C}{\partial V_{EB}}\right|_{V_{CB}},\quad g_{22}=\left.\frac{\partial I_C}{\partial V_{CB}}\right|_{V_{EB}}$$ For the pnp, the same definitions apply with all voltages and currents reversed in polarity (voltages $V_{BE}$, $V_{BC}$; currents $I_E$, $I_C$ taken out of the terminals), which leaves the conductances unchanged. Equations: $$i_e=g_{11}v_{eb}+g_{12}v_{cb},\qquad i_c=g_{21}v_{eb}+g_{22}v_{cb}$$ Circuit: between E and B a conductance $g_{11}$ in parallel with a current source $g_{12}v_{cb}$; between C and B a conductance $g_{22}$ in parallel with a current source $g_{21}v_{eb}$. This is the topology of Fig. 12.1(b) with (B, C, E) relabeled (E, C, B). Active-mode check (no base-width modulation), $I_E\propto e^{qV_{EB}/kT}$, $|I_C|=\alpha_F|I_E|$: $$g_{11}=\frac{qI_E}{kT}=\frac{g_m}{\alpha_F},\qquad |g_{21}|=\alpha_Fg_{11}=g_m,\qquad g_{12}=g_{22}=0$$ So the CB input resistance is $1/g_{11}=\alpha_F/g_m\approx kT/qI_E$ and the current gain is $\alpha_F$. $\blacksquare$ --- ### Problem 12.3 — $g_m$ and $r_\pi$ for the 2N2605 **Problem.** Using the 2N2605 pnp CE characteristics (Fig. 11.5d), $I_B=5\ \mu$A and $V_{EC}=10$ V; find $g_m$ and $r_\pi$. **Assumption (figure not available).** Reading the $I_B=5\ \mu$A curve at $V_{EC}=10$ V, take $I_C\approx1.0$ mA, i.e. $\beta_{dc}\approx200$. If your figure reads differently, use the formulas with your value. $$g_m=\frac{qI_C}{kT}=\frac{1.0\times10^{-3}\ \text{A}}{0.0259\ \text{V}}=3.86\times10^{-2}\ \text{S}$$ $$r_\pi=\frac{\beta_{dc}}{g_m}=\frac{I_C/I_B}{qI_C/kT}=\frac{kT/q}{I_B}=\frac{0.0259}{5\times10^{-6}}=5.2\ \text{k}\Omega$$ (Note $r_\pi$ depends only on $I_B$.) $\blacksquare$ --- ### Problem 12.4 — Confirm Eqs. (12.10) **Problem.** Show that Fig. 12.2(b) gives $g_m=g_{21}-g_{12}$, $r_o=1/(g_{22}+g_{12})$, $r_\pi=1/(g_{11}+g_{12})$, $r_\mu=-1/g_{12}$. **Solution.** KCL at the base node (current splits between $r_\pi$ to E and $r_\mu$ to C): $$i_b=\frac{v_{be}}{r_\pi}+\frac{v_{be}-v_{ce}}{r_\mu}=\left(\frac1{r_\pi}+\frac1{r_\mu}\right)v_{be}-\frac1{r_\mu}v_{ce}$$ KCL at the collector node (source, $r_o$, and the current arriving through $r_\mu$ from B): $$i_c=g_mv_{be}+\frac{v_{ce}}{r_o}+\frac{v_{ce}-v_{be}}{r_\mu}=\left(g_m-\frac1{r_\mu}\right)v_{be}+\left(\frac1{r_o}+\frac1{r_\mu}\right)v_{ce}$$ Match coefficients with $i_b=g_{11}v_{be}+g_{12}v_{ce}$, $i_c=g_{21}v_{be}+g_{22}v_{ce}$ (valid for arbitrary $v_{be},v_{ce}$): $$g_{12}=-\frac1{r_\mu}\ \Rightarrow\ r_\mu=-\frac1{g_{12}}$$ $$g_{11}=\frac1{r_\pi}+\frac1{r_\mu}=\frac1{r_\pi}-g_{12}\ \Rightarrow\ r_\pi=\frac1{g_{11}+g_{12}}$$ $$g_{21}=g_m-\frac1{r_\mu}=g_m+g_{12}\ \Rightarrow\ g_m=g_{21}-g_{12}$$ $$g_{22}=\frac1{r_o}+\frac1{r_\mu}=\frac1{r_o}-g_{12}\ \Rightarrow\ r_o=\frac1{g_{22}+g_{12}}$$ Since $g_{12}<0$ (a larger collector voltage narrows the base and reduces base current), $r_\mu>0$ as required. $\blacksquare$ --- ### Problem 12.5 — Computer exercise: Hybrid-Pi parameters **Problem.** Write a MATLAB program that computes $g_m, r_\pi, r_o, r_\mu$ with and without base-width modulation for given $V_{EB}$, $V_{EC}$; run with $V_{EC}=10$ V and $V_{EB}$ chosen so $I_C\le1$ mA; also compute $g_m$, $r_\pi$ from (12.9); discuss. **Assumption.** The BJT0/BJTplus listings (Appendix M) are not in the supplied text, so the program below is self-contained with a stated pnp model (replace the parameter block with the BJT0 values if desired). Model: narrow-base diffusion current $I_p=qAD_Bn_i^2e^{V_{EB}/V_t}/(N_BW)$; $I_C=I_p(1-W^2/2L_B^2)$; $I_B=$ base recombination plus emitter back-injection; $W=W_{B0}-x_{n,C}$, with $x_{n,C}$ the C–B depletion extension into the base. **Logic.** 1. Define $I_B(V_{EB},V_{EC})$ and $I_C(V_{EB},V_{EC})$ as functions; the flag `bwm` selects base-width modulation (1) or fixed $W=W_{B0}$ (0). 2. Find $V_{EB}$ giving $I_C=1$ mA by bisection ($I_C$ is monotonic in $V_{EB}$). 3. Compute $g_{11},g_{12},g_{21},g_{22}$ by central finite differences (the definition (12.5)). 4. Apply (12.10): $g_m=g_{21}-g_{12}$, $r_\pi=1/(g_{11}+g_{12})$, $r_o=1/(g_{22}+g_{12})$, $r_\mu=-1/g_{12}$. 5. Compare with $g_m=qI_C/kT$ and $r_\pi=\beta_{dc}/g_m$ (12.9). ```matlab % hybrid_pi.m -- Hybrid-Pi parameters of a pnp BJT (self-contained) clear; clc; q=1.602e-19; k=1.381e-23; T=300; Vt=k*T/q; es=11.7*8.854e-14; ni=1e10; NE=1e18; NB=1e16; NC=1e15; % assumed doping (cm^-3) DB=12; DE=3; tauB=1e-7; tauE=1e-8; % cm^2/s, s LB=sqrt(DB*tauB); LE=sqrt(DE*tauE); A=1e-4; WB0=1.5e-4; % cm^2, cm Vbi=Vt*log(NB*NC/ni^2); VEC=10; ICtarget=1e-3; d=1e-4; % V, A, V (finite-difference step) for bwm=[0 1] % W as function of (VEB,VEC); VBC = VEC-VEB is the C-B reverse bias W=@(VEB,VEC) WB0 - bwm*sqrt(2*es*(Vbi+VEC-VEB)/q*NC/(NB*(NB+NC))); Ip=@(VEB,VEC) q*A*DB*ni^2./(NB*W(VEB,VEC)).*exp(VEB/Vt); IEn=@(VEB) q*A*DE*ni^2/(NE*LE)*exp(VEB/Vt); IC=@(VEB,VEC) Ip(VEB,VEC).*(1-W(VEB,VEC).^2/(2*LB^2)); IB=@(VEB,VEC) Ip(VEB,VEC).*W(VEB,VEC).^2/(2*LB^2)+IEn(VEB); lo=0.4; hi=0.8; % bisection for IC = 1 mA for n=1:60 m=(lo+hi)/2; if IC(m,VEC)>ICtarget, hi=m; else lo=m; end end VEB=m; g11=(IB(VEB+d,VEC)-IB(VEB-d,VEC))/(2*d); g12=(IB(VEB,VEC+d)-IB(VEB,VEC-d))/(2*d); g21=(IC(VEB+d,VEC)-IC(VEB-d,VEC))/(2*d); g22=(IC(VEB,VEC+d)-IC(VEB,VEC-d))/(2*d); gm=g21-g12; rpi=1/(g11+g12); ro=1/(g22+g12); rmu=-1/g12; % Inf if g12=g22=0 (no BWM) gm_a=q*IC(VEB,VEC)/(k*T); % Eq. (12.9a) rpi_a=(IC(VEB,VEC)/IB(VEB,VEC))/gm_a; % Eq. (12.9b) fprintf('BWM=%d: VEB=%.4f V, IB=%.3e A, IC=%.3e A\n',bwm,VEB,IB(VEB,VEC),IC(VEB,VEC)); fprintf(' gm=%.4e S, rpi=%.4e, ro=%.4e, rmu=%.4e\n',gm,rpi,ro,rmu); fprintf(' Eq(12.9): gm=%.4e S, rpi=%.4e\n\n',gm_a,rpi_a); end ``` **Sample results (assumed parameters above; yours will differ with the BJT0 values):** | Case | $V_{EB}$ (V) | $I_B$ ($\mu$A) | $g_m$ (mS) | $r_\pi$ | $r_o$ | $r_\mu$ | |---|---|---|---|---|---|---| | No BWM | 0.668 | 43.5 | 38.67 | 595 $\Omega$ | $\infty$ | $\infty$ | | With BWM | 0.642 | 7.29 | 38.65 | 3.55 k$\Omega$ | 67 k$\Omega$ | 12 M$\Omega$ | Eq. (12.9) gives $g_m=38.67$ mS in both cases and $r_\pi=\beta_{dc}/g_m=595\ \Omega$ and $3.55$ k$\Omega$ respectively, matching the finite-difference values to within about $10^{-5}$ relative (the tiny $g_{12}$ correction). **Discussion.** - $g_m$ is essentially identical in all cases and equals $qI_C/kT$: it depends on $I_C$, not on device details. - The two cases give different $r_\pi$ because the fixed wide base ($W=W_{B0}$) has far more recombination current, hence a smaller $\beta_{dc}$ at the same $I_C$. - Without base-width modulation $g_{12}=g_{22}=0$, so $r_o$ and $r_\mu$ are infinite, recovering Fig. 12.2(a). - With modulation, $g_{22}>0$ gives a finite $r_o$ (Early voltage $V_A\approx I_Cr_o\approx67$ V here) and a tiny negative $g_{12}$ gives a huge $r_\mu\gg r_\pi$, which is why $r_\mu$ is often dropped. - Agreement of (12.9) with (12.10) confirms that neglecting $g_{12}$ is justified. --- ### Problem 12.6 — Unity beta frequency $f_T$ **Problem.** (a) With the high-frequency Hybrid-Pi model derive $i_c/i_b$ when $v_{ce}=0$ and from it an expression for $f_T$ (the frequency where $|i_c/i_b|=1$). (b) Evaluate $f_T$ for the 2N3906 at $V_{EB}=0.68$ V, $V_{EC}=10$ V, $I_B=5.57\ \mu$A, $I_C=1.00$ mA, with $g_m=3.86\times10^{-2}$ S, $r_o=2.00\times10^{4}\ \Omega$, $r_\pi=4.65\times10^3\ \Omega$, $r_\mu=3.59\times10^6\ \Omega$, $C_{eb}=23.6$ pF, $C_{cb}=2.32$ pF, $r_b=10\ \Omega$, $r_c=2.8\ \Omega$, $r_e=0$. **(a) Derivation.** With $v_{ce}=0$ the collector is a.c. shorted to the emitter. $r_o$ is shorted; $r_c$ is in series with the short; $r_e=0$; $r_b$ is in series with a driving current $i_b$, so none affects the current ratio. Let $v$ be the voltage across the intrinsic B′–E′ port. Because C′ is at a.c. ground, $r_\pi$, $r_\mu$, $C_{eb}$, $C_{cb}$ all have $v$ across them: $$i_b=v\left[\frac1{r_\pi}+\frac1{r_\mu}+j\omega(C_{eb}+C_{cb})\right]=v\left[g_\pi+j\omega C\right],\quad g_\pi=\frac1{r_\pi}+\frac1{r_\mu},\ C=C_{eb}+C_{cb}$$ The current into the shorted collector is the source current minus the current that flows back from B′ through $C_{cb}$ and $r_\mu$: $$i_c=g_mv-\frac v{r_\mu}-j\omega C_{cb}v\approx\left(g_m-j\omega C_{cb}\right)v\qquad(g_m\gg1/r_\mu)$$ Dividing: $$\frac{i_c}{i_b}=\frac{g_m-j\omega C_{cb}}{g_\pi+j\omega C}$$ Low-frequency limit: $\beta_0=g_m/g_\pi\approx g_mr_\pi$. Magnitude: $$\left|\frac{i_c}{i_b}\right|=\sqrt{\frac{g_m^2+\omega^2C_{cb}^2}{g_\pi^2+\omega^2C^2}}$$ Set equal to 1 at $\omega_T=2\pi f_T$: $$g_m^2+\omega_T^2C_{cb}^2=g_\pi^2+\omega_T^2C^2$$ $$\omega_T^2\left(C^2-C_{cb}^2\right)=g_m^2-g_\pi^2$$ $$\boxed{f_T=\frac1{2\pi}\sqrt{\frac{g_m^2-g_\pi^2}{(C_{eb}+C_{cb})^2-C_{cb}^2}}\ \approx\ \frac{g_m}{2\pi\,(C_{eb}+C_{cb})}}$$ (The approximation neglects $g_\pi\ll g_m$ and $C_{cb}^2$ relative to $C^2$.) Also, for $f\ll f_T$, $|i_c/i_b|\approx\beta_0/\sqrt{1+(f/f_\beta)^2}$ with $f_\beta=g_\pi/[2\pi C]=f_T/\beta_0$: gain falls at 20 dB/decade beyond $f_\beta$ and reaches unity at $f_T\approx\beta_0f_\beta$. **(b) Numbers.** $$g_\pi=\frac1{4.65\times10^{3}}+\frac1{3.59\times10^{6}}=2.151\times10^{-4}+2.8\times10^{-7}=2.154\times10^{-4}\ \text{S}$$ $$\beta_0=\frac{3.86\times10^{-2}}{2.154\times10^{-4}}\approx179$$ $$C=23.6+2.32=25.92\ \text{pF}$$ $$C^2-C_{cb}^2=671.8-5.4=666.4\ \text{pF}^2$$ $$g_m^2-g_\pi^2=1.490\times10^{-3}-4.6\times10^{-8}\approx1.490\times10^{-3}\ \text{S}^2$$ $$\omega_T=\sqrt{\frac{1.490\times10^{-3}}{666.4\times10^{-24}}}=1.495\times10^{9}\ \text{rad/s}$$ $$f_T=\frac{\omega_T}{2\pi}\approx2.4\times10^{8}\ \text{Hz}\approx240\ \text{MHz}$$ Check with the approximation: $g_m/2\pi C=3.86\times10^{-2}/(1.629\times10^{-10})=237$ MHz. Also $f_\beta=f_T/\beta_0\approx1.3$ MHz. So $f_T\approx240$ MHz, a "general purpose" transistor, as the problem statement suggests. $\blacksquare$ --- ### Problem 12.7 — Diffusion component of $C_{eb}$ **Problem.** Show that for an active-mode pnp BJT $$C_{eb}\big|_{diff}=\frac{q^2An_i^2W}{3N_BkT}\,e^{qV_{EB}/kT}=\tfrac23\,\tau_t\,g_m$$ **Solution.** The E–B junction is a narrow-base diode: n-type base of width $W$ whose far boundary ($x=W$) holds the excess carrier density at zero (the collector sweeps carriers away). Write $\Delta p=\Delta p_0(x)+\delta p(x)e^{j\omega t}$ and use hole continuity: $$j\omega\,\delta p=D_B\frac{d^2\delta p}{dx^2}-\frac{\delta p}{\tau_B}\ \Rightarrow\ \frac{d^2\delta p}{dx^2}=\frac{\delta p}{L^{*2}},\qquad L^{*2}=\frac{L_B^2}{1+j\omega\tau_B}$$ Boundary conditions: $\delta p(W)=0$; at $x=0$, $\Delta p(0)=p_{n0}e^{q(V_0+v)/kT}$, so for small $v$ $$\delta p(0)=p_{n0}e^{qV_0/kT}\,\frac{qv}{kT},\qquad p_{n0}=\frac{n_i^2}{N_B}$$ Solution: $$\delta p(x)=\delta p(0)\,\frac{\sinh[(W-x)/L^*]}{\sinh(W/L^*)}$$ AC current at $x=0$ (hole diffusion): $$\tilde i=-qAD_B\left.\frac{d\delta p}{dx}\right|_{x=0}=\frac{qAD_B}{L^*}\,\delta p(0)\coth\!\left(\frac{W}{L^*}\right)$$ Admittance $Y=\tilde i/v$ with $K\equiv\dfrac{q^2AD_Bp_{n0}}{kT}e^{qV_0/kT}$: $$Y=\frac K{L^*}\coth\!\left(\frac W{L^*}\right)$$ Narrow base, $\xi=W/L^*\ll1$, use $\coth\xi\approx\dfrac1\xi+\dfrac\xi3$: $$Y=\frac K{L^*}\left(\frac{L^*}W+\frac W{3L^*}\right)=K\left[\frac1W+\frac W{3L^{*2}}\right]=K\left[\frac1W+\frac W{3L_B^2}+j\omega\frac{W\tau_B}{3L_B^2}\right]$$ With $L_B^2=D_B\tau_B$ the imaginary part is $j\omega KW/(3D_B)$. So $C_{diff}=\mathrm{Im}\,Y/\omega$: $$C_{diff}=\frac{KW}{3D_B}=\frac{q^2A\,p_{n0}W}{3kT}e^{qV_0/kT}=\frac{q^2An_i^2W}{3N_BkT}e^{qV_{EB}/kT}$$ Relation to $g_m$: the narrow-base dc current is $I_C\approx\dfrac{qAD_Bn_i^2}{N_BW}e^{qV_{EB}/kT}$, so $$\frac{q^2An_i^2W}{N_BkT}e^{qV_{EB}/kT}=\frac{W^2}{D_B}\cdot\frac q{kT}I_C=\frac{W^2}{D_B}g_m$$ Hence $$C_{eb}\big|_{diff}=\frac13\frac{W^2}{D_B}g_m=\frac23\left(\frac{W^2}{2D_B}\right)g_m=\frac23\tau_tg_m\qquad\blacksquare$$ (For a wide-base diode $C_D=\tau g_d/2$; here the transit time replaces the lifetime and the factor becomes $2/3$.) --- ### Problem 12.8 — Estimate $I_{CC}\tau_t/I_{BB}\tau_B$ **Problem.** In Fig. 12.3(b), $V_{CC}/R_L=5$ mA, $V_s/R_s=30\ \mu$A, and $I_B$ was stepped in 5 $\mu$A increments. Estimate $I_{CC}\tau_t/I_{BB}\tau_B$ for switching between the "off" and "on" points. **Solution.** Use $\tau_t/\tau_B=1/\beta_{dc}$: $$x=\frac{I_{CC}\tau_t}{I_{BB}\tau_B}=\frac{I_{CC}}{\beta_{dc}I_{BB}}$$ with $I_{CC}=5$ mA and $I_{BB}=30\ \mu$A. **Assumption (curves not available).** The "on" point at $I_C=5$ mA lies near the knee of the $I_B=20\ \mu$A curve, so the base current just needed to reach $I_{CC}$ is $I_{B,sat}=20\ \mu$A and $\beta_{dc}\approx5\ \text{mA}/20\ \mu\text{A}=250$: $$x=\frac{I_{B,sat}}{I_{BB}}=\frac{20\ \mu\text{A}}{30\ \mu\text{A}}\approx0.67$$ Interpretation: the base is overdriven by a factor $1/x=1.5$, so $0<x<1$ as required. For a different $\beta_{dc}$ read from the curves, $x=167/\beta_{dc}$. $\blacksquare$ --- ### Problem 12.9 — Rise and storage-delay time plots **Problem.** Let $x=I_{CC}\tau_t/I_{BB}\tau_B$, $0\le x\le1$. (a) Plot $t_r/\tau_B$ vs $x$ using (12.27). (b) Plot $t_{sd}/\tau_B$ vs $x$ using (12.32) for $\xi=0$ and $\xi=1$. **Solution.** From (12.27) and (12.32): $$\frac{t_r}{\tau_B}=-\ln(1-x),\qquad\frac{t_{sd}}{\tau_B}=\ln\frac{1+\xi}{\xi+x}$$ $$\xi=0:\ \frac{t_{sd}}{\tau_B}=-\ln x,\qquad\xi=1:\ \frac{t_{sd}}{\tau_B}=\ln\frac2{1+x}$$ | $x$ | $t_r/\tau_B$ | $t_{sd}/\tau_B$ ($\xi=0$) | $t_{sd}/\tau_B$ ($\xi=1$) | |---|---|---|---| | 0.05 | 0.051 | 3.00 | 0.646 | | 0.1 | 0.105 | 2.303 | 0.598 | | 0.2 | 0.223 | 1.609 | 0.511 | | 0.3 | 0.357 | 1.204 | 0.431 | | 0.4 | 0.511 | 0.916 | 0.357 | | 0.5 | 0.693 | 0.693 | 0.288 | | 0.6 | 0.916 | 0.511 | 0.223 | | 0.7 | 1.204 | 0.357 | 0.163 | | 0.8 | 1.609 | 0.223 | 0.105 | | 0.9 | 2.303 | 0.105 | 0.051 | | 0.99 | 4.605 | 0.010 | 0.005 | | 1.0 | $\infty$ | 0 | 0 | **Description.** (a) $t_r/\tau_B$ rises slowly from 0 at $x=0$ (large overdrive, fast fill) and diverges logarithmically as $x\to1$: when $I_{BB}$ barely supports $I_{CC}$ the exponential only asymptotically reaches the required charge. (b) $t_{sd}(\xi=0)=t_r(1-x)$, the mirror image. For $\xi=0$ it diverges as $x\to0$ (deep saturation, much stored charge) and vanishes at $x=1$. For $\xi=1$ the curve is far lower everywhere (finite, $\ln2\approx0.69$ at $x=0$): a reverse base current slashes storage delay, especially in deep saturation. ```matlab x = linspace(0.001,0.999,500); tr = -log(1-x); tsd0 = log(1./x); % xi = 0 tsd1 = log(2./(1+x)); % xi = 1 figure; plot(x,tr,'LineWidth',1.5); grid on xlabel('x = I_{CC}\tau_t/(I_{BB}\tau_B)'); ylabel('t_r/\tau_B'); ylim([0 5]) figure; plot(x,tsd0,x,tsd1,'LineWidth',1.5); grid on xlabel('x'); ylabel('t_{sd}/\tau_B'); legend('\xi=0','\xi=1'); ylim([0 3]) ``` $\blacksquare$ --- ### Problem 12.10 — Fall time **Problem.** (a) Derive $t_f$ using the measurement-based definition (Fig. 12.6: $i_C$ falls from $0.9I_{CC}$ to $0.1I_{CC}$) and (12.31b). (b) Plot $t_f/\tau_B$ vs $x$ for $\xi=0,1$ and discuss. **(a) Derivation.** In the active-mode portion, using $I_{BB}\tau_B/\tau_t=I_{CC}/x$: $$i_C(t)=\frac{I_{BB}\tau_B}{\tau_t}\left[(1+\xi)e^{-t/\tau_B}-\xi\right]=\frac{I_{CC}}{x}\left[(1+\xi)e^{-t/\tau_B}-\xi\right]$$ Time at which $i_C=0.9I_{CC}$: $$(1+\xi)e^{-t_1/\tau_B}-\xi=0.9x\ \Rightarrow\ t_1=\tau_B\ln\frac{1+\xi}{\xi+0.9x}$$ Time at which $i_C=0.1I_{CC}$: $$t_2=\tau_B\ln\frac{1+\xi}{\xi+0.1x}$$ (Both lie after $t_{sd}$ because $0.9x<x$.) The fall time is the difference: $$t_f=t_2-t_1=\tau_B\left[\ln\frac{1+\xi}{\xi+0.1x}-\ln\frac{1+\xi}{\xi+0.9x}\right]$$ $$\boxed{t_f=\tau_B\ln\!\left(\frac{\xi+0.9x}{\xi+0.1x}\right)}$$ **(b)** For $\xi=0$: $t_f=\tau_B\ln9=2.197\,\tau_B$, independent of $x$ (pure exponential decay with time constant $\tau_B$). For $\xi=1$: $t_f/\tau_B=\ln\dfrac{1+0.9x}{1+0.1x}$. | $x$ | $t_f/\tau_B$ ($\xi=0$) | $t_f/\tau_B$ ($\xi=1$) | |---|---|---| | 0.1 | 2.197 | 0.076 | | 0.2 | 2.197 | 0.146 | | 0.3 | 2.197 | 0.210 | | 0.4 | 2.197 | 0.268 | | 0.5 | 2.197 | 0.323 | | 0.6 | 2.197 | 0.374 | | 0.7 | 2.197 | 0.421 | | 0.8 | 2.197 | 0.465 | | 0.9 | 2.197 | 0.507 | | 1.0 | 2.197 | 0.547 | ```matlab x = linspace(0.001,1,300); tf0 = log(9)*ones(size(x)); tf1 = log((1+0.9*x)./(1+0.1*x)); plot(x,tf0,x,tf1,'LineWidth',1.5); grid on xlabel('x = I_{CC}\tau_t/(I_{BB}\tau_B)'); ylabel('t_f/\tau_B'); legend('\xi=0','\xi=1') ``` **Discussion.** With $\xi=0$ the active-mode charge leaves only by recombination, so $t_f=2.2\,\tau_B$ regardless of drive. With a reverse base current ($\xi=1$) the fall is at least four times faster. The smaller $x$ is (heavier overdrive, larger $I_{BB}$), the faster the fall: the reverse current drives $Q_B$ toward $-I_{BB}\tau_B$ steeply, so the $0.9\to0.1\,I_{CC}$ window is crossed quickly ($t_f\to0.8x\,\tau_B$ as $x\to0$). As $x\to1$ the benefit shrinks to $0.55\,\tau_B$. $\blacksquare$ --- # Chapter 13 — PNPN Devices > **The big picture.** We have met the two-region, one-junction diode and the three-region, two-junction transistor. Add one more region and you get four alternating regions and three interacting junctions: the **PNPN device**, or **thyristor**. Think of a mousetrap: it sits quietly until a small nudge releases stored energy and it snaps shut, staying shut. Thyristors are electronic mousetraps. They block a high voltage until a tiny gate trigger flips them into a low-resistance state that *latches*. They are used as high-power rectifiers and switches, often large-area devices in metal stud packages for heat dissipation, with ratings above 5 kA and 10 kV. We concentrate on the **silicon controlled rectifier (SCR)**. (The source text lists this chapter as read-only; it has no end-of-chapter problem set or summary section, so none is solved here. A short recap is given at the end.) --- ## 13.1 Silicon Controlled Rectifier (SCR) **Structure (Fig. 13.1a).** Four alternately doped regions P1, N2, P3, N4 and three junctions: $J_{12}$ (P1–N2), $J_{23}$ (N2–P3), $J_{34}$ (P3–N4). - The outer P region (P1) and its contact: **anode (A)**. - The outer N region (N4) and its contact: **cathode (K)**. - The inner regions N2 and P3 are the **bases**. - The third terminal, the **gate (G)**, connects to P3. - Without a gate lead, the device is the **PNPN (Shockley) diode**. Variables: $V_{AK}$ anode-to-cathode voltage, $I_{AK}$ current into the anode, $I_G$ current into the gate. **Fabrication and doping profile (Fig. 13.1b).** Start with a thinned, lightly doped n-type Si wafer ($N_D\sim5\times10^{13}$ cm$^{-3}$), which becomes N2. Diffuse p-type dopant into both sides, giving nearly identical $J_{12}$ and $J_{23}$ junctions. Finally diffuse n-type dopant into one side to form the heavily doped N4 cathode. The base regions N2 and P3 have *moderate electrical widths*, e.g. $W_2\sim250\ \mu$m and $W_3\sim25\ \mu$m, comparable to a minority-carrier diffusion length. That is the secret of the device: unlike a wide-base diode (widths $\gg L$) or a BJT (base $\ll L$), the junctions are *close enough to talk to each other but not so close that they merge*. **Characteristics (Fig. 13.2).** - $V_{AK}<0$: independent of $I_G$, like a reverse-biased pn diode; little current until breakdown at $-V_{BR}$ (the reverse blocking voltage). - $V_{AK}>0$ with $I_G=0$: little current until $V_{AK}$ reaches the **forward blocking voltage** $V_{BF}$. Beyond it the device switches from the high-impedance blocking mode to the low-impedance conducting mode and then follows a diode-like forward characteristic near $V_{AK}\approx1$ V. - $I_G>0$ lowers $V_{BF}$, so the device turns on at a lower $V_{AK}$. - $V_{BR}$ and $V_{BF}$ can be hundreds to thousands of volts; the conducting drop is typically about a volt. --- ## 13.2 SCR Operational Theory Start with $I_G=0$ (then the SCR behaves as the PNPN diode). Strategy: begin with the simplest model and refine it only as far as needed. ### The diode model Model the SCR as three pn diodes in series (Fig. 13.3a). - $V_{AK}<0$: the middle diode ($J_{23}$) is forward biased and replaced by a short (Fig. 13.3b). The remaining diodes $J_{12}$ and $J_{34}$ are reverse biased. From Fig. 13.1(b), the lightly doped side of $J_{12}$ is about $10^{14}$ cm$^{-3}$ (high breakdown voltage), whereas the lightly doped side of $J_{34}$ (the P3 side) is about $10^{16}$ cm$^{-3}$, so $J_{34}$ breaks down at a relatively small reverse voltage and then acts like a short. The reverse characteristic is therefore basically that of the $J_{12}$ junction. - $V_{AK}>0$: $J_{12}$ and $J_{34}$ are forward biased and replaced by shorts (Fig. 13.3c); $J_{23}$ is reverse biased, so the forward blocking characteristic is that of the $J_{23}$ junction. Because $J_{12}$ and $J_{23}$ have nearly identical doping profiles, the predicted characteristic is nearly symmetric about $V_{AK}=0$ (Fig. 13.4), with $V_{BF}\approx V_{BR}$, both near the breakdown voltage of $J_{12}$ or $J_{23}$ (the smaller of the avalanche breakdown voltage and the punch-through voltage that fully depletes the lightly doped N2 region). In reality the observed blocking voltages are always somewhat *smaller* than these upper limits. **What the diode model gets right:** the general form of the reverse and forward blocking regions. **What it gets wrong:** it cannot produce forward switching, because it ignores the **interaction between junctions**, and such an interaction must exist since $W_2$, $W_3$ are comparable to the diffusion lengths. ### The two-transistor model Split the SCR into two interconnected BJTs (Fig. 13.5): P1–N2–P3 is a **pnp** transistor (emitter P1, base N2, collector P3) and N2–P3–N4 is an **npn** transistor (emitter N4, base P3, collector N2). Wires join the shared regions: the base of the pnp is the collector of the npn and vice versa. Both BJTs are intrinsically **low gain** (base widths only comparable to $L$, so transport factors and hence $\alpha$'s are well below unity at normal operation). **Qualitative regeneration (Fig. 13.6).** Assume blocking mode, $I_G=0$, $V_{AK}>0$. Both transistors are active-mode biased. 1. Majority carriers from the P1 and N4 emitters are injected into the adjacent bases. 2. A fraction diffuses across the base. 3. They arrive in the base of the *other* transistor, where they appear as an excess of *majority* carriers that cannot escape through an external terminal. 4. This excess raises base-to-emitter forward bias in the other transistor, increasing its emitter injection, which feeds back around the loop, giving more carriers... a **regenerative** (positive-feedback) process, like the squeal of a microphone near its loudspeaker. At low $V_{AK}$ the loop gain is below one, so the process is stable and the carrier build-up around $J_{23}$ is negligible. As $V_{AK}$ grows, the build-up reduces the reverse bias of $J_{23}$ and finally forward biases it (a junction is forward biased when minority carriers exceed equilibrium at the depletion-region edges). In BJT language both transistors become saturated. The SCR enters the low-impedance conducting mode when all three junctions are forward biased. **Quantitative condition.** Each transistor in blocking mode is active biased, so use the large-signal active-mode Ebers–Moll equivalent circuits (Fig. 13.7a,b): current source $\alpha_FI_E$ (the reverse generator is negligible) and a leakage $I_{RO}$ ($\alpha_R I_R\to$ small, $I_F\approx I_E$, $I_{CO}=I_{RO}$). Connect them as in Fig. 13.5 (Fig. 13.7c), subscript 1 for the pnp and 2 for the npn. Equating the current into and out of node $E_2$ (the cathode)... equivalently the current at the anode: $$I_{AK}=\alpha_1I_{AK}+I_{RO1}+\alpha_2I_{AK}+I_{RO2}\tag{13.1}$$ Collect the $I_{AK}$ terms: $$I_{AK}(1-\alpha_1-\alpha_2)=I_{RO1}+I_{RO2}$$ $$\boxed{I_{AK}=\frac{I_{RO1}+I_{RO2}}{1-(\alpha_1+\alpha_2)}}\tag{13.2}$$ As $\alpha_1+\alpha_2\to1$ the denominator vanishes and the current grows without bound (limited only by the external circuit): **the critical switching condition is $\alpha_1+\alpha_2=1$**. The SCR is built so that $\alpha_1+\alpha_2<1$ at small $V_{AK}>0$. Several mechanisms push $\alpha_1+\alpha_2$ up as $V_{AK}$ increases. Recall $$\alpha_{dc}=M\gamma\alpha_T\tag{13.3}$$ with $\alpha_T$ the base transport factor, $\gamma$ the emitter efficiency, and $M$ the carrier multiplication factor. 1. **Base-width modulation.** $J_{23}$ is reverse biased, and its depletion region extends mostly into the lightly doped N2 base of the pnp. Increasing $V_{AK}$ narrows the quasineutral base, increasing $\alpha_T$. 2. **Emitter efficiency.** Recombination–generation current reduces $\gamma$ at low injection levels. As the current increases with $V_{AK}$, the injection level rises and $\gamma$ increases. 3. **Carrier multiplication.** If the first two effects are insufficient, avalanche multiplication $M>1$ in the reverse-biased $J_{23}$ junction ultimately triggers the SCR. Hence the breakdown voltage of $J_{23}$ is the upper limit of $V_{BF}$. **Why gate current lowers $V_{BF}$.** A positive $I_G$ into P3 supplies majority carriers (holes) to the base of the N2–P3–N4 transistor, enhancing injection from the N4 emitter. The regenerative loop becomes unstable (loop gain reaching one) at a lower $V_{AK}$. A few milliamps of gate current can trigger many amperes of anode current. --- ## 13.3 Practical Turn-On/Turn-Off Considerations ### 13.3.1 Circuit Operation Consider the bias circuit of Fig. 13.8b ($V_{AA}$ in series with $R_L$) and the load line $I_{AK}=(V_{AA}-V_{AK})/R_L$ on the forward characteristics (Fig. 13.8a). The "off" point is the intersection with the blocking branch; the "on" point is the intersection with the conducting branch near $I_{AK}\approx V_{AA}/R_L$. **Turn-on:** - Increase $V_{AA}$: the load line shifts parallel toward larger $V_{AK}$. When it no longer intersects the blocking branch (i.e. $V_{AA}>V_{BF}$), the device switches on. - Or increase $I_G$ from zero: when the high-impedance branch of the $I_G\neq0$ curve (e.g. curve $G_2$) no longer meets the load line, the operating point jumps to the "on" point. - Once latched on, $I_G$ can be removed and the SCR stays on. **Turn-off:** - Reduce $V_{AA}$ so that $I_{AK}$ falls below the **holding current** $I_H$; the conducting state cannot be sustained and the device reverts to blocking. - Theoretically $I_G<0$ (extracting gate current) can also turn it off, but the extraction rate from P3 must exceed the entry rate, which requires special devices carrying large gate currents (e.g., $I_G\approx10$ A to turn off $I_{AK}=100$ A). Specially designed devices are **gate turn-off (GTO) SCRs**. ### 13.3.2 Additional Triggering Mechanisms - **Light:** a sufficiently intense pulse with $h\nu>E_G$ generates carriers, producing the required injection, with a window to the P3 region. Advantage: complete electrical isolation of the input. Such a device is a **light-activated SCR (LASCR)** or light-activated switch (LAS). - **Temperature:** the forward blocking voltage falls with temperature, equivalent to applying $I_G>0$; this limits operation to roughly 125–150 °C. - **Noise:** because of the very high current gain ($I_{AK}(\text{on})/I_G(\text{turn-on})\sim10^4$–$10^6$), noise pulses in the gate circuit can trigger the device inadvertently. This is remedied, at the expense of triggering sensitivity, by the shorted-cathode configuration. Unexpected temperature rise can also cause inadvertent switching; mounting high-power devices on metal studs improves heat dissipation. ### 13.3.3 Shorted-Cathode Configuration Built as described in 13.1, $\alpha_1+\alpha_2$ (and hence $V_{BF}$ and the forward characteristics) are extremely sensitive to the base dopings and widths. The first commercial PNPN diode venture failed financially because the turn-on voltage could not be held within tolerance. The now-standard fix is the **shorted-cathode** structure (Fig. 13.9): the cathode metallization is extended to also contact the P3 region at a point remote from the gate. How it works: at low $I_{AK}$ the P3/N4 short essentially removes injection from N4, so the gain of the N2–P3–N4 transistor is nearly zero. At larger $I_{AK}$, the lateral hole current flowing in P3 beneath N4 produces a forward P3-to-N4 voltage drop that increases from point A (near the short) to point B (far from the short). When the potential at B reaches about 0.7 V, strong N4-to-P3 injection starts near B. In quick succession $I_{AK}$ rises, the injecting region widens from B toward A, the npn gain rises abruptly, injection becomes uniform across the N4 face, $\alpha_1+\alpha_2\to1$, and the SCR latches. Thus the switching characteristic is made reproducible by controlling the *resistance of P3 beneath the cathode*, instead of the delicate gains. ### 13.3.4 di/dt and dv/dt Effects **di/dt.** A large, short gate pulse can minimize turn-on time. But in a shorted-cathode SCR, if $I_G$ rises too fast the injection across P3–N4 is concentrated near the gate before spreading across the cathode face. The resulting local current crowding heats the region and can cause thermal failure: **di/dt burnout**. The di/dt of the turn-on pulse must be limited. **dv/dt.** A rapidly applied anode-to-cathode voltage makes the SCR conduct prematurely: the faster the rise, the lower the voltage required for turn-on. The same occurs under high-frequency ac operation or a high-voltage noise spike. Cause: the depletion width of $J_{23}$ widens as $V_{AK}$ increases, which requires a displacement (charging) current $C_{J23}\,dV/dt$; this majority-carrier current in the bases induces extra injection from P1 and N4, raising the transistor gains. The larger $dv/dt$, the larger the enhanced gain and the lower the blocking voltage. ### 13.3.5 Triggering Time Carriers injected from the anode and cathode must cross the adjacent bases. The average diffusion transit time across a quasineutral base is $W^2/2D$ (Chapter 12). With quasineutral widths $W_2$ (N2) and $W_3$ (P3), the transit times are $$t_1=\frac{W_2^2}{2D_p},\qquad t_2=\frac{W_3^2}{2D_n}$$ (holes cross N2; electrons cross P3). To first order the triggering time is the geometric mean: $$t_{on}\approx\sqrt{t_1t_2}=\sqrt{\frac{W_2^2}{2D_p}\cdot\frac{W_3^2}{2D_n}}=\frac{W_2W_3}{2\sqrt{D_pD_n}}\tag{13.4}$$ Small base widths give fast switching. But smaller widths mean larger transistor gains and lower punch-through voltages, so lower blocking voltage and power capability: a **trade-off between high power handling and rapid switching**. ### 13.3.6 Switching Advantages/Disadvantages (SCR vs BJT) **Advantages** 1. Very little gate current turns on very large anode currents. 2. Can block both polarities of an ac signal. 3. Very high blocking voltage together with low conducting voltage drop. 4. Unlike the BJT, no current crowding in steady-state conduction. **Disadvantages** 1. Cannot be turned off by setting $I_G=0$ (a BJT turns off when base current is removed). 2. Cannot operate at high frequencies. 3. Prone to turning on from noise voltage spikes. 4. Limited temperature range of operation. --- ## 13.4 Other PNPN Devices - **Dual-gate SCR / silicon controlled switch (SCS) (Fig. 13.10):** gates to both bases (G1 at P3... and G2 at N2); can be triggered through either, giving the designer more flexibility. - **Bilateral devices** respond symmetrically to positive and negative applied voltage. - **DIAC (diode AC) (Fig. 13.11):** an integrated pair of oppositely oriented PNPN diodes connected in parallel; the I-V curve switches in both polarities. - **TRIAC (triode AC) (Fig. 13.12):** the bilateral SCR, two oppositely oriented SCRs in parallel sharing a gate; ideal for ac power control such as light dimmers and motor speed control. - **Programmable unijunction transistor (PUT) (Fig. 13.13):** despite its name, not a unijunction transistor. A UJT is a three-lead device with two ohmic contacts at the ends of a bar-like resistive region and a third lead to a pn junction along the bar; it switches somewhat like an SCR. The PUT is actually an *anode-gated PNPN* device: the gate contacts the N2 base instead of P3. Connected as in Fig. 13.13(b) (gate set by a divider $R_1$, $R_2$ from $V_{BB}$, with $R_L$ in the cathode), it mimics a UJT, with the peak voltage $V_P\approx V_{BB}R_2/(R_1+R_2)$ (plus a diode drop) and characteristics like Fig. 13.13(c). Changing the external $R_1$, $R_2$ changes the characteristics: hence "programmable." --- ## Chapter 13 recap - Four layers, three junctions; base widths $\sim$ diffusion length so the junctions interact. - Diode model gives the blocking regions; the two-transistor model (pnp + npn cross-connected) explains switching via positive feedback: $I_{AK}=(I_{RO1}+I_{RO2})/[1-(\alpha_1+\alpha_2)]$, switching when $\alpha_1+\alpha_2=1$. - Triggering: voltage, gate current, light, temperature, $dv/dt$. Turn-off: current below $I_H$ (or gate extraction in a GTO). - Shorted-cathode structure makes the characteristics reproducible; $di/dt$ and $dv/dt$ ratings limit use; $t_{on}\approx W_2W_3/2\sqrt{D_pD_n}$. - Family: SCR, PNPN diode, SCS, DIAC, TRIAC, PUT. # Chapter 14: MS Contacts and Schottky Diodes > **The big question of this chapter.** Every chip you have ever used has to talk to the outside world through metal wires. At the exact spot where a metal touches a semiconductor, something remarkable happens: electrons have to decide whether to cross freely (an *ohmic contact*, a good door) or to face a one-way turnstile (a *rectifying contact*, a Schottky diode). Both behaviours come from one idea: **line up the Fermi levels, and see what the bands do.** Think of two reservoirs of water at different heights joined by a pipe. Water flows until the levels match. Electrons do the same thing: they flow from the material with the higher Fermi level to the one with the lower Fermi level until $E_F$ is flat. The piled-up charge left behind builds the barrier. Almost everything in this chapter is that one sentence, dressed up. The metal-semiconductor (MS) contact appears in every solid-state device. As an **ohmic contact** it is the link between the semiconductor and the outside world. As a **rectifying contact** (the *Schottky diode* or *MS diode*) it is a device in its own right. The MS diode is a close cousin of the asymmetrical ($p^+$-$n$ or $n^+$-$p$) step-junction diode, so a large part of the $pn$ analysis can be borrowed with minor modifications. Plan of the chapter: (14.1) the equilibrium band diagram of an ideal contact, and how it separates rectifying from ohmic; (14.2) the Schottky diode: electrostatics, $I$-$V$, a.c. and transient response; (14.3) practical complications; (14.4) summary; then the problems. --- ## 14.1 Ideal MS Contacts An **ideal MS contact** satisfies three assumptions: 1. Metal and semiconductor are in intimate contact on an atomic scale, with no layer (such as an oxide) between them. 2. There is no interdiffusion or intermixing of metal and semiconductor. 3. There are no adsorbed impurities or surface charges at the MS interface. ### Ingredients: vacuum level, work function, electron affinity Fig. 14.1 shows *surface-included* energy band diagrams of an isolated metal (left) and an isolated $n$-type semiconductor (right), the latter at flat-band (zero field) conditions. The vertical line where the bands end abruptly is the surface. Cross-hatching marks allowed states almost completely filled with electrons. *Figure 14.1 (described).* Left: a metal with a filled band up to $E_{FM}$ and a vertical wall whose top ledge is the vacuum level $E_0$; the gap from $E_{FM}$ up to $E_0$ is labelled $\Phi_M$. Right: a semiconductor with $E_c$, $E_F$ (near $E_c$ because $n$-type), $E_i$, $E_v$ and the same vacuum level; the gap from $E_c$ to $E_0$ is $\chi$; from $E_F$ to $E_0$ is $\Phi_s$. Definitions: - **Vacuum level** $E_0$: the minimum energy an electron needs to free itself completely from the material. Imagine it as the rim of a bathtub: an electron at $E_0$ is just out of the tub. - **Work function** $\Phi$: $E_0-E_F$. It is the energy price for pulling an electron *at the Fermi level* out of the solid. - **Metal work function** $\Phi_M$: an invariant property of the metal, from 3.66 eV (Mg) to 5.15 eV (Ni). - **Semiconductor work function** $\Phi_s$ has two pieces: $$\Phi_s=\chi+(E_c-E_F)_{FB}\qquad(14.1)$$ - **Electron affinity** $\chi=(E_0-E_c)_{\text{surface}}$: invariant property of the semiconductor. $\chi=4.00,\ 4.03,\ 4.07$ eV for Ge, Si, GaAs. - $(E_c-E_F)_{FB}$: computable from the doping (flat-band, bulk value). Why two pieces? Because the metal's Fermi level is fixed by the metal, but the semiconductor's Fermi level *moves with doping*. Dope more heavily $n$-type and $E_F$ climbs toward $E_c$, so $\Phi_s$ falls. ### Building the equilibrium diagram ($\Phi_M>\Phi_s$, $n$-type) *Thought experiment.* Join the two solids so quickly that no electrons have time to move. An instant after contact, the diagram is simply the two isolated diagrams abutted at the interface, aligned to the common $E_0$ (Fig. 14.2(a)). $\Phi_M$ and $\chi$ are material constants and are *not* altered by contacting. But $E_{Fs}\ne E_{FM}$. For $\Phi_M>\Phi_s$, $E_{Fs}$ lies *above* $E_{FM}$. In equilibrium the Fermi level must be position independent (Subsection 3.2.4), so: 1. Electrons begin to transfer from semiconductor to metal (high to low Fermi level). 2. The semiconductor loses electrons near the surface, creating a **depletion region** of uncompensated positive donors. 3. This builds a growing barrier to further transfer. 4. The process stops when the transfer rate is the same both ways and $E_F$ is flat. The result is Fig. 14.2(b). The barrier measured from $E_F$ in the metal up to $E_c$ at the interface is $$\boxed{\Phi_B=\Phi_M-\chi\quad\text{(ideal MS, }n\text{-type)}}\qquad(14.2)$$ $\Phi_B$ is the barrier height seen by metal electrons with $E=E_{FM}$. Mnemonic: *climb from the metal's Fermi level to the vacuum level ($\Phi_M$), then drop down to the conduction-band edge ($\chi$).* *Figure 14.2 (described).* (a) Instant after contact, $\Phi_M>\Phi_s$: $E_{FM}$ sits below $E_{Fs}$, bands flat, interface abrupt. (b) Equilibrium: $E_F$ flat; semiconductor bands bend *upward* toward the interface; $E_c$ at the interface is $\Phi_B$ above $E_F$; far away $E_c-E_F=(E_c-E_F)_{FB}$. (c) Instant after contact, $\Phi_M<\Phi_s$: $E_{FM}$ above $E_{Fs}$. (d) Equilibrium for $\Phi_M<\Phi_s$: electrons flow metal $\to$ semiconductor, an *accumulation* layer forms; bands bend *downward* toward the interface; no barrier for electrons going semiconductor $\to$ metal. ### Biasing the two structures: who is rectifying? Conventions (Fig. 14.3(a)): semiconductor grounded, $V_A$ applied to the metal, and $I>0$ when current flows from metal to semiconductor. **Case $\Phi_M>\Phi_s$ (barrier type).** - $V_A>0$ (Fig. 14.3(b)): $E_{FM}$ is pushed *below* $E_{Fs}$. The barrier seen by semiconductor electrons shrinks to $q(V_{bi}-V_A)$. An exponentially increasing number of electrons can climb over it: a rapidly rising **forward current**. - $V_A<0$ (Fig. 14.3(c)): $E_{FM}$ rises above $E_{Fs}$, the barrier for semiconductor electrons grows, and semiconductor-to-metal flow is nearly blocked. Metal electrons still see the same $\Phi_B$, so a small current flows that is essentially bias independent once $|V_A|$ exceeds a few $kT/q$: **a small, saturating reverse current**. That is rectification, Fig. 14.3(d). The device is the **MS diode**. **Case $\Phi_M<\Phi_s$.** - No barrier for electrons going semiconductor $\to$ metal, so even a small $V_A>0$ gives a large current. - Under reverse bias, metal electrons face a small barrier that vanishes after a few tenths of a volt, so large reverse current appears at small bias, with no saturation. That is **ohmic-like** behaviour. Parallel arguments for a $p$-type semiconductor (Exercise 14.1) give the opposite pattern. **Table 14.1 Electrical nature of ideal MS contacts** | | $n$-type semiconductor | $p$-type semiconductor | |---|---|---| | $\Phi_M>\Phi_s$ | Rectifying | Ohmic | | $\Phi_M<\Phi_s$ | Ohmic | Rectifying | Everything here assumes the contact is *ideal*; Section 14.3 explains real-world twists. > **Anticipated confusion.** "Why is the *larger* metal work function the rectifying one for $n$-type?" A larger $\Phi_M$ means the metal's Fermi level is deeper below vacuum, so the metal is a thirstier acceptor of electrons; it drains the semiconductor surface and leaves a depleted barrier. For $p$-type the carriers are holes, which float *upward* in the diagram, so the logic flips. ### Exercise 14.1 **Problem.** (a) Construct the equilibrium band diagram for an ideal $p$-type semiconductor-to-metal contact with $\Phi_M<\Phi_s$. (b) Repeat for $\Phi_M>\Phi_s$. (c) Verify the contact is rectifying for $\Phi_M<\Phi_s$ and ohmic-like for $\Phi_M>\Phi_s$. (d) Find the barrier height $\Phi_B$ (from $E_{FM}$ to $E_v$ at the interface) of the rectifying $p$-type contact. **Solution.** (a), (b) The recipe: 1. Draw the surface-included diagrams of the isolated parts. 2. Align both to the common $E_0$ and join at the interface. 3. Without changing the interfacial positions of the semiconductor bands, slide the field-free semiconductor bulk up or down until $E_F$ is constant everywhere. 4. Connect $E_c$, $E_i$, $E_v$ at the interface to the bulk positions. 5. Erase the construction lines. Result (described): for $\Phi_M<\Phi_s$ the $p$-type semiconductor has $E_F$ low (near $E_v$) and $E_{FM}$ is higher, so electrons flow from metal into semiconductor, recombine with holes near the surface and leave ionized acceptors: a depletion layer with bands bending *downward* toward the interface (a barrier for holes). For $\Phi_M>\Phi_s$ holes are pulled into the surface layer, the bands bend *upward* toward the interface and $E_v$ moves toward $E_F$ (accumulation). (c) Empty electron states in the metal below $E_{FM}$ (exponentially few) act like holes. - $\Phi_M<\Phi_s$: there is a hole barrier in both directions at equilibrium. Raising $E_{FM}$ relative to $E_{Fs}$ lowers the barrier for holes going semiconductor $\to$ metal, so that current grows exponentially with the separation $E_{FM}-E_{Fs}$. Reversing the bias blocks it, leaving a small saturating metal $\to$ semiconductor hole current. **Rectifying.** - $\Phi_M>\Phi_s$: there is no barrier to holes going semiconductor $\to$ metal, and the small barrier the other way vanishes with a slight downward shift of $E_{FM}$. **Ohmic-like.** (d) Measure the barrier to the valence band at the interface: $$\Phi_B=E_{FM}-E_v|_{\text{interface}}=(E_c-E_v)-\big(E_c|_{\text{interface}}-E_{FM}\big)$$ Vacuum-level alignment gives $E_0-E_c=\chi$ and $E_0-E_{FM}=\Phi_M$, so $E_c|_{\text{interface}}-E_{FM}=\Phi_M-\chi$. Therefore $$\boxed{\Phi_B=E_G+\chi-\Phi_M\quad\text{(ideal MS, }p\text{-type)}}$$ --- ## 14.2 Schottky Diode Here the semiconductor is taken to be uniformly doped $n$-type, with the polarities of Fig. 14.3(a). The $pn$-diode analogies keep the presentation compact. ### 14.2.1 Electrostatics **Built-in voltage.** As in a $pn$ diode there is a voltage drop across the MS diode at equilibrium. From Fig. 14.4(a) (barrier $\Phi_B$ at the interface; bulk $E_c$ sits $(E_c-E_F)_{FB}$ above $E_F$), the band bending is the difference: $$V_{bi}=\frac{1}{q}\Big[\Phi_B-(E_c-E_F)_{FB}\Big]\qquad(14.3)$$ with $\Phi_B=\Phi_M-\chi$ for an ideal $n$-type contact. **Charge density.** Electrons are depleted next to the interface; the depletion region extends a distance $W$ into the semiconductor and carries the positive donor charge. Unlike the $pn$ junction there is no acceptor charge on the other side: the compensating negative charge is an essentially $\delta$-function sheet of extra electrons in the metal right at the interface (picture the charge on a thin conducting plate). Fig. 14.4(b): a rectangle of height $+qN_D$ from $0$ to $W$ and a $\delta$-spike of negative charge at the interface. Depletion approximation: $$\rho=\begin{cases}qN_D & 0\le x\le W & (14.4a)\\ 0 & x>W & (14.4b)\end{cases}$$ Here $x_n\to W$ since the depletion region lies entirely in the semiconductor. Because the charge in the metal is a $\delta$-function, $\mathcal E=0$ and $V=$ const inside the metal, so the metal needs no further attention. **Field.** Poisson's equation in the semiconductor: $$\frac{d\mathcal E}{dx}=\frac{\rho}{K_s\varepsilon_0}=\frac{qN_D}{K_s\varepsilon_0}\quad 0\le x\le W\qquad(14.5)$$ Separate variables and integrate from an arbitrary $x$ to $W$, where $\mathcal E(W)=0$: $$\int_{\mathcal E(x)}^{0}d\mathcal E'=\frac{qN_D}{K_s\varepsilon_0}\int_x^W dx'\qquad(14.6)$$ $$-\mathcal E(x)=\frac{qN_D}{K_s\varepsilon_0}(W-x)$$ $$\boxed{\mathcal E(x)=-\frac{qN_D}{K_s\varepsilon_0}(W-x)\quad 0\le x\le W}\qquad(14.7)$$ The field is negative (it points toward $-x$, from the positive donors toward the metal's negative sheet). Fig. 14.4(c): a straight line from $-qN_DW/K_s\varepsilon_0$ at $x=0$ to zero at $x=W$. This is identical to the $n$-side field of a $p^+$-$n$ step junction with $x_n\to W$ (Eq. 5.21). The maximum magnitude occurs at the interface: $$|\mathcal E|_{\max}=\frac{qN_DW}{K_s\varepsilon_0}$$ **Potential.** Use $dV/dx=-\mathcal E$: $$\frac{dV}{dx}=-\mathcal E=\frac{qN_D}{K_s\varepsilon_0}(W-x)\qquad(14.8)$$ Integrate from $x$ to $W$ with $V(W)\equiv0$ (the surface-side edge of the semiconductor bulk is the natural reference): $$\int_{V(x)}^{0}dV'=\frac{qN_D}{K_s\varepsilon_0}\int_x^W (W-x')\,dx'\qquad(14.9)$$ $$-V(x)=\frac{qN_D}{K_s\varepsilon_0}\cdot\frac{(W-x)^2}{2}$$ $$\boxed{V(x)=-\frac{qN_D}{2K_s\varepsilon_0}(W-x)^2\quad 0\le x\le W}\qquad(14.10)$$ This looks different from the $pn$ result but is not. In the $pn$ analysis $V=0$ was set on the far $p$-side; a solution of identical form results if the metal is taken as the $V=0$ reference. With a single semiconductor region, the bulk edge is the preferred reference. At equilibrium the potential drop across the depletion region is $-V_{bi}$ at $x=0$. With bias, $V_{bi}\to V_{bi}-V_A$ and $V(0)=-(V_{bi}-V_A)$. (This assumes an ohmic back contact and negligible $IR$ drop across the neutral bulk.) **Depletion width.** Evaluate (14.10) at $x=0$: $$-(V_{bi}-V_A)=-\frac{qN_DW^2}{2K_s\varepsilon_0}\qquad(14.11)$$ $$\boxed{W=\left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi}-V_A)\right]^{1/2}}\qquad(14.12)$$ Identical to a $p^+$-$n$ junction. ### Exercise 14.2 **Problem.** Copper is deposited on a carefully prepared $n$-type silicon substrate to make an ideal Schottky diode. $\Phi_M\approx4.65$ eV, $\chi=4.03$ eV, $N_D=10^{16}/\text{cm}^3$, $T=300$ K. Find (a) $\Phi_B$, (b) $V_{bi}$, (c) $W$ at $V_A=0$, (d) $|\mathcal E|_{\max}$ at $V_A=0$. **Solution.** (a) Ideal $n$-type contact (14.2): $$\Phi_B=\Phi_M-\chi=4.65-4.03=0.62\ \text{eV}$$ (b) Flat-band position of the Fermi level ($E_i\approx$ midgap, $n_i=10^{10}\,\text{cm}^{-3}$): $$(E_c-E_F)_{FB}=\frac{E_G}{2}-kT\ln\frac{N_D}{n_i}=0.56-0.0259\ln\!\left(\frac{10^{16}}{10^{10}}\right)=0.56-0.358\approx0.20\ \text{eV}$$ $$V_{bi}=\frac{1}{q}\Big[\Phi_B-(E_c-E_F)_{FB}\Big]=0.62-0.20=0.42\ \text{V}$$ (c) Put $V_A=0$ in (14.12): $$W=\left[\frac{2(11.8)(8.85\times10^{-14})}{(1.6\times10^{-19})(10^{16})}(0.42)\right]^{1/2}$$ $$W=\left[\frac{8.77\times10^{-13}}{1.6\times10^{-3}}\right]^{1/2}=(5.48\times10^{-10})^{1/2}\ \text{cm}=2.34\times10^{-5}\ \text{cm}=0.234\ \mu\text{m}$$ (d) $$|\mathcal E|_{\max}=\frac{qN_DW}{K_s\varepsilon_0}=\frac{(1.6\times10^{-19})(10^{16})(2.34\times10^{-5})}{(11.8)(8.85\times10^{-14})}=3.59\times10^{4}\ \text{V/cm}$$ ### 14.2.2 $I$-$V$ Characteristics The electrostatics and the general shape of the $I$-$V$ curve resemble a $pn$ diode, but the *current mechanism* is quite different. In a $p^+$-$n$ diode (Fig. 14.5(a)) the dominant forward current is recombination in the depletion region at small bias and hole injection from $p^+$ into $n$ at larger bias; electron injection into the heavier-doped $p^+$ side is negligible. In an MS ($n$-type) diode (Fig. 14.5(b)) the recombination and hole-injection currents still exist, but the barrier seen by semiconductor electrons is small, so **electron injection from the semiconductor into the metal routinely dominates**. It gives a huge forward current long before recombination or hole diffusion matter. Under reverse bias, electron flow from metal into semiconductor (Fig. 14.3(c)) totally dominates; the reverse hole diffusion current and R-G generation current are typically negligible. Since minority carriers play an insignificant role, the MS diode is called a **majority carrier device**. The majority-carrier current over the barrier is the **thermionic emission current**. We derive it for $n$-type, treating first electrons going from semiconductor to metal. Take $x$ normal to the interface, directed into the semiconductor. **Step 1: Which electrons can cross?** An electron entering the depletion region from the bulk crosses into the metal only if its kinetic energy along $x$ exceeds the remaining barrier: $$\frac12 m_n^*v_x^2\ \ge\ q(V_{bi}-V_A)\qquad(14.13)$$ $$|v_x|\ \ge\ v_{\min}=\left[\frac{2q(V_{bi}-V_A)}{m_n^*}\right]^{1/2}\qquad(14.14)$$ with $v_x$ directed toward the interface (negative). **Step 2: Current from one velocity group.** If $n(v_x)$ is the number of electrons per cm$^3$ with that $v_x$, then (as in the drift-current derivation of Subsection 3.1.2) $$I_{S\to M,x}=-qA\,v_x\,n(v_x)\qquad(14.15)$$ **Step 3: Sum over all fast enough electrons.** $$I_{S\to M}=-qA\int_{-\infty}^{-v_{\min}}v_x\,n(v_x)\,dv_x\qquad(14.16)$$ **Step 4: The distribution.** For a nondegenerate semiconductor (Maxwell-Boltzmann statistics, integrated over $v_y,v_z$): $$n(v_x)=\left(\frac{4\pi kT\,m_n^{*2}}{h^3}\right)e^{(E_F-E_c)/kT}\,e^{-(m_n^*/2kT)v_x^2}\qquad(14.17)$$ **Step 5: Integrate and simplify.** Let $a=m_n^*/2kT$. The velocity integral is $$\int_{-\infty}^{-v_{\min}}v_xe^{-av_x^2}dv_x=-\frac{1}{2a}e^{-av_{\min}^2}=-\frac{kT}{m_n^*}e^{-m_n^*v_{\min}^2/2kT}$$ and $\tfrac12m_n^*v_{\min}^2=q(V_{bi}-V_A)$. Hence $$I_{S\to M}=qA\left(\frac{4\pi kT\,m_n^{*2}}{h^3}\right)\frac{kT}{m_n^*}\,e^{(E_F-E_c)/kT}e^{-q(V_{bi}-V_A)/kT}=A\left(\frac{4\pi q m_n^*k^2}{h^3}\right)T^2e^{-[(E_c-E_F)_{FB}+qV_{bi}-qV_A]/kT}$$ Using $(E_c-E_F)_{FB}+qV_{bi}=\Phi_B$ (Eq. 14.3): $$\boxed{I_{S\to M}=AA^*T^2e^{-\Phi_B/kT}e^{qV_A/kT}}\qquad(14.18)$$ $$A^*=\frac{4\pi q\,m_n^*k^2}{h^3}=\left(\frac{m_n^*}{m_0}\right)\!\frac{4\pi q\,m_0k^2}{h^3}\qquad(14.19)$$ $$\frac{4\pi q m_0k^2}{h^3}=120\ \text{A/(cm}^2\text{-K}^2)\qquad(14.20)$$ $A^*$ is the **Richardson constant** (first found in the study of electron emission from hot metals). **Step 6: The opposite flow.** Metal electrons always see the same barrier $\Phi_B$, so $$I_{M\to S}(V_A)=I_{M\to S}(V_A=0)\qquad(14.21)$$ At equilibrium the two flows must cancel exactly: $$I_{M\to S}(V_A=0)=-I_{S\to M}(V_A=0)=-AA^*T^2e^{-\Phi_B/kT}\qquad(14.22)$$ **Step 7: Total current.** $$I=I_{S\to M}+I_{M\to S}=I_{S\to M}+I_{M\to S}(V_A=0)\qquad(14.23)$$ Combining (14.18) and (14.22): $$\boxed{I=I_S\big(e^{qV_A/kT}-1\big)}\qquad(14.24)$$ $$\boxed{I_S=AA^*T^2e^{-\Phi_B/kT}}\qquad(14.25)$$ This is the MS version of the ideal diode equation. For forward bias beyond a few $kT/q$, $I\to I_Se^{qV_A/kT}$; for reverse bias beyond a few $kT/q$, $I\to-I_S$ (saturation). *Intuition.* $I_S$ looks different from the $pn$ diode's $qAD_pn_i^2/(L_pN_D)$ because it counts *hot-enough majority electrons* crossing a barrier, not minority carriers diffusing away. The $\Phi_B/kT$ in the exponent is why a 0.1 eV change in barrier height changes the saturation current by a factor $e^{0.1/0.0259}\approx48$. **Experiment versus theory (Fig. 14.6).** *Figure 14.6 (described).* (a) Semilog forward $I$ versus $V_A$ for an MBR040 diode: a straight line of slope $q/kT$ from about 0.1 V to 0.35 V (over many decades of $I$), with the theoretical diffusion ($I_{DIFF}$) and R-G dashed curves far below; at higher bias the curve bends over because of series resistance. (b) Reverse characteristic: $|I|$ grows steadily (microamps) as $V_A$ goes from 0 to about $-50$ V, ending in breakdown. For $V_A\le0.35$ V theory and experiment agree almost perfectly (slope close to $q/kT$). As in the $pn$ diode, the slope decrease at larger forward bias is due to the voltage drop across the bulk series resistance. In reverse bias there are two departures from ideal: 1. **Breakdown** limits the maximum reverse bias. Without edge effects, avalanche breakdown in an MS diode is essentially the same as in an equally doped $p^+$-$n$ or $n^+$-$p$ diode. 2. The reverse current **does not saturate**. Unlike a $pn$ diode this is *not* R-G current; it is mainly **Schottky barrier lowering**: $\Phi_B$ decreases slightly with increasing reverse bias, $$\Phi_B=\Phi_{B0}-\Delta\Phi_B\qquad(14.26)$$ where $\Phi_{B0}$ is the barrier at zero field at the interface and $$\Delta\Phi_B=\left[\frac{q\,\mathcal E_s}{4\pi K_s\varepsilon_0}\right]^{1/2}\qquad(14.27)$$ ($\mathcal E_s$ is the surface field, found from the electrostatic relations; with $q$ in coulombs and $\mathcal E_s$ in V/cm the result is in volts, i.e. eV.) Physically, an escaping electron induces an image charge in the metal, and the image attraction plus the applied field rounds off and lowers the top of the barrier. Because $I_S\propto e^{-\Phi_B/kT}$, even a small $\Delta\Phi_B$ produces a noticeable increase in reverse current. Near $-V_{BR}$ avalanche multiplication may further enhance the current. ### Exercise 14.3 **Problem.** Model the $0.1\ \text{V}\le V_A\le0.35$ V portion of Fig. 14.6(a) as $I=I_0e^{qV_A/n_1kT}$ where $I_0$ and $n_1$ are constants ($n_1$ is the "ideality factor," unity in Eq. 14.24 but usually slightly above 1). Using the data below, perform a least-squares fit for $I_0$ and $n_1$. | $V_A$ (V) | $I$ (A) | |---|---| | 0.10 | $4.047\times10^{-7}$ | | 0.15 | $2.792\times10^{-6}$ | | 0.20 | $1.890\times10^{-5}$ | | 0.25 | $1.263\times10^{-4}$ | | 0.30 | $8.084\times10^{-4}$ | | 0.35 | $4.487\times10^{-3}$ | **Solution logic.** Take the log: $$\ln I=\ln I_0+\frac{q}{n_1kT}V_A$$ This is a straight line $y=c_2+c_1x$ with $x=V_A$, $y=\ln I$, slope $c_1=q/n_1kT$ and intercept $c_2=\ln I_0$. MATLAB's `polyfit(x,y,1)` returns `[c1, c2]` (highest power first), computed by least squares. Then $$n_1=\frac{q}{kT\,c_1}=\frac{1}{0.0259\,c_1},\qquad I_0=e^{c_2}$$ ```matlab % Least-squares fit to MS diode I-V data % ln(I) = ln(Is) + q*VA/(n1*k*T) I = [4.047e-7 2.792e-6 1.890e-5 1.263e-4 8.084e-4 4.487e-3]; VA = [0.10 0.15 0.20 0.25 0.30 0.35]; y = log(I); c = polyfit(VA, y, 1); % least squares; c(1)=slope, c(2)=ln(Is) slope = c(1); format compact n1 = 1/(0.0259*slope) % kT/q = 0.0259 V Is = exp(c(2)) ``` Results: $\boxed{n_1=1.03,\quad I_0=I_s=1.02\times10^{-8}\ \text{A}}$. Quick sanity check by hand: from the end points, $\ln(4.487\times10^{-3}/4.047\times10^{-7})=9.31$ over 0.25 V gives slope 37.3 V$^{-1}$, so $n_1=1/(0.0259\times37.3)\approx1.04$, consistent. ### 14.2.3 a.c. Response A small a.c. signal on a d.c. reverse bias causes a charge fluctuation (Fig. 14.7): the depletion width oscillates about its dc value, the donor charge in the semiconductor changes, and the $\delta$-sheet of electrons at the metal surface changes by the opposite amount. Majority carriers flow quickly in and out of the semiconductor to make this happen. With the charge fluctuation on two planes a distance $W$ apart, it is exactly a parallel-plate capacitor: $$C=\frac{K_s\varepsilon_0A}{W}\qquad(14.28)$$ With (14.12): $$\boxed{C=\frac{K_s\varepsilon_0A}{\left[\dfrac{2K_s\varepsilon_0}{qN_D}(V_{bi}-V_A)\right]^{1/2}}}\qquad(14.29)$$ Inverting and squaring (14.29): $$\frac{1}{C^2}=\frac{2(V_{bi}-V_A)}{qN_DK_s\varepsilon_0A^2}\qquad(14.30)$$ *Figure 14.7 (described).* A semiconductor of width $W$ with $+qN_D$ charge, with the small signal $v_a$ shifting the depletion edge $x=W$ in and out, and the compensating charge fluctuating at $x=0$ in the metal. *Reading Fig. 14.8.* (a) $C$ versus $V_A$ for an MBR040 diode, falling from tens of pF near zero bias to a few pF at $-12$ V, decreasing more slowly with larger reverse bias. (b) $1/C^2$ versus $V_A$ is a nearly straight line, as (14.30) predicts (3.4 pF of package stray capacitance was subtracted first). The slope gives $N_D$, the intercept at $1/C^2=0$ gives $V_{bi}$ and, once both are known, (14.3) gives $\Phi_B$. This is one of the most popular ways to measure barrier heights. **Forward bias is different from a $pn$ diode.** Reverse-bias behaviour mirrors an asymmetric $pn$ diode, but in forward bias the MS diode's diffusion current is negligible, so there is little minority carrier injection or storage. Since stored minority carriers give the diffusion admittance, **the MS diode has no diffusion capacitance or diffusion conductance**. There is still the depletion capacitance, a potentially large parallel conductance $G=dI/dV_A$, and the series resistance $R_S$, but even at frequencies approaching or exceeding GHz, $C$ and $G$ stay frequency independent. ### Exercise 14.4 **Problem.** From the plotted $1/C^2$ data of Fig. 14.8(b), estimate the barrier height $\Phi_B$. Take an $n$-type device, $A=1.5\times10^{-3}$ cm$^2$, room temperature. **Solution.** 1. Intercept of the extrapolated straight line on the $V_A$ axis: $V_{bi}\approx0.6$ V. 2. Slope of the line: $\Delta(1/C^2)\approx6\times10^{-3}\ \text{pF}^{-2}=6\times10^{21}\ \text{F}^{-2}$ over $\Delta V=10.3+0.6=10.9$ V: $$|\text{slope}|=\frac{6\times10^{21}}{10.3+0.6}\approx5.5\times10^{20}\ \text{F}^{-2}\text{V}^{-1}$$ 3. From (14.30), slope $=2/(qN_DK_s\varepsilon_0A^2)$, so $$N_D=\frac{2}{qK_s\varepsilon_0A^2|\text{slope}|}=\frac{2}{(1.6\times10^{-19})(11.8)(8.85\times10^{-14})(1.5\times10^{-3})^2(5.5\times10^{20})}=9.7\times10^{15}\ \text{cm}^{-3}$$ 4. Fermi-level position: $$(E_c-E_F)_{FB}=\frac{E_G}{2}-kT\ln\frac{N_D}{n_i}\approx0.20\ \text{eV}$$ 5. From (14.3): $$\Phi_B=qV_{bi}+(E_c-E_F)_{FB}\approx0.6+0.2=\boxed{0.8\ \text{eV}}$$ ### 14.2.4 Transient Response A very rapid transient response is the signature of the MS diode. In a $pn$ diode, excess minority carriers stored in the quasineutral regions must be removed before it can switch from forward on-state to reverse off-state (the storage delay). In an MS diode there is almost no minority carrier injection or storage, because the diffusion current is negligible. Typical commercial MS diodes have reverse recovery times of just a few nanoseconds. The response can in fact be limited not by stored charge but by the internal $RC$ time constant of the junction capacitance and the bulk series resistance. Small-area devices, often called **hot carrier diodes**, have $C\lesssim1$ pF and sub-nanosecond response times. **The Schottky diode clamp (Fig. 14.9).** Recall the BJT turn-off discussion of Chapter 12. Connect an MS diode between base and collector of an npn BJT (anode to base, cathode to collector). As the transistor tries to enter saturation during turn-on, the base-collector voltage drops until the MS diode (which conducts at a lower forward voltage than a $pn$ junction) turns on and diverts the extra base current. The C-B junction stays at a relatively low forward voltage, so very little charge is stored in the BJT, and very little is stored in the Schottky diode. Less charge to remove means a much shorter turn-off time. *Figure 14.9 (described).* npn BJT with terminals C, B, E and a Schottky diode symbol across B-C. --- ## 14.3 Practical Contact Considerations ### 14.3.1 Rectifying Contacts Real MS structures are never quite ideal, even though progress has been made. Silicon devices usually have a thin (5-25 Å) native oxide between metal and semiconductor, which forms almost instantly in air. Arsenic precipitates are believed to form on GaAs surfaces. Allowed electronic states that charge and discharge exist at essentially all semiconductor surfaces. In extreme cases these nonidealities ruin the diode; in modern structures they primarily change the barrier height $\Phi_B$. In most real $n$-type diodes $\Phi_B\ne\Phi_M-\chi$, and similarly for $p$-type. In Si, GaAs and most other semiconductors, surface charges tend to **pin** the equilibrium Fermi level at a specific energy in the surface gap (**Fermi-level pinning**). Hence the observed $\Phi_B$ varies only slightly with the metal used: for GaAs, regardless of metal, $\Phi_B\approx\tfrac23E_G\approx0.95$ eV ($n$-type) and $\Phi_B\approx\tfrac13E_G\approx0.47$ eV ($p$-type). Since $\Phi_B$ cannot be predicted from $\Phi_M$ and $\chi$, it must be *measured*. The $C$-$V$ procedure of Exercise 14.4 is one popular technique. All formulas of the previous section still hold if the measured $\Phi_B$ is used. ### 14.3.2 Ohmic Contacts Low-impedance contacts, regardless of polarity, are essential in nearly every device. Ideally $\Phi_M<\Phi_s$ ($n$-type) and $\Phi_M>\Phi_s$ ($p$-type) give ohmic contacts, but Fermi-level pinning means these combinations might still be rectifying. For example, because the surface Fermi level in GaAs is pinned about $2E_G/3$ below $E_c$, *any* metal on $n$-type GaAs forms a barrier. So how are ohmic contacts made? **Heavy doping of the surface region** beneath the contact (Fig. 14.10(a)): create an $n^+$ layer on $n$-Si before depositing the metal. The reason (Fig. 14.10(b)): the barrier height is to first order unaffected by doping, but the depletion width, hence the barrier *width*, shrinks with increasing doping ($W\propto N_D^{-1/2}$). When the doping exceeds about $10^{17}$/cm$^3$, carriers can **tunnel** through the thin top of the barrier (thermionic-field emission); for dopings above about $10^{19}$/cm$^3$ the whole barrier is so narrow that even low-energy majority carriers tunnel through (**field emission**). The barrier exists but is effectively transparent. (Think of a wall so thin that a ball can quantum-leak through.) *Figure 14.10 (described).* (a) Metal on $n^+$ layer on $n$-Si, with SiO$_2$ on both sides. (b) Three band diagrams, low, moderate and high doping, with arrows showing the emission changing from solely thermionic (over the top) to mostly field emission (through the base). Heavy doping is seldom the whole story. **Annealing** is typically required to minimize contact resistance. Aluminum is the most widely used contacting metal for discrete Si devices; heating to about 475 C for a few minutes in $N_2$ lets Al penetrate the native oxide and interdiffuse beneficially with Si. In contacts over shallow $p^+$-$n$ or $n$-$p$ junctions, nonuniform Al-Si interdiffusion can drive the Al through the junction and short it: **spiking** (Fig. 14.11(a)). A small percentage of Si is sometimes added to the Al to inhibit Si diffusion from the substrate and thus the spiking. Stability under later processing is another concern. Aluminum is unacceptable in complex ICs where steps above 500 C follow contact formation. High-temperature stability comes from the metal-like **silicides** formed with refractory metals (Mo, Ta, Ti, W). Titanium silicide, TiSi$_2$, is widely used: a Ti film over the Si contact area is converted to TiSi$_2$ by heating in an inert atmosphere. Since Si is consumed, the silicide-silicon interface moves a short distance into the Si, removing surface defects and contamination. The result: a clean, planar, temperature-stable ohmic contact. Fig. 14.11(b) shows the TiSi$_2$ contact/interconnect metallization typical of modern ICs (a stack of Ti/TiN/Al with a tungsten "plug"). *Figure 14.11 (described).* (a) An Al contact on a thin $n$-type layer over Si with an Al spike piercing down through the junction. (b) A cross-section: Si substrate, $n^+$ region with TiSi$_2$ on top, a tungsten plug in an insulator hole, TiN/Ti liner, and Al interconnect on top. --- ## 14.4 Summary The properties of an ideal MS contact were specified and a method was presented for constructing its equilibrium band diagram. Using different combinations of $\Phi_M$ and $\Phi_s$, the basic nature of the contacts was probed. Ideally, $\Phi_s(p)>\Phi_M>\Phi_s(n)$ gives a rectifying contact and $\Phi_s(p)<\Phi_M<\Phi_s(n)$ gives ohmic behaviour. The rectifying contact, called the Schottky diode or MS diode when used as a device, has electrostatics and reverse-bias a.c. response all but identical to a comparably doped $p^+$-$n$ or $n^+$-$p$ junction. Unlike the $pn$ diode, the dominant d.c. current is **thermionic emission** of majority carriers over the barrier, and since the diffusion current is small, the minority carrier injection and storage are also small. Consequently the MS diode has no significant diffusion admittance and switches between on and off states extremely fast. In real Schottky diodes $\Phi_B$ is seldom the ideal value, and the ohmic or rectifying nature may differ from the ideal prediction. An ohmic contact is usually produced in practice by heavy doping of the semiconductor surface region under the contact. --- ## Chapter 14 Problems **Problem information table** (difficulty level 1 = easiest, 4 = hardest) | Problem | Complete after | Difficulty | Point weighting | Short description | |---|---|---|---|---| | 14.1 | 14.4 | 1 | 10 (1 each part) | Quick quiz | | 14.2 | 14.1 | 2 | 10 (5 each part)/comb. | Draw/use band diagram | | 14.3 | 14.2.1 | 2 | 8 (2 each part) | Parameter computation | | 14.4 | 14.2.1 | 2 | 8 | $W$ versus $N_D$ plot | | 14.5 | 14.2.1 | 4 | 30 (a-18, b-8, c-4) | Auto-draw band diagram | | 14.6 | 14.2.2 | 2 | 8 | Confirm Eq. (14.18) | | 14.7 | 14.2.2 | 2-3 | 15 (a-4, b-5, c-4, d-2) | MS photodiode | | 14.8 | 14.2.2 | 1 | 8 | Effect of $R_S$ on $I$-$V$ | | 14.9 | 14.2.2 | 2 | 8 | Injection ratio | | 14.10 | 14.2.2 | 3 | 15 (a-10, b-4, c-1) | Schottky barrier lowering | | 14.11 | 14.2.3 | 2 | 12 | $C$-$V$ data analysis | | 14.12 | 14.2.3 | 4 | 20 (a-12, b-6, c-2) | Linearly graded MS diode | ### Problem 14.1 Quick Quiz **(a) MS diode, Schottky diode, hot carrier diode?** "MS diode" and "Schottky diode" are two names for the same device: a rectifying metal-semiconductor contact. A "hot carrier diode" is a small-area Schottky diode (with $C\lesssim1$ pF) built for sub-nanosecond response; the name refers to the energetic majority carriers that cross the barrier. **(b) Ideal contact with $\Phi_M=\Phi_s$ boundary.** For an $n$-type semiconductor: ohmic-like if $\Phi_M<\Phi_s$, rectifying if $\Phi_M>\Phi_s$. For a $p$-type semiconductor: ohmic-like if $\Phi_M>\Phi_s$, rectifying if $\Phi_M<\Phi_s$ (Table 14.1). **(c) Electrostatics differences?** The formulas for $\rho$, $\mathcal E$, $V$ on the lightly doped side are the same. The differences are: (i) in the MS diode the compensating charge is a $\delta$-function in the metal rather than a depletion layer on the heavy side, so $x_n\to W$ and $x_p\to0$; (ii) the $V=0$ reference is the semiconductor bulk edge; and (iii) $V_{bi}=[\Phi_B-(E_c-E_F)_{FB}]/q$ is computed from the barrier height instead of the two doping levels. **(d) Dominant current component:** thermionic emission of majority carriers over the barrier. **(e) How was $I_{M\to S}$ found without analyzing metal injection?** Metal electrons always see the same barrier $\Phi_B$ regardless of $V_A$, so $I_{M\to S}$ is bias independent; at $V_A=0$ the net current must vanish, so $I_{M\to S}=-I_{S\to M}(V_A=0)$ (detailed balance, Eqs. 14.21-14.22). **(f) No diffusion capacitance or conductance:** these arise from stored minority carriers injected into the quasineutral regions. In an MS diode the diffusion (minority injection) current is negligible, so there is essentially no minority carrier store. **(g) Fast switching:** no minority carrier storage, hence no storage delay; the switching speed is limited only by the $RC$ time constant of the junction capacitance and series resistance. **(h) Schottky diode clamp:** an MS diode connected from base to collector of a BJT. It conducts at a lower forward voltage than the C-B $pn$ junction and so prevents deep saturation, limiting stored charge and speeding up turn-off. **(i) Practical ohmic contact procedure:** heavily dope the semiconductor surface region beneath the contact (making the barrier thin enough for tunneling), deposit the metal (e.g. Al or a silicide such as TiSi$_2$), and anneal. **(j) Spiking:** nonuniform Al-Si interdiffusion at an Al contact during annealing, with Al penetrating through a shallow junction and shorting it. ### Problem 14.2 *Assumptions for garbled data.* The OCR lost the dopant labels for B, C, D and F. I take: B is $p$-type Ge with $N_A=10^{16}$; C is $n$-type Si with $N_D=10^{16}$; D is $p$-type Si $N_A=10^{16}$ (as stated); F is $p$-type GaAs with $N_A=10^{16}$. Use $T=300$ K, $kT=0.0259$ eV, and intrinsic-level approximation $E_i\approx$ midgap: $n_i$(Ge)$=2.4\times10^{13}$, $n_i$(Si)$=10^{10}$, $n_i$(GaAs)$=1.8\times10^{6}$ cm$^{-3}$; $E_G=0.66,\,1.12,\,1.42$ eV. Flat-band formulas ($E_i$ at midgap): $$(E_c-E_F)_{FB}=\frac{E_G}{2}-kT\ln\frac{N_D}{n_i}\ (n\text{-type}),\qquad (E_F-E_v)_{FB}=\frac{E_G}{2}-kT\ln\frac{N_A}{n_i}\ (p\text{-type})$$ $$\Phi_s=\chi+(E_c-E_F)_{FB}\ (n),\qquad \Phi_s=\chi+E_G-(E_F-E_v)_{FB}\ (p)$$ **Combination A** (Ge, $n$, $N_D=10^{16}$, $\Phi_M=4.75$): $$(E_c-E_F)_{FB}=0.33-0.0259\ln\frac{10^{16}}{2.4\times10^{13}}=0.33-0.0259(6.03)=0.174\ \text{eV}$$ $$\Phi_s=4.00+0.174=4.174\ \text{eV}<\Phi_M$$ $n$-type with $\Phi_M>\Phi_s$: **rectifying**; $\Phi_B=4.75-4.00=0.75$ eV. (a) Diagram: bands bend up toward the interface, $E_c-E_F=0.75$ eV at the interface, $0.174$ eV in the bulk, $qV_{bi}=0.75-0.174=0.58$ eV. (b) Argument: $V_A>0$ lowers $E_{FM}$, reducing the barrier for semiconductor electrons, giving exponentially growing forward current; $V_A<0$ raises the barrier and leaves only the small, constant metal-to-semiconductor current $\Rightarrow$ rectifying. **Combination B** (Ge, $p$, $N_A=10^{16}$, $\Phi_M=4.75$): $$(E_F-E_v)_{FB}=0.174\ \text{eV},\qquad\Phi_s=4.00+0.66-0.174=4.486\ \text{eV}<\Phi_M$$ $p$-type with $\Phi_M>\Phi_s$: **ohmic**. (a) Diagram: bands bend up toward the interface and $E_v$ approaches or crosses $E_F$ at the surface (hole accumulation); no barrier for holes. (b) Holes flow from semiconductor to metal freely in forward bias; the small barrier for the reverse direction vanishes after a small shift of $E_{FM}$; a nonsaturating, ohmic-like characteristic. **Combination C** (Si, $n$, $N_D=10^{16}$, $\Phi_M=4.00$): $$(E_c-E_F)_{FB}=0.56-0.0259\ln10^6=0.56-0.358=0.202\ \text{eV}$$ $$\Phi_s=4.03+0.202=4.232\ \text{eV}>\Phi_M$$ $n$-type, $\Phi_M<\Phi_s$: **ohmic**. (a) Electrons flow metal $\to$ semiconductor at contact; accumulation layer; bands bend down toward the interface; $E_c$ at the interface lies *below* $E_F$ by $\Phi_s-\Phi_M-0.202+\dots$ (specifically $E_c-E_{F}=\Phi_M-\chi=-0.03$ eV at the interface). (b) No barrier for $S\to M$ electrons; the small barrier for $M\to S$ electrons disappears with a few tenths of a volt reverse bias $\Rightarrow$ ohmic. **Combination D** (Si, $p$, $N_A=10^{16}$, $\Phi_M=4.25$): $$(E_F-E_v)_{FB}=0.202\ \text{eV},\qquad\Phi_s=4.03+1.12-0.202=4.948\ \text{eV}>\Phi_M$$ $p$-type with $\Phi_M<\Phi_s$: **rectifying**; $\Phi_B=E_G+\chi-\Phi_M=1.12+4.03-4.25=0.90$ eV. (a) Diagram: bands bend *down* toward the interface; $E_F-E_v$ at the interface is 0.90 eV, $0.202$ eV in the bulk; $qV_{bi}=0.90-0.202=0.70$ eV. (b) Forward bias (metal negative relative to the semiconductor, $V_A<0$) lowers the hole barrier giving exponentially growing hole current; the other polarity leaves a small saturated current $\Rightarrow$ rectifying. **Combination E** (GaAs, $n$, $N_D=10^{16}$, $\Phi_M=4.75$): $$(E_c-E_F)_{FB}=0.71-0.0259\ln\frac{10^{16}}{1.8\times10^6}=0.71-0.0259(22.4)=0.13\ \text{eV}$$ $$\Phi_s=4.07+0.13=4.20\ \text{eV}<\Phi_M$$ $n$-type, $\Phi_M>\Phi_s$: **rectifying**; $\Phi_B=4.75-4.07=0.68$ eV. (a), (b) as for A: bands bend up, $qV_{bi}=0.68-0.13=0.55$ eV, forward bias lowers the barrier. **Combination F** (GaAs, $p$, $N_A=10^{16}$, $\Phi_M=4.75$): $$(E_F-E_v)_{FB}=0.13\ \text{eV},\qquad\Phi_s=4.07+1.42-0.13=5.36\ \text{eV}>\Phi_M$$ $p$-type, $\Phi_M<\Phi_s$: **rectifying**; $\Phi_B=1.42+4.07-4.75=0.74$ eV. (a), (b) as for D: bands bend down, $qV_{bi}=0.74-0.13=0.61$ eV. (Note: these are *ideal-contact* predictions. Section 14.3 explains why measured barriers in GaAs would be pinned near $0.95$ eV ($n$) and $0.47$ eV ($p$).) ### Problem 14.3 *Assumption:* $N_D=10^{16}/\text{cm}^3$ (the OCR shows "$10^1$/cm"; Exercise 14.2 uses the same value). Gold on Si, $\Phi_M=5.10$ eV. (a) $\Phi_B=\Phi_M-\chi=5.10-4.03=1.07$ eV. (b) $(E_c-E_F)_{FB}=0.56-0.0259\ln10^6=0.202$ eV, so $$V_{bi}=\frac{1}{q}[\Phi_B-(E_c-E_F)_{FB}]=1.07-0.202=0.868\ \text{V}$$ (c) $$W=\left[\frac{2K_s\varepsilon_0V_{bi}}{qN_D}\right]^{1/2}=\left[\frac{2(1.0443\times10^{-12})(0.868)}{(1.6\times10^{-19})(10^{16})}\right]^{1/2}=\left[1.133\times10^{-9}\right]^{1/2}=3.37\times10^{-5}\ \text{cm}=0.337\ \mu\text{m}$$ (d) $$|\mathcal E|_{\max}=\frac{qN_DW}{K_s\varepsilon_0}=\frac{(1.6\times10^{-19})(10^{16})(3.37\times10^{-5})}{1.0443\times10^{-12}}=5.16\times10^4\ \text{V/cm}$$ ### Problem 14.4 **Problem.** Plot the equilibrium $W$ versus $N_D$ for Si MS diodes at 300 K, with $10^{14}\le N_D\le10^{17}$ cm$^{-3}$ (the OCR shows "$10^7$"; I assume $10^{17}$), for $\Phi_B=0.5,0.6,0.7$ eV. **Solution.** At equilibrium $$W=\left[\frac{2K_s\varepsilon_0V_{bi}}{qN_D}\right]^{1/2},\quad V_{bi}=\Phi_B-\left[0.56-kT\ln\frac{N_D}{n_i}\right]$$ Because $V_{bi}$ increases only logarithmically with $N_D$, $W$ falls nearly as $N_D^{-1/2}$: a straight line of slope about $-\tfrac12$ on a log-log plot, with the curves ordered by $\Phi_B$ (larger $\Phi_B$ gives larger $W$). Numerically, $2K_s\varepsilon_0/q=1.305\times10^{7}$ (cm$^{-3}$ cm$^2$/V), so $W^2=1.305\times10^{7}\,V_{bi}/N_D$ in cm$^2$ with $V_{bi}$ in volts: | $N_D$ (cm$^{-3}$) | $(E_c-E_F)_{FB}$ (eV) | $W$ (μm), $\Phi_B=0.5$ | $\Phi_B=0.6$ | $\Phi_B=0.7$ | |---|---|---|---|---| | $10^{14}$ | 0.322 | 1.53 | 1.91 | 2.22 | | $10^{15}$ | 0.262 | 0.56 | 0.66 | 0.76 | | $10^{16}$ | 0.202 | 0.20 | 0.23 | 0.26 | | $10^{17}$ | 0.143 | 0.068 | 0.077 | 0.085 | ```matlab % Equilibrium depletion width vs ND in Si Schottky diodes q=1.6e-19; eps=11.8*8.85e-14; kT=0.0259; ni=1e10; EG=1.12; ND=logspace(14,17,200); PhiB=[0.5 0.6 0.7]; for k=1:3 Vbi = PhiB(k) - (EG/2 - kT*log(ND/ni)); % volts W = sqrt(2*eps*Vbi./(q*ND))*1e4; % micrometers loglog(ND,W); hold on end xlabel('N_D (cm^{-3})'); ylabel('W (\mum)'); legend('\Phi_B=0.5 eV','\Phi_B=0.6 eV','\Phi_B=0.7 eV'); grid on ``` ### Problem 14.5 Auto-draw band diagram **Logic.** Place $E_F=0$. For $n$-type with barrier $\Phi_B$ and bulk $(E_c-E_F)_{FB}=E_G/2-kT\ln(N_D/n_i)$: $V_{bi}=\Phi_B-(E_c-E_F)_{FB}$. If $V_{bi}\le0$ the contact is not rectifying (ohmic-like), so the program must reject it. The depletion width is $W=[2K_s\varepsilon_0V_{bi}/(qN_D)]^{1/2}$, and the potential energy relative to the bulk is $-qV(x)=qV_{bi}(1-x/W)^2$ (from (14.10), parabola). So for $0\le x\le W$: $$E_c(x)=(E_c-E_F)_{FB}+V_{bi}\left(1-\frac xW\right)^2\ \text{(eV)},\quad E_v=E_c-E_G,\quad E_i=E_c-\Big[\tfrac{E_G}{2}\Big]$$ and flat beyond $x=W$. Metal side ($x<0$) is just $E_F=0$. For $p$-type: $(E_F-E_v)_{FB}=E_G/2-kT\ln(N_A/n_i)$, $V_{bi}=\Phi_B-(E_F-E_v)_{FB}$ with $\Phi_B$ the barrier to $E_v$ at the interface; the bands bend *down*: $$E_v(x)=-\left[(E_F-E_v)_{FB}+V_{bi}\left(1-\frac xW\right)^2\right]\ \text{(eV)}\quad\text{with the sign: bands lower near surface.}$$ Careful with signs: for $p$-type, bulk $E_v=-(E_F-E_v)_{FB}$ and at the surface $E_v=-\Phi_B$; so $E_v(x)=-(E_F-E_v)_{FB}-V_{bi}(1-x/W)^2$. (a)-(c) combined program (silicon default; GaAs by changing the material; $n$ or $p$ by a flag): ```matlab function mscon(PhiB, Ndop, type, mat) % Equilibrium band diagram of an MS diode. % PhiB : barrier height (eV) [n-type: Ec-EF at interface; p-type: EF-Ev] % Ndop : doping (cm^-3); type = 'n' or 'p'; mat = 'Si' or 'GaAs' q=1.6e-19; e0=8.85e-14; kT=0.0259; switch mat case 'Si', EG=1.12; ni=1e10; Ks=11.8; case 'GaAs', EG=1.42; ni=1.8e6; Ks=13.1; end dEF = EG/2 - kT*log(Ndop/ni); % (Ec-EF) for n, (EF-Ev) for p Vbi = PhiB - dEF; % volts if Vbi <= 0 error('Vbi<=0: contact is not rectifying. Choose a larger PhiB.'); end W = sqrt(2*Ks*e0*Vbi/(q*Ndop)); % cm xm = linspace(-0.5*W, 0, 20); % metal side xs = linspace(0, W, 100); % depletion region xb = linspace(W, 1.5*W, 20); % neutral bulk x = [xm xs xb]*1e4; % micrometers Vs = Vbi*(1-xs/W).^2; % band bending (eV) if type=='n' Ec = [PhiB*ones(size(xm)), dEF+Vs, dEF*ones(size(xb))]; Ev = Ec - EG; Ei = Ec - EG/2; else Ev = -[PhiB*ones(size(xm)), dEF+Vs, dEF*ones(size(xb))]; Ec = Ev + EG; Ei = Ev + EG/2; end % inside the metal the semiconductor bands do not exist: blank them Ec(x<0)=NaN; Ev(x<0)=NaN; Ei(x<0)=NaN; plot(x,Ec,'b',x,Ev,'b',x,Ei,'b--',[min(x) max(x)],[0 0],'r'); hold on xlabel('x (\mum)'); ylabel('Energy (eV), E_F = 0'); title(sprintf('%s %s-type MS diode, \\Phi_B=%.2f eV, V_{bi}=%.3f V, W=%.3f \\mum',mat,type,PhiB,Vbi,W*1e4)); text(min(x),0.05,'E_F'); end ``` Example calls: `mscon(0.62,1e16,'n','Si')` reproduces Exercise 14.2 ($V_{bi}=0.42$ V, $W=0.234\ \mu$m); `mscon(0.95,1e16,'n','GaAs')` gives a GaAs diode with pinned barrier. ### Problem 14.6 Derivation of Eq. (14.18) **Problem.** Accept (14.16) and (14.17); fill in the steps leading to (14.18). **Solution.** Start: $$I_{S\to M}=-qA\int_{-\infty}^{-v_{\min}}v_x\,n(v_x)\,dv_x,\qquad n(v_x)=\left(\frac{4\pi kTm_n^{*2}}{h^3}\right)e^{(E_F-E_c)/kT}e^{-(m_n^*/2kT)v_x^2}$$ Pull out the constants: $$I_{S\to M}=-qA\left(\frac{4\pi kTm_n^{*2}}{h^3}\right)e^{(E_F-E_c)/kT}\int_{-\infty}^{-v_{\min}}v_xe^{-(m_n^*/2kT)v_x^2}dv_x$$ Write $a=m_n^*/2kT$. The velocity integral is an exact differential: $\int v_xe^{-av_x^2}dv_x=-\dfrac{1}{2a}e^{-av_x^2}$, so $$\left[-\frac{1}{2a}e^{-av_x^2}\right]_{-\infty}^{-v_{\min}}=-\frac{1}{2a}e^{-av_{\min}^2}$$ With $a=m_n^*/2kT$: $\dfrac1{2a}=\dfrac{kT}{m_n^*}$. Then $$I_{S\to M}=+qA\left(\frac{4\pi kTm_n^{*2}}{h^3}\right)\frac{kT}{m_n^*}e^{(E_F-E_c)/kT}e^{-m_n^*v_{\min}^2/2kT}$$ The two minus signs combine to a positive result. Simplify the prefactor: $$qA\frac{4\pi k^2T^2m_n^*}{h^3}=A\left(\frac{4\pi qm_n^*k^2}{h^3}\right)T^2=AA^*T^2$$ Now the exponent. From (14.14), $\tfrac12m_n^*v_{\min}^2=q(V_{bi}-V_A)$, so $e^{-m_n^*v_{\min}^2/2kT}=e^{-q(V_{bi}-V_A)/kT}$. Combining exponentials: $$e^{(E_F-E_c)/kT}e^{-qV_{bi}/kT}e^{qV_A/kT}=e^{-[(E_c-E_F)_{FB}+qV_{bi}]/kT}e^{qV_A/kT}$$ By (14.3), $(E_c-E_F)_{FB}+qV_{bi}=\Phi_B$. Therefore $$\boxed{I_{S\to M}=AA^*T^2e^{-\Phi_B/kT}e^{qV_A/kT}}$$ with $A^*=4\pi qm_n^*k^2/h^3=(m_n^*/m_0)\cdot120$ A/cm$^2$K$^2$. This is (14.18). ### Problem 14.7 MS Photodiode **(a) Short circuit.** With the terminals shorted, $V_A=0$ and the band diagram equals the equilibrium diagram (bands bent up toward the interface, $qV_{bi}$ of bending). Electron-hole pairs created in the depletion region, and minority holes created within a diffusion length of its edge that diffuse into it, are swept apart by the built-in field: electrons slide down the potential into the neutral bulk (toward $+x$), holes slide toward the metal (toward the interface), where they recombine with the metal's electrons. Net conventional current flows from the semiconductor toward the metal. With the polarity convention of Fig. 14.3(a) ($I>0$ from metal to semiconductor), the **short-circuit photocurrent is negative**: $I_{SC}=I=-I_L<0$. **(b) Open circuit.** Now the total current must be zero. The photocurrent (holes to the metal, electrons to the bulk) charges the semiconductor positively relative to the metal, i.e. it builds a photovoltage $V_{OC}>0$ across the contact (metal positive relative to semiconductor in the sense of forward bias). That reduces the band bending to $q(V_{bi}-V_{OC})$, so the semiconductor Fermi level sits $qV_{OC}$ *above* the metal's, and the barrier for semiconductor electrons shrinks until the forward thermionic emission current exactly cancels the photocurrent. Diagram: $\Phi_B$ at the interface unchanged (set by the metal and surface), bulk $E_c$ lowered by $qV_{OC}$ relative to equilibrium, $E_{FM}$ below the semiconductor quasi-Fermi level $E_{Fn}$ by $qV_{OC}$; bands partially flattened. The form follows from zero total current: forward injected current $=$ photocurrent. **(c) Photocurrent.** By the simplified procedure of Subsection 9.2.1, the photocurrent is the generation rate times the volume of the "collecting" region: carriers generated within the depletion region and within one minority-carrier diffusion length of its edge are collected. Minority carriers in the $n$-type semiconductor are holes, and the metal side contributes none, so (low-level injection, uniform generation $G_L$, area $A$): $$I_L=-qAG_L\,(W+L_p),\qquad W=\left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi}-V_A)\right]^{1/2}$$ where the negative sign follows the Fig. 14.3(a) polarity. Total current: $$I=I_S\left(e^{qV_A/kT}-1\right)-qAG_L(W+L_p)$$ (Compared with the $pn$ photodiode, the term $L_n$ is absent because there is no quasi-neutral minority-carrier region on the metal side.) **(d) Sketch.** The illuminated $I$-$V$ curve is the dark diode curve shifted *downward* by $qAG_L(W+L_p)$ (slightly bias dependent through $W$): it crosses the $V_A$ axis at $V_{OC}>0$ (I=0), and passes through $I=-I_L$ at $V_A=0$. The fourth quadrant ($V_A>0$, $I<0$) is where power is delivered, as in a solar cell. This agrees with (a) (negative $I_{SC}$ at $V_A=0$) and (b) (positive $V_{OC}$ where $I=0$): $$V_{OC}=\frac{kT}{q}\ln\!\left(1+\frac{I_L}{I_S}\right)$$ ### Problem 14.8 Effect of series resistance **Problem.** Plot the forward $I$-$V$ for $R_S=0,0.1,1,10\ \Omega$, $0\le V_A\le0.6$ V, $10^{-9}\le I\le10^{-1}$ A. *Assumption:* the OCR shows "$I=10^{-?}$ A", the saturation current's exponent is lost. I take $I_S=10^{-10}$ A, which makes the $R_S=0$ curve span the requested ranges (at $I=0.1$ A: $V=0.0259\ln10^9=0.54$ V). **Solution.** The applied voltage is the junction voltage plus the resistor drop: $$V_A=V_j+IR_S,\qquad V_j=\frac{kT}{q}\ln\!\left(\frac{I}{I_S}+1\right)$$ So solve explicitly for $V_A$ as a function of $I$: $$V_A(I)=\frac{kT}{q}\ln\!\left(\frac I{I_S}+1\right)+IR_S$$ ```matlab Is=1e-10; Vt=0.0259; I=logspace(-9,-1,400); Rs=[0 0.1 1 10]; for k=1:4 VA = Vt*log(I/Is+1) + I*Rs(k); semilogy(VA,I); hold on end axis([0 0.6 1e-9 1e-1]); grid on xlabel('V_A (V)'); ylabel('I (A)'); legend('R_S=0','0.1 \Omega','1 \Omega','10 \Omega') ``` Result: for $R_S=0$ a straight line of slope $q/kT$ (one decade per 59.6 mV). The series resistance pushes the curves to the right by $IR_S$, which becomes significant only when $IR_S\gtrsim kT/q$, i.e. $I\gtrsim0.0259/R_S$. With $R_S=10\ \Omega$ the curve bends over beyond $\sim3$ mA and at $I=10$ mA is displaced by 0.1 V; with $R_S=1\ \Omega$ the bend is near 30 mA; with $0.1\ \Omega$ near 0.3 A (beyond the plotted range, so almost indistinguishable from ideal). At high current the characteristic becomes an ohmic straight line $I\approx(V_A-V_{j})/R_S$. ### Problem 14.9 Minority carrier injection ratio **Problem.** Estimate $I_{DIFF}/I_{TE}$ for Si with $A^*=140$ A/cm$^2$K$^2$, $\Phi_B=0.72$ eV, $N_D=10^{16}$ cm$^{-3}$, $\tau_p=10^{-6}$ s, $T=300$ K. **Solution.** Both currents scale as $e^{qV_A/kT}$, so the ratio is bias independent and equals the ratio of the saturation current densities. Thermionic emission: $$J_{TE,S}=A^*T^2e^{-\Phi_B/kT}=(140)(300)^2e^{-0.72/0.0259}=(1.26\times10^{7})e^{-27.8}$$ $$e^{-27.8}=8.46\times10^{-13}\ \Rightarrow\ J_{TE,S}=1.07\times10^{-5}\ \text{A/cm}^2$$ Diffusion (equated to a wide-base $p^+$-$n$ diode): $$J_{DIFF,S}=q\frac{D_p}{L_p}\frac{n_i^2}{N_D}$$ Take $\mu_p\approx400$ cm$^2$/V-s at $N_D=10^{16}$ (assumption; the problem does not give it): $$D_p=0.0259(400)=10.4\ \text{cm}^2/\text{s},\quad L_p=\sqrt{D_p\tau_p}=\sqrt{(10.4)(10^{-6})}=3.22\times10^{-3}\ \text{cm}$$ Evaluate the factors: $\dfrac{D_p}{L_p}=\dfrac{10.4}{3.22\times10^{-3}}=3.2\times10^{3}$ cm/s and $\dfrac{n_i^2}{N_D}=\dfrac{10^{20}}{10^{16}}=10^{4}$ cm$^{-3}$; so $$J_{DIFF,S}=(1.6\times10^{-19})(3.2\times10^{3})(10^{4})=5.2\times10^{-12}\ \text{A/cm}^2$$ Ratio: $$\frac{I_{DIFF}}{I_{TE}}=\frac{5.2\times10^{-12}}{1.07\times10^{-5}}\approx\boxed{5\times10^{-7}}$$ So for roughly every two million majority electrons injected into the metal, only about one hole is injected into the semiconductor: the MS diode is truly majority-carrier. ### Problem 14.10 Schottky barrier lowering *Assumptions:* $N_D=10^{16}$ cm$^{-3}$ (OCR garbled; consistent with the Fig. 14.6 diode, $N_D\approx9.7\times10^{15}$) and $V_{bi}=0.6$ V (the value found in Exercise 14.4). **(a)** Surface field (from (14.7) at $x=0$ with $V_{bi}\to V_{bi}-V_A$): $$\mathcal E_s=\left[\frac{2qN_D(V_{bi}-V_A)}{K_s\varepsilon_0}\right]^{1/2}$$ $$\Delta\Phi_B=\left[\frac{q\mathcal E_s}{4\pi K_s\varepsilon_0}\right]^{1/2}\ \text{(V)}$$ Selected values: | $V_A$ (V) | $\mathcal E_s$ (V/cm) | $\Delta\Phi_B$ (V) | $I_S(V_A)/I_S(0)$ | |---|---|---|---| | 0 | $4.29\times10^4$ | 0.0229 | 1 | | $-1$ | $7.0\times10^4$ | 0.0292 | 1.27 | | $-10$ | $1.80\times10^5$ | 0.0469 | 2.5 | | $-50$ | $3.94\times10^5$ | 0.0693 | 6.0 | Sample computation at $V_A=-50$ V: $\mathcal E_s=\{3.2\times10^{-3}(50.6)/1.0443\times10^{-12}\}^{1/2}=3.94\times10^5$ V/cm; $\Delta\Phi_B=\{(1.6\times10^{-19})(3.94\times10^5)/[4\pi(1.0443\times10^{-12})]\}^{1/2}=0.0693$ V. ```matlab q=1.6e-19; eps=11.8*8.85e-14; ND=1e16; Vbi=0.6; Vt=0.0259; VA=linspace(-50,0,300); Es=sqrt(2*q*ND*(Vbi-VA)/eps); % V/cm dPhi=sqrt(q*Es/(4*pi*eps)); % V figure(1); plot(VA,dPhi); xlabel('V_A (V)'); ylabel('\Delta\Phi_B (V)'); ratio=exp((dPhi-dPhi(end))/Vt); % I_S(VA)/I_S(0) figure(2); plot(VA,ratio); xlabel('V_A (V)'); ylabel('I(V_A)/I(0)'); ``` **(b)** Since $I_S\propto e^{-\Phi_B/kT}=e^{-(\Phi_{B0}-\Delta\Phi_B)/kT}$, the reverse saturation current relative to its zero-bias value is $$\frac{I(V_A)}{I(V_A=0)}=\exp\!\left[\frac{\Delta\Phi_B(V_A)-\Delta\Phi_B(0)}{kT}\right]$$ (the factor $1-e^{qV_A/kT}\to1$ once $|V_A|\gtrsim0.1$ V). The values are in the table column above, from 1 at $V_A=0$ to 6.0 at $-50$ V. **(c)** $\Delta\Phi_B$ is small (tens of millivolts, a few percent of $\Phi_B$) but because of the exponential dependence on $\Phi_B/kT$ it makes the reverse current rise by a factor of several over the reverse range: the non-saturation seen in Fig. 14.6(b). The lowering grows as $(V_{bi}-V_A)^{1/4}$, so it rises fast at first and then slowly, consistent with the slow steady creep of the measured reverse current. ### Problem 14.11 $C$-$V$ data analysis **Problem.** Least-squares fit of Table P14.11 for $V_{bi}$, $N_D$, $\Phi_B$, with $A=1.5\times10^{-3}$ cm$^2$. In the table the $1/C^2$ column is in units of $10^{21}\ \text{F}^{-2}$ (this scaling is the one consistent with the slope used in Exercise 14.4). | $-V_A$ (V) | $1/C^2$ ($10^{21}$ F$^{-2}$) | |---|---| | 1.09 | 0.953 | | 2.08 | 1.494 | | 3.07 | 2.035 | | 4.06 | 2.579 | | 5.05 | 3.125 | | 6.04 | 3.673 | | 7.03 | 4.217 | | 8.02 | 4.763 | | 9.01 | 5.320 | | 10.00 | 5.890 | **Solution.** Per (14.30), $1/C^2=m(V_{bi}-V_A)=mV_{bi}+m(-V_A)$: a line $y=a+bx$ with $x=-V_A$, $b=m=2/(qK_s\varepsilon_0A^2N_D)$, $a=mV_{bi}$. Least-squares sums ($n=10$): $\bar x=5.545$, $\bar y=3.405$, $S_{xx}=\sum(x-\bar x)^2=80.86$, $S_{xy}=\sum(x-\bar x)(y-\bar y)=44.71$. $$b=\frac{S_{xy}}{S_{xx}}=0.5529\times10^{21}\ \text{F}^{-2}\text{V}^{-1}$$ $$a=\bar y-b\bar x=3.405-0.5529(5.545)=0.339\times10^{21}\ \text{F}^{-2}$$ $$V_{bi}=\frac ab=\frac{0.339}{0.5529}=0.613\ \text{V}$$ $$N_D=\frac{2}{qK_s\varepsilon_0A^2b}=\frac{2}{(1.6\times10^{-19})(1.0443\times10^{-12})(2.25\times10^{-6})(5.529\times10^{20})}=9.6\times10^{15}\ \text{cm}^{-3}$$ $$(E_c-E_F)_{FB}=0.56-0.0259\ln\frac{9.6\times10^{15}}{10^{10}}=0.56-0.357=0.203\ \text{eV}$$ $$\Phi_B=qV_{bi}+(E_c-E_F)_{FB}=0.613+0.203=\boxed{0.82\ \text{eV}}$$ Results: $V_{bi}=0.61$ V, $N_D=9.6\times10^{15}$ cm$^{-3}$, $\Phi_B=0.82$ eV, agreeing within a few percent with the graphical estimates of Exercise 14.4 ($0.6$ V, $9.7\times10^{15}$, $0.8$ eV). ```matlab VA = -[1.09 2.08 3.07 4.06 5.05 6.04 7.03 8.02 9.01 10.00]; y = [0.953 1.494 2.035 2.579 3.125 3.673 4.217 4.763 5.320 5.890]*1e21; c = polyfit(VA, y, 1); % y = c(1)*VA + c(2); slope is negative m = -c(1); Vbi = c(2)/m q=1.6e-19; eps=11.8*8.85e-14; A=1.5e-3; kT=0.0259; ND = 2/(q*eps*A^2*m) PhiB = Vbi + (0.56 - kT*log(ND/1e10)) ``` ### Problem 14.12 Linearly graded MS diode **Problem.** $N_D(x)=ax$ in the semiconductor ($a$ constant). (a) Derive $\rho,\mathcal E,V,W$. (b) How is $V_{bi}$ determined? (c) Find the junction capacitance. **(a) Solution.** Depletion approximation: $\rho=qN_D=qax$ for $0\le x\le W$, zero beyond. Poisson, with $\mathcal E(W)=0$: $$\frac{d\mathcal E}{dx}=\frac{qax}{K_s\varepsilon_0}$$ $$\int_{\mathcal E(x)}^{0}d\mathcal E'=\frac{qa}{K_s\varepsilon_0}\int_x^Wx'dx'$$ $$\boxed{\mathcal E(x)=-\frac{qa}{2K_s\varepsilon_0}\left(W^2-x^2\right)\quad 0\le x\le W}$$ (Parabolic, maximum magnitude $qaW^2/2K_s\varepsilon_0$ at the interface.) The potential with $V(W)=0$: $$\frac{dV}{dx}=-\mathcal E=\frac{qa}{2K_s\varepsilon_0}(W^2-x^2)$$ $$-V(x)=\frac{qa}{2K_s\varepsilon_0}\int_x^W(W^2-x'^2)dx'=\frac{qa}{2K_s\varepsilon_0}\left[W^2(W-x)-\frac{W^3-x^3}{3}\right]$$ $$\boxed{V(x)=-\frac{qa}{6K_s\varepsilon_0}\left(2W^3-3W^2x+x^3\right)=-\frac{qa}{6K_s\varepsilon_0}(W-x)^2(2W+x)}$$ Check: $(W-x)^2(2W+x)=2W^3-3W^2x+x^3$. At $x=0$, $V(0)=-(V_{bi}-V_A)$: $$V_{bi}-V_A=\frac{qaW^3}{3K_s\varepsilon_0}\ \Rightarrow\ \boxed{W=\left[\frac{3K_s\varepsilon_0}{qa}(V_{bi}-V_A)\right]^{1/3}}$$ **(b) Built-in voltage.** $V_{bi}=\dfrac1q[\Phi_B-(E_c-E_F)]$ where $(E_c-E_F)$ is the Fermi-level position in the neutral semiconductor *at the edge of the depletion region*, where $N_D=aW$ (the doping is not uniform, so there is no single bulk value): $$(E_c-E_F)\Big|_{x=W}=kT\ln\frac{N_c}{aW}\quad\text{(or }\tfrac{E_G}2-kT\ln\tfrac{aW}{n_i})$$ $$V_{bi}=\frac1q\left[\Phi_B-kT\ln\frac{N_c}{aW_{eq}}\right],\quad W_{eq}=\left[\frac{3K_s\varepsilon_0V_{bi}}{qa}\right]^{1/3}$$ This is a transcendental pair in $(V_{bi},W_{eq})$ (the same self-consistency as in the linearly graded $pn$ junction, Subsection 5.2.5); solve it by iteration: guess $V_{bi}$, compute $W_{eq}$, update $V_{bi}$ with the logarithm, repeat until convergence. (The log varies slowly, so it converges in a few passes.) **(c) Capacitance.** $C=K_s\varepsilon_0A/W$ per (14.28), so $$\boxed{C=K_s\varepsilon_0A\left[\frac{qa}{3K_s\varepsilon_0(V_{bi}-V_A)}\right]^{1/3}=A\left[\frac{qa(K_s\varepsilon_0)^2}{3(V_{bi}-V_A)}\right]^{1/3}}$$ Thus $1/C^3\propto(V_{bi}-V_A)$: a plot of $1/C^3$ versus $V_A$ is a straight line, whose slope gives $a$ and whose intercept gives $V_{bi}$. --- # Part II Supplement and Review > Part II was a long tour of junction devices: the $pn$ diode (static, a.c., transient), its special relatives (solar cells, LEDs, $p$-$i$-$n$, varactors), the BJT, the PNPN family, heterojunctions, and finally the MS contact. This closing section gives a reading list, a glossary to test yourself against, and three problem sets that require *integrated* knowledge of Chapters 5-11, with fully worked solutions. ## Alternative / Supplemental Reading List | Author(s) | Type (A = alternative, S = supplemental) | Level | Relevant chapters | |---|---|---|---| | Streetman | A | Undergraduate | 5-7, 11 | | Neamen | A | Undergraduate | 7-10 | | Tyagi | A/S | Advanced undergraduate, introductory graduate | 6-10, 12-14, 18 | | Navon | A | Undergraduate | 12 (PNPN) | | Sah | S | Undergraduate to graduate (variable) | 7 (BJT, HBT and PNPN) | | Yang | A/S | Undergraduate | 3 ($pn$ electrostatics), 6 (BJT technology), 7 (MS diodes) | Books: (1) D. H. Navon, *Semiconductor Microdevices and Materials*, Holt, Rinehart and Winston, 1986. (2) D. A. Neamen, *Semiconductor Physics and Devices, Basic Principles*, Irwin, 1992. (3) C. T. Sah, *Fundamentals of Solid-State Electronics*, World Scientific, 1991. (4) B. G. Streetman, *Solid State Electronic Devices*, 4th ed., Prentice Hall, 1995. (5) M. S. Tyagi, *Introduction to Semiconductor Materials and Devices*, Wiley, 1991. (6) E. S. Yang, *Microelectronic Devices*, McGraw-Hill, 1988. ## Figure Sources / Cited References (1) S. M. Sze, *Physics of Semiconductor Devices*, 2nd ed., Wiley, 1981. (2) C. T. Sah, R. N. Noyce and W. Shockley, "Carrier Generation and Recombination in P-N Junctions and P-N Junction Characteristics," *Proc. IRE*, 45, 1228 (1957). (3) D. K. Schroder, *Semiconductor Material and Device Characterization*, Wiley, 1990. (4) R. H. Kingston, "Switching Times in Junction Diodes and Junction Transistors," *Proc. IRE*, 42, 829-834 (1954). (5) R. H. Dean and C. J. Neuse, "A Refined Step-Recovery Technique for Measuring Minority Carrier Lifetimes and Related Parameters in Asymmetric p-n Junction Diodes," *IEEE Trans. Electron Devices*, ED-18, 151-158 (March 1971). (6) E. S. Yang, *Microelectronic Devices*, McGraw-Hill, 1988, p. 383. (7) M. A. Green et al., "Characterization of 23-Percent Efficient Silicon Solar Cells," *IEEE Trans. Electron Devices*, 37, 331-336 (Feb. 1990). (8) *Photovoltaics: Program Overview Fiscal Year 1992*, NREL, DOE/CH10093-190, March 1993. (9) M. G. Craford, "LEDs Challenge the Incandescents," *IEEE Circuits and Devices*, 8, 24-29 (Sept. 1992). (10) M. G. Craford, "Recent Developments in Light-Emitting-Diode Technology," *IEEE Trans. Electron Devices*, ED-24, 935-943 (July 1977). (11) BiCMOS technology, *Solid State Technology*, 35, 33 (Oct. 1992); see also the special series on BiCMOS processing edited by D. Gupta and M. T. Bohr, begun in *Solid State Technology*, June 1992. (12) M. Lundstrom, "III-V Heterojunction Bipolar Transistors," Ch. 1 in *Heterojunction Transistors and Small Size Effects in Devices*, ed. M. Willander, Studentlitteratur and Chartwell Bratt, 1992. (13) I. R. C. Post, P. Ashburn and G. R. Wolstenholme, "Polysilicon Emitters for Bipolar Transistors: A Review and Re-Evaluation of Theory and Experiment," *IEEE Trans. Electron Devices*, 39, 1717 (July 1992). (14) For MBE, consult the device fabrication references in the R1 mini-chapter. (15) S. Tiwari and D. J. Frank, "Empirical Fit to Band Discontinuities and Barrier Heights in III-V Alloy Systems" (title partly illegible in the OCR). (16) Navon, 1986, pp. 410-411. (17) Sze, 2nd ed., 1981, pp. 255-256. ## Review List of Terms Defining these in your own words is a rapid review of Part II. Concise definitions follow so you can check yourself. **Junction diode basics (1-17)** | # | Term | Meaning | |---|---|---| | 1 | wide-base diode | Quasineutral widths $\gg$ minority-carrier diffusion lengths; the classic ideal diode. | | 2 | narrow-base diode | Quasineutral widths $\ll$ diffusion lengths; minority profiles are straight lines, $I_0\propto1/W_n$ not $1/L_p$. | | 3 | ideal diode | Obeys $I=I_0(e^{qV_A/kT}-1)$: no R-G in depletion region, low-level injection, no other processes. | | 4 | law of the junction | Minority concentration at the depletion edge: $p_n(x_n)=p_{n0}e^{qV_A/kT}$. | | 5 | quasineutral region | Region outside the depletion region where $\rho\approx0$, $\mathcal E$ small. | | 6 | breakdown | Sudden large reverse current at $V_A=-V_{BR}$. | | 7 | impact ionization | A carrier gains enough energy from the field to knock a bound electron free, creating an electron-hole pair. | | 8 | avalanching | Repeated impact ionization chain reaction. | | 9 | multiplication factor | $M=I/I_0$, ratio of the multiplied current to the unmultiplied current. | | 10 | Zener process | Breakdown by direct band-to-band tunneling in very narrow junctions. | | 11 | tunneling | Quantum penetration of a thin barrier. | | 12 | conductivity modulation | Large injected carrier densities raise the conductivity of the lightly doped region (high-level injection). | | 13 | punch-through (narrow-base diode) | Depletion region reaches the end of the narrow region, so breakdown or current limiting occurs earlier. | | 14 | high-level injection | Injected minority concentration comparable to or exceeding the majority doping. | | 15 | quasistatic | Signals vary slowly enough that the device follows the d.c. characteristics. | | 16 | hyperabrupt | Doping profile with doping decreasing away from the junction, giving a steeper $C$-$V$. | | 17 | varactor | Diode designed as a voltage-variable capacitor. | **Capacitance and transients, optoelectronics (18-37)** | # | Term | Meaning | |---|---|---| | 18 | abrupt | Step doping profile. | | 19 | profiling | Deducing $N(x)$ from $C$-$V$ data ($1/C^2$ slope). | | 20 | junction capacitance | Depletion-layer capacitance, $C_J=K_s\varepsilon_0A/W$. | | 21 | diffusion admittance | Small-signal admittance from stored minority carriers in forward bias ($G_0$ and $C_D$). | | 22 | turn-off transient | Switching from forward to reverse bias. | | 23 | storage delay time ($t_s$) | Time during which stored charge is being removed and the current stays at $-I_R$. | | 24 | reverse recovery time | Total time until the current falls to a specified fraction of $I_R$. | | 25 | reverse injection | Minority carrier injection in the reverse direction when the stored charge flows back. | | 26 | step recovery diode | Diode with a very abrupt end of the storage phase, used for pulse sharpening. | | 27 | photodetector | Device that converts light to an electrical signal. | | 28 | solar cell | $pn$ junction delivering power from light in the fourth quadrant. | | 29 | LED | Forward-biased direct-gap diode emitting photons by radiative recombination. | | 30 | $p$-$i$-$n$ diode | Diode with an intrinsic layer between $p$ and $n$, wide depletion region, fast, low capacitance. | | 31 | avalanche photodiode | Photodiode operated near breakdown to multiply photocurrent. | | 32 | fill factor | $FF=P_{max}/(I_{SC}V_{OC})$, how "square" the solar cell's $I$-$V$ is. | | 33 | shadowing | Loss from the metal grid blocking light. | | 34 | texturing | Roughened surface that reduces reflection and traps light. | | 35 | concentrator solar cells | Cells used with lenses or mirrors to focus sunlight. | | 36 | isoelectronic trap | Impurity of the same valence that traps carriers and enhances radiative recombination in indirect-gap LED materials (e.g. N in GaP). | | 37 | total internal reflection | Light striking a boundary beyond the critical angle stays inside the high-index semiconductor, limiting LED extraction. | **BJT (38-68)** | # | Term | Meaning | |---|---|---| | 38 | BJT | Bipolar junction transistor. | | 39 | emitter, base, collector | The three regions: emitter injects carriers, base transports them, collector collects. | | 40 | common base | Base is common to input and output. | | 41 | common emitter | Emitter is common; $\beta_{dc}$ amplifier configuration. | | 42 | active mode | E-B forward, C-B reverse. | | 43 | saturation mode | Both junctions forward. | | 44 | cutoff mode | Both junctions reverse. | | 45 | inverted mode | E-B reverse, C-B forward (roles reversed). | | 46 | buried layer | Heavily doped layer beneath the collector to lower collector resistance in ICs. | | 47 | quasineutral base width | $W_B$ minus depletion encroachment. | | 48 | emitter efficiency | $\gamma$, fraction of emitter current that is carriers injected into the base. | | 49 | base transport factor | $\alpha_T$, fraction of injected carriers reaching the collector. | | 50 | common base d.c. gain | $\alpha_{dc}=\gamma\alpha_T$ (to first order). | | 51 | common emitter d.c. gain | $\beta_{dc}=\alpha_{dc}/(1-\alpha_{dc})$. | | 52 | performance parameters | $\gamma,\alpha_T,\alpha_{dc},\beta_{dc}$ and related figures of merit. | | 53 | Ebers-Moll equations/model | Superposition of two coupled diodes describing all modes. | | 54 | forward gain | Gain in the normal (active) direction. | | 55 | base-width modulation | Change of $W_B$ with collector voltage. | | 56 | Early effect | Output conductance caused by base-width modulation. | | 57 | punch-through (BJT) | Base fully depleted so the C-B depletion region touches the E-B region. | | 58 | regenerative | Positive feedback process (as in the SCR two-transistor model). | | 59 | phototransistor | BJT whose base current comes from light. | | 60 | intrinsic transistor | The active part of the BJT excluding parasitics. | | 61 | base series resistance | Resistance of the lateral base path. | | 62 | current crowding | Emitter current concentrates at the edge due to the base resistance drop. | | 63 | graded base | Base doping decreasing from emitter to collector, producing a built-in field that speeds carriers. | | 64 | Gummel plot | $\log I_C$ and $\log I_B$ versus $V_{BE}$. | | 65 | BiCMOS | Technology integrating bipolar and CMOS. | | 66 | shallow emitter | Very thin emitter for high-speed devices. | | 67 | polysilicon emitter | Polycrystalline Si emitter giving high $\gamma$. | | 68 | HBT | Heterojunction bipolar transistor. | **Heterojunctions, BJT dynamics, PNPN, MS contacts (69-100)** | # | Term | Meaning | |---|---|---| | 69 | heterojunction | Junction between two different semiconductors. | | 70 | band alignment | The relative positions of the band edges across a heterojunction. | | 71 | lattice-matched | Materials with equal lattice constants, giving defect-free interfaces. | | 72 | graded junction (HBT) | Composition gradually changed across the junction to remove band spikes. | | 73 | Hybrid-pi equivalent circuit | Small-signal BJT model with $r_\pi$, $g_m$, $C_\pi$, $C_\mu$, $r_o$. | | 74 | transconductance | $g_m=dI_C/dV_{BE}=qI_C/kT$. | | 75 | base transit time | Average time for carriers to cross the base, $\tau_B\approx W_B^2/2D_B$. | | 76 | rise, storage delay, fall times | The three stages of the BJT switching waveform. | | 77 | Schottky diode clamp | MS diode from base to collector preventing saturation. | | 78 | thyristors | Four-layer switching devices. | | 79 | SCR | Silicon controlled rectifier. | | 80 | anode, cathode, gate | The three SCR terminals. | | 81 | blocking voltages | Maximum forward and reverse voltages without conduction. | | 82 | two-transistor model | An SCR as a $pnp$ and an $npn$ connected in a regenerative loop. | | 83 | GTO SCR | Gate turn-off thyristor. | | 84 | LAS | Light-activated switch (light-triggered SCR). | | 85 | $di/dt$ burnout | Overheating from too fast a current rise at turn-on. | | 86 | $dv/dt$ effect | False triggering from fast voltage rise via capacitive charging current. | | 87 | DIAC, TRIAC | Bidirectional triggering diode and bidirectional triode thyristor. | | 88 | PUT | Programmable unijunction transistor. | | 89 | Schottky diode | Rectifying MS diode. | | 90 | vacuum level | Energy of a free electron at rest just outside the solid. | | 91 | metal work function | $\Phi_M=E_0-E_{FM}$. | | 92 | electron affinity | $\chi=E_0-E_c$ at the semiconductor surface. | | 93 | thermionic emission | Emission of carriers over a barrier due to thermal energy. | | 94 | Richardson's constant | $A^*=4\pi qm^*k^2/h^3$. | | 95 | Schottky barrier lowering | Reduction of $\Phi_B$ with field due to image force. | | 96 | hot carrier diode | Small-area, very fast Schottky diode. | | 97 | Fermi-level pinning | Surface states fix $E_F$ at the surface, making $\Phi_B$ nearly independent of the metal. | | 98 | field emission | Tunneling through a thin barrier. | | 99 | Al spiking | Al penetration through a shallow junction. | | 100 | silicide | Metal-silicon compound used for stable ohmic contacts. | --- ## Part II: Review Problem Sets and Answers These sets assume knowledge, at times integrated, of Chapters 5-11. Set A is adapted from a one-hour open-book exam; Sets B and C combine select problems from closed-book exams. Answers to Sets A and B are given at the end; Set C has no published answers (so it can serve as homework), but a worked solution is provided here for completeness. --- ## Problem Set A ### Problem A1 A $pn$ junction has the doping profile sketched: $N_D-N_A=-N_B$ for $x<0$ (equal magnitude on both sides), $0$ for $0\le x\le x_i$, and $+N_B$ for $x>x_i$. In the entire problem the carrier concentrations may be neglected ($n=0$, $p=0$) in the $0\le x\le x_i$ region. *(Figure described: $N_D-N_A$ versus $x$: a step down to $-N_B$ on the left, zero between $0$ and $x_i$, and $+N_B$ for $x\ge x_i$. This is effectively a $p$-$i$-$n$ structure with equal-magnitude doping and a thin "intrinsic" spacer.)* **(a) Built-in voltage?** Following Subsection 5.1.4, $V_{bi}=\frac{kT}{q}\ln\!\big[n(x_n)/n(-x_p)\big]$, the ratio of electron concentrations deep in the two neutral regions. Deep in the $n$-side, $n=N_B$; deep in the $p$-side, $n=n_i^2/N_B$. So $$V_{bi}=\frac{kT}{q}\ln\!\left[\frac{N_B}{n_i^2/N_B}\right]=\frac{kT}{q}\ln\frac{N_B^2}{n_i^2}=\frac{2kT}{q}\ln\frac{N_B}{n_i}$$ Justification: the built-in voltage depends only on the carrier concentrations in the neutral regions on each side (equivalently the Fermi-level position relative to the band edges there), not on the profile in between. **(b) Charge density (depletion approximation).** $\rho=-qN_B$ for $-x_p\le x\le0$; $\rho=0$ for $0\le x\le x_i$ (no dopants, no carriers); $\rho=+qN_B$ for $x_i\le x\le x_n$; $0$ elsewhere. The sketch is two equal-height rectangles of opposite sign separated by a gap of width $x_i$. **(c) Electric field.** Poisson: $d\mathcal E/dx=\rho/K_s\varepsilon_0$. With $\mathcal E(-x_p)=0$: For $-x_p\le x\le0$: $$\frac{d\mathcal E}{dx}=-\frac{qN_B}{K_s\varepsilon_0}\ \Rightarrow\ \mathcal E(x)=-\frac{qN_B}{K_s\varepsilon_0}(x+x_p)$$ At $x=0$: $\mathcal E(0)=-qN_Bx_p/K_s\varepsilon_0$. For $0\le x\le x_i$: $d\mathcal E/dx=0$, so the field is constant: $$\mathcal E(x)=\mathcal E(0)=-\frac{qN_Bx_p}{K_s\varepsilon_0}$$ For $x_i\le x\le x_n$ (integrate from $x$ to $x_n$ where $\mathcal E(x_n)=0$): $$\int_{\mathcal E(x)}^{0}d\mathcal E'=\frac{qN_B}{K_s\varepsilon_0}\int_x^{x_n}dx'\ \Rightarrow\ \mathcal E(x)=-\frac{qN_B}{K_s\varepsilon_0}(x_n-x)$$ The sketch: a negative-going straight line from zero at $-x_p$ down to $-qN_Bx_p/K_s\varepsilon_0$ at $x=0$, a flat plateau up to $x_i$, then a straight line rising back to zero at $x_n$. **(d) Relation between $x_n$ and $x_p$.** The positive and negative charge areas must be equal: $$(qN_B)x_p=(qN_B)(x_n-x_i)\ \Rightarrow\ x_p=x_n-x_i$$ Equivalently, continuity of $\mathcal E$ at $x=x_i$: $\dfrac{qN_Bx_p}{K_s\varepsilon_0}=\dfrac{qN_B(x_n-x_i)}{K_s\varepsilon_0}$, the same result. (In a standard step junction with $N_A=N_D$ this gives $x_p=x_n$; here the central region merely adds the spacer.) **(e) Equilibrium band diagram.** $E_F$ is flat. Far left: $E_F$ below $E_i$ by $kT\ln(N_B/n_i)$ ($p$-side); far right: $E_F$ above $E_i$ by the same amount. Bands curve (parabola, since $\mathcal E$ is linear) on $-x_p\le x\le0$, run as a *straight line* (constant slope, constant field) for $0\le x\le x_i$, and curve again (opposite concavity) on $x_i\le x\le x_n$, with total bending $qV_{bi}$. The only difference from a simple step junction ($N_A=N_D=N_B$) is the constant-slope straight-line segment in $0\le x\le x_i$. The diagram is symmetric about the midpoint of the spacer, since $(E_i-E_F)_{p\text{-side}}=(E_F-E_i)_{n\text{-side}}$ for $N_A=N_D=N_B$. ### Problem A2 Two Si $p^+$-$n$ diodes at 300 K are identical except for $n$-side doping: diode #1 has $N_D=10^{14}$ cm$^{-3}$, diode #2 has $N_D=10^{16}$ cm$^{-3}$. **(a) Larger $V_{bi}$? Diode #2.** $qV_{bi}=(E_i-E_F)_{p^+}+(E_F-E_i)_n$. The $p^+$ side term is the same for both; $(E_F-E_i)_n=kT\ln(N_D/n_i)$ is larger for the heavier-doped diode, so $V_{bi,2}>V_{bi,1}$. (Heavy doping on the $p^+$ side is a hint that $V_{bi}=(kT/q)\ln(N_AN_D/n_i^2)$ cannot be used blindly: the $p^+$ side may be degenerate, so the Fermi-level construction is the safe route.) **(b) Larger $V_{BR}$? Diode #1.** Avalanche breakdown on the lightly doped side scales roughly as $1/N_D$: $V_{BR}>1000$ V for $N_D=10^{14}$ cm$^{-3}$, but about 60 V for $10^{16}$ cm$^{-3}$ (typical Si values). **(c) Larger $C_J$ at a given reverse bias $|V_A|>V_{bi}$? Diode #2.** $C_J=K_s\varepsilon_0A/W$, with $$W=\left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi}-V_A)\right]^{1/2}\approx\left[\frac{2K_s\varepsilon_0}{qN_D}(-V_A)\right]^{1/2}\quad(|V_A|\gg V_{bi})$$ Thus $W_1>W_2$ (smaller $N_D$) and $C_2>C_1$. **(d) Larger $|I|$ if ideal, reverse bias beyond a few volts? Diode #1.** The ideal reverse current is $$I=-qA\frac{D_p}{L_p}\frac{n_i^2}{N_D}$$ Now $1/N_{D1}>1/N_{D2}$. Also $D_p=(kT/q)\mu_p$ and $\mu_p$ decreases with doping, so $D_{p1}>D_{p2}$; with equal $\tau_p$, $L_p=\sqrt{D_p\tau_p}$ so $D_p/L_p=\sqrt{D_p/\tau_p}$ is also larger for #1. Both factors favour $|I_1|>|I_2|$. **(e) Larger $|I|$ if the diodes are not ideal (real Si at 300 K, reverse bias beyond a few volts)? Diode #1.** The reverse current is dominated by R-G generation in the depletion region: $$I_{R-G}=-qA\frac{n_iW}{2\tau_0}$$ With $\tau_0\propto1/N_T$ (trap density, same in both diodes since only the doping differs), $\tau_0$ is the same. But $W_1>W_2$ from (c), so $|I_{R-G,1}|>|I_{R-G,2}|$. **(f) Larger diffusion capacitance at a given forward bias? Diode #1.** In the ideal region, $C_D\propto G_0\tau_p\propto(I+I_0)\tau_p/\dots$; more precisely $C_D=\tfrac12G_0\tau_p$ with $G_0=qI/kT$ at a *given forward bias*, and $G_0\propto I_0$, which is larger for diode #1 by (d). With equal $\tau_p$, $C_{D1}>C_{D2}$. **(g) Larger storage delay time $t_s$ for the same $I_F/I_R$? Same for both.** The approximate solution (Eq. 8.8) $t_s\approx\tau_p\ln(1+I_F/I_R)$ (or the more accurate (8.9)) depends only on $\tau_p$ and $I_F/I_R$. The $I_F/I_R$ ratio is given as the same; $N_T$, and hence $\tau_p$, is inferred to be the same in both diodes since only $N_D$ is different. Hence $t_{s1}=t_{s2}$. ### Problem A3 **(a) Inverted-mode carrier activity in a $pnp$ BJT.** The given diagram (active mode) shows holes injected from the heavily doped emitter ($p$) across the base to the collector. In the inverted mode the junction biases are interchanged: the original collector-base junction is *forward* biased and the original emitter-base junction is *reverse* biased. The diagram: Fermi level flat apart from the junction offsets; the "collector" (now acting as emitter, lightly doped $p$) injects holes into the base; they diffuse across the base, and are collected by the reverse-biased original emitter. The injected hole flow is therefore from the *right* ($C$) to the *left* ($E$) in the drawing. A caution to avoid misinterpretation: the carriers still diffuse "downhill" in concentration, but the roles are exchanged, and because the lightly doped collector is a poor injector, a large share of the forward-biased junction current is *electron* injection from base into the collector (a wasted, base-current component). **(b)(i) Why lower gain in backward connection.** The gain is limited by emitter efficiency $\gamma$. In the proper connection the E-B junction has $N_E\gg N_B$, so hole injection into the base dominates over electron injection into the emitter, $\gamma\approx1$. In the backward connection the injecting junction is the C-B junction, where $N_C<N_B$: electron injection from base into collector becomes significant, so the injection efficiency is far below that of the real emitter, degrading $\gamma$ and with it $\alpha_{dc}=\gamma\alpha_T$. **(b)(ii) Why greater sensitivity to base-width modulation.** The larger depletion region sits at the reverse-biased junction, extending mainly into the more lightly doped side. In the proper connection the reverse-biased C-B junction has $N_B\gg N_C$, so most of the depletion layer extends into the collector, only slightly altering $W_B$. In the backward connection, the reverse-biased junction is the E-B one with $N_E\gg N_B$, so most of the depletion region extends into the *base*. Much larger changes of $W_B$ result for equivalent bias changes. --- ## Problem Set B ### Problem B1 A $pn$ diode has $N_D-N_A=N_0\left[e^{\alpha x}-1\right]$ with constants $N_0,\alpha$ (negative for $x<0$, positive for $x>0$, vanishing at $x=0$ and saturating at $-N_0$ on the left). **(a) Depletion approximation, in a sentence.** (i) $n$ and $p$ are much less than $|N_D-N_A|$ in a region $-x_p\le x\le x_n$ about the metallurgical junction, so $\rho=q(N_D-N_A)$ there; (ii) $\rho=0$ elsewhere (neutral regions). **(b) Charge density.** $\rho=qN_0(e^{\alpha x}-1)$ in $-x_p\le x\le x_n$: negative for $x<0$ (value $-qN_0(1-e^{-\alpha x_p})$ at $x=-x_p$, saturating toward $-qN_0$ for large $\alpha x_p$), zero at $x=0$, and growing exponentially positive up to $x_n$; zero beyond $x_n$ and below $-x_p$ (abrupt edges). **(c) Electric field.** With $\mathcal E(-x_p)=0$: $$\frac{d\mathcal E}{dx}=\frac{qN_0}{K_s\varepsilon_0}\left(e^{\alpha x}-1\right)\quad-x_p\le x\le x_n$$ $$\mathcal E(x)=\frac{qN_0}{K_s\varepsilon_0}\int_{-x_p}^{x}\left(e^{\alpha x'}-1\right)dx'=\frac{qN_0}{K_s\varepsilon_0}\left[\frac{e^{\alpha x'}}{\alpha}-x'\right]_{-x_p}^{x}$$ $$\boxed{\mathcal E(x)=\frac{qN_0}{K_s\varepsilon_0}\left[\frac{e^{\alpha x}-e^{-\alpha x_p}}{\alpha}-(x+x_p)\right]}$$ **(d) Completing the electrostatics (organized steps).** *Step 1.* Solve for the potential in the depletion region using $dV/dx=-\mathcal E(x)$ with $V=0$ at $x=-x_p$: $$V(x)=-\int_{-x_p}^{x}\mathcal E(x')\,dx'$$ *Step 2.* Impose two boundary conditions at the right edge: $\mathcal E(x_n)=0$ and $V(x_n)=V_{bi}-V_A$. These two equations contain three unknowns, $x_n$, $x_p$ and $V_{bi}$ (for a given $V_A$); once $V_{bi}$ is known they can be solved simultaneously for $x_n$ and $x_p$, giving $W=x_n+x_p$. In this problem the solution must be obtained numerically (transcendental equations). *Step 3.* Obtain $V_{bi}$ by a procedure paralleling the linearly graded junction analysis of Subsection 5.2.5: $V_{bi}$ follows from the Fermi-level position in the neutral regions at the depletion-region edges (using $N_D-N_A$ evaluated at $x_n$ and $-x_p$), which depends on the unknown $x_n,x_p$, so Steps 2 and 3 are solved self-consistently. ### Problem B2 The band diagram of a Si step junction at room temperature (described, because it is not legible): the quasi-Fermi levels in the far regions are labelled so that $E_{Fp}-E_{Fn}$ is the applied reverse voltage, $E_i-E_{Fp}=E_G/4$ on the $p$-side and $E_{Fn}-E_i=E_G/4$ on the $n$-side in the neutral regions. Given: $x_n+x_p=2\times10^{-4}$ cm, $A=10^{-3}$ cm$^2$, $\tau_n=\tau_p=\tau_0=10^{-6}$ s, $\mu_n=1352$, $\mu_p=459$ cm$^2$/V-s, $K_s=11.8$, $\varepsilon_0=8.85\times10^{-14}$ F/cm, $n_i=10^{10}$ cm$^{-3}$, $kT/q=0.0259$ V. *Reading of the data:* the ratio of the OCR-ed quantities (e.g. the R-G current of $1.6\times10^{-10}$ A in the key) requires $\tau_0=10^{-6}$ s, which agrees with the lifetime used in part (d). **(a) Reverse bias magnitude.** The difference of the Fermi levels in the neutral regions equals $qV_A$: $$V_A=-\frac1q(E_{Fp}-E_{Fn})=-\frac{E_G}{2q}=-0.56\ \text{V}$$ Explanation: the diagram shows the $p$-side Fermi level above the $n$-side Fermi level by $E_G/2$, the separation of Fermi levels being the applied voltage across the diode ($E_{Fn}-E_{Fp}=qV_A$). The reverse bias is 0.56 V. **(b) Built-in voltage.** From the equilibrium positions, $qV_{bi}=(E_i-E_F)_p+(E_F-E_i)_n=\dfrac{E_G}{4}+\dfrac{E_G}{4}=\dfrac{E_G}{2}$: $$V_{bi}=\frac{E_G}{2q}=0.56\ \text{V}$$ **(c) R-G current (reverse bias, $W=x_n+x_p$):** $$I_{R-G}=-qA\frac{n_iW}{2\tau_0}=-\frac{(1.6\times10^{-19})(10^{-3})(10^{10})(2\times10^{-4})}{2(10^{-6})}=-1.6\times10^{-10}\ \text{A}$$ **(d) Diffusion current.** First the doping: $N_A=N_D=n_ie^{E_G/4kT}$: $$N_A=N_D=(10^{10})e^{1.12/[(0.0259)(4)]}=(10^{10})e^{10.81}=4.96\times10^{14}\ \text{cm}^{-3}$$ Diffusion constants and lengths: $$D_N=\frac{kT}{q}\mu_n=(0.0259)(1352)=35.02\ \text{cm}^2/\text{s},\quad L_N=\sqrt{D_N\tau_n}=\sqrt{(35.02)(10^{-6})}=5.92\times10^{-3}\ \text{cm}$$ $$D_P=(0.0259)(459)=11.89\ \text{cm}^2/\text{s},\quad L_P=\sqrt{(11.89)(10^{-6})}=3.45\times10^{-3}\ \text{cm}$$ (The widths $x_n,x_p\ll$ the quasineutral extent in a wide-base diode, so the wide-base formula applies.) Under reverse bias the ideal-diode current is the saturation current with a negative sign: $$I_{DIFF}=-qA\,n_i^2\left[\frac{D_N}{L_NN_A}+\frac{D_P}{L_PN_D}\right]$$ $$=-(1.6\times10^{-19})(10^{-3})(10^{20})\left[\frac{35.02}{(5.92\times10^{-3})(4.96\times10^{14})}+\frac{11.89}{(3.45\times10^{-3})(4.96\times10^{14})}\right]$$ Evaluate the bracket: $\dfrac{35.02}{2.936\times10^{12}}=1.193\times10^{-11}$ and $\dfrac{11.89}{1.711\times10^{12}}=6.95\times10^{-12}$. The prefactor is $qAn_i^2=(1.6\times10^{-19})(10^{-3})(10^{20})=1.6\times10^{-2}$. Combining: $$I_{DIFF}=-(1.6\times10^{-2})(1.888\times10^{-11})\ \text{A}=\boxed{-3.02\times10^{-13}\ \text{A}}$$ The diffusion current is some 500 times smaller than the R-G current, as is typical of Si at room temperature. **(e) Junction capacitance.** $$C_J=\frac{K_s\varepsilon_0A}{x_n+x_p}=\frac{(11.8)(8.85\times10^{-14})(10^{-3})}{2\times10^{-4}}=5.22\ \text{pF}$$ **(f) Minority carrier plot (linear scale).** Under reverse bias the minority concentrations at the depletion edges fall to essentially zero (law of the junction with $V_A\ll-kT/q$): $n_p(-x_p)\approx0$, $p_n(x_n)\approx0$. Far away they recover to the equilibrium values $n_p0=n_i^2/N_A=2.0\times10^5$ cm$^{-3}$ and $p_n0=n_i^2/N_D=2.0\times10^5$ cm$^{-3}$ over distances of order $L_N$ (left) and $L_P$ (right). Since $L_N>L_P$, the electron deficit on the $p$-side extends farther than the hole deficit on the $n$-side. The sketch: two exponential rises from near zero at the edges to the (tiny) equilibrium level, the left one stretched out longer. **(g) Significant diffusion capacitance?** No. $C_D$ comes from stored excess minority carriers adjacent to the depletion region; under reverse bias there is a *deficit*, not a store. **(h) Storage delay for a pulse to larger reverse bias?** No. The diode is already reverse biased before the pulse and has no stored minority-carrier charge to remove; the current merely adjusts quickly (with the $RC$ charging of $C_J$). ### Problem B3 Two Si $pnp$ BJTs are identical except $W_{B1}>W_{B2}$. $N_E\gg N_B>N_C$ and $W_B\le L_B$ in both, active mode at the same bias. **(a) Larger emitter efficiency? BJT #2.** With the quasineutral base width $W_{B1}>W_{B2}$ under the same biasing, $$\gamma=\frac{1}{1+\dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{W_B}{L_E}}$$ so $\gamma_1<\gamma_2$: a wider base gives a larger electron back-injection term, hence lower efficiency. **(b) Larger base transport factor? BJT #2.** $$\alpha_T=\frac{1}{\cosh(W_B/L_B)}\approx\frac{1}{1+\tfrac12(W_B/L_B)^2}$$ so $\alpha_{T1}<\alpha_{T2}$. Physically, fewer carriers recombine while crossing the narrower base. **(c) Larger $\beta_{dc}$? BJT #2.** $\beta_{dc}=\alpha_{dc}/(1-\alpha_{dc})$ increases monotonically with $\alpha_{dc}=\gamma\alpha_T$, and both factors are larger for #2. **(d) Greater sensitivity to base-width modulation? BJT #2.** A given bias change produces the same $\Delta W_B$ in both. Sensitivity is the *fractional* change $\Delta W_B/W_B$, which is larger for the narrower base: $\Delta W/W_{B2}>\Delta W/W_{B1}$. **(e) Larger punch-through voltage? BJT #1.** Punch-through occurs when the C-B depletion region consumes the whole base. For the same $\Delta W$ per volt, completely depleting the wider base of #1 requires a larger $V_{CB}$. If the maximum output current is limited by carrier multiplication and avalanching: **(f) Larger $V_{CBO}$? Same for both.** $V_{CBO}$ equals the avalanche breakdown voltage of the C-B junction, which depends on the junction dopings (identical in the two transistors). **(g) Larger $V_{CEO}$? BJT #1.** From Eq. (11.54), $$V_{CEO}=V_{CBO}\left(\frac{1}{\beta_{dc}+1}\right)^{1/m},\quad3\le m\le6$$ $\beta_{dc1}<\beta_{dc2}$ by (c), so $V_{CEO,1}>V_{CEO,2}$. --- ## Problem Set C (no published answers; worked solutions supplied) ### Problem C1 **(a) Which assumption is *not* invoked in deriving the ideal diode equation?** (iii) Narrow-base diode. The derivation assumes a *wide-base* diode (quasineutral widths much *greater* than the diffusion lengths); (i) no R-G in the depletion region, (ii) low-level injection, and (iv) no other processes are all invoked. **(b) Dominant current in most Si diodes at room temperature under reverse and small forward bias?** (ii) The R-G current. (The ideal diffusion current is far smaller in Si at 300 K because $n_i$ is small.) **(c) Which statement is incorrect?** (iv). Breakdown in a $pn$ diode is reversible provided the power is limited (so (i) is true), the Zener effect needs very narrow depletion regions $\lesssim10^{-6}$ cm (ii true), and $V_{BR}\propto1/N$ roughly for avalanching (iii true). Statement (iv) as printed, that avalanching dominates when $V_{BR}\le4.5$ V, is reversed: *below about 4.5 V (heavy doping, narrow depletion) the Zener (tunneling) process dominates*; avalanching dominates for larger breakdown voltages. (The OCR of (iv) is garbled; I reconstructed it from the usual textbook statement.) **(d) Which statement about the junction capacitance $C_J$ is correct?** The OCR dropped (i) and (ii). Statements (iii) "$C_J$ vanishes under forward biasing" and (iv) "$C_J$ is associated physically with fluctuations in the *minority* carrier concentrations at the edges of the depletion region" are both false: $C_J$ does not vanish in forward bias (it grows), and it is associated with fluctuations in the *majority*-carrier charge at the depletion edges. The correct statement is therefore one of the first two, which in the usual version reads "the magnitude of $C_J$ decreases with increasing reverse bias" (assumption, since it is the standard property: $C_J\propto(V_{bi}-V_A)^{-1/2}$). ### Problem C2 *Figure described and assumption stated.* The linear plot shows $p_n(x)$ in two $p^+$-$n$ diodes with the same $N_D$ and $A$: both curves start at the depletion-edge value $p_n(x_n)$ (marked $p_n(x_n)=n_i^2/N_D\,e^{qV_A/kT}$, above the equilibrium level $n_i^2/N_D$) and decay with the same diffusion length to the same equilibrium level; Diode A has the higher edge value. Low-level injection prevails. **(a)** (i) forward biased: the minority concentration at the edge exceeds the equilibrium value $p_{n0}$ (law of the junction), which happens only under forward bias. **(b)** (iii) smaller than: $p_n(x_n)=p_{n0}e^{qV_A/kT}$, and Diode B's edge value is lower, so $V_{AB}<V_{AA}$. **(c)** (iii) significantly smaller than: the current is proportional to the slope at the edge, $I\propto(p_n(x_n)-p_{n0})$, which is lower for B (exponentially smaller for a slightly smaller bias). **(d)** (ii) roughly the same as: $V_{BR}$ depends mostly on the (equal) $N_D$ on the lightly doped side, not on the applied forward bias. **(e)** (iii) essentially the same for both: with identical $\tau_p$ (same decay length $L_p$ implies same $\tau_p$) and the same $I_F/I_R$, $t_s\approx\tau_p\ln(1+I_F/I_R)$ is the same. ### Problem C3 *Assumed figure.* A log plot of carrier concentrations versus $x$: flat majority levels $p_p$ and $n_n$ in the neutral regions; the minority levels $n_p$ and $p_n$ *rise above* their equilibrium values $n_i^2/N_A$ and $n_i^2/N_D$ at the depletion-region edges $-x_p$ and $x_n$ and decay away from the junction; all minority values remain far below the majority levels. **(a) Forward or reverse biased?** Forward biased: the minority concentrations at the depletion edges are *larger* than the equilibrium values (excess carriers, piled up), as predicted by the law of the junction $p_n(x_n)=p_{n0}e^{qV_A/kT}$ with $V_A>0$. In reverse bias they would fall below equilibrium. **(b) Low-level injection?** Yes: even the elevated minority concentrations $p_n(x_n)$, $n_p(-x_p)$ are far below the majority levels $n_n$, $p_p$ on the log plot, so the majority concentrations are not perturbed. **(c) Link between minority pile-up and $C_D$.** The diffusion capacitance measures the *change in stored minority carrier charge* with applied voltage, $C_D=dQ_{stored}/dV_A$. A larger pile-up near the edge means more stored charge, and since the pile-up grows exponentially with $V_A$, $C_D$ is proportional to the stored charge (about $\tfrac12G_0\tau_p$). **(d) Link between the pile-up and $t_s$.** At turn-off, the stored excess minority carriers must be removed (swept out by the reverse current and by recombination) before the junction can become reverse biased. The larger the store, the longer the current stays pinned at $-I_R$: $t_s$ grows with the stored charge (with $\tau_p$ and $I_F/I_R$). ### Problem C4 *Setup.* Given sketched characteristics of a $p^+$-$n$ Si step junction at room temperature: reverse $I$-$V_A$ (small, slowly rising magnitude, R-G dominated), $C$-$V_A$ (decreasing as $(V_{bi}-V_A)^{-1/2}$), and the turn-off current $i(t)$ (plateau at $-I_R$ for $t_s$ then decay). Draw dashed curves on each (verbal description of what to draw): **(a-c) $N_D$ doubled.** (a) $I$-$V_A$: reverse current is dominated by R-G, $|I_{R-G}|\propto W\propto N_D^{-1/2}$, so the dashed curve lies lower by a factor of about $1/\sqrt2\approx0.71$ and keeps the same slow rise. (The ideal saturation component would also fall, roughly by 2, but it is negligible.) (b) $C$-$V_A$: $C\propto N_D^{1/2}$, so the dashed curve is higher by about $\sqrt2$, with the same shape (a slight additional change through $V_{bi}$). (c) $i(t)$: **no effect** to first order; $t_s\approx\tau_p\ln(1+I_F/I_R)$ depends on $\tau_p$ and $I_F/I_R$, which are unchanged. **(d-f) $\tau_p$ and $\tau_0$ doubled.** (d) $I$-$V_A$: $I_{R-G}=-qAn_iW/2\tau_0$, so the dashed reverse-current curve is **lower by a factor of 2** (the diffusion current changes only as $1/\sqrt{\tau}$, negligible in comparison). (e) $C$-$V_A$: **no effect** ($C_J$ depends only on $N_D$, $V_{bi}$ and $V_A$). (f) $i(t)$: the storage plateau at $-I_R$ becomes longer: $t_s\approx\tau_p\ln(1+I_F/I_R)$ doubles (the plateau level $-I_R$ and $I_F$ are set by the circuit, so they stay the same), and the subsequent recovery tail also lasts longer. # Part III: Field Effect Devices (Opener) Parts I and II of the book built the physics of semiconductors and of junctions. Part III now turns to a different way of controlling current: instead of injecting carriers across a junction (as in a BJT), we reach into a slab of semiconductor with an **electric field** and squeeze or widen the conducting path. Think of a garden hose and your foot. The hose is the semiconductor, the water is current, and your foot, pressing from the outside without entering the hose, is the gate. This is the entire idea of the field effect, and every MOSFET in every processor you have ever used is a refinement of it. --- # Chapter 15: Field Effect Introduction: The J-FET and MESFET ## 15.1 General Introduction ### The idea is older than the transistor you know The **field effect** is the modulation of the conductivity of a semiconductor by an electric field applied normal (perpendicular) to the semiconductor surface. Here is the surprising history. Field effect transistors predate the bipolar junction transistor by about 20 years. In patents filed in the 1920s and 1930s, J. E. Lilienfeld (USA) and O. Heil (Germany) independently conceived the structure of Fig. 15.1: a semiconductor (Cu$_2$S) bar with ohmic contacts A and B at the ends and a metal plate (aluminum) lying over it. A voltage on the plate changes the conductance of the bar underneath, and therefore the current between A and B. **Figure 15.1 (described):** Lilienfeld's idealization. A slab of semiconductor with two metal end contacts A and B, and a metal plate C resting on top of the slab. The plate-to-semiconductor voltage modulates the conductance between A and B. Why did it not work then? It was an idea ahead of its time. Good semiconductor material did not exist, and the technology was immature. Practical field-effect devices had to wait for the success of the bipolar transistor in the late 1940s and early 1950s. ### A timeline of field-effect devices | Year | Event | |---|---| | 1920s-30s | Lilienfeld and Heil propose the field-effect transistor | | 1952 | Shockley proposes and analyzes the junction FET (J-FET) | | 1953 | Dacey and Ross build operational J-FETs | | 1960 | Kahng and Atalla report the planar MOSFET (thermal SiO$_2$ gate insulator, inversion-layer channel) | | 1964 | Commercial MOSFETs from Fairchild and RCA | | late 1960s | Dennard invents the one-transistor DRAM cell | | early 1970s | Charge-coupled device (CCD) introduced | | 1980s | CMOS becomes the integrated-circuit technology of choice | | later | MODFET (modulation-doped FET), fastest three-terminal semiconductor device | **The J-FET.** Shockley's J-FET replaced the Lilienfeld metal plate by $pn$ junctions (Fig. 15.2). The end contacts A and B are renamed **source** and **drain**, and the field-effect electrode is the **gate**. Here it is the depletion regions of the $pn$ junctions that directly modulate the conductance between source and drain. **The MOSFET.** The planar MOSFET (Fig. 15.3) has a thermally grown SiO$_2$ gate insulator, a surface-inversion channel, and source/drain islands doped opposite to the substrate. **The DRAM cell.** A 1-T DRAM cell (Fig. 15.4) is a storage element (a capacitor or $pn$ junction) plus a MOSFET acting as a switch. A review article quoted in the book says the 1-T DRAM cell "is now the most abundant man-made object on this planet earth." Fig. 15.4 shows a planar cell: a polysilicon capacitor plate, an access transistor with gate oxide, an $n^+$ drain, field oxide, and a field implant in $p$-silicon. **The CCD.** A CCD can be viewed as a MOSFET with a segmented gate (Fig. 15.5, three-phase gates $P_1, P_2, P_3$, input and output gates and diodes on $p$-silicon). Biasing the gates in sequence pushes stored charge packets along the surface channel to the output. The image sensor in a camcorder is a CCD with a two-dimensional gate array: charge is generated in proportion to the optical image, then shifted out and converted to an electrical signal. **CMOS.** *Complementary* MOS puts both $n$-channel (electron-carrying) and $p$-channel (hole-carrying) MOSFETs on the same chip. Fig. 15.6 shows a cut-away CMOS inverter: an $n$-channel MOSFET (with $n^+$ source/drain in a $p$ region) beside a $p$-channel MOSFET ($p^+$ source/drain in an $n$ well), with thermal and CVD SiO$_2$ and Al interconnect. CMOS won in the 1980s because of low power dissipation. The change was evolutionary: CMOS existed in the 1960s and was even mass produced in LED watches by 1972. **The MODFET.** GaAs has better electron transport than Si but lacks a good native insulator, and deposited insulators on GaAs give poor devices. Molecular beam epitaxy (MBE) can grow lattice-matched layers of different semiconductors. The alloy AlGaAs has a larger band gap than GaAs and takes over the role of "insulator." In the MODFET (Fig. 15.7) an $n$-AlGaAs layer sits on undoped GaAs, and a two-dimensional electron gas (2 DEG) forms at the interface. It is touted as the fastest three-terminal semiconductor device and is used at high frequencies. ### Roadmap of Part III (Part III is five chapters) - **Chapter 15** (this chapter): the J-FET and the closely related MESFET. They act as conceptual bridges between the junction devices of Part II and the "pure" field-effect devices. They also serve as a first pass through terminology and analytical methods. - **Chapters 16-17:** MOS fundamentals. The MOS capacitor (MOS-C) is the structural heart of all MOS devices. Chapter 16 covers the ideal MOS-C and its $C$-$V$ behavior. Chapter 17 covers the long-channel enhancement-mode MOSFET. - **Chapters 18-19:** deviations from the ideal (Ch. 18) and small-dimension effects plus a survey of other FETs (Ch. 19). > **Intuition check.** Why is the J-FET a "bridge"? It still contains $pn$ junctions (like Part II) but the *current* is controlled by a field acting across a depletion region (like Part III). --- ## 15.2 J-FET ### 15.2.1 Introduction The cross section of the basic structure is Fig. 15.8. A central $n$ region (the channel) connects the source (S) and drain (D). Heavily doped $p^+$ gates sit on top and bottom (both labelled G). A gate voltage changes the $pn$-junction depletion widths and the associated electric field normal to the semiconductor surface. The width changes modulate the conductance between the ohmic source and drain contacts. **Why "unipolar"?** The J-FET was first called the *unipolar transistor*. Normal operation involves only one type of carrier, here electrons in the $n$ region moving from source to drain. (A BJT needs both electrons and holes.) **Naming the terminals.** - **Source:** carriers enter the semiconductor from the external circuit here. - **Drain:** carriers leave the semiconductor ("are drained") here. - **Gate:** gets its name from its controlling, "gating" action. The modern epilayer J-FET (Fig. 15.9) has an $n$-epi layer on a $p$-Si substrate, $p^+$ gate diffusions, $n^+$ source and drain contacts, SiO$_2$ passivation and isolation. Its appearance differs, but it is functionally equivalent to Shockley's structure. ### 15.2.2 Qualitative Theory of Operation **Standard biasing.** For the symmetrical Shockley structure (Fig. 15.10) with the top and bottom gates tied together: $$V_G \le 0, \qquad V_D \ge 0$$ - $V_G\le 0$ means the $p^+$-$n$ junctions are always zero or reverse biased, so the gate draws almost no current. - $V_D\ge 0$ pulls electrons from source to drain, consistent with the names S and D. Our plan: **change the terminal voltages one at a time and look inside the device.** #### Case 1: $V_G=0$, raise $V_D$ from 0 (Fig. 15.11) **(a) Equilibrium, $V_D=0$.** The device is in thermal equilibrium. All one sees are small depletion regions around the two $p^+$-$n$ junctions, extending mainly into the lightly doped $n$ side. **(b) Small $V_D$.** A current $I_D$ flows through the nondepleted $n$ region between the two junctions. That conducting region is the **channel**. For small $V_D$ it looks like a plain resistor, so $I_D$ varies linearly with $V_D$ (Fig. 15.12a). **(c)-(d) Moderate $V_D$ (a few tenths of a volt and up).** Imagine $V_D=5$ V, source grounded. Along the channel the potential must pass through 1, 2, 3, 4 V going from source to drain. The $p^+$ gate sides are held at 0 V. So the junction reverse bias grows from the source end toward the drain end. The depletion regions widen progressively along the channel and the channel funnels down (Fig. 15.11d). This is still a resistor, but no longer a simple one: less conducting volume means more resistance, so the slope of the $I_D$-$V_D$ curve decreases (Fig. 15.12b). > **Analogy.** A river flowing between two banks that slowly squeeze inward. The more the banks pinch near the end, the harder it is to push additional water through. **(e) Pinch-off.** Keep raising $V_D$ and the top and bottom depletion regions finally touch near the drain (Fig. 15.11e). This is **pinch-off**: complete depletion of the channel at one point. The slope of $I_D$-$V_D$ becomes approximately zero (Fig. 15.12c). The drain voltage at which this happens is named $V_{Dsat}$. For $V_D>V_{Dsat}$ the characteristic **saturates** at $I_{Dsat}$. **(f) Post pinch-off, $V_D>V_{D sat}$.** A short depleted section of length $\Delta L$ appears next to the drain. #### Two paradoxes that confuse everyone **Paradox 1: Doesn't pinch-off kill the current?** Suppose $I_D=0$ at pinch-off. Then no current flows anywhere in the channel, so there is no voltage drop along it, so the channel potential is zero everywhere. Then the $pn$ junctions are zero biased and the channel is wide open, contradicting the assumption of a pinched channel. So **current must flow to create and maintain pinch-off.** The hidden misconception is that a large current cannot pass through a depletion region. Depletion regions are *not* empty of carriers. Their carrier densities are merely small relative to the doping and can still reach $\sim 10^{12}$ cm$^{-3}$ or more. Large currents pass through depletion regions all the time: the forward-biased diode and both depletion regions of a BJT. **Paradox 2: Why doesn't $I_D$ change for $V_D>V_{Dsat}$?** As $V_D$ rises above $V_{Dsat}$, a section $\Delta L$ next to the drain becomes depleted. The voltage on its drain side is $V_D$ and on its source side is $V_{Dsat}$. So the extra voltage $V_D-V_{Dsat}$ is dropped across the depleted section. If $\Delta L\ll L$ (the usual case), the source-to-pinch-off portion has the **same shape and the same end-point voltages** (0 and $V_{Dsat}$) as at the onset of saturation. Same shape plus same applied voltage means same current. So $I_D$ stays about constant. If $\Delta L$ is comparable to $L$, the same $V_{Dsat}$ appears across a shorter section $L-\Delta L$, and $I_D$ rises noticeably with $V_D$. This is a short-channel effect (channel-length modulation). > **The waterfall analogy (Fig. 15.13).** The flow over a waterfall is set not by the height of the fall but by the flow rate in the rapids leading to it. Two falls of different heights fed by identical rapids deliver identical flow. Rapids = source side of the channel. Fall = the pinched $\Delta L$ section. Height of the fall = the potential drop $V_D-V_{Dsat}$. #### Case 2: $V_G<0$ (Fig. 15.14) Similar to $V_G=0$ with three modifications: 1. The junctions are reverse biased even at $V_D=0$, so depletion regions are wider and the channel is thinner. Channel resistance is higher and the linear-region slope is smaller (Fig. 15.14a). 2. A thinner channel pinches off at a smaller drain bias, so $V_{Dsat}$ and $I_{Dsat}$ are smaller than for $V_G=0$ (Fig. 15.14b). 3. For sufficiently negative $V_G$ the entire channel is depleted even at $V_D=0$ (Fig. 15.14c). The gate voltage where this first happens is the **pinch-off gate voltage**, $V_G=V_P$. For $V_G\le V_P$, $I_D=0$ for all drain biases. *Footnote from the text:* at very large $V_D$ the junctions near the drain break down and $I_D$ rises abruptly. This is omitted from all sketched characteristics. ### 15.2.3 Quantitative $I_D$-$V_D$ Relationships **Wanted:** $I_D=I_D(V_D,V_G)$. **Device specification (Fig. 15.15).** - $y$-axis runs down the channel from source ($y=0$) to drain ($y=L$). - $x$-axis is normal to the $p^+$-$n$ metallurgical junctions. - $L$ = channel length, $Z$ = lateral width, $2a$ = separation between the top and bottom metallurgical junctions. - $V(y)$ = channel potential at $y$. $W(y)=W_{top}(y)=W_{bottom}(y)$ = depletion width, lying almost entirely in the $n$ side because the junctions are $p^+$-$n$. **Basic assumptions.** 1. $p^+$-$n$ step junctions, uniformly doped $n$ region with donor concentration $N_D$. 2. Structure symmetric about the plane $x=a$ (the same $V_G$ on both gates). 3. Current flows only in the nondepleted $n$ region, exclusively in the $y$ direction. 4. $W(y)$ can reach at least $a$ without junction breakdown. 5. $L\gg a$. Valid range: $0\le V_D\le V_{Dsat}$ and $0\ge V_G\ge V_P$. #### Step 1: Current density in the channel The general electron current density is $$J_N = q\mu_n n\,\mathcal{E} + qD_N\nabla n \tag{15.1}$$ Inside the channel $n\approx N_D$ (uniform), so the diffusion term is negligible, and the field is $\mathcal{E}_y=-dV/dy$: $$J_{N_y}=q\mu_n N_D\,\mathcal{E}_y=-q\mu_n N_D\frac{dV}{dy} \tag{15.2}$$ #### Step 2: Integrate over the cross section No carriers are created or lost, so the current through any cross section equals $I_D$. $I_D$ is defined positive in the $-y$ direction, hence the minus sign. Using the symmetry about $x=a$ (two halves, $x$ from $W$ to $a$): $$I_D=-\int J_{N_y}\,dA = -Z\int_{W(y)}^{2a-W(y)}J_{N_y}\,dx = 2Zq\mu_nN_D\frac{dV}{dy}\int_{W(y)}^{a}dx \tag{15.3a}$$ $$I_D=2qZ\mu_nN_D\,[a-W(y)]\frac{dV}{dy} \tag{15.3b}$$ #### Step 3: Separate variables and integrate over the channel length $I_D$ is independent of $y$, so write $I_D\,dy=2qZ\mu_nN_D[a-W]\,dV$ and integrate from $y=0$ ($V=0$) to $y=L$ ($V=V_D$): $$\int_0^L I_D\,dy = I_DL = 2qZ\mu_nN_D\int_{V(0)=0}^{V_D}[a-W(V)]\,dV \tag{15.4}$$ $$I_D=\frac{2qZ\mu_nN_Da}{L}\int_0^{V_D}\left[1-\frac{W(V)}{a}\right]dV \tag{15.5}$$ #### Step 4: The gradual channel approximation The true electrostatics are two-dimensional (Poisson in $x$ and $y$). Because $L\gg a$, variations along $y$ are far more gradual than those along $x$. The **gradual channel approximation** neglects $\partial/\partial y$ in Poisson's equation, giving a one-dimensional $x$-direction solution at each $y$. Thus, using the one-sided step-junction result (Eq. 5.38), $$W(V)=\left[\frac{2K_S\varepsilon_0}{qN_D}\,(V_{bi}-V_A)\right]^{1/2}=\left[\frac{2K_S\varepsilon_0}{qN_D}\,(V_{bi}+V-V_G)\right]^{1/2} \tag{15.6}$$ Here $V_A=V_G-V(y)$ is the applied voltage across the junction at $y$ (negative means reverse bias). #### Step 5: Pinch-off voltage in terms of device parameters $W\to a$ when $V_D=0$ and $V_G=V_P$ (that is how $V_P$ is defined). Putting $V=0$, $V_G=V_P$ in (15.6): $$a=\left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}-V_P)\right]^{1/2} \tag{15.7}$$ Equivalently $V_P=V_{bi}-\dfrac{qN_Da^2}{2K_S\varepsilon_0}$. Dividing (15.6) by (15.7): $$\frac{W(V)}{a}=\left[\frac{V_{bi}+V-V_G}{V_{bi}-V_P}\right]^{1/2} \tag{15.8}$$ #### Step 6: Do the integral Let $c\equiv V_{bi}-V_P$. Substitute (15.8) into (15.5): $$\int_0^{V_D}\left[1-\frac{(V_{bi}-V_G+V)^{1/2}}{c^{1/2}}\right]dV = V_D-\frac{1}{c^{1/2}}\cdot\frac{2}{3}\Big[(V_{bi}-V_G+V)^{3/2}\Big]_0^{V_D}$$ $$= V_D-\frac{2}{3}\,\frac{(V_D+V_{bi}-V_G)^{3/2}-(V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}$$ So, with $G_0\equiv 2qZ\mu_nN_Da/L$: $$\boxed{I_D=G_0\left\{V_D-\frac{2}{3}\,\frac{(V_D+V_{bi}-V_G)^{3/2}-(V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}\right\},\quad 0\le V_D\le V_{Dsat},\ V_P\le V_G\le 0} \tag{15.9}$$ $G_0$ has a clear physical meaning: it is the channel conductance you would have if there were **no depletion regions** (full thickness $2a$, so $G=q\mu_nN_D\,(2a)Z/L$). Check: for small $V_D$ expand $(V_D+x)^{3/2}\approx x^{3/2}+\tfrac32x^{1/2}V_D$ with $x=V_{bi}-V_G$, giving $I_D\approx G_0V_D\left[1-\sqrt{(V_{bi}-V_G)/(V_{bi}-V_P)}\right]$. That is a resistor whose conductance shrinks as $V_G$ becomes more negative, and vanishes at $V_G=V_P$. This is the linear region of Fig. 15.14a. #### Step 7: Saturation Equation (15.9) is valid only **below** pinch-off. If you stubbornly plug in $V_D>V_{Dsat}$, the computed $I_D$ actually *decreases*. Since $I_D$ is really about constant above pinch-off, model it as $$I_D\big|_{V_D>V_{Dsat}}=I_{Dsat}=I_D\big|_{V_D=V_{Dsat}} \tag{15.10a, b}$$ At the drain end pinch-off means $W\to a$ at $V(L)=V_{Dsat}$. From (15.6): $$a=\left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}+V_{Dsat}-V_G)\right]^{1/2} \tag{15.11}$$ Comparing with (15.7), $V_{bi}+V_{Dsat}-V_G=V_{bi}-V_P$, so $$\boxed{V_{Dsat}=V_G-V_P} \tag{15.12}$$ (Remember $V_P<0$, so $V_{Dsat}>0$ for $V_G\ge V_P$. At $V_G=V_P$, $V_{Dsat}=0$: no current. At $V_G=0$, $V_{Dsat}=|V_P|$.) Substituting $V_D=V_G-V_P$ into (15.9): $(V_D+V_{bi}-V_G)^{3/2}=(V_{bi}-V_P)^{3/2}$, so $$I_{Dsat}=G_0\left\{(V_G-V_P)-\frac{2}{3}(V_{bi}-V_P)\left[1-\left(\frac{V_{bi}-V_G}{V_{bi}-V_P}\right)^{3/2}\right]\right\} \tag{15.13}$$ where the bracket arises because $(V_{bi}-V_P)^{3/2}/(V_{bi}-V_P)^{1/2}=V_{bi}-V_P$ and $(V_{bi}-V_G)^{3/2}/(V_{bi}-V_P)^{1/2}=(V_{bi}-V_P)\left(\tfrac{V_{bi}-V_G}{V_{bi}-V_P}\right)^{3/2}$. **Comparison with experiment.** Fig. 15.16 plots normalized theory ($V_{bi}=1$ V, $V_P=-2.5$ V; $V_G=0,-0.5,-1.0,-1.5,-2.0$ V; the locus $I_{Dsat}$ vs. $V_{Dsat}$ is a dashed line through the knee of every curve). Fig. 15.17 shows measured curves for a TI 2N3823 $n$-channel J-FET ($I_D$ up to 5 mA, $V_D$ to 5 V). Theory models experiment reasonably well. Better agreement follows from including the series resistance of the regions between the active channel and the contacts (Fig. 15.18, $R_S$ and $R_D$; see Exercise 15.3). **Square-law approximation.** Most J-FETs in saturation obey $$I_{Dsat}=I_{D0}\left(1-\frac{V_G}{V_P}\right)^2,\qquad I_{D0}=I_{Dsat}\big|_{V_G=0} \tag{15.14}$$ It looks completely different from (15.13) but gives similar numbers and is easier for first-order circuit work (device as a "black box"). Eq. (15.13) remains indispensable for studying dependence on temperature, doping, or other fundamental parameters. --- ### Exercise 15.1: Maximum half-width $a$ **Problem.** An $n$-channel Si J-FET has $N_D=10^{16}$ cm$^{-3}$. At room temperature, find the maximum half-width $a$. **Solution.** For saturating characteristics the channel must pinch off before the junctions break down, i.e. $a\le W_{BR}$ (depletion width at the breakdown voltage). From Fig. 6.11, $V_{BR}\approx 55$ V for $N_D=10^{16}$ cm$^{-3}$ $p^+$-$n$ junctions, and $V_{bi}=0.92$ V (from the Exercise 5.1 figure). $$a_{max}=W_{BR}=\left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}+V_{BR})\right]^{1/2}$$ $$=\left[\frac{(2)(11.8)(8.85\times10^{-14})}{(1.6\times10^{-19})(10^{16})}(0.92+55)\right]^{1/2}\ \text{cm}$$ $$=\left[7.30\times10^{-8}\right]^{1/2}\ \text{cm}=2.70\times10^{-4}\ \text{cm}$$ $$\boxed{a_{max}=2.70\ \mu\text{m}}$$ ### Exercise 15.2: Phenomenological versus square-law $I_{Dsat}/I_{D0}$ **Problem.** Plot $I_{Dsat}/I_{D0}$ vs. $V_G/V_P$ using (15.13) and (15.14) for $V_{bi}/V_P=-0.4$ (i.e. $V_{bi}=1$ V, $V_P=-2.5$ V). How does the comparison vary with $V_{bi}/V_P$? **Solution.** Setting $V_G=0$ in (15.13): $$I_{D0}=G_0\left\{-V_P-\frac{2}{3}(V_{bi}-V_P)\left[1-\left(\frac{V_{bi}}{V_{bi}-V_P}\right)^{3/2}\right]\right\}$$ Divide numerator and denominator of the ratio by $V_P$ (which is negative, but cancels in the ratio). Define $x=V_G/V_P$ and $z=V_{bi}/V_P$: $$\frac{I_{Dsat}}{I_{D0}}=\frac{(x-1)-\frac{2}{3}(z-1)\left[1-\left(\frac{z-x}{z-1}\right)^{3/2}\right]}{-1-\frac{2}{3}(z-1)\left[1-\left(\frac{z}{z-1}\right)^{3/2}\right]}$$ The square-law ratio is simply $$\frac{I_{Dsat}}{I_{D0}}=(1-x)^2$$ Both equal 1 at $x=0$ and 0 at $x=1$. The two curves agree well in between. Changing $z$ in the program shows that **agreement improves as $|V_{bi}/V_P|$ decreases** (the book states agreement improves with increasing $|V_P/V_{bi}|$: the $V_{bi}$ corrections become small and the exact expression approaches the square law). MATLAB program: ```matlab % Exercise 15.2 -- Comparison of IDsat Relationships % Computational parameters clear z=-0.4; % z = Vbi/VP x=linspace(0,1); % x = VG/VP % Eq. (15.13) ("P-theory"); y = IDsat/ID0 Num=(x-1)-(2/3)*(z-1)*(1-((z-x)./(z-1)).^(1.5)); Den=-1-(2/3)*(z-1)*(1-(z/(z-1))^(1.5)); yP=Num./Den; % Square-law theory yS=(1-x).^2; % Plotting result close plot(x,yP,x,yS,'--');grid xlabel('VG/VP');ylabel('IDsat/ID0'); text(0.38,0.4,'Square-Law') text(0.27,0.3,'Eq.(15.13)') text(0.8,0.83,'Vbi/VP=-0.4') ``` Logic: `x` sweeps $V_G/V_P$ from 0 to 1; the numerator and denominator are evaluated from the normalized formula above (element-wise `./` and `.^` because `x` is a vector); `plot` overlays the solid exact curve and dashed square-law curve (Fig. above, both falling monotonically from 1 to 0 with the square law slightly different in the middle). ### Exercise 15.3: Series source and drain resistance **Problem.** Show that $R_S$ and $R_D$ (Fig. 15.18) are accounted for by replacing $V_D\to V_D-I_D(R_S+R_D)$ and $V_G\to V_G-I_DR_S$ in Eqs. (15.9), (15.12), (15.13). **Solution.** Voltage drops at the channel ends change the channel limits: $$V(0)=I_DR_S,\qquad V(L)=V_D-I_DR_D$$ Putting these limits into (15.4), (15.5): $$I_D=\frac{2qZ\mu_nN_Da}{L}\int_{I_DR_S}^{V_D-I_DR_D}\left[1-\frac{W(V)}{a}\right]dV \tag{15.5'}$$ $W(V)/a$ from (15.8) is unchanged since the gate is still at $V_G$ and the junction still sees $V_{bi}+V-V_G$. Evaluating the integral with the new limits: $$I_D=G_0\left\{[V_D-I_DR_D-I_DR_S]-\frac{2}{3}\frac{(V_D-I_DR_D+V_{bi}-V_G)^{3/2}-(I_DR_S+V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}\right\} \tag{15.9'}$$ Now check the replacement rule. With $V_D'=V_D-I_D(R_S+R_D)$ and $V_G'=V_G-I_DR_S$: $$V_D'+V_{bi}-V_G'=V_D-I_DR_D+V_{bi}-V_G,\qquad V_{bi}-V_G'=I_DR_S+V_{bi}-V_G$$ so (15.9) with primed voltages is exactly (15.9'). Pinch-off: $W\to a$ when $V(L)=V_{Dsat}-I_{Dsat}R_D$. From (15.6), $$a=\left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}+V_{Dsat}-I_{Dsat}R_D-V_G)\right]^{1/2} \tag{15.11'}$$ and from (15.7) $a=[\ldots(V_{bi}-V_P)]^{1/2}$. Equating, The gate-to-channel voltage at the drain end involves $V_G$ itself, so the condition is $V_{bi}+V_{Dsat}-I_{Dsat}R_D-V_G=V_{bi}-V_P$, i.e. $$V_{Dsat}-I_{Dsat}R_D=V_G-V_P \tag{15.12'}$$ Hence, in terms of the intrinsic drain-end channel potential, $V_{D}'$ with $V_G'$: $V_{Dsat}'=V_G'-V_P$ gives $V_{Dsat}-I_{Dsat}(R_S+R_D)=V_G-I_{Dsat}R_S-V_P$, which is identical to (15.12'). Setting $V_D=V_{Dsat}$, $I_D=I_{Dsat}$ in (15.9) and using (15.12'): $$I_{Dsat}=G_0\left\{(V_G-I_{Dsat}R_S-V_P)-\frac{2}{3}(V_{bi}-V_P)\left[1-\left(\frac{I_{Dsat}R_S+V_{bi}-V_G}{V_{bi}-V_P}\right)^{3/2}\right]\right\} \tag{15.13'}$$ which is (15.13) with $V_G\to V_G-I_{Dsat}R_S$. So the rule works. **Remarks.** (i) The primed equations are implicit in $I_D$; no closed form exists, but numerical iteration solves them easily (Problem 15.7). (ii) A shortcut: change the integration variable to $V'=V-I_DR_S$ and require the same form as (15.5); this works but teaches little. (iii) The saturated current is still flat in $V_D$, but at a lower value, because $R_S$ makes the gate bias seen by the channel less favorable (**source degeneration**, negative feedback). --- ### 15.2.4 a.c. Response Treat the J-FET as a two-port (Fig. 15.19a): input gate-source, output drain-source. First consider low frequencies where capacitances are negligible. **Input.** The gate-source port sees a reverse-biased diode, which is (to first order) an **open circuit** at low frequencies. **Output.** The d.c. drain current is $I_D(V_D,V_G)$. Superimposing small a.c. voltages $v_d$ and $v_g$, the drain current becomes $I_D(V_D,V_G)+i_d$, quasi-statically: $$I_D(V_D,V_G)+i_d=I_D(V_D+v_d,\,V_G+v_g) \tag{15.15a}$$ $$i_d=I_D(V_D+v_d,V_G+v_g)-I_D(V_D,V_G) \tag{15.15b}$$ Taylor expand about the operating point, keeping first-order terms: $$I_D(V_D+v_d,V_G+v_g)=I_D(V_D,V_G)+\frac{\partial I_D}{\partial V_D}\bigg|_{V_G}v_d+\frac{\partial I_D}{\partial V_G}\bigg|_{V_D}v_g \tag{15.16}$$ Substituting in (15.15b): $$i_d=\frac{\partial I_D}{\partial V_D}\bigg|_{V_G}v_d+\frac{\partial I_D}{\partial V_G}\bigg|_{V_D}v_g \tag{15.17}$$ The partial derivatives are conductances. Define $$g_d\equiv\frac{\partial I_D}{\partial V_D}\bigg|_{V_G=\text{const}}\quad\text{(drain or channel conductance)} \tag{15.18a}$$ $$g_m\equiv\frac{\partial I_D}{\partial V_G}\bigg|_{V_D=\text{const}}\quad\text{(transconductance or mutual conductance)} \tag{15.18b}$$ $$i_d=g_dv_d+g_mv_g \tag{15.19}$$ This is the a.c. KCL at the drain node: a conductance $g_d$ from drain to source in parallel with a voltage-controlled current source $g_mv_g$ (Fig. 15.19b, with an open circuit at the gate). The parameter $g_m$ plays a role analogous to $\alpha$ and $\beta$ in BJT analysis. **Table 15.1: J-FET small-signal parameters** (by direct differentiation of (15.9) and (15.13); $G_0=2qZ\mu_nN_Da/L$): | | Below pinch-off ($V_D\le V_{Dsat}$) | Above pinch-off ($V_D\ge V_{Dsat}$) | |---|---|---| | $g_d$ | $G_0\left[1-\left(\dfrac{V_D+V_{bi}-V_G}{V_{bi}-V_P}\right)^{1/2}\right]$ | $0$ | | $g_m$ | $G_0\,\dfrac{(V_D+V_{bi}-V_G)^{1/2}-(V_{bi}-V_G)^{1/2}}{(V_{bi}-V_P)^{1/2}}$ | $G_0\left[1-\left(\dfrac{V_{bi}-V_G}{V_{bi}-V_P}\right)^{1/2}\right]$ | Derivations: $\partial/\partial V_D$ of (15.9) gives $G_0\{1-(V_D+V_{bi}-V_G)^{1/2}/(V_{bi}-V_P)^{1/2}\}$. $\partial/\partial V_G$ of (15.9) gives $G_0\{-\tfrac23\cdot(-\tfrac32)[(V_D+V_{bi}-V_G)^{1/2}-(V_{bi}-V_G)^{1/2}]/(V_{bi}-V_P)^{1/2}\}$. For saturation differentiate (15.13) with respect to $V_G$: $G_0\{1-\tfrac23(V_{bi}-V_P)\cdot(-\tfrac32)\ldots\}$; more simply, $I_{Dsat}=G_0\{V_G-V_P-\tfrac23(V_{bi}-V_P)+\tfrac23(V_{bi}-V_G)^{3/2}(V_{bi}-V_P)^{-1/2}\}$, whose $V_G$ derivative is $G_0\{1-(V_{bi}-V_G)^{1/2}(V_{bi}-V_P)^{-1/2}\}$. The result $g_d=0$ in saturation matches the zero slope of the characteristic. **High frequency.** The gate $pn$ junctions are reverse biased, so each is represented adequately by its depletion capacitance. That capacitance is split between gate-source ($C_{gs}$) and gate-drain ($C_{gd}$), giving Fig. 15.19(c): $C_{gs}$ and $C_{gd}$ added to the low-frequency circuit. ### Exercise 15.4: Effective transconductance with $R_S$, $R_D$ **Problem.** Using the low-frequency circuit of Fig. 15.19(b) for a saturation-biased J-FET, show that $R_S$ and $R_D$ give a reduced effective transconductance $g_m'=g_m/(1+g_mR_S)$. **Solution.** In saturation $g_d=0$, so replace $g_d$ by an open circuit and insert $R_S$, $R_D$ at the proper nodes (Fig. E15.4). The controlling voltage is the gate-to-*intrinsic* source voltage $v_g-i_dR_S$: $$i_d=g_m\,(v_g-i_dR_S)$$ $$i_d(1+g_mR_S)=g_mv_g$$ $$i_d=\frac{g_m}{1+g_mR_S}\,v_g\equiv g_m'v_g,\qquad g_m'=\frac{g_m}{1+g_mR_S}$$ $R_D$ carries the current $i_d$ without affecting the controlling voltage, so it does not change $g_m'$ when $g_d=0$. $R_S$ reduces $g_m'$ appreciably when $g_mR_S$ is comparable to or larger than 1. This is negative feedback. --- ## 15.3 MESFET ### 15.3.1 General Information A rectifying metal-semiconductor (Schottky) contact has a bias-controllable depletion region just like a $pn$ junction (Chapter 14). So one can build FETs with Schottky gates: **MESFETs** (metal-semiconductor FETs). Fig. 15.20 shows a simplified perspective of a GaAs MESFET (source and drain on $n^+$ regions, gate metal on $n$-GaAs, on a semi-insulating GaAs substrate) and an idealized cross section. Commercial MESFETs are GaAs-based $n$-channel devices, chosen for GaAs's superior electron transport. Metal-insulator-GaAs FETs were historically hard to make, so MESFET fabrication became the most mature GaAs technology. GaAs MESFETs are the heart of **MMICs** (monolithic microwave integrated circuits) and are sold as discrete devices for amplifiers and oscillators operating above 5 GHz. **Two types (Fig. 15.21).** - **D-MESFET (depletion mode):** a gate bias further depletes the subgate region and reduces channel conductance (like the J-FET). It is normally on at $V_G=0$. - **E-MESFET (enhancement mode):** built so the built-in voltage of the metal-semiconductor contact alone totally depletes the channel. A forward gate bias is needed to shrink the depletion width and obtain current. It is normally off. Fig. 15.21 shows implanted $n$ and $n^+$ regions, a Ti/Pt/Au Schottky gate contact, AuGe/Ni ohmic contacts, and a passivation insulator. Analog circuits (including MMICs) almost exclusively use D-MESFETs. Digital logic uses both D- and E-MESFETs. High frequency demands extremely short channels: gate (active channel) length is routinely 1 $\mu$m, and discrete GaAs MESFETs in the 1993 Hewlett-Packard catalog have nominal gate length of 0.25 $\mu$m. ### 15.3.2 Short-Channel Considerations The J-FET theory applies to the MESFET if all assumptions hold. For short channels two are suspect: 1. $L\gg a$ (and hence the gradual channel approximation). 2. Low field: the drift velocity is assumed $v=\mu_0\mathcal{E}$. With $V_D=1$ V across $L=1\ \mu$m, the average field is $\bar{\mathcal{E}}=V_D/L=10^4$ V/cm. The field near the drain is larger. At $10^4$ V/cm the Si electron drift velocity (Fig. 3.4) departs markedly from the linear extrapolation, and in GaAs the failure sets in at an even lower field. Three modified models follow. #### Variable mobility model Use Eq. (3.3) with $\beta=1$: $$v=\frac{\mu_0\mathcal{E}}{1+\mu_0\mathcal{E}/v_{sat}} \tag{15.20}$$ $$\mu(\mathcal{E})=\frac{|v|}{|\mathcal{E}|}=\frac{\mu_0}{1+\mu_0|\mathcal{E}|/v_{sat}} \tag{15.21}$$ Substituting $\mu(\mathcal{E})$ for $\mu_n$ in (15.1) with $\mathcal{E}=-dV/dy$ and redoing the derivation gives (derived in Problem 15.12) $$I_D=\frac{I_D(\text{long-channel})}{1+\dfrac{\mu_0V_D}{v_{sat}L}},\qquad 0\le V_D\le V_{Dsat} \tag{15.22}$$ where $I_D$(long-channel) is from (15.9) (with $\mu_0$). Above pinch-off, $I_D$(long)$\to I_{Dsat}$(long) and $V_D\to V_G-V_P$. For a single-gate MESFET (Fig. 15.20) replace $2a$ by $a$ in the long-channel equations. $I_D$ is always reduced relative to the long-channel case since $\mu(\mathcal{E})\le\mu_0$; when $v\ll v_{sat}$, $\mu_0V_D/(v_{sat}L)\ll1$ and the long-channel result returns. This model suits moderate-length FETs ($L\gtrsim10\ \mu$m) but cannot capture the failure of the gradual channel approximation. #### Saturated velocity model In long channels, current saturation is caused by pinch-off near the drain. In short channels another mechanism exists. Suppose $v=v_{sat}$ at a point $y_1<L$. The current there is $$I(y_1)=qv_{sat}N_DZ[a-W(y_1)] \tag{15.23}$$ Current continuity requires $I(y>y_1)=I(y_1)$. Since $v$ cannot exceed $v_{sat}$ beyond $y_1$, (15.23) forces $W(y)$ to stay constant for $y_1\le y\le L$ (Fig. 15.22). This needs an appreciable field in the subgate depletion region with some field lines ending on charge outside the gated region, the opposite of the gradual channel assumption. Though the channel is only partly constricted, $I_D$ has saturated. A larger $V_D$ just moves $y_1$ toward the source. In general, saturation occurs when the channel pinches off or when $v\to v_{sat}$ at $y=L$. For channels of a few microns or less, $W$ for $y_1\le y\le L$ differs little from $W(0)$: $$I_{Dsat}=I(y_1)\approx qv_{sat}N_DZ[a-W(0)] \tag{15.24}$$ $$W(0)=\left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}-V_G)\right]^{1/2}=a\left(\frac{V_{bi}-V_G}{V_{bi}-V_P}\right)^{1/2} \tag{15.25}$$ so $I_{Dsat}=qv_{sat}N_DZa\left[1-\sqrt{(V_{bi}-V_G)/(V_{bi}-V_P)}\right]$. This fits $L\sim1\ \mu$m GaAs MESFET data fairly well. Note that $I_{Dsat}$ depends **linearly** on $\sqrt{V_{bi}-V_G}$, and the velocity, not mobility, sets the current: $g_m$ is nearly constant for $V_G$ well above $V_P$. #### Two-region model The saturated-velocity model gives only $I_{Dsat}$. The two-region model gives a complete characteristic: - For $0\le y\le y_1$: gradual channel approximation and long-channel theory with $v=\mu_0\mathcal{E}$. - For $y_1\le y\le L$: $v=v_{sat}$. - The transition $y_1$ is where $|\mathcal{E}|=\mathcal{E}_{sat}=v_{sat}/\mu_0$. Hence the characteristics for $V_D\le V_{Dsat}$ come from (15.9) (or the single-gate equivalent), and for $V_D>V_{Dsat}$ set $I_D=I_{Dsat}=I_D|_{V_D=V_{Dsat}}$. In general $V_{Dsat}\ne V_G-V_P$, and (15.13) is not used. $V_{Dsat}$ is the drain voltage at which $|\mathcal{E}|$ reaches $\mathcal{E}_{sat}$ at $y=L$ (see Problem 15.13). Solving the long-channel relations for $V(y)$ (Problem 15.3) and differentiating with respect to $y$ at $y=L$ yields $$\mathcal{E}_{sat}L=\frac{(V_{bi}-V_P)\left[3\left(u^2-w^2\right)-2\left(u^3-w^3\right)\right]}{3\,(u-1)},\quad u=\left(\frac{V_{Dsat}+V_{bi}-V_G}{V_{bi}-V_P}\right)^{1/2},\ w=\left(\frac{V_{bi}-V_G}{V_{bi}-V_P}\right)^{1/2} \tag{15.26}$$ with $\mathcal{E}_{sat}<0$ in the book's sign convention. This is solved numerically for $V_{Dsat}$ (the derivation is Problem 15.13). **Fig. 15.23:** normalized two-region characteristics for $V_{bi}=1$ V, $V_P=-2.5$ V, $\mathcal{E}_{sat}=-5\times10^3$ V/cm, $L=1\ \mu$m, normalized to the $V_G=0$ saturation current $I_{D0}$ of an equivalent long-channel FET. Compared to Fig. 15.16: $V_{Dsat}$ is much lower and $I_{Dsat}/I_{D0}$ is reduced. $I_{Dsat}$ varies with $V_G$ roughly as (15.24). There is a slope discontinuity at $V_D=V_{Dsat}$, so the curves look crude, but they give a reasonable first-order match to observed short-channel curves. Accurate short-channel modeling needs a numerical two-dimensional solution of the electrostatic and current equations. --- ## 15.4 Summary The modulation of the conductivity of a semiconductor by an electric field normal to its surface is the **field effect**. The chapter began with the history of the field-effect family. The rest was devoted to the J-FET and MESFET. J-FET operation was developed qualitatively (linear region, slope-over, pinch-off, saturation, gate pinch-off) and quantitatively (classic long-channel $I_D$-$V_D$ derivation using the gradual channel approximation). The channel conductance $g_d$ and transconductance $g_m$ are the key parameters for the small-signal a.c. model. The MESFET discussion stressed short-channel physics: variable mobility, velocity saturation, and the two-region model, as modifications to the long-channel J-FET theory. **Key formulas.** $$I_D=G_0\left\{V_D-\tfrac23\frac{(V_D+V_{bi}-V_G)^{3/2}-(V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}\right\},\quad V_{Dsat}=V_G-V_P,\quad G_0=\frac{2qZ\mu_nN_Da}{L}$$ --- ## Problems ### Problem 15.1: Quick quiz The OCR lost the wording of parts (b), (d), (e), (g). I reconstruct them from the chapter content and state the reconstruction each time. **(a) Define "field effect."** The modulation of the conductivity of a semiconductor by an electric field applied normal to its surface. **(b) (Reconstructed) How do the source, drain and gate get their names? Why was the J-FET called unipolar?** Source: carriers enter the semiconductor from the external circuit. Drain: carriers leave. Gate: controlling or gating action. Unipolar: only one carrier type (electrons in an $n$-channel device) carries the current. **(c) For a $p$-channel J-FET ($n^+$-$p$ gates, $p$ region between source and drain), does $I_D$ flow into or out of the drain under normal operation?** Holes are the carriers; they move from source to drain, so positive current flows out of the drain contact into the external circuit. (Opposite of the $n$-channel case, where conventional current flows into the drain.) Normal biasing is $V_G\ge0$ and $V_D\le0$ so the $n^+p$ gate junctions stay reverse biased. **(d) (Reconstructed) Why does $I_D$ saturate for $V_D>V_{Dsat}$, and why is $I_D\neq0$ at pinch-off?** The voltage in excess of $V_{Dsat}$ is dropped across a short depleted section $\Delta L$; the source-to-pinch-off region keeps the same shape and same end-point voltages, so the same current flows. $I_D$ cannot be zero at pinch-off, because zero current would give zero voltage drop along the channel, which would remove the pinch-off. **(e) What is meant by "pinch-off"?** Complete depletion of the channel (top and bottom depletion regions touching) at some point. In the drain-end sense it occurs at $V_D=V_{Dsat}=V_G-V_P$. In the gate sense, $V_G=V_P$ pinches off the whole channel at $V_D=0$. **(f) Mathematical definition of drain conductance and transconductance.** $$g_d=\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G},\qquad g_m=\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D}$$ **(g) (Reconstructed) Sketch the low-frequency small-signal equivalent circuit of a J-FET in saturation.** Open circuit between $G$ and $S$ (reverse-biased gate). Between $D$ and $S$: a current source $g_mv_g$ in parallel with $g_d$, with $g_d\approx0$ in saturation (from the flat characteristics of Fig. 15.16), so only the current source $g_mv_{gs}$ remains. Adding $C_{gs}$ and $C_{gd}$ gives the high-frequency model. **(h) What do MESFET, D-MESFET, E-MESFET stand for?** Metal-semiconductor field-effect transistor; depletion-mode MESFET (normally on); enhancement-mode MESFET (normally off). **(i) Why is the magnitude of the electric field in the channel of concern in modeling short-channel MESFETs?** In a short channel the field $\sim V_D/L$ becomes large (e.g. $10^4$ V/cm), so the drift velocity is no longer $\mu_0\mathcal{E}$: it saturates, which changes current saturation and invalidates the long-channel theory. **(j) Primary difference between the long-channel and two-region short-channel $I_D$-$V_D$ theories?** Long-channel: saturation occurs only by pinch-off ($V_{Dsat}=V_G-V_P$) with $v=\mu_0\mathcal{E}$ everywhere. Two-region: the channel is split into a low-field region ($v=\mu_0\mathcal{E}$) and a velocity-saturated region ($v=v_{sat}$); saturation occurs when $\mathcal{E}$ at the drain end reaches $\mathcal{E}_{sat}$, at a lower $V_{Dsat}$ and lower $I_{Dsat}$. --- ### Problem 15.2: The composite transistor (Fig. P15.2) An $n$-type bar has ohmic contacts D (top) and B (bottom), and $p^+$-$n$ step junctions on its two sides with contacts E and C. $d$ is the separation of the two $p^+$ regions and $L$ their lateral length. (OCR damaged some conditions; I use the plausible original: $d\ll L_p$ in (a),(b); I assume (a) asks for $I_C$ versus $V_{CB}$.) **(a) D-B tied, $d\ll L_p$, $V_{EB}>0$, $V_{CB}<0$. Sketch $I_C$ (out of C) versus the collector-base voltage at several constant $I_E$.** *Reasoning.* E-$n$-C is a $p^+nn$... a lateral $p^+$-$n$-$p^+$ **BJT** with E as emitter (forward biased, $V_{EB}>0$), the $n$ bar (D-B) as base, and C as collector (reverse biased). Since $d\ll L_p$, holes injected from E cross the narrow base without recombining, so the transport factor is about 1, and the injection efficiency is about 1 ($p^+$ emitter). Hence $\alpha\approx1$: $$I_C\approx\alpha I_E\approx I_E$$ Sketch: common-base output characteristics, flat horizontal lines, spaced by the chosen $I_E$ values, nearly independent of $V_{CB}$ in reverse bias, with a slight upward tilt (base-width modulation). As $V_{CB}$ moves toward zero and becomes positive (collector junction forward biased, about 0.4 to 0.5 V), the collector starts injecting holes back and $I_C$ drops through zero and goes negative. At large reverse $V_{CB}$ the curves rise abruptly (avalanche breakdown, multiplication). **(b) E-C tied, D-B tied, $d\ll L_p$. Sketch the current into the E-C leads versus the voltage from E-C to D-B.** *Reasoning.* The two $p^+$-$n$ junctions are in parallel, with the same voltage. The structure is simply a diode of about twice the single-junction area. Forward bias ($V>0$): exponential $I\approx I_0(e^{qV/kT}-1)$ (current rising sharply near 0.6 V for Si). Reverse bias: small, nearly constant saturation current $-I_0$ (with slight generation increase), then abrupt breakdown at $V=-V_{BR}$. The two junctions at the same potential inject equally into the region between them, so there is no hole current across the gap $d$ and no transistor action. **(c) E-C tied, $d<2W_{BR}$, $L\approx$ the bar length between D and B, $V_{DB}>0$, $V_{EB}<0$. Sketch $I_D$ versus $V_{DB}$ at several $V_{EB}$.** *Reasoning.* Now the $p^+$ regions are reverse-biased gates. B acts as source, D as drain, the gap $d$ is the channel ($2a=d$): this is a **J-FET**. The condition $d<2W_{BR}$ guarantees pinch-off ($W\to a=d/2$) occurs before junction breakdown. The sketch is the J-FET family: for each gate bias a linear rise, slope-over, and saturation at $V_{DBsat}=V_{EB}-V_P$ with $I_{Dsat}$ falling as $V_{EB}$ becomes more negative; for $V_{EB}\le V_P$, $I_D=0$. At large $V_{DB}$ all curves rise abruptly (breakdown near the drain end). --- ### Problem 15.3: Voltage along the channel **Problem.** (a) Derive an expression for the position $y/L$ where a given channel voltage $V(y)$ occurs. (b) For $V_G=0$, $V_D=5$ V, $V_{bi}=1$ V, $V_P=-8$ V, find $y/L$ for $V(y)=1,2,3,4$ V and compare with Fig. 15.11(c). **(a) Solution.** Following the hint, put $L\to y$, $V_D\to V(y)$ in (15.4): $$I_Dy=2qZ\mu_nN_D\int_0^{V(y)}[a-W(V)]\,dV$$ Divide by (15.4) itself: $$\frac{y}{L}=\frac{\int_0^{V(y)}[1-W/a]\,dV}{\int_0^{V_D}[1-W/a]\,dV}$$ Evaluate the integrals as in (15.9) with $F(V)\equiv V-\tfrac23\left[(V+V_{bi}-V_G)^{3/2}-(V_{bi}-V_G)^{3/2}\right]/(V_{bi}-V_P)^{1/2}$: $$\boxed{\frac{y}{L}=\frac{F(V(y))}{F(V_D)}=\frac{V(y)-\frac23\frac{[V(y)+V_{bi}-V_G]^{3/2}-(V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}}{V_D-\frac23\frac{[V_D+V_{bi}-V_G]^{3/2}-(V_{bi}-V_G)^{3/2}}{(V_{bi}-V_P)^{1/2}}}}$$ **(b) Numbers.** $V_{bi}-V_P=9$ V so $(V_{bi}-V_P)^{1/2}=3$, and with $V_G=0$: $$F(V)=V-\frac{2}{9}\left[(V+1)^{3/2}-1\right]$$ (Check that we are below pinch-off: $V_{Dsat}=V_G-V_P=8$ V $>5$ V.) $$F(5)=5-\tfrac29(6^{3/2}-1)=5-\tfrac29(13.697)=5-3.044=1.956$$ | $V(y)$ (V) | $F(V)$ | $y/L$ | |---|---|---| | 1 | $1-\frac29(2.828-1)=0.594$ | 0.304 | | 2 | $2-\frac29(5.196-1)=1.068$ | 0.546 | | 3 | $3-\frac29(8-1)=1.444$ | 0.738 | | 4 | $4-\frac29(11.180-1)=1.738$ | 0.888 | **Comparison.** Fig. 15.11(c) draws 1, 2, 3, 4 V at equal spacing, i.e. $y/L=0.2, 0.4, 0.6, 0.8$. The true positions are $0.30, 0.55, 0.74, 0.89$. The voltage rises slowly near the source and steeply near the drain (the last volt occurs in the last 11% of the channel). Reason: the channel is narrowest near the drain, so the resistance per unit length is largest there, and for the same current the voltage drop per unit length is largest there. --- ### Problem 15.4: $V_{Dsat}$ from the maximum of $I_D$ **Problem.** If (15.9) is evaluated for $V_D>V_{Dsat}$, $I_D$ is a peaked function of $V_D$ with its maximum at $V_{Dsat}$. Use the maximum condition to derive (15.12) directly from (15.9). **Solution.** Set $dI_D/dV_D=0$: $$\frac{dI_D}{dV_D}=G_0\left\{1-\frac{(V_D+V_{bi}-V_G)^{1/2}}{(V_{bi}-V_P)^{1/2}}\right\}=0$$ $$(V_D+V_{bi}-V_G)^{1/2}=(V_{bi}-V_P)^{1/2}$$ $$V_D+V_{bi}-V_G=V_{bi}-V_P$$ $$\boxed{V_{Dsat}=V_G-V_P}$$ It is a maximum since $d^2I_D/dV_D^2=-\dfrac{G_0}{2}\dfrac{1}{(V_D+V_{bi}-V_G)^{1/2}(V_{bi}-V_P)^{1/2}}<0$. --- ### Problem 15.5: J-FET with the bottom gate grounded (Fig. P15.5) Long-channel J-FET, bottom gate tied to the source and grounded ($V_{GB}=0$), top gate at $V_G\equiv V_{GT}\le0$, $V_D\ge0$. Let $A=2K_S\varepsilon_0/(qN_D)$, so $W=\sqrt{A(V_{bi}-V_A)}$ and $a=\sqrt{A(V_{bi}-V_P)}$ where $V_P$ is the pinch-off voltage with the two gates tied. **(a) Sketch the depletion regions when $V_{GT}$ pinches off the channel at $V_D=0$.** The bottom junction is at equilibrium: a thin depletion layer of width $W_B(0)=\sqrt{AV_{bi}}$ at the bottom. The top depletion layer is much wider and must extend down to touch it, so $W_T=2a-W_B(0)$: a deep depletion region from the top, a thin one from the bottom, meeting at a plane closer to the bottom than the midplane $x=a$. **(b) $V_P=-8$ V (gates tied), $V_{bi}=1$ V. Find the top-gate pinch-off voltage $V_{PT}$ with $V_{GB}=0$, $V_D=0$.** $$a=\sqrt{A\cdot 9}=3\sqrt A\quad(\text{since }V_{bi}-V_P=9\text{ V}),\qquad W_B(0)=\sqrt{A\cdot1}=\sqrt A$$ Pinch-off condition: $$W_T=2a-W_B(0)=6\sqrt A-\sqrt A=5\sqrt A$$ $$\sqrt{A(V_{bi}-V_{PT})}=5\sqrt A$$ $$V_{bi}-V_{PT}=25$$ $$\boxed{V_{PT}=1-25=-24\ \text{V}}$$ Consistent with (a): the top depletion layer must span 5/6 of the thickness $2a$ (more than $a$), requiring much more reverse bias than the symmetric $-8$ V where each layer spans only $a$ (depletion $\propto\sqrt{V}$, so $25/9$ more voltage-equivalent). Assumes no breakdown at $-24$ V. **(c) $V_{PT}<V_{GT}<0$, drain raised to pinch-off. Sketch.** Both junctions get more reverse biased toward the drain. The top is always more reverse biased ($V-V_{GT}$) than the bottom ($V-0$), so the top layer is thicker at every $y$. The wedges meet only at the drain end, at a point shifted toward the bottom gate. **(d) Expression for $V_{Dsat}$ using only voltages.** Pinch-off at $y=L$ where $V=V_{Dsat}$: $W_T+W_B=2a$: $$\sqrt{A(V_{bi}+V_{Dsat}-V_{GT})}+\sqrt{A(V_{bi}+V_{Dsat}-0)}=2\sqrt{A(V_{bi}-V_P)}$$ $$\boxed{\sqrt{V_{bi}+V_{Dsat}-V_{GT}}+\sqrt{V_{bi}+V_{Dsat}}=2\sqrt{V_{bi}-V_P}}$$ **(e) Larger or smaller than the tied-gate value $V_{Dsat}=V_{GT}-V_P$?** Let $u=V_{bi}+V_{Dsat}$ and let $u'$ be the tied-gate value, $u'=V_{bi}-V_P+V_{GT}$. The left side of (d), $f(u)=\sqrt{u-V_{GT}}+\sqrt u$, increases with $u$. At $u'$: $f(u')=\sqrt{V_{bi}-V_P}+\sqrt{u'}$. Because $V_{GT}<0$, $u'<V_{bi}-V_P$, so $f(u')<2\sqrt{V_{bi}-V_P}$. Hence the true root satisfies $u>u'$: $$V_{Dsat}(V_{GB}=0)>V_{Dsat}(V_{GB}=V_{GT})$$ Physically, the bottom junction is less reverse biased, its depletion layer is thinner, and a larger drain voltage is needed to close the channel. **(f) $I_D(V_D,V_{GT})$.** Channel thickness $2a-W_T-W_B$: $$I_DL=qZ\mu_nN_D\int_0^{V_D}\left[2a-W_T(V)-W_B(V)\right]dV$$ with $W_T=\sqrt{A(V_{bi}+V-V_{GT})}$, $W_B=\sqrt{A(V_{bi}+V)}$. Integrating, $$I_D=\frac{qZ\mu_nN_D}{L}\left\{2aV_D-\tfrac23\sqrt A\Big[(V_D+V_{bi}-V_{GT})^{3/2}-(V_{bi}-V_{GT})^{3/2}+(V_D+V_{bi})^{3/2}-V_{bi}^{3/2}\Big]\right\}$$ Using $\sqrt A=a/\sqrt{V_{bi}-V_P}$ and $G_0=2qZ\mu_nN_Da/L$: $$\boxed{I_D=G_0\left\{V_D-\frac{1}{3}\frac{(V_D+V_{bi}-V_{GT})^{3/2}-(V_{bi}-V_{GT})^{3/2}+(V_D+V_{bi})^{3/2}-V_{bi}^{3/2}}{(V_{bi}-V_P)^{1/2}}\right\}}$$ valid for $V_D\le V_{Dsat}$ of part (d). Check: if both gates were at $V_{GT}$ the bracket doubles and $\tfrac13\to\tfrac23$, recovering (15.9). --- ### Problem 15.6: Linearly graded channel (Fig. P15.6) The net donor doping rises linearly from the $p^+$-$n$ junction at $x=0$ (zero) to a maximum at the center $x=a$ and falls symmetrically to zero at $x=2a$ (assumed reading of the figure). Let $N_D-N_A=\kappa x$ for $0\le x\le a$ and $\kappa(2a-x)$ for $a\le x\le2a$, $\kappa$ = grade constant (cm$^{-4}$). **(a) Depletion width.** One side is $p^+$, so depletion lies in the $n$ side. Poisson on $0\le x\le W$ (junction at $x=0$): $$\frac{d\mathcal{E}}{dx}=\frac{q\kappa x}{K_S\varepsilon_0}\Rightarrow\mathcal{E}(x)=-\frac{q\kappa}{2K_S\varepsilon_0}(W^2-x^2)\quad(\mathcal{E}(W)=0)$$ Voltage across the junction: $$V_{bi}-V_A=\int_0^W|\mathcal{E}|dx=\frac{q\kappa}{2K_S\varepsilon_0}\left(W^3-\frac{W^3}{3}\right)=\frac{q\kappa W^3}{3K_S\varepsilon_0}$$ $$\boxed{W=\left[\frac{3K_S\varepsilon_0}{q\kappa}(V_{bi}-V_A)\right]^{1/3}}$$ (The symmetric linearly graded junction of Section 5.2.5 has 12 instead of 3; here only one side of the junction is graded.) **(b) $I_D$-$V_D$.** With gates tied, $V_A=V_G-V(y)$, so $W/a=[(V_{bi}+V-V_G)/(V_{bi}-V_P)]^{1/3}$, where $a^3=\frac{3K_S\varepsilon_0}{q\kappa}(V_{bi}-V_P)$ defines $V_P$. The caution: the carrier density is not constant. The conducting cross-section carries $$\int_{W}^{2a-W}N(x)dx=2\int_W^a\kappa x\,dx=\kappa(a^2-W^2)$$ (ignore the $\mu_n$ doping dependence). Then $$I_D=qZ\mu_n\frac{dV}{dy}\kappa(a^2-W^2)\Rightarrow I_DL=qZ\mu_n\kappa a^2\int_0^{V_D}\left[1-\left(\frac{W}{a}\right)^2\right]dV$$ $$\left(\frac{W}{a}\right)^2=\left(\frac{V_{bi}-V_G+V}{V_{bi}-V_P}\right)^{2/3}$$ $$\int_0^{V_D}\left[1-\left(\frac{V_{bi}-V_G+V}{V_{bi}-V_P}\right)^{2/3}\right]dV=V_D-\frac{3}{5}\frac{(V_D+V_{bi}-V_G)^{5/3}-(V_{bi}-V_G)^{5/3}}{(V_{bi}-V_P)^{2/3}}$$ $$\boxed{I_D=G_0'\left\{V_D-\frac35\,\frac{(V_D+V_{bi}-V_G)^{5/3}-(V_{bi}-V_G)^{5/3}}{(V_{bi}-V_P)^{2/3}}\right\},\quad G_0'=\frac{qZ\mu_n\kappa a^2}{L}}$$ $G_0'$ is the no-depletion conductance. $V_{Dsat}=V_G-V_P$ still holds because pinch-off is still $W=a$. --- ### Problem 15.7: Effect of $R_S$ and $R_D$ (MATLAB) **Task.** Plot normalized $I_D/I_{D0}$ vs. $V_D$ for several $V_G$, with $G_0R_S=G_0R_D=0,0.1,0.5$, $V_{bi}=1$ V, $V_P=-2.5$ V, with $I_{D0}=I_{Dsat}|_{V_G=0}$ from (15.13) (the intrinsic device). **Method.** Let $F(V_1,V_2;V_G)=\int_{V_1}^{V_2}[1-W/a]dV=(V_2-V_1)-\tfrac23\left[(V_2+V_{bi}-V_G)^{3/2}-(V_1+V_{bi}-V_G)^{3/2}\right]/\sqrt{V_{bi}-V_P}$. Then $I_D=G_0F$ with $$V_1=I_DR_S,\qquad V_2=\min\{V_D-I_DR_D,\ V_G-V_P\}$$ (the min enforces saturation: the drain-end channel potential cannot exceed $V_G-V_P$). Let $j=I_D/I_{D0}$ and $f_0=I_{D0}/G_0$ (volts). Then $I_DR_S=j\,f_0\,(G_0R_S)$, so only $G_0R_S$, $G_0R_D$ appear. The equation to solve per bias point: $$h(j)=jf_0-F\big(jf_0\,g_{rs},\ \min\{V_D-jf_0\,g_{rd},V_G-V_P\};V_G\big)=0$$ with $g_{rs}=G_0R_S$, $g_{rd}=G_0R_D$. $h$ is strictly increasing in $j$ (increasing $j$ raises $V_1$ and lowers $V_2$, both reducing $F$), $h(0)\le0$, $h(1)\ge0$. So bisection on $[0,1]$ always finds a unique root. ```matlab % Problem 15.7 -- J-FET ID-VD with series resistances Rs, Rd clear; close all Vbi=1; Vp=-2.5; c=Vbi-Vp; VGlist=[0 -0.5 -1 -1.5 -2]; VD=linspace(0,5,101); grs=[0 0.1 0.5]; % G0*Rs = G0*Rd values % F(V1,V2;VG): integral of (1-W/a) dV F=@(V1,V2,VG) (V2-V1)-(2/3)*((V2+Vbi-VG).^1.5-(V1+Vbi-VG).^1.5)/sqrt(c); f0=F(0,-Vp,0); % ID0/G0 (intrinsic, VG=0) for k=1:length(grs) g=grs(k); % G0*Rs=G0*Rd=g figure(k); hold on for m=1:length(VGlist) VG=VGlist(m); j=zeros(size(VD)); for n=1:length(VD) lo=0; hi=1; for it=1:60 % bisection on j=ID/ID0 mid=(lo+hi)/2; V1=mid*f0*g; V2=min(VD(n)-mid*f0*g, VG-Vp); h=mid*f0-F(V1,V2,VG); if h>0, hi=mid; else lo=mid; end end j(n)=(lo+hi)/2; end plot(VD,j) end grid on; xlabel('V_D (volts)'); ylabel('I_D/I_{D0}') title(['G_0R_S=G_0R_D=' num2str(g)]) end ``` **Results** (from running the same algorithm; $V_D=5$ V; columns $V_G=0,-0.5,-1,-1.5,-2$ V): | $G_0R$ | $I_D/I_{D0}$ at $V_D=5$ V | |---|---| | 0 | 1.000, 0.614, 0.334, 0.144, 0.035 | | 0.1 | 0.956, 0.594, 0.326, 0.142, 0.035 | | 0.5 | 0.820, 0.527, 0.298, 0.134, 0.034 | For $V_G=0$: $I_D/I_{D0}$ at $V_D=0.5,1,2,5$ is 0.386, 0.666, 0.965, 1.000 ($G_0R=0$), 0.357, 0.622, 0.916, 0.956 (0.1) and 0.275, 0.493, 0.763, 0.820 (0.5). **Comment.** $G_0R=0$ reproduces Fig. 15.16. With resistance: (i) the linear-region slope drops (resistances add in series with the channel); (ii) $I_{Dsat}$ is lower, most for $V_G=0$ (high current) and negligibly for near-pinch-off curves (low current, little $IR$ drop); (iii) the knee moves to higher terminal $V_D$ (from (15.12'), $V_{Dsat}=V_G-V_P+I_{Dsat}R_D$), the saturated curves remain flat, and they bunch together because $R_S$ reduces the effective gate swing (reduced $g_m$, consistent with Exercise 15.4). --- ### Problem 15.8: The pinch resistor A pinch resistor is a J-FET with its gate(s) shorted to the source, a two-terminal nonlinear resistor with $V_G=0$ and applied voltage $V=V_D$. **(a) Conductances.** Below pinch-off ($0\le V\le V_{Dsat}=-V_P$), from (15.9) with $V_G=0$: $$I=G_0\left\{V-\frac23\frac{(V+V_{bi})^{3/2}-V_{bi}^{3/2}}{(V_{bi}-V_P)^{1/2}}\right\}$$ d.c. conductance: $$\boxed{G=\frac IV=G_0\left\{1-\frac23\frac{(V+V_{bi})^{3/2}-V_{bi}^{3/2}}{V\,(V_{bi}-V_P)^{1/2}}\right\}}$$ a.c. conductance (this is $g_d$ at $V_G=0$): $$\boxed{g=\frac{dI}{dV}=G_0\left[1-\left(\frac{V+V_{bi}}{V_{bi}-V_P}\right)^{1/2}\right]}$$ For $V>-V_P$: $I=I_{D0}$ is constant, so $G=I_{D0}/V$ and $g=0$. **(b) Numbers at $V=V_{Dsat}/2$.** $V_P=-2$ V so $V_{Dsat}=2$ V and $V=1$ V. $Z/L=1$, $a=0.5\ \mu$m, $N_D=10^{16}$ cm$^{-3}$, $V_{bi}=1$ V. Assumption: electron mobility is not given; I take $\mu_n\approx1250$ cm$^2$/V s (Si at $10^{16}$ cm$^{-3}$, about 300 K). Also, the given $a$ and $N_D$ imply $V_{bi}-V_P=1.9$ V, not 3 V; I use the stated $V_P=-2$ V for the voltage dependence. $$G_0=2q\frac ZL\mu_nN_Da=2(1.6\times10^{-19})(1)(1250)(10^{16})(0.5\times10^{-4})=2.0\times10^{-4}\ \text{S}$$ With $V_{bi}-V_P=3$ V: $$\frac{I}{G_0}=1-\frac23\frac{2^{3/2}-1}{\sqrt3}=1-\frac23(1.0556)\Rightarrow\frac I{G_0}=0.2963\ \text{V}\quad(\text{since }V=1\text{ V})$$ $$I=(2.0\times10^{-4})(0.2963)=5.93\times10^{-5}\ \text{A}$$ $$G=\frac IV=5.93\times10^{-5}\ \text{S}\Rightarrow R=\frac1G\approx16.9\ \text{k}\Omega$$ $$g=2.0\times10^{-4}\left[1-\sqrt{\tfrac23}\right]=2.0\times10^{-4}(0.1835)=3.67\times10^{-5}\ \text{S}\Rightarrow r=\frac1g\approx27.2\ \text{k}\Omega$$ $$\boxed{R\approx16.9\ \text{k}\Omega,\quad r\approx27.2\ \text{k}\Omega}$$ (Dynamic resistance exceeds static: the I-V curve bends over.) --- ### Problem 15.9: Cutoff frequency $$f_{max}=\frac{g_m}{2\pi C_G}$$ **(a) Derivation.** Treat the gate input as the capacitor $C_G$ (the open circuit is replaced by the junction capacitance). An a.c. gate voltage $v_g$ draws input current $i_g=j\omega C_Gv_g$ with magnitude $\omega C_G|v_g|$. The output (saturation, $g_d=0$) delivers $i_d=g_mv_g$. The current gain $|i_d/i_g|=g_m/(\omega C_G)$ falls to unity at the maximum useful frequency: $$\omega_{max}C_G=g_m\Rightarrow f_{max}=\frac{g_m}{2\pi C_G}$$ (analogous to the MOSFET cutoff analysis of Section 17.3.2). **(b) J-FET expression.** The largest $g_m$ is the no-depletion limit $G_0=2qZ\mu_nN_Da/L$ (Table 15.1: saturation $g_m\to G_0$ as $V_G\to V_{bi}$). Estimate $C_G$ with both gates in parallel at pinch-off width $a$: $C_G=2\dfrac{K_S\varepsilon_0ZL}{a}$. $$f_{max}=\frac{G_0}{2\pi C_G}=\frac{2qZ\mu_nN_Da/L}{2\pi\cdot2K_S\varepsilon_0ZL/a}$$ $$\boxed{f_{max}=\frac{q\mu_nN_Da^2}{2\pi K_S\varepsilon_0L^2}}$$ Short channels ($f\propto1/L^2$), high mobility and heavy channel doping give higher speed. **(c) Numbers.** Si, $N_D=10^{16}$, $a=0.5\ \mu$m, $L=5\ \mu$m, assuming $\mu_n=1250$ cm$^2$/V s: $$f_{max}=\frac{(1.6\times10^{-19})(1250)(10^{16})(0.5\times10^{-4})^2}{2\pi(11.8)(8.85\times10^{-14})(5\times10^{-4})^2}=\frac{5.0\times10^{-9}}{1.64\times10^{-18}}\approx3.0\times10^{9}\ \text{Hz}$$ $$\boxed{f_{max}\approx3\ \text{GHz}}$$ --- ### Problem 15.10: Temperature dependence of $g_m$ **Setup.** Saturation, $V_G=0$: $g_m=G_0\left[1-\sqrt{V_{bi}/(V_{bi}-V_P)}\right]$. Neglecting dimensional changes, $V_{bi}-V_P=qN_Da^2/(2K_S\varepsilon_0)$ is temperature independent: $$V_{bi}-V_P=\frac{(1.6\times10^{-19})(10^{16})(0.6\times10^{-4})^2}{2(11.8)(8.85\times10^{-14})}=2.76\ \text{V}$$ $G_0\propto\mu_n(T)$, and $V_{bi}(T)=\dfrac{kT}{q}\ln\dfrac{N_AN_D}{n_i^2(T)}$ with $N_A=5\times10^{18}$, $N_D=10^{16}$: $$\frac{g_m(T)}{g_m(300)}=\frac{\mu_n(T)}{\mu_n(300)}\cdot\frac{1-\sqrt{V_{bi}(T)/2.76}}{1-\sqrt{V_{bi}(300)/2.76}}$$ **Model used (stated assumptions).** - $n_i(T)=10^{10}(T/300)^{3/2}\exp\left[-\left(\frac{E_G(T)}{2kT}-\frac{E_G(300)}{2k\cdot300}\right)\right]$, $E_G=1.17-4.73\times10^{-4}T^2/(T+636)$. - $\mu_n$: Matthiessen with lattice $1430(T/300)^{-2.42}$ and ionized-impurity $9900(T/300)^{3/2}$ cm$^2$/V s ($\mu_n(300)\approx1250$). **(a) Results.** | $T$ (K) | $V_{bi}$ (V) | $\mu_n(T)/\mu_n(300)$ | $g_m(T)/g_m(300)$ | |---|---|---|---| | 225 | 0.972 | 1.588 | 1.477 | | 250 | 0.940 | 1.374 | 1.309 | | 275 | 0.908 | 1.174 | 1.146 | | 300 | 0.875 | 1.000 | 1.000 | | 325 | 0.841 | 0.853 | 0.874 | | 350 | 0.806 | 0.730 | 0.768 | | 375 | 0.771 | 0.629 | 0.679 | | 400 | 0.734 | 0.545 | 0.604 | | 425 | 0.698 | 0.475 | 0.541 | | 450 | 0.660 | 0.417 | 0.487 | | 475 | 0.622 | 0.368 | 0.442 | ```matlab % Problem 15.10 -- gm(T)/gm(300) for saturation-biased n-channel Si J-FET clear; close all k=8.617e-5; q=1.6e-19; eps=11.8*8.85e-14; ND=1e16; NA=5e18; a=0.6e-4; c=q*ND*a^2/(2*eps); % Vbi-Vp (T independent) Eg=@(T) 1.17-4.73e-4*T.^2./(T+636); ni=@(T) 1e10*(T/300).^1.5.*exp(-(Eg(T)./(2*k*T)-Eg(300)/(2*k*300))); mu=@(T) 1./(1./(1430*(T/300).^-2.42)+1./(9900*(T/300).^1.5)); Vbi=@(T) k*T.*log(NA*ND./ni(T).^2); gm=@(T) mu(T).*(1-sqrt(Vbi(T)/c)); T=225:50:475; T=[225 250 275 300 325 350 375 400 425 450 475]; loglog(T,gm(T)/gm(300),'o-',T,mu(T)/mu(300),'s--'); grid xlabel('T (K)'); ylabel('normalized'); legend('g_m','\mu_n') p=polyfit(log(T/300),log(gm(T)/gm(300)),1); n=-p(1) ``` **(b) Exponent.** Fit $g_m\propto T^{-n}$ to the log-log data: $n\approx1.65$. The mobility alone behaves as $T^{-2.0}$ over this range with this model. So $g_m$ falls more slowly than $\mu_n$. Reason: as $T$ rises, $V_{bi}$ falls (intrinsic carriers grow), so the zero-bias depletion shrinks and the channel opens a little, partly compensating the mobility drop. Therefore the mobility is the dominant effect, with $V_{bi}$ (and $V_P$) providing a partial offset. (Numbers depend on the assumed mobility model.) --- ### Problem 15.11: Back-gated MESFET, two-region model Top MS gate (built-in $V_{biT}$, voltage $V_{GT}$) and bottom $p^+$-$n$ gate ($V_{biB}$, $V_{GB}$), thickness $2a$ between the junctions, channel length $L$ set by the shorter MS gate. Let $V_{PT}$ be the top-gate voltage that pinches off the channel when $V_D=0$ and $V_{GB}=0$ (reading of the garbled definition). **Solution.** Below saturation the two-region model uses the long-channel theory ($v=\mu_0\mathcal{E}$) up to $V_{Dsat}$. With $A=2K_S\varepsilon_0/(qN_D)$: $$I_DL=qZ\mu_0N_D\int_0^{V_D}\left[2a-W_T-W_B\right]dV,\quad W_T=\sqrt{A(V_{biT}+V-V_{GT})},\ W_B=\sqrt{A(V_{biB}+V-V_{GB})}$$ Integrating: $$I_D=\frac{qZ\mu_0N_D}{L}\left\{2aV_D-\frac23\sqrt A\left[\Sigma_T+\Sigma_B\right]\right\}$$ $$\Sigma_T=(V_D+V_{biT}-V_{GT})^{3/2}-(V_{biT}-V_{GT})^{3/2},\quad\Sigma_B=(V_D+V_{biB}-V_{GB})^{3/2}-(V_{biB}-V_{GB})^{3/2}$$ Eliminate $\sqrt A$ with the pinch-off definition ($W_T+W_B=2a$ at $V=0$, $V_{GT}=V_{PT}$, $V_{GB}=0$): $$2a=\sqrt A\left[(V_{biT}-V_{PT})^{1/2}+V_{biB}^{1/2}\right]\Rightarrow\frac{\sqrt A}{a}=\frac{2}{(V_{biT}-V_{PT})^{1/2}+V_{biB}^{1/2}}$$ With $G_0=2qZ\mu_0N_Da/L$, $qZ\mu_0N_D/L=G_0/(2a)$, so $\frac{G_0}{2a}\cdot\frac23\sqrt A=\frac{2G_0}{3}\frac{1}{D}$ with $D\equiv(V_{biT}-V_{PT})^{1/2}+V_{biB}^{1/2}$: $$\boxed{I_D=G_0\left\{V_D-\frac{2}{3}\,\frac{\Sigma_T+\Sigma_B}{(V_{biT}-V_{PT})^{1/2}+V_{biB}^{1/2}}\right\},\quad0\le V_D\le V_{Dsat}}$$ Above $V_{Dsat}$, $I_D=I_{Dsat}=I_D|_{V_{Dsat}}$, where $V_{Dsat}$ is found from the field condition $|\mathcal{E}(L)|=\mathcal{E}_{sat}$ (analogue of 15.26). Check: for a symmetric device $V_{biT}=V_{biB}$, $V_{GT}=V_{GB}$, $D=2\sqrt{V_{bi}-V_P}$ and $\Sigma_T=\Sigma_B$, giving (15.9). (If the single-gate variant is wanted, the same algebra with $2a\to a$ applies.) --- ### Problem 15.12: Derive Eq. (15.22) **Solution.** Use magnitudes with $\mathcal{E}=dV/dy>0$ along the channel. The mobility is $\mu(\mathcal{E})=\mu_0/(1+\mu_0\mathcal{E}/v_{sat})$. Current through the channel: $$I_D=2qZN_D\,\mu(\mathcal{E})\,\mathcal{E}\,[a-W]=\frac{2qZN_D\mu_0[a-W]\,dV/dy}{1+(\mu_0/v_{sat})\,dV/dy}$$ Multiply out: $$I_D+I_D\frac{\mu_0}{v_{sat}}\frac{dV}{dy}=2qZ\mu_0N_D[a-W]\frac{dV}{dy}$$ Multiply by $dy$: $$I_D\,dy=2qZ\mu_0N_D[a-W]\,dV-\frac{I_D\mu_0}{v_{sat}}\,dV$$ Integrate over $0\le y\le L$ ($I_D$ constant), $0\le V\le V_D$: $$I_DL=2qZ\mu_0N_D\int_0^{V_D}[a-W]dV-\frac{I_D\mu_0V_D}{v_{sat}}$$ $$I_D\left(L+\frac{\mu_0V_D}{v_{sat}}\right)=2qZ\mu_0N_D\int_0^{V_D}[a-W]dV=I_D(\text{long})\,L$$ $$\boxed{I_D=\frac{I_D(\text{long-channel})}{1+\mu_0V_D/(v_{sat}L)}}$$ (For single-gate MESFET use $a$ in place of $2a$ in the long-channel factor.) --- ### Problem 15.13: Derive Eq. (15.26) **Solution.** From Problem 15.3(a), $I_Dy=G_0L\,F(V(y))$ with $F(V)=\int_0^V[1-W/a]dV$ and $I_D=G_0F(V_D)$. Differentiate with respect to $y$: $$I_D=G_0L\left[1-\frac{W(V)}{a}\right]\frac{dV}{dy}$$ At the drain end for $V_D=V_{Dsat}$, the transition to the velocity-saturated region occurs when the field magnitude there equals $|\mathcal{E}_{sat}|$ (with $\mathcal{E}=-dV/dy$, so $\mathcal{E}_{sat}<0$): $$\left.\frac{dV}{dy}\right|_{L}=-\mathcal{E}_{sat}=\frac{I_D}{G_0L[1-W(V_{Dsat})/a]}=\frac{F(V_{Dsat})}{L[1-u]}$$ where $u=W(V_{Dsat})/a=\left[(V_{Dsat}+V_{bi}-V_G)/(V_{bi}-V_P)\right]^{1/2}$. Write with $c=V_{bi}-V_P$, $w=\sqrt{(V_{bi}-V_G)/c}$ so that $V_{Dsat}=c(u^2-w^2)$: $$F(V_{Dsat})=c(u^2-w^2)-\frac23c(u^3-w^3)=\frac c3\left[3(u^2-w^2)-2(u^3-w^3)\right]$$ $$\mathcal{E}_{sat}L=-\frac{F(V_{Dsat})}{1-u}=\frac{F(V_{Dsat})}{u-1}$$ $$\boxed{\mathcal{E}_{sat}L=\frac{(V_{bi}-V_P)\left[3(u^2-w^2)-2(u^3-w^3)\right]}{3(u-1)}}$$ The same result in the unsimplified form is $\mathcal{E}_{sat}L=\left\{V_{Dsat}-\tfrac23[(V_{Dsat}+V_{bi}-V_G)^{3/2}-(V_{bi}-V_G)^{3/2}]/\sqrt c\right\}\Big/\left[\sqrt{(V_{Dsat}+V_{bi}-V_G)/c}-1\right]$, solved numerically for $V_{Dsat}$. --- ### Problem 15.14: MATLAB two-region model **Logic.** Only $\mathcal{E}_{sat}L$ matters. For each $V_G$: 1. $V_{Dsat}$ is the root of $\Phi(V_D)=F(V_D)/(1-u(V_D))-|\mathcal{E}_{sat}|L=0$ on $(0,V_G-V_P)$. $\Phi$ rises monotonically from $-|\mathcal{E}_{sat}|L<0$ to $+\infty$ as $u\to1$, so there is exactly one root (found with `fzero`/bisection). 2. For $V_D\le V_{Dsat}$: $I_D/I_{D0}=F(V_D)/f_0$. For $V_D>V_{Dsat}$: $I_{Dsat}/I_{D0}=F(V_{Dsat})/f_0$. 3. $f_0=F(V_G=0,V_D=-V_P)=-V_P-\tfrac23(V_{bi}-V_P)\left[1-\left(\tfrac{V_{bi}}{V_{bi}-V_P}\right)^{3/2}\right]$. Normalizing by $f_0$ cancels the single/double-gate factor and $G_0$. ```matlab % Problem 15.14 -- two-region model ID-VD characteristics clear; close all Vbi=1; Vp=-2.5; Esat=5e3; L=1e-4; % V, V, V/cm, cm (Fig. 15.23 values) EL=Esat*L; c=Vbi-Vp; F =@(V,VG) V-(2/3)*((V+Vbi-VG).^1.5-(Vbi-VG)^1.5)/sqrt(c); f0=-Vp-(2/3)*c*(1-(Vbi/c)^1.5); % ID0/G0 VGlist=[0 -0.5 -1 -1.5 -2]; hold on for VG=VGlist Phi=@(V) F(V,VG)./(1-sqrt((V+Vbi-VG)/c))-EL; VDsat=fzero(Phi,[1e-9, VG-Vp-1e-9]); % root in (0, VG-Vp) IDsat=F(VDsat,VG)/f0; VD=linspace(0,2,201); ID=zeros(size(VD)); for n=1:length(VD) if VD(n)<=VDsat, ID(n)=F(VD(n),VG)/f0; else, ID(n)=IDsat; end end plot(VD,ID) end grid on; xlabel('V_D (volts)'); ylabel('I_D/I_{D0}') ``` **(a) Check with the Fig. 15.23 parameters** ($V_{bi}=1$, $V_P=-2.5$, $|\mathcal{E}_{sat}|=5\times10^3$ V/cm, $L=1\ \mu$m, so $\mathcal{E}_{sat}L=0.5$ V). Output: $V_G=0$: $V_{Dsat}=0.437$ V, $I_{Dsat}/I_{D0}=0.343$; $V_G=-0.5$: 0.429 V, 0.246; $-1.0$: 0.412 V, 0.162; $-1.5$: 0.377 V, 0.089; $-2.0$: 0.291 V, 0.029. This reproduces the shape of Fig. 15.23: low $V_{Dsat}$ (under 0.5 V), $I_{Dsat}/I_{D0}$ maximum about 0.34 (the plot axis tops at 0.5), slope discontinuity at $V_{Dsat}$. **(b) (Assumed reading.)** Use $\mathcal{E}_{sat}=-10^4$ V/cm with a long channel (for example $L=100\ \mu$m, $\mathcal{E}_{sat}L=100$ V), and compare with Fig. 15.16. Setting `Esat=1e4; L=100e-4` and extending `VD` to 5 V, the curves converge to the long-channel set of Fig. 15.16: $V_{Dsat}\to V_G-V_P$ (2.5 V at $V_G=0$) and $I_{Dsat}/I_{D0}\to$ the (15.13) values (1.00, 0.61, 0.33, 0.14, 0.035 at $V_G=0,-0.5,-1,-1.5,-2$ V). Decreasing $L$ progressively lowers $V_{Dsat}$ and $I_{Dsat}$ (e.g. $\mathcal{E}_{sat}L=5$ V gives $V_{Dsat}=1.85$ V, ratio 0.94; 20 V gives 2.32 V, ratio 0.996). **(c) Where long-channel theory fails.** Define failure as $I_{Dsat}/I_{D0}<0.95$ at $V_G=0$. Solving $F(V_{Dsat})/f_0=0.95$ together with the $V_{Dsat}$ equation numerically: $$V_{Dsat}\approx1.90\ \text{V},\qquad\boxed{\mathcal{E}_{sat}L\approx5.6\ \text{V}}$$ The long-channel theory therefore holds if $|\mathcal{E}_{sat}|L\gtrsim5.6$ V. Examples: $L=5.6/(5\times10^3)=1.1\times10^{-3}$ cm $=11\ \mu$m for $|\mathcal{E}_{sat}|=5\times10^3$ V/cm, and $5.6/10^4$ cm $=5.6\ \mu$m for $10^4$ V/cm. So channels shorter than about 11 $\mu$m (for $5\times10^3$ V/cm) or 5.6 $\mu$m (for $10^4$ V/cm) begin to deviate from long-channel behavior. Values of $I_{Dsat}/I_{D0}$ at $V_G=0$ versus $\mathcal{E}_{sat}L$ from the program: 0.5 V: 0.343; 1 V: 0.551; 2 V: 0.767; 3 V: 0.864; 5 V: 0.940; 10 V: 0.983; 20 V: 0.996. # Chapter 16: MOS Fundamentals > Constants used throughout (as the book instructs): $T=300\ \text{K}$, $kT/q=0.0259\ \text{V}$, $n_i=10^{10}\ \text{cm}^{-3}$, $K_S=11.8$, $K_O=3.9$, $\varepsilon_0=8.85\times10^{-14}\ \text{F/cm}$, $q=1.6\times10^{-19}\ \text{C}$. Imagine a sandwich: a metal slice, a whisper-thin sheet of glass, and a slab of silicon. That sandwich, the **metal-oxide-semiconductor (MOS) structure**, sits at the heart of nearly every chip made today. A processor with billions of transistors is, at bottom, billions of these sandwiches. The idea of a "field effect" device dates to the 1920s, but modern history begins with Kahng and Atalla, who filed patents on the Si-SiO$_2$ field-effect transistor in 1960. "MOS" is reserved for the metal-SiO$_2$-Si system; the generic name **MIS** (metal-insulator-semiconductor) covers other insulators or semiconductors. This chapter studies the simplest MOS device, the two-terminal **MOS capacitor (MOS-C)**. The plan: 1. Define precisely an *ideal* MOS-C. 2. Draw energy-band and block-charge pictures of what happens inside under a static bias. 3. Turn the pictures into equations (charge, field, potential), and link them to the gate voltage. 4. Explain the capacitance-voltage ($C$-$V$) curve, the "window" through which engineers look inside the sandwich. --- ## 16.1 Ideal Structure Definition The MOS-C (Fig. 16.1) is a thin SiO$_2$ layer (0.01 $\mu$m to 1.0 $\mu$m) between a Si substrate and a metallic **field plate**. Common field-plate materials are aluminum and heavily doped polycrystalline silicon (polysilicon, deposited by chemical vapor deposition and then heavily n- or p-doped by diffusion or ion implantation). A second metallic layer on the back gives electrical contact to the silicon. The field plate and its terminal are the **gate**; the silicon terminal, normally grounded, is the **back** or **substrate contact**. $x$ is measured from the oxide-semiconductor interface into the silicon; $x_o$ is the oxide thickness; $V_G$ is the gate voltage. **Eight idealizations** define the ideal MOS-C: 1. The gate is thick enough to be an **equipotential region**, under a.c. as well as d.c. bias. 2. The oxide is a **perfect insulator**: zero current through it under all static biases. 3. There are **no charge centers** in the oxide or at the oxide-semiconductor interface. 4. The semiconductor is **uniformly doped**. 5. The semiconductor is thick enough that, whatever the gate potential, a **field-free region** (the "bulk") is reached before the back contact. 6. The back contact is **ohmic**. 7. The structure is **one-dimensional**; everything depends only on $x$. 8. $\Phi_M=\Phi_S=\chi+(E_C-E_F)_{FB}$: the metal and semiconductor work functions are equal. **Are these realistic?** Largely yes: SiO$_2$ resistivity can reach $10^{18}\ \Omega\cdot$cm, so d.c. leakage is negligible; thin gates are equipotential; ohmic backs are easy. Idealization 8 is the exception. It is only a convenience, and Chapter 18 removes it. Its purpose here is to avoid early clutter in the static description. --- # Chapter 16 — MOS Fundamentals (continued) ## 16.2 Electrostatics — Mostly Qualitative Before we crunch numbers, we want *pictures*. The MOS capacitor (metal–oxide–semiconductor, "MOS-C") is a sandwich, and its behaviour is easiest to grasp with two kinds of cartoon: the **energy band diagram** (what the electrons' energy landscape looks like) and the **block charge diagram** (where the charge sits). ### 16.2.1 Visualization Aids #### Energy Band Diagram The energy band diagram is an indispensable aid for visualizing the internal status of the MOS structure under static bias. Our first task is to build the diagram for the **ideal MOS structure in equilibrium (zero bias)**. Figure 16.2 shows the band diagrams of the three components *on their own*, each terminated abruptly by a vertical line that represents a surface. The "ledge" at the top of the vertical line is the **vacuum level** $E_0$: the minimum energy an electron must have to escape the material completely. - **Metal:** the energy gap between the vacuum level and the Fermi level is the **metal work function** $\Phi_M$. - **Semiconductor:** the surface barrier is described by the **electron affinity** $\chi$, the energy difference between the vacuum level and the conduction band edge *at the surface*. We use $\chi$ rather than "vacuum level minus $E_F$" because $E_C-E_F$ is not constant in a semiconductor: it changes with doping and with band bending near the surface. $\chi$, by contrast, is invariant with band bending. The quantity $(E_C-E_F)_{FB}$ is the $E_C$–$E_F$ separation in the flat-band (field-free) part of the semiconductor. - **Insulator:** modelled essentially as an intrinsic, wide-gap semiconductor whose surface barrier is again specified through an electron affinity $\chi_i$. > **Intuition:** Think of each material as a lake whose water level is the Fermi energy $E_F$, sitting in a basin whose rim is the vacuum level. The work function says "how far below the rim is the water"; the electron affinity says "how far below the rim is the bottom of the empty upper floor (the conduction band)". **Forming the zero-bias diagram** is a two-step conceptual process: 1. Bring the metal and semiconductor together to a distance $x_o$ apart (the eventual oxide thickness) and let the two-component system equilibrate. In equilibrium the two Fermi levels line up (Subsection 3.2.4). Also, because the ideal MOS-C is defined with $\Phi_M=\Phi_S$, the *vacuum levels of M and S are automatically aligned too*. Hence there is no charge and no electric field anywhere in the metal–gap–semiconductor system. 2. Insert the insulator of thickness $x_o$ into the empty gap. Because the field in the gap is zero, the only effect is to lower slightly the barrier between M and S. The result is Fig. 16.3: flat $E_C$, $E_i$, $E_V$ in the semiconductor, a flat $E_F$ throughout, and a rectangular insulator barrier between the metal and the semiconductor. #### Block Charge Diagrams A block charge diagram is the complement of the band diagram: it shows the *approximate* charge distribution. In the ideal structure at equilibrium there is no charge anywhere. When a bias is applied, charge appears in the metal and in the semiconductor near the metal–oxide and oxide–semiconductor interfaces. A sample is Fig. 16.4: a $+Q$ block on the metal side and a $-Q$ block on the semiconductor side of the oxide (position $x$ along the horizontal axis; M, oxide, S labelled). No attempt is made to depict exact charge distributions. A squared-off ("block") approximation is used, so magnitudes and spatial extents are only qualitative. Nevertheless, because the electric field vanishes deep inside both the metal and the semiconductor (idealization 5), **Gauss's law demands that the total charge in the structure is zero.** > **Rule:** in a block charge diagram, the area of the positive charge blocks always equals the area of the negative charge blocks. ### 16.2.2 Effect of an Applied Bias #### General Observations Assume normal operating conditions: the back of the MOS-C is grounded and $V_G$ is the d.c. bias on the gate. With $V_G\neq 0$: 1. The **semiconductor Fermi level is unaffected** and stays level (independent of position). This follows from the assumption of zero current under all static biasing: the semiconductor is always in equilibrium, whatever the gate voltage. 2. As in a pn junction, the applied bias separates the Fermi energies at the two ends of the structure by $qV_G$: $$E_F(\text{metal})-E_F(\text{semiconductor})=-qV_G \tag{16.1}$$ > **Intuition (the handles):** the metal and semiconductor Fermi levels are "handles connected to the outside world". The back contact is grounded, so the semiconductor handle is bolted in place. The gate handle is moved *down* if $V_G>0$ and *up* if $V_G<0$. Because the barrier heights are fixed quantities, moving the metal handle bends everything in between, like twisting a rubber doll out of shape. Equivalently, $V_G\neq0$ causes potential drops and band bending inside the structure: - No band bending in the metal (equipotential region). - In the oxide and semiconductor the bands slope **upward** (increasing $E$ going from gate toward back contact) when $V_G>0$, and **downward** when $V_G<0$. - Poisson's equation in the oxide (an ideal insulator: no carriers, no charge centers) gives $d\mathcal{E}_{ox}/dx=0$, so $\mathcal{E}_{ox}$ is constant. The oxide band edges $E_C$, $E_V$ are therefore **straight lines**. - In the semiconductor the bending has a more complex shape, but by idealization 5 it must vanish before the back contact is reached. #### Specific Biasing Regions Take an **n-type Si substrate** first. **Accumulation ($V_G>0$).** Positive $V_G$ lowers $E_F$ in the metal relative to the semiconductor and makes the bands slope upward in insulator and semiconductor (Fig. 16.5a). The electron concentration $n=n_i e^{(E_F-E_i)/kT}$ *increases* as one approaches the interface (because $E_i$ moves farther below $E_F$ there). A situation where the majority-carrier concentration is greater near the interface than in the bulk is called **accumulation**. From the charge view: $V_G>0$ puts positive charge on the gate; charge balance draws negative electrons to the interface (Fig. 16.5b). **Depletion (small $V_G<0$).** A small negative bias raises $E_F$ in the metal and gives the bands a small downward slope (Fig. 16.5c). Majority electrons are *depleted* near the interface. In charge language, the negative gate repels electrons from the interface and exposes positively charged donor sites (Fig. 16.5d). Here both $n$ and $p$ at the interface are less than the background doping ($N_D$ or $N_A$), hence "depletion". **Inversion (large negative $V_G$).** As $V_G$ becomes more and more negative, the semiconductor surface bands bend upward more and more. The surface hole concentration $p_s$ climbs: from less than $n_i$ when $E_i(\text{surface})<E_F$; to $n_i$ when $E_i(\text{surface})=E_F$; to greater than $n_i$ when $E_i(\text{surface})$ exceeds $E_F$. Eventually (Fig. 16.5e,f) we reach $$E_i(\text{surface})-E_i(\text{bulk})=2\,[E_F-E_i(\text{bulk})] \tag{16.2}$$ and then $$p_s=n_i\,e^{[E_i(\text{surface})-E_F]/kT}=n_i\,e^{[E_F-E_i(\text{bulk})]/kT}=n_{bulk}=N_D \tag{16.3}$$ So at this special bias $V_G=V_T$ the surface hole concentration equals the bulk electron concentration: the surface is no longer "depleted". For $V_G<V_T$, $p_s$ exceeds $n_{bulk}=N_D$, and the surface region has changed character from n-type to p-type. The condition in which the *minority*-carrier concentration at the surface exceeds the bulk *majority*-carrier concentration is **inversion** (Fig. 16.5g,h). > **Intuition (a see-saw of populations):** Imagine a crowded dance floor (n-type: full of electron-dancers). A negative gate is a bouncer that first pushes dancers away from the door (depletion). Shove hard enough and the few "hole-guests" gather at the door until they outnumber the original crowd (inversion). The door region now behaves like p-type material sitting on n-type material, a pn junction created purely by an electric field! For an **ideal p-type device** the results are those of Fig. 16.6, with reversed polarities: flat band ($V_G=0$), accumulation ($V_G<0$; a block of holes at the interface), depletion ($V_G>0$ small; exposed acceptors) and inversion ($V_G>V_T>0$; an electron sheet plus a depletion block of exposed acceptors). **Summary.** There are three physically distinct bias regions: | | Accumulation | Depletion | Inversion | |---|---|---|---| | ideal **n-type** | $V_G>0$ | $V_T<V_G<0$ | $V_G<V_T$ | | ideal **p-type** | $V_G<0$ | $0<V_G<V_T$ | $V_G>V_T$ | Flat band ($V_G=0$, no band bending) marks the boundary between accumulation and depletion. The boundary $V_G=V_T$ is the **depletion–inversion transition point**, with Eq. (16.2) quantitatively specifying the onset of inversion for both n- and p-type devices. ### Exercise 16.1 **Problem.** Construct line plots (with $V_G$ along the $x$-axis) that identify the voltage ranges for accumulation, depletion and inversion in ideal n- and p-type MOS devices. **Solution.** The "plots" graphically restate the summary above. On a $V_G$ number line: $$\begin{aligned} \text{n-type:}\quad & \underbrace{V_G<V_T}_{\text{INV}}\ \Big|\ \underbrace{V_T<V_G<0}_{\text{DEPL}}\ \Big|\ \underbrace{V_G>0}_{\text{ACC}}\\ \text{p-type:}\quad & \underbrace{V_G<0}_{\text{ACC}}\ \Big|\ \underbrace{0<V_G<V_T}_{\text{DEPL}}\ \Big|\ \underbrace{V_G>V_T}_{\text{INV}} \end{aligned}$$ The n-type axis reads (left to right) INV | $V_T$ | DEPL | $0$ | ACC; the p-type axis reads ACC | $0$ | DEPL | $V_T$ | INV. Note $V_T<0$ for n-type and $V_T>0$ for p-type. $$\boxed{\text{n: INV}<V_T<\text{DEPL}<0<\text{ACC};\qquad \text{p: ACC}<0<\text{DEPL}<V_T<\text{INV}}$$ --- ## 16.3 Electrostatics — Quantitative Formulation ### 16.3.1 Semiconductor Electrostatics #### Preparatory Considerations We now want analytic expressions for the charge density $\rho(x)$, electric field $\mathcal{E}(x)$ and electrostatic potential $\phi(x)$ inside an ideal MOS-C under static bias. The problem simplifies because: - The metal is an equipotential region. Charge near the metal–oxide interface sits only a few Ångstroms (1 Å $=10^{-8}$ cm) deep, so it is modelled as a **$\delta$-function of charge** at the M–O interface. Since the oxide holds no charge (idealization 3), the metal charge equals (in magnitude) the total charge in the semiconductor. - With no oxide charge, $\mathcal{E}_{ox}$ is constant and the oxide potential is linear in position. So solving the MOS-C essentially reduces to solving the semiconductor part. Starting from Poisson's equation we use the **depletion approximation** for a first-order closed form (closely paralleling the pn junction, Chapter 5). An exact solution exists (because the semiconductor is always in equilibrium regardless of bias); it is given in Appendix B. **Definitions.** $\phi(x)$ is the electrostatic potential at depth $x$ into the semiconductor, measured from the oxide–semiconductor interface (Fig. 16.7a). We use $\phi$ rather than $V$ to avoid confusion with externally applied voltages. The field $\mathcal{E}=-d\phi/dx$ vanishes deep in the substrate. By convention $\phi=0$ in the field-free bulk. The value of $\phi$ at $x=0$ is the **surface potential** $\phi_s$. Relation to the bands (Fig. 16.7b, consistent with Eq. 3.12): $$\phi(x)=\frac{1}{q}\left[E_i(\text{bulk})-E_i(x)\right] \tag{16.4}$$ $$\phi_s=\frac{1}{q}\left[E_i(\text{bulk})-E_i(\text{surface})\right] \tag{16.5}$$ and a key material parameter, the **Fermi potential**: $$\phi_F=\frac{1}{q}\left[E_i(\text{bulk})-E_F\right] \tag{16.6}$$ $\phi_F>0$ for p-type; $\phi_F<0$ for n-type; its magnitude is tied to the doping. For nondegenerate Si near room temperature (Chapter 2): $$p_{bulk}=n_i e^{[E_i(\text{bulk})-E_F]/kT}=N_A\quad(N_A\gg N_D) \tag{16.7a}$$ $$n_{bulk}=n_i e^{[E_F-E_i(\text{bulk})]/kT}=N_D\quad(N_D\gg N_A) \tag{16.7b}$$ Combining (16.6) and (16.7): $$\phi_F=+\frac{kT}{q}\ln\frac{N_A}{n_i}\quad\text{(p-type)} \tag{16.8a}$$ $$\phi_F=-\frac{kT}{q}\ln\frac{N_D}{n_i}\quad\text{(n-type)} \tag{16.8b}$$ > **Sign convention:** with the book's definitions, positive $\phi_s$ means the bands bend *downward* at the surface ($E_i$ at the surface is lower than in the bulk). In a p-type device positive $\phi_s$ therefore attracts electrons. Under flat band $\phi_s=0$. Substituting (16.5) and (16.6) into (16.2): $$\phi_s=2\phi_F\quad\text{at the depletion–inversion transition point} \tag{16.9}$$ For $\phi_F>0$ (p-type): $\phi_s<0$ is accumulation, $0<\phi_s<2\phi_F$ is depletion, and $\phi_s>2\phi_F$ is inversion. For an n-type semiconductor the inequalities are reversed (accumulation $\phi_s>0$, depletion $2\phi_F<\phi_s<0$, inversion $\phi_s<2\phi_F$). ### Exercise 16.2 **Problem.** (a) Construct line plots (with $\phi_s$ on the axis) identifying the surface-potential ranges for accumulation, depletion and inversion in ideal n- and p-type devices. (b) For each of the $(\phi_F,\phi_s)$ sets below (in units of $kT/q$), give the doping type and biasing condition, and draw the energy band and block charge diagrams. (i) $\phi_F=12,\ \phi_s=12$; (ii) $\phi_F=-9,\ \phi_s=3$; (iii) $\phi_F=-9,\ \phi_s=-18$; (iv) $\phi_F=15,\ \phi_s=36$; (v) $\phi_F=-15,\ \phi_s=0$. **Solution (a).** From (16.9) and the inequalities above: $$\begin{aligned} \text{p-type }(\phi_F>0):\quad & \underbrace{\phi_s<0}_{\text{ACC}}\ \Big|\ \underbrace{0<\phi_s<2\phi_F}_{\text{DEPL}}\ \Big|\ \underbrace{\phi_s>2\phi_F}_{\text{INV}}\\ \text{n-type }(\phi_F<0):\quad & \underbrace{\phi_s<2\phi_F}_{\text{INV}}\ \Big|\ \underbrace{2\phi_F<\phi_s<0}_{\text{DEPL}}\ \Big|\ \underbrace{\phi_s>0}_{\text{ACC}} \end{aligned}$$ **Solution (b).** The doping type follows from the sign of $\phi_F$. The condition follows from comparing $\phi_s$ with $0$ and $2\phi_F$: | Set | $\phi_F$ | $\phi_s$ | $2\phi_F$ | Doping | Condition | |---|---|---|---|---|---| | (i) | 12 | 12 | 24 | p | $0<12<24$: **Depletion** (middle, $\phi_s=\phi_F$) | | (ii) | $-9$ | 3 | $-18$ | n | $\phi_s>0$: **Accumulation** | | (iii) | $-9$ | $-18$ | $-18$ | n | $\phi_s=2\phi_F$: **Depletion/inversion transition** | | (iv) | 15 | 36 | 30 | p | $36>30$: **Inversion** | | (v) | $-15$ | 0 | $-30$ | n | $\phi_s=0$: **Flat band** | Diagrams: - **(i) p-type, depletion:** $E_F$ flat; $E_C,E_i,E_V$ bend downward toward the interface by $12\,kT$ (so $E_i$ is still above $E_F$ at the surface, i.e. the surface is still p-type but depleted of holes). Charge: $+Q$ on the metal; $-Q$ as a wide block of exposed acceptors. - **(ii) n-type, accumulation:** bands bend downward by only $3\,kT$ toward the interface, so $E_C$ moves closer to $E_F$ (more electrons). Charge: $+Q$ on metal; $-Q$ as a thin sheet of electrons at the interface. - **(iii) n-type, transition:** bands bend upward by $18\,kT$; at the surface $E_i$ lies as far *above* $E_F$ as it lies below it in the bulk, so $p_s=N_D$. Charge: $-Q$ on the metal; $+Q$ as a wide block of ionized donors. - **(iv) p-type, inversion:** bands bend downward by $36\,kT$, beyond $2\phi_F=30\,kT$. Charge: $+Q$ on metal; $-Q$ as a thin electron sheet at the interface plus a block of exposed acceptors of width about $W_T$. - **(v) n-type, flat band:** all bands flat, $E_F$ level across the structure, no charge anywhere. --- ### Delta-Depletion Solution The approximate closed-form solution based on the depletion approximation is divided into three segments matching the three bias regions. The exact-solution plots of Fig. 16.8 (a p-type device with $\phi_F=12\,kT/q$, $T=300$ K, $kT/q=0.0259$ V) guide the approximation. - **Fig. 16.8(a), accumulation** ($\phi_s=-6kT/q$): the accumulation-layer charge is squeezed into an extremely narrow region at the interface. - **Fig. 16.8(b), middle of depletion** ($\phi_s=\phi_F=12kT/q$): the depletion region reaches deep (a fraction of a micron). - **Fig. 16.8(c), onset of inversion** ($\phi_s=2\phi_F=24kT/q$): wider depletion region with charge density $-qN_A$. - **Fig. 16.8(d), deep inversion** ($\phi_s=2\phi_F+6kT/q=30kT/q$): a very thin, very dense electron layer at the interface (hundreds of $qN_A$) on top of a depletion region only slightly wider than in (c). (The $\rho$ plots use a linear scale with the $x$ axis pointing downward to match Fig. 16.6; dashed lines in (b)–(d) outline the depletion-approximation charge.) **Accumulation.** Since the accumulation charge sits in an extremely narrow region while the depleted region is far deeper, replace the accumulation charge by a $\delta$-function of equal charge at $x=0$. Then automatically $\mathcal{E}=0$ and $\phi=0$ for all $x>0$. This is somewhat inaccurate but acceptable to first order. **Inversion.** Likewise the inversion charge is confined to an extremely thin layer, so it is modelled by a $\delta$-function at the interface. In addition, comparing $\phi_s=\phi_F$, $2\phi_F$ and $2\phi_F+6kT/q$, the depletion width grows a lot up to $2\phi_F$ but only slightly afterwards. So in the delta-depletion model the $\delta$-function added in inversion exactly balances the extra gate charge; the depletion width, the field and the potential for $x>0$ stay frozen at their $\phi_s=2\phi_F$ values. The inversion solution = end-of-depletion solution + a $\delta$-function surface charge. > **Intuition (the pinned surface):** once inversion begins, electrons are so cheap to supply that any extra gate charge is answered by extra electrons in a razor-thin layer. It is like a reservoir with an overflow pipe: the water level ($\phi_s$) stops rising, and extra input spills into the inversion layer. **Depletion.** In the depletion approximation the true charge is replaced by a squared-off distribution ending abruptly at $x=W$. For p-type: $$\rho=q(p-n+N_D^+-N_A^-)=-qN_A\qquad(0\le x\le W) \tag{16.10}$$ Poisson's equation reduces to $$\frac{d\mathcal{E}}{dx}=\frac{\rho}{K_s\varepsilon_0}=-\frac{qN_A}{K_s\varepsilon_0}\qquad(0\le x\le W) \tag{16.11}$$ Integrating with $\mathcal{E}=0$ at $x=W$: $$\mathcal{E}(x)=-\frac{d\phi}{dx}=\frac{qN_A}{K_s\varepsilon_0}(W-x)\qquad(0\le x\le W) \tag{16.12}$$ A second integration with $\phi=0$ at $x=W$: $$\phi(x)=\frac{qN_A}{2K_s\varepsilon_0}(W-x)^2\qquad(0\le x\le W) \tag{16.13}$$ Applying $\phi(0)=\phi_s$ determines $W$: $$\phi_s=\frac{qN_AW^2}{2K_s\varepsilon_0} \tag{16.14}$$ $$W=\left[\frac{2K_s\varepsilon_0}{qN_A}\,\phi_s\right]^{1/2} \tag{16.15}$$ Equations (16.10), (16.12), (16.13), (16.15) constitute the depletion-bias solution. For an n-bulk device replace $N_A$ by $-N_D$ (and use the magnitude of $\phi_s$ inside the square root). **Maximum depletion width.** In the delta-depletion picture $W$ stops growing once the surface inverts ($\phi_s=2\phi_F$), so it reaches its maximum equilibrium value $$W_T=\left[\frac{2K_s\varepsilon_0}{qN_A}\,(2\phi_F)\right]^{1/2} \tag{16.16}$$ Fig. 16.9 plots $W_T$ against doping for Si at 300 K over the typical MOS range (log–log): it falls monotonically with doping, from several micrometres near $N_A\sim10^{14}\ \text{cm}^{-3}$ to a few hundredths of a micrometre near $10^{18}\ \text{cm}^{-3}$. > **Sanity check:** for $N_A=10^{16}\ \text{cm}^{-3}$, $\phi_F=0.0259\ln(10^{6})=0.358$ V and $W_T=\sqrt{2(11.8)(8.854\times10^{-14})(0.716)/[(1.6\times10^{-19})(10^{16})]}\approx0.31\ \mu\text{m}$. ### 16.3.2 Gate Voltage Relationship The results above are cast in terms of the surface potential $\phi_s$, an internal quantity. The knob we actually turn is $V_G$, so we need a relation between $V_G$ and $\phi_s$. The gate voltage in the ideal structure is dropped partly across the oxide and partly across the semiconductor: $$V_G=\Delta\phi_{semi}+\Delta\phi_{ox} \tag{16.17}$$ Since $\phi=0$ in the bulk, the semiconductor drop is just $$\Delta\phi_{semi}=\phi(x=0)=\phi_s \tag{16.18}$$ So we must express $\Delta\phi_{ox}$ in terms of $\phi_s$. In the ideal insulator $$\frac{d\mathcal{E}_{ox}}{dx}=0 \tag{16.19}$$ $$\mathcal{E}_{ox}=-\frac{d\phi_{ox}}{dx}=\text{constant} \tag{16.20}$$ so integrating across the oxide of thickness $x_o$, $$\Delta\phi_{ox}=\int_{-x_o}^{0}\mathcal{E}_{ox}\,dx=\mathcal{E}_{ox}x_o \tag{16.21}$$ Relating $\mathcal{E}_{ox}$ to the semiconductor field uses the boundary condition on normal displacement at an interface: $$(D_{semi}-D_{ox})\big|_{O\text{-}S}=Q_{O\text{-}S} \tag{16.22}$$ where $D=K\varepsilon_0\mathcal{E}$ and $Q_{O\text{-}S}$ is the interface charge per unit area. Since $Q_{O\text{-}S}=0$ in the ideal structure (idealization 3): $$D_{ox}=D_{semi}\big|_{x=0} \tag{16.23}$$ $$\mathcal{E}_{ox}=\frac{K_s}{K_o}\mathcal{E}_s \tag{16.24}$$ $$\Delta\phi_{ox}=\frac{K_s}{K_o}\mathcal{E}_sx_o \tag{16.25}$$ Here $K_s$ is the semiconductor dielectric constant, $K_o$ that of the oxide, and $\mathcal{E}_s$ the semiconductor field at the interface. Substituting (16.18) and (16.25) into (16.17): $$V_G=\phi_s+\frac{K_s}{K_o}\mathcal{E}_sx_o \tag{16.26}$$ with $\mathcal{E}_s$ a known (or readily computed) function of $\phi_s$. In the delta-depletion solution, (16.12) at $x=0$ combined with (16.15) gives $$\mathcal{E}_s=\left[\frac{2qN_A\phi_s}{K_s\varepsilon_0}\right]^{1/2} \tag{16.27}$$ $$V_G=\phi_s+\frac{K_s}{K_o}x_o\left[\frac{2qN_A\phi_s}{K_s\varepsilon_0}\right]^{1/2}\qquad(0\le\phi_s\le2\phi_F) \tag{16.28}$$ > **Footnote from the book:** strictly, the $\delta$-function layers of carriers at the O–S interface in the delta-depletion model would contribute a $Q_{O\text{-}S}$ under accumulation and inversion. However, in that model $\phi_s=0$ for all accumulation biases and $\phi_s=2\phi_F$ for all inversion biases, so the $V_G$–$\phi_s$ relationship above is only used for *depletion-bias* calculations. > > **Handy fact:** since $K_s=11.8$ for Si and $K_o=3.9$ for SiO$_2$, Eq. (16.24) says $\mathcal{E}_{ox}\approx3\,\mathcal{E}_s$ in an MOS system with no charge at the Si–SiO$_2$ interface. All energy band diagrams in this chapter should be consistent with this (the oxide band edges are about three times as steep as the semiconductor bands at the interface). **Reading Fig. 16.10** (the $V_G$ vs. $\phi_s$ curve for $x_o=0.1\ \mu$m, $N_A=10^{15}\ \text{cm}^{-3}$, $T=300$ K; delta-depletion vs. exact): - $\phi_s$ varies quite rapidly with $V_G$ in *depletion*: the gate voltage is shared proportionally between oxide and semiconductor. - In *accumulation* ($\phi_s<0$) and *inversion* ($\phi_s>2\phi_F$) it takes a large change in $V_G$ to produce a small change in $\phi_s$. Changes in the applied potential are then dropped almost entirely across the oxide. - The depletion-bias region is only slightly more than 1 V wide. Since the semiconductor's character changes drastically from one side of this region to the other, we expect a significant variation in electrical characteristics over a rather narrow voltage range. > **Intuition:** think of $\phi_s$ as the water level in a tank that is easy to move while the tank has a floor that is "soft" (depletion) but becomes a rigid wall in accumulation (a layer of majority carriers that answers any push instantly) and in inversion (a layer of minority carriers that answers any push instantly). In those two regimes the semiconductor is a nearly ideal metal-like plate and the oxide takes the voltage. ### Exercise 16.3 **Problem.** An MOS-C at $T=300$ K has $x_o=0.1\ \mu$m and Si doping $N_A=10^{15}\ \text{cm}^{-3}$. Compute (a) $\phi_F$ in $kT/q$ units and in volts, (b) $W$ when $\phi_s=\phi_F$, (c) $\mathcal{E}_s$ when $\phi_s=\phi_F$, (d) $V_G$ when $\phi_s=\phi_F$. **Constants:** $kT/q=0.0259$ V, $n_i=10^{10}\ \text{cm}^{-3}$, $K_s=11.8$, $K_o=3.9$, $\varepsilon_0=8.85\times10^{-14}$ F/cm, $q=1.6\times10^{-19}$ C, $x_o=0.1\ \mu\text{m}=10^{-5}$ cm. **(a)** p-type, so use (16.8a): $$\begin{aligned} \frac{\phi_F}{kT/q}&=\ln\frac{N_A}{n_i}=\ln\frac{10^{15}}{10^{10}}=\ln 10^{5}=11.51\\ \phi_F&=11.51\,(0.0259\ \text{V})=0.298\ \text{V} \end{aligned}$$ $$\boxed{\phi_F=11.51\,kT/q=0.298\ \text{V}}$$ **(b)** Use (16.15) with $\phi_s=\phi_F=0.298$ V: $$\begin{aligned} W&=\left[\frac{2K_s\varepsilon_0\phi_s}{qN_A}\right]^{1/2} =\left[\frac{2(11.8)(8.85\times10^{-14})(0.298)}{(1.6\times10^{-19})(10^{15})}\right]^{1/2}\\ &=\left[\frac{6.224\times10^{-13}}{1.6\times10^{-4}}\right]^{1/2} =\left[3.89\times10^{-9}\ \text{cm}^2\right]^{1/2}\\ &=6.24\times10^{-5}\ \text{cm} \end{aligned}$$ $$\boxed{W=0.624\ \mu\text{m}}$$ **(c)** Evaluate (16.12) at $x=0$: $$\begin{aligned} \mathcal{E}_s&=\frac{qN_A}{K_s\varepsilon_0}W=\frac{(1.6\times10^{-19})(10^{15})(6.24\times10^{-5})}{(11.8)(8.85\times10^{-14})}\\ &=\frac{9.98\times10^{-9}}{1.044\times10^{-12}}\ \text{V/cm} \end{aligned}$$ $$\boxed{\mathcal{E}_s\approx9.56\times10^{3}\ \text{V/cm}}$$ **(d)** Use (16.26): $$\begin{aligned} V_G&=\phi_s+\frac{K_s}{K_o}\,\mathcal{E}_s\,x_o\\ &=0.298+\frac{11.8}{3.9}\,(9.56\times10^{3})(10^{-5})\\ &=0.298+(3.026)(0.0956)\\ &=0.298+0.289 \end{aligned}$$ $$\boxed{V_G\approx0.587\ \text{V}}$$ Check via (16.28): $\mathcal{E}_s=\sqrt{2qN_A\phi_s/(K_s\varepsilon_0)}=\sqrt{2(1.6\times10^{-4})(0.298)/(1.044\times10^{-12})}=9.56\times10^3$ V/cm. Same. **Comment.** These results are representative. In $kT/q$ units $|\phi_F|$ typically lies between 9 and 18 at 300 K, and for nondegenerate dopings $|\phi_F|$ is less than half the Si band gap in volts ($<0.56$ V). The computed $W$ is less than $W_T$ for this doping, as it must be (Fig. 16.9); indeed $W_T=\sqrt{2}\,W=0.88\ \mu$m here. The only possible surprise is the size of the surface field, $\sim10^4$ V/cm. The device parameters are identical to those of Fig. 16.10, and the computed $V_G$ agrees with the value read from that plot. --- ## 16.4 Capacitance–Voltage Characteristics Since the oxide blocks d.c. current, the main observable of an MOS-C is **capacitance**. It varies with gate voltage, and the measured $C$–$V$ characteristic is of great practical importance: to the device specialist it is a picture window into the internal nature of the structure. It is a powerful diagnostic tool for finding deviations from the ideal in both the oxide and the semiconductor, and therefore is routinely monitored during MOS fabrication. **Measurement.** Typical laboratories use automated equipment like that of Subsection 7.2.3 / Fig. 7.6. The MOS-C sits on a probing station, usually in a light-tight box (to exclude room light), connected by shielded cables to a $C$–$V$ meter. The meter superimposes a small a.c. signal (typically $\lesssim15$ mV rms, commonly at 1 MHz) on a preselected d.c. voltage and detects the resulting a.c. current. The d.c. voltage is swept slowly to obtain a continuous or quasi-continuous $C$–$V$ curve, displayed on a monitor and printed. More sophisticated equipment (Fig. 16.11a) combines a high-frequency (1 MHz) $C$–$V$ meter with low-frequency (quasistatic) capability. This section models the $C$–$V$ characteristic in the **low-frequency** and **high-frequency** limits. (These names originally referred to the a.c. signal frequency used in the measurement.) The theory is restricted to the ideal structure; practical measurement considerations come at the end of the section. ### 16.4.1 Theory and Analysis #### Qualitative Theory Fig. 16.11(b) shows representative high- and low-frequency data for a device on $N_D=9.1\times10^{14}\ \text{cm}^{-3}$ (100) Si with $x_o=0.119\ \mu$m. Capacitance (normalized to $C_o$, vertical axis 0 to 1.0) is plotted against $V_G$ from about $-5$ V to $+2$ V: it is flat near $1$ at positive $V_G$ (accumulation), falls through depletion at about $-1$ to $-3$ V, and then, at large negative $V_G$, the low-frequency curve climbs back up toward 1 while the high-frequency curve stays at its minimum. To explain this, follow an **n-type** MOS-C from accumulation through depletion to inversion, and ask how the charge responds to the small a.c. signal. **Accumulation.** The d.c. state is a pile-up of majority carriers right at the O–S interface. The system can change very rapidly: majority carriers equilibrate with time constants of order $10^{-10}$ to $10^{-13}$ s. Thus at probing frequencies $\le1$ MHz the device follows the a.c. signal quasistatically: the a.c. signal adds or subtracts a small $\Delta Q$ on the two sides of the oxide (Fig. 16.12a). This is exactly a parallel-plate capacitor, for low *or* high probing frequency: $$C(\text{acc})=C_o=\frac{K_o\varepsilon_0A_G}{x_o} \tag{16.29}$$ where $A_G$ is the gate area. **Depletion.** For n-type, the d.c. state is $-Q$ on the gate and $+Q$ of depletion-layer charge in the semiconductor (majority carriers withdrawn from a width $W$ next to the interface). Again only majority carriers participate, so the charge changes very rapidly. When the a.c. signal puts more negative charge on the gate, the depletion layer widens almost instantaneously (Fig. 16.12b): $W$ quasistatically fluctuates about its d.c. value. Remove the stationary d.c. charge conceptually and what remains is a small fluctuating charge on the two sides of a *double-layer insulator*. For all probing frequencies this is two parallel-plate capacitors in **series**: $$C_o=\frac{K_o\varepsilon_0A_G}{x_o}\quad\text{(oxide capacitance)} \tag{16.30a}$$ $$C_s=\frac{K_s\varepsilon_0A_G}{W}\quad\text{(semiconductor capacitance)} \tag{16.30b}$$ $$C(\text{depl})=\frac{C_oC_s}{C_o+C_s}=\frac{C_o}{1+\dfrac{K_oW}{K_sx_o}} \tag{16.31}$$ Because $W$ grows with depletion bias, $C(\text{depl})$ *decreases* as the d.c. bias moves from flat band to the onset of inversion. > **Intuition:** Picture the capacitor as two pillows stacked. The oxide pillow has fixed thickness. The semiconductor pillow is *inflatable*: the more you deplete, the thicker it is, and the harder it is for the plates to "feel" each other, so total capacitance falls. **Inversion.** Now minority carriers pile up near the interface and the d.c. depletion width is maximum, $W_T$. How does the a.c. charge respond? Two extreme options: the inversion charge fluctuates (Fig. 16.12c), or the depletion edge fluctuates (Fig. 16.12d); a combination is also conceivable. It turns out the answer depends on frequency. - **Very low frequency ($\omega\to0$):** minority carriers can be generated or annihilated in response to the a.c. signal; the a.c. state is a succession of d.c. states. As in accumulation, charge is added/removed near the edges of a single-layer insulator (Fig. 16.12c): $$C(\text{inv})\approx C_o\quad\text{for }\omega\to0 \tag{16.32}$$ - **Very high frequency ($\omega\to\infty$):** the sluggish generation–recombination process cannot supply or remove minority carriers in step with the a.c. signal. The inversion layer charge remains fixed at its d.c. value and only the depletion edge wiggles about $W_T$ (Fig. 16.12d), just like depletion biasing: two capacitors in series with $W=W_T$ constant: $$C(\text{inv})=\frac{C_oC_s}{C_o+C_s}=\frac{C_o}{1+\dfrac{K_oW_T}{K_sx_o}}\quad\text{for }\omega\to\infty \tag{16.33}$$ Because $W_T$ is independent of the d.c. inversion bias, the high-frequency inversion capacitance is a constant, the minimum: $C(\text{depl})_{min}$. At intermediate frequencies, where only part of the inversion charge can follow the signal, an intermediate inversion capacitance between the two limits is observed. > **Intuition (the slow mailman):** minority carriers in inversion are created by thermal generation, a slow process compared to a 1 MHz signal. A high-frequency signal is like shouting to a person who needs a second to respond: the "inversion layer" can't respond, so the only thing that moves is the depletion edge. A very slow signal gives thermal generation time to keep up, so the electrons obey and the capacitor looks like the oxide alone. **Overall theory.** For an n-type device the accumulating gate voltages ($C\approx C_o$) are positive, the inverting gate voltages are negative, and the decreasing-capacitance depletion region is of the order of a volt wide. $C\approx C_o$ under accumulation, decreases through depletion, and is again approximately constant under inversion: $\approx C_o$ if $\omega\to0$ or $C(\text{depl})_{min}$ if $\omega\to\infty$. This is in good agreement with the experimental curves of Fig. 16.11(b). ### Exercise 16.4 **Problem.** Complete the table using the ideal-structure $C$–$V$ characteristic (points a–g) and the block charge diagrams (1–5) of Figure E16.4. For each biasing condition give the letter(s) of the bias point(s) on the $C$–$V$ curve and the number of the block charge diagram. The figure shows an n-bulk $C$–$V$ curve vs. $V_G$: a flat high-$C$ plateau at positive $V_G$ (point g), a falling depletion portion (points f at flat band, e in mid-depletion, d at the depletion/inversion transition), and the inverted region on the left (points c, b, a; the high-frequency curve stays at minimum, the low-frequency curve rises to $C_o$). Charge diagrams: (1) $-Q$ on M, wide ionized-donor block in S (depletion); (2) no charge (flat band); (3) $+Q$ on M, thin electron sheet in S (accumulation); (4) $-Q$ on M, hole sheet plus ionized-donor block (inversion); (5) $-Q$ on M, wide ionized-donor block with the holes just appearing (transition). **Solution.** Reason from the polarity (n-type: positive $V_G$ = accumulation) and from what happens to $C$: - **Accumulation:** large positive $V_G$, $C=C_o$, electrons at the interface: point **g**, diagram **3**. - **Flat band:** $V_G=0$, boundary between accumulation and depletion, no charge: point **f**, diagram **2**. - **Depletion:** negative $V_G$ of modest size, $C$ falling, depletion block: point **e**, diagram **1**. - **Depletion/inversion transition:** maximum depletion width just reached: point **d**, diagram **5**. - **Inversion:** $V_G$ more negative than $V_T$, $W=W_T$, hole layer present: points **a/b/c** (a, b, c, depending on frequency: $C_{min}$ at high frequency, rising toward $C_o$ at low frequency), diagram **4**. | Bias condition | Capacitance point | Charge diagram | |---|---|---| | Accumulation | g | 3 | | Depletion | e | 1 | | Inversion | a / b / c | 4 | | Flat band | f | 2 | | Depl/inv transition | d | 5 | This matches the book's answer key. --- #### Delta-Depletion Analysis Building on the previous subsection, a first-order quantitative theory follows from the delta-depletion formulation: the charge blocks representing accumulation and inversion layers in Fig. 16.12 are formally replaced by $\delta$-functions at the O–S interface. Then $C$ equals $C_o$ *exactly* for accumulation biases, and for inversion biases in the low-frequency limit. The depletion relationship (16.31) and the high-frequency inversion relationship (16.33) are used unmodified (the block modelling in Fig. 16.12 matches the depletion approximation exactly). Thus: $$C=C_o\quad\text{accumulation} \tag{16.34a}$$ $$C=\frac{C_o}{1+\dfrac{K_oW}{K_sx_o}}\quad\text{depletion} \tag{16.34b}$$ $$C=C_o\quad\text{inversion }(\omega\to0) \tag{16.34c}$$ $$C=\frac{C_o}{1+\dfrac{K_oW_T}{K_sx_o}}\quad\text{inversion }(\omega\to\infty) \tag{16.34d}$$ Given device parameters, $C_o$ and $W_T$ follow from earlier relationships. To complete the analytical solution, the depletion-bias $W$ in (16.34b) must be expressed as a function of $V_G$. Inverting (16.28) to get $\sqrt{\phi_s}$ as a function of $V_G$ and substituting into (16.15) gives $$W=\frac{K_s}{K_o}x_o\left[-1+\sqrt{1+\frac{V_G}{V_\delta}}\,\right] \tag{16.35}$$ where $$V_\delta\equiv\frac{qN_A\,K_s\,x_o^2}{2K_o^2\varepsilon_0}\quad\text{(p-bulk device; for n-bulk }N_A\to-N_D) \tag{16.36}$$ (The derivation: (16.28) is quadratic in $\sqrt{\phi_s}$, $V_G=\phi_s+2\sqrt{V_\delta\phi_s}$, whose positive root is $\sqrt{\phi_s}=\sqrt{V_\delta}\big(-1+\sqrt{1+V_G/V_\delta}\big)$; then $W=\frac{K_s}{K_o}x_o\sqrt{\phi_s/V_\delta}$.) Note that if (16.35) is substituted into (16.34b) one obtains a very simple result: $$C=\frac{C_o}{\sqrt{1+\dfrac{V_G}{V_\delta}}}\qquad\text{(depletion biases)} \tag{16.37}$$ (Check: $1+\frac{K_oW}{K_sx_o}=1+\big[-1+\sqrt{1+V_G/V_\delta}\big]=\sqrt{1+V_G/V_\delta}$.) A sample set of low- and high-frequency $C$–$V$ curves from the delta-depletion analysis is shown in Fig. 16.13 ($x_o=0.1\ \mu$m, $N_D=10^{15}\ \text{cm}^{-3}$, $T=300$ K, $V_G$ from $-2$ to $+1$ V): both curves equal $C_o$ (=1.0 normalised) on the accumulation side, both fall along the same curve (16.37) through depletion, and at the depletion/inversion transition the low-frequency curve returns to 1 while the high-frequency curve levels off at its minimum (about 0.25). ### 16.4.2 Computations and Observations #### Exact Computations The delta-depletion characteristics (Fig. 16.13) are a rather crude representation of reality. The first-order theory is credible for gate voltages comfortably inside a biasing region but fails badly near the transition points (accumulation to depletion, and depletion to inversion). Practical applications often need more accuracy, obtained from the exact charge distribution (results in Appendix C; the derivation is beyond the text's scope, but the results are tractable, so highly accurate ideal-structure $C$–$V$ curves are easily constructed). Sample exact-charge curves appear in Figs. 16.14–16.16, showing the effect of doping (16.14), oxide thickness (16.15) and temperature (16.16). Observations: - **Doping (Fig. 16.14, $x_o=0.1\ \mu$m, n-type $N_D=10^{14},10^{15},10^{16}$ and p-type analogues):** higher doping causes a significant *increase* in the high-frequency inversion capacitance and a substantial *widening* of the depletion bias region. At very high doping the capacitance approaches a bias-independent constant. This is no surprise: as doping increases the semiconductor looks more and more like a metal, and the MOS-C behaves more and more like an ordinary capacitor. - **Oxide thickness (Fig. 16.15, $x_o=0.1,0.2,0.3\ \mu$m):** thicker oxide also widens the depletion-bias region (a proportionately larger voltage is dropped across the oxide) and affects the high-frequency inversion capacitance (thicker oxide gives a *larger* $C/C_o$ minimum). - **Temperature (Fig. 16.16, $T=0$ to $100^\circ$C, $x_o=0.1\ \mu$m, $N_D=5\times10^{14}\ \text{cm}^{-3}$):** a moderate sensitivity of the inversion-bias capacitance and a near insensitivity of the depletion-bias capacitance to temperature. ### Exercise 16.5 **Problem.** (a) Using the exact-charge relationships of Appendix C, write a computer (MATLAB) program that computes and plots low-frequency $C/C_o$ vs. $V_G$ for $U=\phi_s/(kT/q)$ stepped in unit increments from $U=U_F-21$ to $U_F+21$. Use $K_s=11.8$, $K_o=3.9$, $T=300$ K, inputs $N_A$ and $x_o$ only. Set $N_A=10^{15}\ \text{cm}^{-3}$ and plot for $x_o=0.1,0.2,0.3\ \mu$m; compare with Fig. 16.15. (b) Do the same for high-frequency $C/C_o$ with $x_o=0.1\ \mu$m and $N_A=10^{14},10^{15},10^{16}\ \text{cm}^{-3}$ and compare with Fig. 16.14. (The high-frequency calculation is much more involved.) **Method.** Appendix C is outside this excerpt, so I use the standard exact (Boltzmann, equilibrium) MOS results for a p-substrate, with $U=\phi_s/(kT/q)$, $U_F=\phi_F/(kT/q)=\ln(N_A/n_i)$, $L_D=\sqrt{K_s\varepsilon_0kT/(q^2N_A)}$: $$\begin{aligned} F(U,U_F)&=\sqrt{e^{-U}+U-1+e^{-2U_F}\left(e^{U}-U-1\right)}\\ \mathcal{E}_s&=\operatorname{sgn}(U)\,\frac{\sqrt2\,(kT/q)}{L_D}\,F\\ V_G&=\phi_s+\frac{K_s}{K_o}x_o\,\mathcal{E}_s\ \ \ (\text{Eq. (16.26) with the exact }\mathcal{E}_s)\\ C_s&=\frac{\sqrt2\,K_s\varepsilon_0}{L_D}\,\frac{\left|1-e^{-U}+e^{-2U_F}\left(e^{U}-1\right)\right|}{2F}\\ \frac{C_{LF}}{C_o}&=\frac{C_s}{C_o+C_s}\quad\text{with}\quad C_o=\frac{K_o\varepsilon_0}{x_o} \end{aligned}$$ Step through $U$ from $U_F-21$ to $U_F+21$ (skipping $U=0$, where $F=0$ and a limit is used), compute $V_G$ and $C_{LF}/C_o$ at each point, and plot $C/C_o$ against $V_G$. In MATLAB: ```matlab NA=1e15; xo=[0.1 0.2 0.3]*1e-4; ni=1e10; kT=0.0259; q=1.6e-19; e0=8.854e-14; Ks=11.8; Ko=3.9; LD=sqrt(Ks*e0*kT/(q*NA)); UF=log(NA/ni); U=(UF-21):0.25:(UF+21); U(U==0)=1e-6; F=sqrt(exp(-U)+U-1+exp(-2*UF)*(exp(U)-U-1)); Cs=sqrt(2)*Ks*e0/LD*abs(1-exp(-U)+exp(-2*UF)*(exp(U)-1))./(2*F); for k=1:3 Co=Ko*e0/xo(k); VG=U*kT+(Ks/Ko)*xo(k)*sign(U).*sqrt(2)*(kT/LD).*F; plot(VG, Cs./(Co+Cs)); hold on end ``` **Numerical results I computed (p-type, $N_A=10^{15}$ cm$^{-3}$, $T=300$ K):** | $x_o$ | flat-band $C/C_o$ | LF exact dip: min $C/C_o$ (at $V_G$) | exact $V_T$ (at $\phi_s=2\phi_F$) | |---|---|---|---| | 0.1 $\mu$m | 0.70 | 0.28 (at 0.87 V) | 1.005 V | | 0.2 $\mu$m | 0.82 | 0.44 (at 1.24 V) | 1.414 V | | 0.3 $\mu$m | 0.87 | 0.54 (at 1.60 V) | 1.823 V | Sample points for $x_o=0.1\ \mu$m: $U=-9.5 o V_G=-10.0$ V, $C/C_o=0.995$; $U=-0.5 o V_G=-0.045$ V, $0.73$; $U=+1.5 o0.11$ V, $0.60$; $U=11.5\ (\phi_F) o0.57$ V, $0.34$; $U=21.5 o0.95$ V, $0.31$; $U=23.5 o1.03$ V, $0.47$; $U=26.5 o1.34$ V, $0.88$; $U=32.5 o10.6$ V, $0.995$. (the $V_T$ values agree with the delta-depletion result: $2\phi_F+\frac{K_s}{K_o}x_o\sqrt{2qN_A(2\phi_F)/(K_s\varepsilon_0)}$ with $2\phi_F=0.596$ V; the exact and delta-depletion $V_T$ coincide because $\phi_s$ is pinned there.) The curves show the depletion region widening and the dip becoming shallower as $x_o$ grows, matching the trends of Fig. 16.15, and the LF curve returning to 1.0 in inversion ($V_G$ beyond about $V_T$) and in accumulation (negative $V_G$ for p-type). For (b), the high-frequency inversion plateau is the depletion capacitance at $W_T$. Delta-depletion values for $x_o=0.1\ \mu$m are, using (16.33): $$\begin{aligned} N_A=10^{14}:&\ W_T=2.50\ \mu\text{m},\ \ C_{min}/C_o=\frac{1}{1+\frac{3.9}{11.8}\frac{2.50}{0.1}}=0.11\\ N_A=10^{15}:&\ W_T=0.88\ \mu\text{m},\ \ C_{min}/C_o=0.26\\ N_A=10^{16}:&\ W_T=0.31\ \mu\text{m},\ \ C_{min}/C_o=0.50 \end{aligned}$$ The exact $V_T$ values ($\phi_s=2\phi_F$) are 0.59, 1.01 and 2.13 V respectively, so the depletion region widens with doping as Fig. 16.14 shows. The exact-charge high-frequency curves land close to these plateaus and reproduce the trend of Fig. 16.14: higher doping gives higher $C_{min}$ and a wider depletion region. The book points the reader to the `MOS_CV` MATLAB file (Appendix M); results should compare favorably with Figs. 16.14 and 16.15. $$\boxed{\text{HF }C_{min}/C_o\approx0.11,\ 0.26,\ 0.50\ \text{for }N_A=10^{14},10^{15},10^{16}\ \text{cm}^{-3}}$$ #### Practical Observations So far we have avoided saying what "low" and "high" frequency mean in real measurements. Would a 100 Hz signal give a low-frequency $C$–$V$ curve? Surprisingly, *no*. In modern MOS-Cs, with long carrier lifetimes and low generation rates, even 10 Hz (the practical limit of bridge-type measurements) gives high-frequency-type characteristics. A genuine low-frequency curve requires indirect means such as the **quasistatic technique**: a slow (typically 10–100 mV/s) linear voltage ramp is applied to the gate while the current into the gate is monitored. The quasistatic displacement current is directly proportional to the low-frequency capacitance, so with calibration the current-vs-voltage data replicates the low-frequency $C$–$V$ curve. On the high side, one cannot let $\omega\to\infty$. Measurement frequencies seldom exceed 1 MHz: at higher frequencies the bulk resistance of the semiconductor lowers the observed capacitance, and at $\gtrsim1$ GHz the response time of majority carriers itself matters. The near-universal standard is **1 MHz**. Even so, high-frequency curves need care. If the d.c. voltage is ramped from accumulation into inversion, the result depends on the ramp rate $R$ (Fig. 16.17: curves for $R=0,0.035,0.096,0.26,2.6$ V/s; the vertical axis is $C/C_o$, the horizontal axis $V_G$ from about $-12$ to $+4$ V). Even at the slowest ramp the inversion part of the true HF curve is not recorded. One must stop the ramp in inversion and let the device equilibrate, or sweep slowly backward from inversion toward accumulation. #### Deep Depletion Let us look at the ramped measurement more closely. In accumulation or depletion only majority carriers are involved and the charge follows the gate bias instantly. On entering inversion, a significant number of *minority* carriers is required for equilibrium. They were not there before, cannot arrive from the distant back contact or across the oxide, and must be *generated* in the near-surface region, a sluggish process. The semiconductor is driven into nonequilibrium (Fig. 16.18a): to balance the charge added to the gate, the depletion width grows beyond $W_T$ to make up for the missing minority carriers. This nonequilibrium state, with a deficit of minority carriers and a depletion width exceeding the equilibrium value, is **deep depletion**. $W>W_T$ explains the *reduced* capacitance in the ramp measurement, and the decrease with faster ramp is logical: the faster the ramp, the fewer minority carriers are generated before a given inversion bias is reached. > **Intuition (the bathtub):** the gate keeps pouring "charge demand" into the tub. Minority carriers are a slow tap. If you pour faster than the tap can deliver, the water level (depletion width) rises above its equilibrium mark, the tub overflows into deep depletion. Stop pouring and the tap catches up, the level falls back. The limiting case is a semiconductor totally devoid of minority carriers, **total deep depletion** (Fig. 16.18b). Apart from the larger depletion width it is exactly the simple depletion state of Fig. 16.12(b), so by the delta-depletion result the capacitance is the depletion-form (16.37) extended to all $V_G$ beyond $V_T$: $$C=\frac{C_o}{\sqrt{1+\dfrac{V_G}{V_\delta}}}\quad\text{total deep depletion}\qquad(V_G>V_T\ \text{p-type};\ V_G<V_T\ \text{n-type}) \tag{16.38}$$ This agrees excellently with experiment and with the exact-charge analysis; the 2.6 V/s curve of Fig. 16.17 is an example. **Applications.** Deep depletion is central to **DRAMs** (deeply depleted MOS-Cs are the storage elements) and **CCDs** (optically generated carrier charge is temporarily stored in partially deep-depleted potential wells under an array of MOS-C gates). ### Exercise 16.6 **Problem.** The experimental $C$–$V$ curve of Fig. E16.6 was observed as follows: the d.c. bias was changed very slowly from point 1 to point 2. At point 2 the $V_G$ sweep rate was increased substantially. On arriving at point 3 the sweep was stopped and the capacitance decayed (rose) to point 4. Qualitatively explain the observed characteristic. **Solution.** 1. **1 to 2 (slow ramp):** the semiconductor equilibrates at each d.c. point, so one traces the standard high-frequency $C$–$V$ curve down to the equilibrium high-frequency minimum $C_{min}$ (at point 2, around the depletion/inversion transition). 2. **2 to 3 (fast ramp):** the semiconductor can no longer equilibrate. It is driven into **deep depletion**: $W>W_T$, so $C$ continues to fall *below* the equilibrium minimum, following the total-deep-depletion curve (16.38). 3. **3 to 4 (sweep stopped):** with $V_G$ fixed, minority carriers (holes for an n-type device) are gradually generated thermally in the near-surface region. They add to the inversion layer, the depletion width shrinks back toward $W_T$, and $C$ increases back to the equilibrium high-frequency value. Equilibrium is systematically restored inside the device through minority-carrier generation. --- ## 16.5 Summary and Concluding Comments This chapter introduced basic MOS terminology, concepts, visualization aids and analytical procedures. We described the MOS capacitor and defined the **ideal MOS structure**. The ideal MOS-C is both a convenient vehicle for MOS fundamentals and the *reference* for analysing more complex real structures; its reference role becomes clearer in Chapter 18, where some idealizations are removed and the resulting perturbations examined. The internal state of the MOS-C under static bias was described qualitatively with energy band and block charge diagrams. With them the terms received a physical meaning: - **Accumulation:** pile-up of majority carriers at the oxide–semiconductor interface. - **Flat band:** no bending of the semiconductor bands (equivalently, no charge in the semiconductor). - **Depletion:** repulsion of majority carriers from the interface, leaving uncompensated impurity ions. - **Inversion:** pile-up of minority carriers at the oxide–semiconductor interface. A remark on terminology: some analyses split the depletion bias region defined here into two sub-regions. In some MOS publications, "depletion" applies only to band bending $0<\phi_s<\phi_F$; **weak inversion** describes $\phi_F<\phi_s<2\phi_F$; and **strong inversion** (more inversion than weak inversion) replaces "inversion" as defined here. Quantitative expressions for charge density, electric field and electrostatic potential were established by solving Poisson's equation. A first-order solution based on the depletion approximation and $\delta$-function modelling of carrier charges was presented; the exact solution is in Appendix B. The formalism was then applied to model the MOS-C $C$–$V$ characteristics in the low- and high-frequency limits. Sample ideal-structure characteristics were examined for parametric dependencies. Finally, the practical meaning of "low" and "high" frequency was clarified, and deep depletion was noted to occur when the MOS-C is not allowed to equilibrate under inversion biasing: a nonequilibrium condition with a deficit of minority carriers and a depletion width in excess of the equilibrium value. --- ## Chapter 16 Problems The book's Problem Information Table lists, for each problem, the completion point, difficulty level, suggested weighting and a short description. The entries for the first problems are: | Problem | Complete after | Difficulty | Weighting | Short description | |---|---|---|---|---| | 16.1 | 16.3.1 | 2 | 10 (one each diagram) | $\phi_F,\phi_s$ sets / infer diagrams | | 16.2 | 16.3.1 | 2 | 8 (a-2, b-3, c-3) | Examine Fig. 16.8(c) | | 16.3 | 16.3.1 | 2 | 8 | GaAs $W_T$ vs. doping plot | # Chapter 17: MOSFETs — The Essentials > **Source slice:** Pierret, *Semiconductor Device Fundamentals*, Chapter 17 (book pp. 611–644). The OCR was noisy; garbled symbols have been repaired from physics knowledge. Where a figure is missing, its content is described. Imagine a garden hose. You do not push water by hand through the hose. You step on it, and the squeeze controls the flow. A MOSFET is the same idea in silicon. The **gate** is your foot. The **source** is the tap. The **drain** is the open end. The "hose" is a thin sheet of electrons that the gate itself creates from nothing. Billions of these tiny taps, switched on and off, make up every phone and laptop processor. This chapter explains how one of them works. Historical note on names: the device was once called MOST, IGFET, MOSFET, and even PIGFET and MISFET. The name **MOSFET** (metal-oxide-semiconductor field-effect transistor) won. We keep assuming an **ideal MOS structure** (no oxide charge, no work-function difference, no interface traps). We study the basic **long-channel enhancement-mode** device. Small-dimension effects come in Chapter 19, and non-idealities in Chapter 18. Plan of the chapter: 1. Qualitative theory of operation (what happens inside, in pictures). 2. Quantitative $I_D$–$V_D$ relationships (square-law theory and bulk-charge theory, and a glance at charge-sheet and exact-charge theories). 3. a.c. response (small-signal circuits, cutoff frequency, characteristics). --- ## 17.1 Qualitative Theory of Operation ### The structure A MOSFET is, physically, **an MOS capacitor with two pn junctions placed immediately next to the gated region**. Figure 17.1 shows it. Take a p-type silicon substrate (an *n-channel* device): - Two heavily doped $n^+$ islands, the **source (S)** and the **drain (D)**, are diffused or implanted in the surface. - Between them sits the **gate (G)**: metal (or polysilicon) on top of a thin $\text{SiO}_2$ layer. - The substrate (back, B) contacts the bulk. Unless stated otherwise, **source and back are grounded**. An $n$-bulk device is the mirror image, with $p^+$ islands. Terminal conventions: - $V_G$ is the gate voltage and $V_D$ the drain voltage, both relative to ground. - The drain current $I_D$ is the main d.c. observable. - Electrons always **enter at the source, leave at the drain, and are controlled by the gate**. That is the origin of the names. - In normal operation the drain pn junction is **reverse biased** ($V_D \ge 0$ for the p-bulk device). - For the p-bulk device $I_D$ is positive when flowing from the external circuit **into** the drain terminal. ### Step 1: what does the gate alone do? (set $V_D=0$) Let $V_T$ be the depletion-to-inversion transition voltage. - If $V_G \le V_T$ (accumulation or depletion), the region between the $n^+$ islands holds an excess or deficit of holes and almost no electrons. Looking along the surface from source to drain you see **an open circuit** (two back-to-back diodes). - If $V_G > V_T$ (inversion), a layer of mobile electrons forms at the Si surface. Looking along the surface now you see an induced "$n$-type" strip, the **inversion layer** or **conducting channel**, joining source to drain (Fig. 17.2a). The more you invert, the more electrons pile up and the larger the channel conductance. So the gate **creates** the channel and sets its **maximum conductance**. ### Step 2: what does the drain voltage do? (fix $V_G > V_T$, raise $V_D$ from 0) **Thought experiment.** The channel is a thin conducting sheet lying on a hill whose height is set by the gate. The channel's electrons exist only because the gate voltage exceeds the local channel potential. As you drive current down the channel, the channel potential rises from 0 at the source to $V_D$ at the drain. The gate's "pull" at the drain end is therefore weaker, $V_G - V_D$ instead of $V_G$. The sheet gets thinner towards the drain. The behavior unfolds in phases (Figs. 17.2 and 17.3): 1. **$V_D = 0$ (Fig. 17.2a).** Thermal equilibrium, $I_D = 0$. 2. **Small $V_D$ (line O to A in Fig. 17.3).** The channel is a plain resistor, so $I_D \propto V_D$. The reverse-biased drain junction adds a junction leakage current, but in a good device it is negligible compared with the channel current as long as $V_D$ is below junction breakdown. 3. **Moderate $V_D$ (Fig. 17.2b).** The voltage drop along the channel starts to cancel the gate's inverting effect. The depletion region widens going from source to drain, and the number of inversion carriers falls. Lower carrier density means lower conductance, so the slope of the $I_D$–$V_D$ curve decreases (the curve "slopes over"). 4. **Pinch-off (Fig. 17.2c, point B).** The inversion layer disappears completely right at the drain end: the surface carrier concentration there falls to the bulk doping level. This onset of surface depletion at the drain end is called **pinch-off**. The slope of the characteristic becomes about zero. The drain voltage at which this happens is $V_{Dsat}$. 5. **Post pinch-off, $V_D > V_{Dsat}$ (Fig. 17.2d).** The pinched-off point grows into a depleted section of length $\Delta L$. This section has few carriers, hence low conductance, so it **absorbs nearly all of the extra voltage** $V_D - V_{D\text{sat}}$. **Why does the current stop growing after pinch-off?** If $\Delta L \ll L$, the part of the channel from source to pinch-off point has the same shape and the same end-point voltages (0 and $V_{Dsat}$) for every $V_D \ge V_{Dsat}$. Same shape and same voltage across it means the same current. So $I_D \approx$ constant. (Water analogy: once the hose is kinked at the end, a bigger pressure just pushes the water over a "waterfall" through the kink. The flow rate is set by the pipe upstream.) If $\Delta L$ is comparable to $L$ (short channels), the same voltage $V_{Dsat}$ drops across a shorter effective channel $L-\Delta L$, so $I_D$ creeps upward with $V_D$ after pinch-off (see the dashed part of Fig. 17.3, "channel-length modulation"). ### Step 3: assembling the family of curves (Fig. 17.4) - For $V_G \le V_T$: no channel, $I_D \approx 0$ for all $V_D$ below junction breakdown. - For each $V_G > V_T$: a curve like Fig. 17.3. A larger $V_G$ means higher channel conductance, hence a **steeper initial slope**, and more carriers at pinch-off, hence a **larger saturation current**, achieved at a larger $V_{Dsat}$. This is **transistor action**: the output-circuit current $I_D$ is modulated by the input voltage $V_G$. **Terminology.** - **Saturation region:** the portion of the characteristic with $V_D > V_{Dsat}$. - **Linear (triode) region:** the portion with $V_D < V_{Dsat}$. - **n-channel device:** channel carriers are electrons. **p-channel device:** channel carriers are holes. ### Exercise 17.1: rewrite the section for a p-channel device **Problem.** How must the figures change if the illustrative MOSFET is p-channel? **Solution (modifications).** - Fig. 17.1(a): p-type substrate becomes n-type substrate; $n^+$ source/drain becomes $p^+$ source/drain. - Fig. 17.1(b): holes replace electrons as channel carriers; p-Si bulk becomes n-Si; $V_D \ge 0$ becomes $V_D \le 0$; arrows reversed so that positive $I_D$ flows from source to drain and out of the drain contact. (IEEE convention would make $I_D \le 0$ for a p-channel; the book prefers positive drain currents.) - Fig. 17.2: source/drain become $p^+$; the inequalities in the caption flip. - Fig. 17.3: $V_D \to -V_D$ and $V_{Dsat} \to -V_{Dsat}$ on the horizontal axis (or redraw along the negative axis). - Fig. 17.4: "$V_G$ increasing" becomes "$-V_G$ increasing", and $V_D$ becomes $-V_D$. Since $V_T<0$ for the p-channel, no current flows if $V_G > V_T$ and current flows if $V_G < V_T$. --- ## 17.2 Quantitative $I_D$–$V_D$ Relationships Over the history of MOSFET modeling a **hierarchy** of long-channel formulas appeared, each more accurate but more complex. We study two: **square-law** theory (simple, qualitatively right) and **bulk-charge** theory (accurate). All but the last derivational steps are identical. ### 17.2.1 Preliminary Considerations #### Threshold voltage The transistor starts to carry current (turns on) at the onset of inversion, $\phi_s = 2\phi_F$ (here $\phi_F = (kT/q)\ln(N_A/n_i)$ for p-bulk). Substituting $\phi_s = 2\phi_F$ in the MOS-C result, Eq. (16.28), gives for an ideal device: $$V_T = 2\phi_F + \frac{K_S}{K_O}\,x_o\sqrt{\frac{4qN_A\phi_F}{K_S\,\epsilon_0}} \qquad \text{ideal n-channel (p-bulk)} \tag{17.1a}$$ $$V_T = 2\phi_F - \frac{K_S}{K_O}\,x_o\sqrt{\frac{4qN_D(-\phi_F)}{K_S\,\epsilon_0}} \qquad \text{ideal p-channel (n-bulk)} \tag{17.1b}$$ Using $C_o = K_O\epsilon_0/x_o$ this is $V_T = 2\phi_F + \sqrt{4qK_S\epsilon_0N_A\phi_F}/C_o$. Interpretation: $2\phi_F$ bends the bands, and the second term pays for the depletion charge $qN_AW_{max}$ that must be supported across the oxide. #### Effective mobility In the bulk, mobility is set by lattice and ionized-impurity scattering. In a MOSFET the gate field presses the carriers against the Si–SiO$_2$ interface (Fig. 17.5), adding **surface scattering** (collisions with the rough interface). The carriers closest to the surface suffer most. The average mobility of inversion-layer carriers is the **effective mobility** $\bar\mu_n$ (or $\bar\mu_p$). With the geometry of Fig. 17.6 ($x$ = depth from the interface, $y$ = distance along the channel from the source, $x_c(y)$ = channel depth, $z$ = width direction with width $Z$, length $L$): $$\bar\mu_n(y)=\frac{\displaystyle\int_0^{x_c(y)}\mu_n(x,y)\,n(x,y)\,dx}{\displaystyle\int_0^{x_c(y)}n(x,y)\,dx} \tag{17.2}$$ The channel electronic charge per cm$^2$ at position $y$ is $$Q_N(y) = -q\int_0^{x_c(y)} n(x,y)\,dx \tag{17.3}$$ Dividing the numerator of (17.2) by the denominator written as $-Q_N/q$ gives the alternative form $$\bar\mu_n = \frac{-q}{Q_N(y)}\int_0^{x_c(y)}\mu_n(x,y)\,n(x,y)\,dx \tag{17.4}$$ Simplification: for long channels the $y$-dependence of $\bar\mu_n$ is small and is neglected, so $\bar\mu_n$ is taken independent of $y$ and of $V_D$. Dependence on $V_G$: larger inversion bias means a stronger $x$-directed field, carriers squeezed closer to the interface, more surface scattering, so $\bar\mu_n$ **falls** with $V_G$ (Fig. 17.7). The first-order empirical model is $$\bar\mu_n = \frac{\mu_0}{1+\theta\,(V_G-V_T)} \tag{17.5}$$ with fit constants $\mu_0$ and $\theta$. In Fig. 17.7: $\mu_{bulk} = 1340$ cm$^2$/V-s, $\mu_0=847$ cm$^2$/V-s, $\theta = 0.0446$ /V. The effective mobility can be far below the bulk value. ### 17.2.2 Square-Law Theory Device: long-channel, Fig. 17.6. Valid for $V_G \ge V_T$ and $0 \le V_D \le V_{Dsat}$. **Step 1: current density.** The general electron current density is $$\mathbf J_N = q\mu_n n\,\boldsymbol{\mathcal E} + qD_N\nabla n \tag{17.6}$$ **Step 2: simplify.** In the channel the current is almost entirely along $y$; diffusion of the majority-type carrier is usually negligible. So, with $\mathcal E_y=-d\phi/dy$ ($\phi(y)$ is the channel potential): $$J_N \equiv J_{Ny} = q\mu_n n\,\mathcal E_y = -q\mu_n n\,\frac{d\phi}{dy} \tag{17.7}$$ ($\mu_n$, $n$, $J_{Ny}$ all depend on $x$ and $y$; $J_{Ny}$ is large at $x=0^+$ and drops quickly into the bulk.) **Step 3: current continuity.** Current through any cross-section equals $I_D$. Integrate $J_{Ny}$ over the channel cross-section ($Z$ wide, $x_c(y)$ deep). With $I_D$ positive in the $-y$ direction (hence the minus sign) and $d\phi/dy$ taken constant across $x$ because the channel is thin: $$I_D = -Z\int_0^{x_c(y)} J_{Ny}\,dx \tag{17.8a}$$ $$I_D = (-Z)\left(-\frac{d\phi}{dy}\right)\left(q\int_0^{x_c}\mu_n n\,dx\right) \tag{17.8b}$$ The last bracket equals $-\bar\mu_nQ_N$ by Eq. (17.4), so $I_D = (-Z)(-\tfrac{d\phi}{dy})(-\bar\mu_nQ_N)$, i.e. $$I_D = -Z\,\bar\mu_n\,Q_N(y)\,\frac{d\phi}{dy} \tag{17.9}$$ Sign check: $Q_N<0$ and $d\phi/dy>0$ give $I_D>0$. Good. **Step 4: integrate along the channel.** $I_D$ does not depend on $y$, and $\phi(0)=0$, $\phi(L)=V_D$: $$\int_0^L I_D\,dy = I_DL = -Z\bar\mu_n\int_0^{V_D}Q_N\,d\phi \tag{17.10}$$ $$I_D = -\frac{Z\bar\mu_n}{L}\int_0^{V_D}Q_N\,d\phi \tag{17.11}$$ **Step 5: charge in the channel (capacitor model).** In an MOS-C above threshold, the added gate charge is balanced by the inversion charge: $$\Delta Q_{gate} = -\Delta Q_{semi} = -Q_N \qquad (V_G\ge V_T) \tag{17.12}$$ Because the charges sit immediately against the oxide edges: $$\Delta Q_{gate} \cong C_o\,\Delta V_G = C_o(V_G-V_T) \tag{17.13}$$ $$Q_N = -C_o(V_G-V_T) \qquad (V_G\ge V_T) \tag{17.14}$$ $$C_o = \frac{K_O\epsilon_0}{x_o} \tag{17.15}$$ In a MOSFET the "bottom plate" potential varies, from 0 at the source to $V_D$ at the drain (Fig. 17.8: a **resistive-plate capacitor**). The plate-to-plate potential difference is $V_G$ at the source, $V_G-V_D$ at the drain, $V_G-\phi$ at point $y$. So the uniform drop $V_G-V_T$ is replaced by $V_G-V_T-\phi$: $$Q_N(y) = -C_o\,\big(V_G-V_T-\phi\big) \tag{17.16}$$ **Step 6: integrate.** $$\begin{aligned} I_D &= \frac{Z\bar\mu_nC_o}{L}\int_0^{V_D}\big(V_G-V_T-\phi\big)\,d\phi\\ &= \frac{Z\bar\mu_nC_o}{L}\left[(V_G-V_T)\phi-\frac{\phi^2}{2}\right]_0^{V_D} \end{aligned}$$ $$\boxed{I_D = \frac{Z\bar\mu_nC_o}{L}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right]}\qquad 0\le V_D\le V_{Dsat},\; V_G\ge V_T \tag{17.17}$$ Equation (17.17) applies **only below pinch-off**. If you plug in $V_D>V_{Dsat}$, the computed $I_D$ actually decreases. To first order, after pinch-off: $$I_D\big|_{V_D>V_{Dsat}} = I_D\big|_{V_D=V_{Dsat}}\equiv I_{Dsat} \tag{17.18}$$ so that $$I_{Dsat} = \frac{Z\bar\mu_nC_o}{L}\left[(V_G-V_T)V_{Dsat}-\frac{V_{Dsat}^2}{2}\right] \tag{17.19}$$ **Finding $V_{Dsat}$.** Pinch-off means $Q_N(L)\to0$ when $\phi(L)=V_D\to V_{Dsat}$: $$Q_N(L) = -C_o(V_G-V_T-V_{Dsat}) = 0 \tag{17.20}$$ $$V_{Dsat} = V_G-V_T \tag{17.21}$$ Substitute into (17.19): $$I_{Dsat} = \frac{Z\bar\mu_nC_o}{L}\left[(V_G-V_T)^2-\frac{(V_G-V_T)^2}{2}\right]$$ $$\boxed{I_{Dsat} = \frac{Z\bar\mu_nC_o}{2L}\,(V_G-V_T)^2} \tag{17.22}$$ This is why it is called the **square-law** theory: saturation current grows as the square of the gate overdrive (neglecting the $V_G$-dependence of $\bar\mu_n$). ### Exercise 17.2: the "universal" square-law characteristics **Problem.** Plot $I_D/(Z\bar\mu_nC_o/L)$ versus $V_D$ for $V_G-V_T=1,2,3,4$ V. Because the normalized current is $(V_G-V_T)V_D-V_D^2/2$, the plot is device independent: the same curves hold for any $Z,L,\bar\mu_n,x_o,N_A$. **Solution.** Let $V_{GT}=V_G-V_T$. Below pinch-off, normalized current $= V_{GT}V_D-V_D^2/2$ for $0\le V_D\le V_{GT}$. Above, it is constant $=V_{GT}^2/2$. Peak values are $0.5, 2, 4.5, 8$. MATLAB logic: loop over $V_{GT}$ from 4 down to 1; compute the parabola on 100 points up to $V_D=V_{GT}$ (default `linspace(0,VGT)`); append the point $V_D=9$ with the saturated value so the curve continues flat; plot with `hold on` and label each curve. ```matlab % "Universal" ID-VD Characteristics -- Square-Law Theory close clear % Let VGT = VG - VT for VGT = 4:-1:1 VD = linspace(0,VGT); % below pinch-off ID = VGT.*VD - VD.*VD./2; % ID/(Z*mu*Co/L) IDsat = VGT*VGT/2; VD = [VD, 9]; % extend flat part ID = [ID, IDsat]; if VGT == 4 plot(VD,ID); grid; axis([0 10 0 10]); xlabel('VD (volts)'); ylabel('ID/(ZCo/L)'); text(8,IDsat+0.2,'VG-VT=4V'); hold on else plot(VD,ID); if VGT == 3 text(8,IDsat+0.2,'VG-VT=3V'); elseif VGT == 2 text(8,IDsat+0.2,'VG-VT=2V'); else text(8,IDsat+0.2,'VG-VT=1V'); end end end hold off ``` (Book Fig. E17.2: four curves, each a rising parabola that flattens at $V_D=V_{GT}$; saturation heights $8, 4.5, 2, 0.5$ on a $0$–$10$ axis. The vertex of each parabola lies on the dashed locus $I=V_D^2/2$.) ### 17.2.3 Bulk-Charge Theory **The flaw in square-law theory.** The capacitor model balanced changes in gate charge only by $Q_N$. That silently assumes the depletion width stays at $W_0$ everywhere along the channel even when $V_D\neq0$. In reality (Figs. 17.2b–d) the depletion width **grows** from source to drain. This point-to-point variation of the "bulk" (depletion) charge must be included. **Step 1: charge balance with $W(y)$.** Gate charge $Q_G = C_o(V_G-V_T-\phi)+qN_AW_0$ (the threshold condition has $Q_N=0$, $W=W_0$). The semiconductor charge is $Q_N-qN_AW(y)=-Q_G$. Therefore $$Q_N(y) = -C_o\,(V_G-V_T-\phi) + qN_A\big[W(y)-W_0\big] \tag{17.23}$$ The bracket is positive (depletion widens), which **reduces** $|Q_N|$. **Step 2: depletion widths** (delta-depletion, surface potential $2\phi_F+\phi$ at point $y$): $$W(y) = \left[\frac{2K_S\epsilon_0\,(2\phi_F+\phi)}{qN_A}\right]^{1/2} \tag{17.24}$$ $$W_0 = \left[\frac{2K_S\epsilon_0\,(2\phi_F)}{qN_A}\right]^{1/2} \tag{17.25}$$ **Step 3: combine.** Define $$V_W \equiv \frac{\sqrt{4qK_S\epsilon_0N_A\phi_F}}{C_o}\qquad(\text{so } V_T=2\phi_F+V_W) \tag{17.26}$$ $$\begin{aligned} qN_A(W-W_0) &= \sqrt{2qK_S\epsilon_0N_A}\left[\sqrt{2\phi_F+\phi}-\sqrt{2\phi_F}\right]\\ &=\sqrt{4qK_S\epsilon_0N_A\phi_F}\left[\sqrt{1+\phi/2\phi_F}-1\right]\\ &=C_oV_W\left[\sqrt{1+\phi/2\phi_F}-1\right] \end{aligned}$$ Hence $$Q_N(y) = -C_o\left\{V_G-V_T-\phi-V_W\left[\sqrt{1+\frac{\phi}{2\phi_F}}-1\right]\right\} \tag{17.27}$$ **Step 4: integrate** (17.27) in (17.11): $$\begin{aligned} I_D &= \frac{Z\bar\mu_nC_o}{L}\int_0^{V_D}\left\{V_G-V_T+V_W-\phi-V_W\sqrt{1+\tfrac{\phi}{2\phi_F}}\right\}d\phi\\ &=\frac{Z\bar\mu_nC_o}{L}\left\{(V_G-V_T+V_W)V_D-\frac{V_D^2}{2}-V_W\cdot\frac{2}{3}(2\phi_F)\left[\left(1+\tfrac{V_D}{2\phi_F}\right)^{3/2}-1\right]\right\} \end{aligned}$$ using $\int\sqrt{1+\phi/2\phi_F}\,d\phi=\tfrac23(2\phi_F)(1+\phi/2\phi_F)^{3/2}$. So $$\boxed{I_D=\frac{Z\bar\mu_nC_o}{L}\left\{\Big(V_G-V_T+V_W\Big)V_D-\frac{V_D^2}{2}-\frac{4\phi_FV_W}{3}\left[\Big(1+\frac{V_D}{2\phi_F}\Big)^{3/2}-1\right]\right\}}\tag{17.28}$$ for $0\le V_D\le V_{Dsat}$, $V_G\ge V_T$. (This is algebraically equivalent to the book's printed form $\{(V_G-V_T-\tfrac{V_D}{2})V_D-\tfrac{2}{3}V_W(2\phi_F)^{1/2}\ldots\}$ rearranged. As a check, for small $V_D/2\phi_F$ expand $(1+x)^{3/2}\approx 1+\tfrac32x+\tfrac38x^2$: $I_D\approx\frac{Z\bar\mu_nC_o}{L}\{(V_G-V_T)V_D-\tfrac12(1+\tfrac{V_W}{4\phi_F})V_D^2\}$, which is the square law plus a small correction that vanishes as $V_W\to0$.) As before, post-pinch-off, $I_D\approx I_D(V_D=V_{Dsat})$. **Step 5: $V_{Dsat}$.** Set $Q_N(L)=0$ with $\phi(L)=V_{Dsat}$ in (17.27). Let $s=\sqrt{1+V_{Dsat}/2\phi_F}$, so $V_{Dsat}=2\phi_F(s^2-1)$: $$\begin{aligned} &V_G-V_T-2\phi_F(s^2-1)-V_W(s-1)=0\\ &2\phi_Fs^2+V_Ws-(V_G-V_T+V_W+2\phi_F)=0\\ &V_G-V_T+V_W+2\phi_F=V_G\quad(\text{since }V_T=2\phi_F+V_W)\\ &2\phi_Fs^2+V_Ws-V_G=0\\ &s=\frac{-V_W+\sqrt{V_W^2+8\phi_FV_G}}{4\phi_F} \end{aligned}$$ Then $V_{Dsat}=V_G-2\phi_F-V_Ws$: $$\boxed{V_{Dsat}=V_G-2\phi_F+\frac{V_W^2}{4\phi_F}-\frac{V_W}{4\phi_F}\sqrt{V_W^2+8\phi_FV_G}}\tag{17.29}$$ (Equivalently $V_{Dsat}=V_G-V_T-V_W\{[\ldots]^{1/2}-1\}$ in the book's grouping; the added term is always negative.) This is the content of Problem 17.14. **Comparing the theories (Fig. 17.9).** - Square law: simple, good for trends, qualitative reasoning, and quick algebra. - Bulk charge: agrees well with long-channel experiments. - The extra terms in (17.28) and (17.29) are **always negative**: they reduce $I_D$ and $V_{Dsat}$. - The square-law accuracy improves as substrate doping decreases. As $N_A\to0$ (and $x_o\to0$ in the sense $V_W\to0$) the bulk-charge theory reduces to the square-law theory. Fig. 17.9 used $x_o=0.1\ \mu$m, $T=300$ K, and compares $N_A=10^{14}$ and $10^{16}$ cm$^{-3}$ with the square law. ### Exercise 17.3: diode-connected MOSFET **Problem.** Gate and drain tied together. $x_o=500$ Å, $N_A=10^{16}$ cm$^{-3}$, $Z/L=10$, $\bar\mu_n=625$ cm$^2$/V-s, $T=300$ K. Using bulk-charge theory find $I_D$ for (a) $V_G=V_D=1$ V; (b) $V_G=V_D=3$ V. **Solution.** $$\phi_F=0.0259\ln\!\frac{10^{16}}{10^{10}}=0.358\ \text{V}$$ $$V_T=2\phi_F+\frac{K_S}{K_O}x_o\sqrt{\frac{4qN_A\phi_F}{K_S\epsilon_0}} =0.716+\frac{11.8}{3.9}(5\times10^{-6})\sqrt{\frac{4(1.6\times10^{-19})(10^{16})(0.358)}{(11.8)(8.85\times10^{-14})}}=1.42\ \text{V}$$ **(a)** $V_G=1\ \text{V}<V_T$, so the transistor is off: $I_D=0$. **(b)** Do **not** just put $V_D=3$ V into (17.28). In the bulk-charge theory $V_{Dsat}$ is below the square-law value $V_G-V_T$. Since $V_D=V_G>V_G-V_T>V_{Dsat}$, the device is saturated (gate tied to drain keeps it in saturation for all $V_G>V_T$). So $I_D=I_{Dsat}$. $$C_o=\frac{(3.9)(8.85\times10^{-14})}{5\times10^{-6}}=6.90\times10^{-8}\ \text{F/cm}^2$$ $$W_0=\left[\frac{2K_S\epsilon_0(2\phi_F)}{qN_A}\right]^{1/2}=3.06\times10^{-5}\ \text{cm}$$ $$V_W=\frac{qN_AW_0}{C_o}=0.71\ \text{V}$$ From (17.29) with $V_G=3$ V: $V_{Dsat}\approx1.15$ V. Then, from (17.28) at $V_D=V_{Dsat}$: $$\frac{Z\bar\mu_nC_o}{L}=10(625)(6.90\times10^{-8})=4.31\times10^{-4}\ \text{A/V}^2$$ $$\{\ldots\}\approx0.885\ \text{V}^2\quad\Rightarrow\quad I_D\approx 0.382\ \text{mA}$$ (The braces: $1.80-0.65+0.81-1.08\approx0.89$ with $V_D=1.14$ V.) ### 17.2.4 Charge-Sheet and Exact-Charge Theories Both earlier theories have two built-in limitations: 1. $Q_N$ is taken as **exactly zero** at and below $V_T$. In a real device the channel charge becomes small but never vanishes, so a residual **subthreshold current** flows. Its value is often important (leakage, low-power circuits). 2. They do **not self-saturate**: the post-pinch-off part must be added by hand. The **charge-sheet** and **exact-charge** theories remove both flaws: either can compute the subthreshold current and both saturate by themselves. (Their computational formulas are in Appendix D, which is outside this slice. Section "Problem 17.16" below uses the standard equivalent formulation.) The exact-charge result uses the exact charge distribution and involves integrals; the charge-sheet model is a simplified version (all inversion charge assumed to sit in an infinitely thin sheet at the surface) and offers the best accuracy/complexity trade-off. Figures 17.10 and 17.11 show: (i) theoretical $I_D$–$V_D$ curves for an n-channel device with $x_o=0.05\ \mu$m, $N_A=10^{15}$ cm$^{-3}$, $\bar\mu_n=550$, $L=7\ \mu$m, $Z=70\ \mu$m, $T=23^\circ$C, $V_G=2,3,4,5$ V (solid exact-charge, dashed charge-sheet, nearly coincident); (ii) subthreshold transfer characteristics, $\log I_D$ versus $V_G$ at $V_D=1$ V, for $N_A=10^{14}$ and $10^{15}$ cm$^{-3}$, with $I_D$ falling roughly exponentially from about $10^{-4}$–$10^{-5}$ A down to $10^{-10}$ A as $V_G$ goes from about $+0.4$ V to $-0.6$ V, and experimental points following the theory. --- ## 17.3 a.c. Response ### 17.3.1 Small-Signal Equivalent Circuits View the MOSFET as a **two-port** (Fig. 17.12a): input between gate and source, output between drain and source (source common). First restrict attention to **low frequencies**, where capacitive effects are negligible. (This mirrors the J-FET treatment of Subsection 15.2.4.) **Input port.** Looking into the gate you see a capacitor. A capacitor is an open circuit at low frequency, so the low-frequency input is modeled by an **open circuit**. **Output port.** The d.c. current is a function $I_D=I_D(V_D,V_G)$. Add small a.c. voltages $v_d$ and $v_g$ to the d.c. biases. If the device can follow (low frequency): $$I_D+i_d=I_D(V_D+v_d,\;V_G+v_g) \tag{17.30a}$$ $$i_d=I_D(V_D+v_d,\;V_G+v_g)-I_D(V_D,V_G) \tag{17.30b}$$ Taylor-expand about the d.c. operating point and keep first-order terms: $$I_D(V_D+v_d,V_G+v_g)=I_D(V_D,V_G)+\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G}v_d+\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D}v_g \tag{17.31}$$ Substitute into (17.30b): $$i_d=\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G}v_d+\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D}v_g \tag{17.32}$$ The partial derivatives have dimensions of conductance. Define $$g_d\equiv\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G=\text{const}}\quad\text{(drain or channel conductance)}\tag{17.33a}$$ $$g_m\equiv\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D=\text{const}}\quad\text{(transconductance or mutual conductance)}\tag{17.33b}$$ $$i_d=g_dv_d+g_mv_g \tag{17.34}$$ Equation (17.34) is the a.c. current-node equation at the drain: a conductance $g_d$ from drain to source in parallel with a current source $g_mv_g$ (Fig. 17.12b). Together with the open-circuit input, this is the **low-frequency small-signal equivalent circuit**. For FETs, $g_m$ plays the role that $\alpha$ and $\beta$ play for BJTs. Explicit $g_d$, $g_m$ come from differentiating (17.17), (17.22), (17.28) per the definitions (17.33). Write $\xi=Z\bar\mu_nC_o/L$ (the $V_G$-dependence of $\bar\mu_n$ is neglected). **Table 17.1 MOSFET small-signal parameters** | | Below pinch-off ($V_D\le V_{Dsat}$) | Above pinch-off ($V_D>V_{Dsat}$) | |---|---|---| | Square law $g_d$ | $\xi\,(V_G-V_T-V_D)$ | $0$ | | Square law $g_m$ | $\xi\,V_D$ | $\xi\,(V_G-V_T)$ | | Bulk charge $g_d$ | $\xi\left[V_G-V_T-V_D+V_W\left(1-\sqrt{1+V_D/2\phi_F}\right)\right]$ | $0$ | | Bulk charge $g_m$ | $\xi\,V_D$ | $\xi\,V_{Dsat}$, with $V_{Dsat}$ from (17.29) | (The bulk-charge below-pinch-off $g_d$ equals $-Z\bar\mu_nQ_N(L)/L$, the channel charge at the drain end. Each entry is verified in Problem 17.17.) **High frequencies.** The circuit must include capacitive coupling between terminals (Fig. 17.12c). A drain-to-source capacitor is omitted because it is usually negligible. Two capacitors appear: - $C_{gd}$: the **overlap capacitance**, from the part of the gate overlapping the drain island. It is minimized with a thicker oxide at the overlap, or better with **self-aligned gate** processing: poly-Si is deposited first, the gate defined, then source and drain islands formed by diffusion or implantation abutting the gate edges. - $C_{gs}$: essentially the capacitance of the MOS gate, $C_{gs}\approx C_G$. ### 17.3.2 Cutoff Frequency Define $\omega_{max}$ ($f_{max}$) as the frequency where the output-to-input current ratio, with the output **short-circuited**, has magnitude one: the transistor no longer amplifies. With $C_{gd}$ small and $C_{gs}\approx C_G$ ($=C_oZL$): $$i_{in}=j\omega\,(C_{gs}+C_{gd})\,v_g\approx j(2\pi f)\,C_G\,v_g\qquad(j=\sqrt{-1}) \tag{17.35}$$ $$i_{out}=g_mv_g \tag{17.36}$$ Setting $|i_{out}/i_{in}|=1$: $$\frac{g_m}{2\pi f_{max}C_G}=1\;\Rightarrow\; f_{max}=\frac{g_m}{2\pi C_G}$$ Using $g_m=\xi V_D$ (below pinch-off) and $C_G=C_oZL$: $$f_{max}=\frac{\bar\mu_nV_D}{2\pi L^2}\qquad(V_D\le V_{Dsat}) \tag{17.37}$$ The key message: **channel length $L$** controls $f_{max}$, and it enters squared. Shrinking $L$ is the route to fast transistors. (Intuition: $f_{max}\sim1/\text{transit time}$, and the transit time is $L/v=L^2/(\mu V_D)$.) ### 17.3.3 Small-Signal Characteristics Fig. 17.13 shows two typical characteristics. **(a) $g_d$ versus $V_G$ at $V_D=0$.** At $V_D=0$ both theories give $$g_d=\frac{Z\bar\mu_nC_o}{L}\,(V_G-V_T)\qquad(V_D=0) \tag{17.38}$$ This is linear in $V_G$ and goes to zero at $V_G=V_T$. So extrapolating the linear part of $g_d$–$V_G$ to the $V_G$ axis gives an accurate estimate of $V_T$. The measured curve does not vanish sharply at $V_T$ because a small minority-carrier concentration exists at the transition, which the theory neglects. Since $g_d\propto\bar\mu_n$, the same data give $\bar\mu_n$ versus $V_G$ directly. This works well if the interface-trap density is low. A moderate-to-large trap density smears the $g_d$–$V_G$ curve and yields a falsely low $\bar\mu_n$ (see Subsection 18.2.4). **(b) Gate capacitance $C_G$ versus $V_G$ at $V_D=0$** (drain grounded). It is used for diagnostics just like the MOS-C $C$–$V$ curve and can be modeled to first order by the **low-frequency** MOS-C curve, even when probed above 1 MHz. Why? The source and drain islands supply the minority carriers. Electrons just run laterally through the surface channel into and out of the gate area to follow the a.c. signal. In the MOS-C, minorities must be thermally generated and are too slow. The curve goes from $C_{Gmax}=C_oZL$ in accumulation, dips to a minimum in depletion, and returns to $C_{Gmax}$ in inversion. ### Exercise 17.4: $g_m$–$V_G$ characteristics **Problem.** (a) Deduce the general form of $g_m$ versus $V_G$ with $V_D$ fixed. Using the square-law entries and $V_T=2$ V, plot $g_m/(Z\bar\mu_nC_o/L)$ versus $V_G$ ($0\le V_G\le10$ V) for $V_D=2,4,6$ V, neglecting $\bar\mu_n(V_G)$. (b) Compare with experimental curves and speculate on differences. **Solution (a).** With $\xi=Z\bar\mu_nC_o/L$, Table 17.1 gives $$\frac{g_m}{\xi}=\begin{cases}0 & V_G<V_T\\ V_G-V_T & 0\le V_G-V_T\le V_D\ \text{(saturated)}\\ V_D & V_G-V_T\ge V_D\ \text{(below pinch-off)}\end{cases}$$ For each $V_D$: zero until $V_G=V_T=2$ V; then it rises linearly with slope 1; at $V_G=2+V_D$ the device leaves saturation and $g_m/\xi$ stays flat at $V_D$. Plateaus of 2, 4, 6 are reached at $V_G=4,6,8$ V. ```matlab VT = 2; VG = linspace(0,10,501); hold on for VD = [2 4 6] gm = zeros(size(VG)); sat = (VG>VT) & (VG-VT<=VD); % saturation: gm/xi = VG-VT lin = (VG-VT>VD); % below pinch-off: gm/xi = VD gm(sat) = VG(sat)-VT; gm(lin) = VD; plot(VG,gm); text(10.1,VD,['VD=' num2str(VD) 'V']); end xlabel('VG (volts)'); ylabel('gm/(Z mu Co/L)'); grid; hold off ``` **(b)** Experimental $g_m$ ($V_D$ from 2 to 6 V, $g_m$ up to about 2.5 mS) shows the same rise but **falls off** for $V_G>V_D+V_T$ instead of staying flat. The cause is the decrease of $\bar\mu_n$ with $V_G$ (surface scattering, Eq. 17.5), which the theory neglected. --- ## 17.4 Summary The chapter introduced MOSFET terminology, operation and analysis for an ideal MOS structure and the basic transistor. - **Qualitative picture.** Inversion creates a surface channel between source and drain. The larger the gate voltage above turn-on, the more conductive the channel at a given $V_D$. A nonzero $V_D$ drives current: proportional to $V_D$ at small $V_D$, sloping over as the channel narrows, and **saturating** when the channel pinches off near the drain. - **Two complications in the quantitative analysis.** (1) Surface scattering reduces mobility, requiring an **effective mobility**. (2) The carrier concentration and current density in the channel vary strongly with position and drop rapidly into the bulk. Even so, the first-order result is surprisingly simple. - **Square-law theory:** Eqs. (17.17), (17.21), (17.22). **Bulk-charge theory** (more accurate, more complex): Eqs. (17.28), (17.29). More exacting theories (charge-sheet, exact-charge) are in Appendix D. - **a.c. response.** Low- and high-frequency small-signal circuits (Figs. 17.12b, c). A **short channel** is required for high-frequency operation, since $f_{max}\propto1/L^2$. Useful information (threshold voltage, effective mobility, interface quality) is extracted from $g_d$, $g_m$, $C_G$ versus d.c. bias. - The treatment parallels the J-FET of Chapter 15; comparing the two is a worthwhile exercise (Problem 17.22). Chapter 18 treats nonidealities; Chapter 19 small-dimension effects and structural variations. --- # Chapter 17 Problems The book's **Problem Information Table** classifies the problems as follows (difficulty level 1 to 4): | Problem | After section | Level | Points | Description | |---|---|---|---|---| | 17.1 | 17.4 | 1 | 10 | Quick quiz | | 17.2 | 17.4 | 2 | 10 | Sample computations | | 17.3 | 17.1 | 2 | 10 | Sketch diagrams | | 17.4 | 17.2.1 | 2 | 10 | Plot $V_T$ vs. $N_A$ for select $x_o$ | | 17.5 | 17.2.1 | 3 | 10 | $V_T$ temperature dependence | | 17.6 | 17.2.2 | 2 | 5 | $V_{Dsat}$ from $I_D$ maximum | | 17.7 | 17.2.2 | 2–3 | 10 | n to p in Subsection 17.2.2 | | 17.8 | 17.2.2 | 1 | 5 | Plot $I_D$–$V_D$, $\bar\mu_n$ per Eq. (17.5) | | 17.9 | 17.2.2 | 3 | 10 | $V_D=V_G-V_B$ | | 17.10 | 17.2.2 | 3–4 | 10 | $I_D$ versus $V_G-V_T$ | | 17.11 | 17.2.2 | 3–4 | 10 | Circular MOSFET | | 17.12 | 17.2.2 | 3 | 12 | $I_D$ temperature dependence | | 17.13 | 17.2.2 | 3–4 | 12 | Include $R_S$, $R_D$ in theory | | 17.14 | 17.2.3 | 3–4 | 10 | Derive Eq. (17.29) | | 17.15 | 17.2.3 | 3 | 10 | Plot bulk-charge $I_D$–$V_D$ | | 17.16 | 17.2.4 | 4 | 25 | Plot sheet, exact $I_D$–$V_D$ | | 17.17 | 17.3.1 | 2 | 5 | Verify Table 17.1 entries | | 17.18 | 17.3.1 | 2 | 5 | MOSFET $Q_N$, $g_d$ | | 17.19 | 17.3.2 | 3 | 10 | Deduce $Z$, $L$ for match | | 17.20 | 17.3.2 | 2–3 | 15 | General MOSFET review | | 17.21 | 17.3.3 | 3 | 12 | Small-signal characteristics | | 17.22 | 17.4 | 2 | 12 | Compare FETs | Constants used below unless stated: $T=300$ K, $kT/q=0.0259$ V, $n_i=10^{10}$ cm$^{-3}$, $K_S=11.8$, $K_O=3.9$, $\epsilon_0=8.85\times10^{-14}$ F/cm, $q=1.6\times10^{-19}$ C. Write $\xi\equiv Z\bar\mu_nC_o/L$. ## Problem 17.1: Quick quiz **(a) Why are the current-carrying contacts called source and drain?** Carriers (electrons in an n-channel device) enter the structure through the **source** (the origin of the carriers) and leave through the **drain** (where they are drained out of the channel). **(b) What is the channel?** The channel is the **surface inversion layer** induced by the gate between the source and drain islands: a thin conducting sheet of mobile carriers (electrons in an n-channel device, holes in a p-channel device). **(c) What is the saturation region?** It is the part of the $I_D$–$V_D$ characteristic with $V_D>V_{Dsat}$, i.e. beyond pinch-off, where $I_D$ is (nearly) independent of $V_D$. **(d) Difference between the depletion-inversion transition voltage and the threshold voltage?** (The item is cut off in the OCR; it asks to compare the "inversion point voltage" of the MOS-C discussion with the MOSFET threshold.) In an ideal device they are the **same quantity**: $V_T$ is the gate voltage at which $\phi_s=2\phi_F$. In the MOS-C it marks the onset of inversion; in the MOSFET it is called the turn-on voltage because the channel (conducting path) first appears there. **(e) Why is the surface-channel mobility different from the bulk mobility?** Besides lattice and ionized-impurity scattering, channel carriers are pressed against the Si–SiO$_2$ interface by the gate field and suffer **surface scattering**. Hence $\bar\mu_n<\mu_n^{bulk}$. **(f) Why "square-law" theory?** Because the saturation current is proportional to the **square** of the gate overdrive, $I_{Dsat}=\frac{Z\bar\mu_nC_o}{2L}(V_G-V_T)^2$ (and $I_D$ involves $V_D^2$ below pinch-off). **(g) Why "bulk-charge" theory?** Because the point-to-point variation of the **depletion ("bulk") charge** along the channel, $qN_A[W(y)-W_0]$, is included in the charge balance (Eq. 17.23). **(h) What is plotted in the subthreshold transfer characteristics?** $\log I_D$ (vertical axis) versus $V_G$ (horizontal), with $V_D$ held constant. **(i) Definitions of $g_d$ and $g_m$.** $$g_d=\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G},\qquad g_m=\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D}$$ **(j) Why is the measured $C$–$V_G$ ($V_D=0$) a low-frequency curve even at 1 MHz?** The source/drain islands directly supply the minority carriers needed to follow the a.c. signal; they flow laterally through the channel, so no slow generation process is involved. ## Problem 17.2: Sample calculations **Problem.** Ideal n-channel MOSFET, $T=300$ K, $Z=50\ \mu$m, $L=5\ \mu$m, $x_o=0.05\ \mu$m, $N_A=10^{16}$ cm$^{-3}$, $\bar\mu_n=800$ cm$^2$/V-s (independent of $V_G$). Find: (a) $V_T$; (b) $I_{Dsat}$ (square law) if $V_G-V_T=2$ V; (c) $I_{Dsat}$ (bulk charge) if $V_G-V_T=2$ V; (d) $g_d$ if $V_G-V_T=2$ V, $V_D=0$; (e) $g_m$ (square law) if $V_G-V_T=2$ V, $V_D=2$ V; (f) $g_m$ (bulk charge), same bias; (g) $f_{max}$ if $V_G-V_T=2$ V, $V_D=1$ V. **Solution.** Common quantities: $$\phi_F=0.0259\ln\frac{10^{16}}{10^{10}}=0.358\ \text{V}$$ $$C_o=\frac{K_O\epsilon_0}{x_o}=\frac{(3.9)(8.85\times10^{-14})}{5\times10^{-6}}=6.90\times10^{-8}\ \text{F/cm}^2$$ $$\xi=\frac{Z}{L}\bar\mu_nC_o=(10)(800)(6.90\times10^{-8})=5.52\times10^{-4}\ \text{A/V}^2$$ **(a)** $$V_W=\frac{\sqrt{4qK_S\epsilon_0N_A\phi_F}}{C_o}=\frac{\sqrt{4(1.6\times10^{-19})(11.8)(8.85\times10^{-14})(10^{16})(0.358)}}{6.90\times10^{-8}}$$ Numerator: $\sqrt{2.39\times10^{-15}}\approx4.88\times10^{-8}$ C/cm$^2$, so $V_W=0.708$ V. $$V_T=2\phi_F+V_W=0.716+0.708=1.42\ \text{V}$$ **(b)** Square law, (17.22): $$I_{Dsat}=\frac{\xi}{2}(V_G-V_T)^2=\frac{5.52\times10^{-4}}{2}(2)^2=1.10\ \text{mA}$$ **(c)** Bulk charge: $V_G=V_T+2=3.42$ V. From (17.29): $$\sqrt{V_W^2+8\phi_FV_G}=\sqrt{0.501+9.795}=3.209$$ $$s=\frac{-0.708+3.209}{4(0.358)}=1.747\ ;\quad V_{Dsat}=V_G-2\phi_F-V_Ws=3.42-0.716-1.237=1.47\ \text{V}$$ From (17.28) at $V_D=1.47$ V: $$\{\ldots\}=(2+0.708)(1.47)-\frac{(1.47)^2}{2}-\frac{4(0.358)(0.708)}{3}\left[\left(1+\frac{1.47}{0.716}\right)^{3/2}-1\right]$$ $$=3.981-1.080-0.338\,(5.34-1)\;\Rightarrow\;\{\ldots\}\approx1.435\ \text{V}^2$$ $$I_{Dsat}=\xi(1.435)=0.79\ \text{mA}$$ (The bulk-charge value is about 28% below the square-law value.) **(d)** At $V_D=0$ (Table 17.1 or (17.38)): $$g_d=\xi(V_G-V_T)=(5.52\times10^{-4})(2)=1.10\ \text{mS}$$ **(e)** Square law: $V_{Dsat}=V_G-V_T=2$ V $=V_D$, so the device sits at the pinch-off edge; both below- and above-pinch-off formulas agree: $$g_m=\xi V_D=(5.52\times10^{-4})(2)=1.10\ \text{mS}$$ **(f)** Bulk charge: $V_{Dsat}=1.47$ V $<V_D=2$ V, so the device is saturated, and $g_m=\xi V_{Dsat}$: $$g_m=(5.52\times10^{-4})(1.47)=0.81\ \text{mS}$$ **(g)** Below pinch-off ($V_D=1$ V $<V_{Dsat}$ in either theory): $$f_{max}=\frac{\bar\mu_nV_D}{2\pi L^2}=\frac{(800)(1)}{2\pi(5\times10^{-4})^2}=\frac{800}{1.571\times10^{-6}}=5.1\times10^{8}\ \text{Hz}\approx509\ \text{MHz}$$ ## Problem 17.3: Sketches for an ideal p-channel MOSFET **(a) Energy-band diagram at threshold, $V_D=0$ (n-bulk, p-channel).** Here $\phi_F<0$ ($E_F$ above $E_i$ in the bulk) and threshold is $V_G=V_T<0$. - Far from the surface: $E_F$ lies $q|\phi_F|$ **above** $E_i$. - Toward the surface the bands bend **upward** by $2q|\phi_F|$, so at the interface $E_i$ lies $q|\phi_F|$ **above** $E_F$ (the surface is as p-type as the bulk is n-type: inversion onset). A depletion region of width $W_{max}$ spans the bend. - Oxide: bands slope across it (a linear potential drop $V_{ox}$), with the gate Fermi level $E_{FM}$ **above** $E_F$ by $q|V_T|$ because $V_G<0$. - Ideal structure: no flat-band offset, so at $V_G=0$ everything would be flat. **(b) Block charge diagram at threshold.** - Gate: charge per unit area $-qN_DW_{max}$ (negative sheet at the gate/oxide interface; a block drawn below the axis at the oxide edge). - Semiconductor: a block of height $+qN_D$ and width $W_{max}$ (ionized donors, positive, depletion) starting at the oxide-semiconductor interface. Inversion charge is zero at threshold. - Total charge is zero: $-qN_DW_{max}+qN_DW_{max}=0$. **(c) Inversion layer and depletion region at pinch-off.** Draw a p-channel device: n-type substrate, $p^+$ source on the left and $p^+$ drain on the right, oxide and gate above, $V_D<0$. The hole inversion layer is a thin strip under the oxide, thick at the source and tapering to zero thickness exactly at the drain edge. Beneath it the n-type depletion region has constant width at the source and widens toward the drain, merging with the drain-substrate junction depletion region. Label: gate, oxide, $p^+$ source/drain, n-substrate, inversion layer, depletion region, channel length $L$, pinch-off point at the drain end ($\Delta L=0$). ## Problem 17.4: $V_T$ versus $N_A$ **Problem.** Plot $V_T$ vs. $N_A$ for ideal n-channel MOSFETs at room temperature, with curves for $x_o=0.0,0.02,0.05,0.1\ \mu$m, $10^{14}\le N_A\le10^{18}$ cm$^{-3}$, plotted $V_T$ limited to 0 to 3 V. **Solution.** Use (17.1a): $V_T=2\phi_F+\sqrt{4qK_S\epsilon_0N_A\phi_F}\,x_o/(K_O\epsilon_0)$, with $\phi_F=(kT/q)\ln(N_A/n_i)$. For $x_o=0$ the second term vanishes: $V_T=2\phi_F$ (only the band bending term). For finite $x_o$ the curve lies above it and rises faster with $N_A$ (the depletion charge $\propto\sqrt{N_A}$ is paid across the oxide). ```matlab % VT versus NA for ideal n-channel MOSFETs close; clear q = 1.6e-19; e0 = 8.85e-14; Ks = 11.8; Ko = 3.9; kT = 0.0259; ni = 1e10; NA = logspace(14,18,200); % cm^-3 phiF = kT*log(NA/ni); xo_list = [0.0 0.02 0.05 0.1]*1e-4; % um -> cm for xo = xo_list Co = Ko*e0/max(xo,eps); % F/cm^2 (xo=0: second term set to 0) if xo == 0, VT = 2*phiF; else, VT = 2*phiF + sqrt(4*q*Ks*e0*NA.*phiF)/Co; end semilogx(NA,VT); hold on end axis([1e14 1e18 0 3]); grid xlabel('N_A (cm^{-3})'); ylabel('V_T (volts)'); legend('x_o=0','0.02 \mum','0.05 \mum','0.1 \mum','Location','NorthWest'); hold off ``` Representative values (computed): at $N_A=10^{16}$: $V_T=0.72$ V ($x_o=0$), $1.00$ V (0.02 $\mu$m), $1.42$ V (0.05 $\mu$m), $2.13$ V (0.1 $\mu$m). The $x_o=0$ curve runs from about $0.48$ V at $10^{14}$ to $0.95$ V at $10^{18}$; thick-oxide curves leave the plotted window ($V_T>3$ V) at high $N_A$. ## Problem 17.5: Temperature dependence of $V_T$ **Problem.** Ideal n-channel MOSFET, $x_o$ and $N_A$ inputs. Write a program to plot $V_T$ vs. $T$, $200\le T\le400$ K. Record the output for $x_o=0.1\ \mu$m and $N_A=10^{16}$ cm$^{-3}$ (the OCR shows "$10^6$"; $10^{16}$ is the intended value). Describe the dependence. **Solution (logic).** $T$ enters through $kT/q$ and through $n_i(T)$ inside $\phi_F$: $$\phi_F(T)=\frac{kT}{q}\ln\frac{N_A}{n_i(T)},\qquad n_i(T)=n_i(300)\left(\frac{T}{300}\right)^{3/2}\exp\!\left[-\frac{E_G}{2k}\left(\frac1T-\frac1{300}\right)\right]$$ **Assumptions:** $E_G=1.12$ eV independent of $T$, $n_i(300)=10^{10}$ cm$^{-3}$ (Exercise 2.4 style). The depletion term also depends on $T$ through $\phi_F$. Then use (17.1a). ```matlab % VT versus temperature, ideal n-channel MOSFET close; clear q = 1.6e-19; e0 = 8.85e-14; Ks = 11.8; Ko = 3.9; k = 8.617e-5; % eV/K xo = input('xo in um = ')*1e-4; NA = input('NA in cm^-3 = '); Co = Ko*e0/xo; Eg = 1.12; T = 200:5:400; ni = 1e10*(T/300).^1.5.*exp(-(Eg/(2*k))*(1./T-1/300)); phiF = k*T.*log(NA./ni); % kT/q in volts = k*T VT = 2*phiF + sqrt(4*q*Ks*e0*NA*phiF)/Co; plot(T,VT); grid; xlabel('T (K)'); ylabel('V_T (volts)'); disp([T(1:10:end)' VT(1:10:end)']) ``` Output for $x_o=0.1\ \mu$m, $N_A=10^{16}$ cm$^{-3}$ (computed with the assumptions above): | $T$ (K) | 200 | 250 | 300 | 350 | 400 | |---|---|---|---|---|---| | $V_T$ (V) | 2.43 | 2.29 | 2.13 | 1.96 | 1.79 | **Description.** $V_T$ **decreases almost linearly** with temperature, about $-3.2$ mV/K (a total drop of roughly 0.64 V over 200 K). Reason: as $T$ rises, $n_i$ rises sharply, the Fermi level moves toward midgap, $\phi_F$ decreases, and both terms of $V_T$ shrink. ## Problem 17.6: $V_{Dsat}$ from the $I_D$ maximum **Problem.** If (17.17) is used beyond $V_{Dsat}$, $I_D$ peaks at $V_{Dsat}$. Use this to derive (17.21) from (17.17). **Solution.** A maximum requires $\partial I_D/\partial V_D=0$ with $\partial^2I_D/\partial V_D^2<0$. $$\frac{\partial I_D}{\partial V_D}=\xi\left[(V_G-V_T)-V_D\right]=0$$ $$V_D=V_G-V_T\equiv V_{Dsat}$$ $$\frac{\partial^2I_D}{\partial V_D^2}=-\xi<0\quad(\text{a maximum, as required})$$ This reproduces (17.21). (Physically the derivative $g_d=\xi(V_G-V_T-V_D)=-Z\bar\mu_nQ_N(L)/L$ vanishes exactly when the channel charge at the drain end vanishes.) ## Problem 17.7: Subsection 17.2.2 for a p-channel device **Problem.** Indicate how the equations of Subsection 17.2.2 change for a p-channel MOSFET. **Solution.** Setup: n-bulk, holes in the channel, $V_G<V_T<0$, $V_D\le0$, source grounded, $\phi(0)=0$, $\phi(L)=V_D$, $\phi\le0$. $I_D$ is taken positive for holes flowing from source to drain. Step by step: $$J_P=q\mu_pp\,\mathcal E-qD_P\nabla p\;\to\; J_{Py}=q\mu_pp\,\mathcal E_y=-q\mu_pp\,\frac{d\phi}{dy}\qquad(17.7\text{ with }\mu_n\to\mu_p,\ n\to p)$$ $$Q_P(y)=+q\int_0^{x_c}p\,dx>0\qquad(17.3)$$ $$I_D=Z\int_0^{x_c}J_{Py}\,dx=-Z\bar\mu_pQ_P\frac{d\phi}{dy}\qquad(17.9)$$ $$I_DL=-Z\bar\mu_p\int_0^{V_D}Q_P\,d\phi\qquad(17.10,\,17.11)$$ Charge: for $V_G\le V_T$, the added gate charge is negative and balanced by $+Q_P$: $$Q_P=-C_o(V_G-V_T)\qquad(17.14),\qquad Q_P(y)=-C_o(V_G-V_T-\phi)\qquad(17.16)$$ This is positive since $V_G-V_T-\phi<0$. Integrate: $$I_D=\frac{Z\bar\mu_pC_o}{L}\int_0^{V_D}(V_G-V_T-\phi)\,d\phi$$ $$I_D=\frac{Z\bar\mu_pC_o}{L}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right]\qquad(V_D\ge V_{Dsat},\ V_G\le V_T)\qquad(17.17)$$ Both terms keep the same form, but note the sign bookkeeping: $(V_G-V_T)V_D>0$ since both factors are negative, and the inequality on $V_D$ reverses ($|V_D|\le|V_{Dsat}|$). Pinch-off ($Q_P(L)=0$): $$V_{Dsat}=V_G-V_T\ (<0)\qquad(17.21)$$ $$I_{Dsat}=\frac{Z\bar\mu_pC_o}{2L}(V_G-V_T)^2\qquad(17.22)$$ Also: $V_T$ per (17.1b), $\bar\mu_n\to\bar\mu_p$ everywhere, and "$V_G\ge V_T$" becomes "$V_G\le V_T$", "$0\le V_D\le V_{Dsat}$" becomes "$V_{Dsat}\le V_D\le0$". ## Problem 17.8: Effect of $\bar\mu_n(V_G)$ **Problem.** Modify the Exercise 17.2 program, replacing $\bar\mu_n$ with Eq. (17.5). Normalize by $Z\mu_0C_o/L$ and superimpose plots for $\theta=0$ and $\theta=0.05$/V. **Solution.** With $\bar\mu_n=\mu_0/[1+\theta(V_G-V_T)]$: $$\frac{I_D}{Z\mu_0C_o/L}=\frac{V_{GT}V_D-V_D^2/2}{1+\theta V_{GT}}\ (V_D\le V_{GT}),\qquad\frac{I_{Dsat}}{Z\mu_0C_o/L}=\frac{V_{GT}^2/2}{1+\theta V_{GT}}$$ For $\theta=0.05$/V the saturation heights become $0.5/1.05=0.476$, $2/1.1=1.82$, $4.5/1.15=3.91$, $8/1.2=6.67$ (instead of $0.5,2,4.5,8$). The curves are progressively compressed at high $V_{GT}$, spacing between curves shrinks (the $V_G$ dependence becomes closer to linear than quadratic). ```matlab close; clear styles = {'-','--'}; theta_list = [0 0.05]; for it = 1:2 theta = theta_list(it); for VGT = 4:-1:1 VD = linspace(0,VGT); ID = (VGT.*VD - VD.*VD./2)/(1+theta*VGT); IDsat = VGT*VGT/2/(1+theta*VGT); VD = [VD, 9]; ID = [ID, IDsat]; plot(VD,ID,styles{it}); hold on end end grid; axis([0 10 0 10]); xlabel('VD (volts)'); ylabel('ID/(Z mu_0 Co/L)'); legend('\theta=0'); hold off % solid: theta=0, dashed: theta=0.05/V ``` ## Problem 17.9: Gate-drain battery **Problem.** A battery $V_B\ge0$ between gate and drain gives $V_D=V_G-V_B$, i.e. $V_G=V_D+V_B$. Using the square-law results, sketch $I_D$ vs. $V_D$ ($V_D\ge0$) for (a) $V_B=V_T/2$; (b) $V_B=2V_T$. **Solution.** The gate overdrive is $V_G-V_T=V_D+V_B-V_T$. Pinch-off condition: $V_D\ge V_G-V_T$, i.e. $V_D\ge V_D+V_B-V_T$, i.e. $V_B\le V_T$. **(a) $V_B=V_T/2\ (<V_T)$.** The pinch-off condition is satisfied for every $V_D$ at which the device is on, so it is always in saturation. It is on when $V_G\ge V_T$: $V_D\ge V_T/2$. $$I_D=\begin{cases}0 & V_D<V_T/2\\[2pt] \dfrac{\xi}{2}\left(V_D-\dfrac{V_T}{2}\right)^2 & V_D\ge V_T/2\end{cases}$$ Sketch: zero out to $V_D=V_T/2$, then a parabola starting with zero slope and growing quadratically (a diode-like "knee"). **(b) $V_B=2V_T\ (>V_T)$.** The device is on already at $V_D=0$ ($V_G=2V_T$), and the pinch-off condition $V_B\le V_T$ is never satisfied, so it stays in the linear region: $$I_D=\xi\left[(V_D+V_T)V_D-\frac{V_D^2}{2}\right]=\xi\left[V_TV_D+\frac{V_D^2}{2}\right]$$ Sketch: starts at the origin with slope $\xi V_T$, then curves **upward** (concave up) as the gate rises together with the drain. (This is the physical consequence of increasing both together.) ## Problem 17.10: $I_D$ versus $V_G-V_T$ at constant $V_D$ **Problem.** Sketch $I_D$ versus $V_G-V_T$ for $V_D=1,2,3,4$ V (the OCR reads "$,2,3,$ and 4 V"; 1 V is the missing value) and explain. **Solution.** Let $x=V_G-V_T$. For $x<0$: $I_D=0$. For $0\le x\le V_D$ the device is **saturated** (since $V_D\ge x=V_{Dsat}$): $$I_D=\frac{\xi}{2}x^2$$ For $x\ge V_D$ the device is **below pinch-off**: $$I_D=\xi\left(xV_D-\frac{V_D^2}{2}\right)$$ This is a straight line with slope $\xi V_D$. At $x=V_D$ both give $\xi V_D^2/2$, and the slopes agree ($\xi x=\xi V_D$): the line is **tangent** to the parabola. So each curve is zero until $x=0$, follows the common parabola $\xi x^2/2$ up to $x=V_D$, and then continues as a straight line of slope $\xi V_D$ (steeper for larger $V_D$). The line extrapolates back to the $x$-axis at $x=V_D/2$. All four curves share the same parabola and peel off it at $x=1,2,3,4$ V. ## Problem 17.11: Circular MOSFET **Problem.** Gate is an annulus (Fig. P17.11). With $r_1,r_2$ the inside and outside radii, show $I_D=\dfrac{2\pi\bar\mu_nC_o}{\ln(r_2/r_1)}\left[(V_G-V_T)V_D-\dfrac{V_D^2}{2}\right]$ (the printed formula uses diameters whose ratio equals $r_2/r_1$). (b) Show it reduces to (17.17) for $r_2=r_1+L$, $Z=2\pi r_1$, $L/r_1\ll1$. **Solution (a).** Use cylindrical coordinates, source inside at $r_1$, drain outside at $r_2$ (the result is the same if reversed). Current flows radially. Replace $y\to r$ and the width $Z\to2\pi r$ (circumference at radius $r$). $$J_{Nr}=-q\mu_nn\frac{d\phi}{dr}\qquad(\text{17.7 in }r)$$ $$I_D=-(2\pi r)\bar\mu_nQ_N(r)\frac{d\phi}{dr}\qquad(\text{17.9 with }Z\to2\pi r)$$ $I_D$ is independent of $r$; separate variables: $$I_D\frac{dr}{r}=-2\pi\bar\mu_nQ_N\,d\phi$$ $$I_D\int_{r_1}^{r_2}\frac{dr}{r}=-2\pi\bar\mu_n\int_0^{V_D}Q_N\,d\phi$$ $$I_D\ln\frac{r_2}{r_1}=2\pi\bar\mu_nC_o\int_0^{V_D}(V_G-V_T-\phi)\,d\phi\qquad(Q_N=-C_o(V_G-V_T-\phi))$$ $$\boxed{I_D=\frac{2\pi\bar\mu_nC_o}{\ln(r_2/r_1)}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right]}$$ **(b)** With $r_2=r_1+L$: $\ln(r_2/r_1)=\ln(1+L/r_1)\approx L/r_1$ for $L/r_1\ll1$ (series $\ln(1+u)\approx u$). Then $$\frac{2\pi}{\ln(r_2/r_1)}\approx\frac{2\pi r_1}{L}=\frac{Z}{L}$$ which gives $I_D=\frac{Z\bar\mu_nC_o}{L}[(V_G-V_T)V_D-V_D^2/2]$, i.e. (17.17). ## Problem 17.12: Temperature dependence of the saturation current **Problem.** Ideal n-channel MOSFET, $x_o=0.1\ \mu$m, $N_A=10^{16}$ cm$^{-3}$ (OCR: "$10^1$"; taken as $10^{16}$, consistent with Problem 17.5). Assume the channel $\bar\mu_n$ has the same temperature dependence as the bulk $\mu_n$. (a) Find $I_{Dsat}(T)/I_{Dsat}(300\text{ K})$ with square-law theory. (b) For $V_G=3$ V compute and plot it for $200\le T\le400$ K with $\mu_n(T)/\mu_n(300)$ superimposed; repeat for $V_G=10$ V; discuss. **Solution (a).** $$I_{Dsat}(T)=\frac{Z\bar\mu_n(T)C_o}{2L}\,[V_G-V_T(T)]^2$$ $Z$, $L$, $C_o$ do not depend on $T$. Taking the ratio: $$\boxed{\frac{I_{Dsat}(T)}{I_{Dsat}(300)}=\frac{\mu_n(T)}{\mu_n(300)}\left[\frac{V_G-V_T(T)}{V_G-V_T(300)}\right]^2}$$ **(b)** Assumption: $\mu_n(T)\propto T^{-3/2}$ (lattice scattering), so $\mu_n(T)/\mu_n(300)=(T/300)^{-3/2}$. $V_T(T)$ is from the Problem 17.5 program ($V_T=2.43,\,2.29,\,2.13,\,1.96,\,1.79$ V at $T=200,250,300,350,400$ K). ```matlab close; clear q=1.6e-19; e0=8.85e-14; Ks=11.8; Ko=3.9; k=8.617e-5; Eg=1.12; xo=0.1e-4; NA=1e16; Co=Ko*e0/xo; T=200:5:400; ni=1e10*(T/300).^1.5.*exp(-(Eg/(2*k))*(1./T-1/300)); phiF=k*T.*log(NA./ni); VT=2*phiF+sqrt(4*q*Ks*e0*NA*phiF)/Co; VT300=VT(T==300); mu=(T/300).^(-1.5); for VG=[3 10] figure ratio=mu.*((VG-VT)/(VG-VT300)).^2; plot(T,ratio,T,mu,'--'); grid xlabel('T (K)'); ylabel('ratio'); title(['VG = ' num2str(VG) ' V']) legend('IDsat(T)/IDsat(300)','mu(T)/mu(300)') end ``` Computed values: | $T$ (K) | $\mu_n(T)/\mu_n(300)$ | $I_{Dsat}$ ratio, $V_G=3$ V | $I_{Dsat}$ ratio, $V_G=10$ V | |---|---|---|---| | 200 | 1.84 | 0.78 | 1.70 | | 300 | 1.00 | 1.00 | 1.00 | | 400 | 0.65 | 1.25 | 0.71 | **Discussion.** Two competing effects: the mobility falls with $T$, but $V_T$ also falls with $T$, which **raises** the overdrive $V_G-V_T$. At $V_G=3$ V the overdrive is small (0.87 V at 300 K), so a given change in $V_T$ is a large *fractional* change in overdrive: the $V_T$ effect wins and $I_{Dsat}$ **increases** with $T$. At $V_G=10$ V the overdrive (7.9 V) is large and barely changes, so the mobility effect wins, the current follows the mobility curve (slightly flattened) and **decreases** with $T$. At a particular intermediate gate voltage the two effects cancel and $I_{Dsat}$ is temperature independent (a "zero-temperature-coefficient" bias point, useful in circuit design). Large-signal power MOSFETs run at high overdrive, so they have a negative temperature coefficient, which prevents thermal runaway. ## Problem 17.13: Source and drain series resistances **Problem.** Resistances $R_S$ and $R_D$ (contact plus island resistance) lie between the terminals and the channel. Show the square-law theory accounts for them by replacing $V_D\to V_D-I_D(R_S+R_D)$ and $V_G\to V_G-I_DR_S$ in (17.17), (17.21), (17.22). **Solution.** The intrinsic channel potentials are (Fig. P17.13): at the intrinsic source ($y=0$) $\phi(0)=I_DR_S$; at the intrinsic drain ($y=L$) $\phi(L)=V_D-I_DR_D$. The gate-to-channel voltage is $V_G-\phi$ as before, so $Q_N=-C_o(V_G-V_T-\phi)$ still holds. Integrate (17.11) between the new limits: $$I_D=\frac{Z\bar\mu_nC_o}{L}\int_{I_DR_S}^{V_D-I_DR_D}(V_G-V_T-\phi)\,d\phi$$ $$=\frac{Z\bar\mu_nC_o}{L}\left[(V_G-V_T)(\phi_L-\phi_0)-\frac{\phi_L^2-\phi_0^2}{2}\right]\qquad\phi_0=I_DR_S,\ \phi_L=V_D-I_DR_D$$ Define $V_D'\equiv\phi_L-\phi_0=V_D-I_D(R_S+R_D)$. Then $\phi_L+\phi_0=V_D'+2I_DR_S$, so $$\frac{\phi_L^2-\phi_0^2}{2}=\frac{V_D'(V_D'+2I_DR_S)}{2}=V_D'\left(\frac{V_D'}{2}+I_DR_S\right)$$ $$I_D=\frac{Z\bar\mu_nC_o}{L}\,V_D'\left[(V_G-V_T)-I_DR_S-\frac{V_D'}{2}\right]$$ With $V_G'\equiv V_G-I_DR_S$: $$\boxed{I_D=\frac{Z\bar\mu_nC_o}{L}\left[(V_G'-V_T)V_D'-\frac{V_D'^2}{2}\right]}$$ which is (17.17) with $V_D\to V_D'$, $V_G\to V_G'$. Pinch-off: $Q_N(L)=0$ gives $\phi_L=V_G-V_T$, i.e. $V_D'+I_DR_S=V_G-V_T$, i.e. $$V_{Dsat}'=V_G'-V_T\quad(17.21),\qquad I_{Dsat}=\frac{Z\bar\mu_nC_o}{2L}(V_G'-V_T)^2\quad(17.22).$$ (Because $I_D$ appears on both sides, the equations are implicit; the resistances reduce the current and the effect grows as $L$ shrinks, since the channel resistance becomes comparable to $R_S+R_D$.) ## Problem 17.14: Derive Eq. (17.29) **Problem.** Derive $V_{Dsat}$ for the bulk-charge theory. **Solution.** Pinch-off: $Q_N(L)=0$ with $\phi(L)=V_{Dsat}$ in (17.27): $$V_G-V_T-V_{Dsat}-V_W\left[\sqrt{1+\frac{V_{Dsat}}{2\phi_F}}-1\right]=0$$ Let $s=\sqrt{1+V_{Dsat}/2\phi_F}$, so $V_{Dsat}=2\phi_F(s^2-1)$: $$V_G-V_T-2\phi_Fs^2+2\phi_F-V_Ws+V_W=0$$ $$2\phi_Fs^2+V_Ws-(V_G-V_T+V_W+2\phi_F)=0$$ Using $V_T=2\phi_F+V_W$ (ideal), $V_G-V_T+V_W+2\phi_F=V_G$: $$2\phi_Fs^2+V_Ws-V_G=0$$ $$s=\frac{-V_W+\sqrt{V_W^2+8\phi_FV_G}}{4\phi_F}\qquad(\text{positive root})$$ Then from the quadratic, $2\phi_Fs^2=V_G-V_Ws$, so $V_{Dsat}=V_G-V_Ws-2\phi_F$: $$\boxed{V_{Dsat}=V_G-2\phi_F+\frac{V_W^2}{4\phi_F}-\frac{V_W}{4\phi_F}\sqrt{V_W^2+8\phi_FV_G}}$$ Check: $V_G=3$, $\phi_F=0.358$, $V_W=0.71$ gives $1.14$ V (Exercise 17.3). Small-$V_W$ check: $\sqrt{V_W^2+8\phi_FV_G}\approx\sqrt{8\phi_FV_G}$, expanding for $V_W\to0$ gives $V_{Dsat}\to V_G-2\phi_F-\ldots\to V_G-V_T$, the square-law result. ## Problem 17.15: Bulk-charge $I_D$–$V_D$ program **Problem.** Write a program that plots $I_D/(Z\bar\mu_nC_o/L)$ versus $V_D$ from the bulk-charge theory and verify the Fig. 17.9 curves ($x_o=0.1\ \mu$m, $T=300$ K, $N_A=10^{14}$ and $10^{16}$ cm$^{-3}$). **Solution (logic).** For each $N_A$ and each overdrive $V_{GT}=V_G-V_T$ ($=1,2,3,4$ V): compute $\phi_F$, $C_o$, $V_W$, $V_T=2\phi_F+V_W$, $V_G=V_T+V_{GT}$; compute $V_{Dsat}$ from (17.29); evaluate (17.28) (normalized by $Z\bar\mu_nC_o/L$) on $0\le V_D\le V_{Dsat}$; append a flat tail at $I_{Dsat}$ out to $V_D=12$ V. Compare against the square law curves of Exercise 17.2. ```matlab close; clear q=1.6e-19; e0=8.85e-14; Ks=11.8; Ko=3.9; kT=0.0259; ni=1e10; xo=0.1e-4; Co=Ko*e0/xo; NAs=[1e14 1e16]; sty={'--',':'}; for ia=1:2 NA=NAs(ia); phiF=kT*log(NA/ni); Vw=sqrt(4*q*Ks*e0*NA*phiF)/Co; VT=2*phiF+Vw; for VGT=1:4 VG=VT+VGT; VDsat=VG-2*phiF+Vw^2/(4*phiF)-(Vw/(4*phiF))*sqrt(Vw^2+8*phiF*VG); VD=linspace(0,VDsat); ID=(VGT+Vw)*VD-VD.^2/2-(4*phiF*Vw/3)*((1+VD/(2*phiF)).^1.5-1); VD=[VD 12]; ID=[ID ID(end)]; plot(VD,ID,sty{ia}); hold on end end for VGT=1:4 % square law for comparison VD=[linspace(0,VGT) 12]; ID=[VGT*VD(1:end-1)-VD(1:end-1).^2/2, VGT^2/2]; plot(VD,ID,'-') end grid; xlabel('VD (volts)'); ylabel('ID/(Z mu Co/L)'); hold off ``` Expected result: the bulk-charge curves lie below the square-law curves (reduced $I_D$ and $V_{Dsat}$); the $N_A=10^{14}$ curves nearly coincide with the square law (small $V_W$), while the $10^{16}$ curves (for $x_o=0.1\ \mu$m: $V_Wpprox1.42$ V, versus $V_Wpprox0.12$ V at $10^{14}$) are noticeably lower, e.g. at $V_{GT}=4$ the saturation level is clearly under 8, as in Fig. 17.9. ## Problem 17.16: Charge-sheet and exact-charge $I_D$–$V_D$ **Problem.** Using the Appendix D relationships, write programs for $I_D$–$V_D$ from (a) charge-sheet theory, (b) exact-charge theory, and compare with Figs. 17.10 and 17.11. **Note on the data.** Appendix D is not in the supplied text. The programs below implement the standard equivalent formulation, which gives the same physics: gradual-channel approximation with the electron quasi-Fermi potential $V_c(y)$ (0 at the source, $V_D$ at the drain) as the independent variable. Drift plus diffusion combine into $$I_D=\frac{Z\bar\mu_n}{L}\int_0^{V_D}\left[-Q_N(V_G,V_c)\right]dV_c$$ The integrand must be found at each $V_c$. **Step 1: surface potential.** With $V_t=kT/q$, $A=\sqrt{2qK_S\epsilon_0N_A}$ and (ideal, $V_{FB}=0$), $\psi_s$ satisfies charge balance $C_o(V_G-\psi_s)=A\,F(\psi_s,V_c)$ where - *Exact-charge* (full Poisson solution): $$F^2=\psi_s+V_t\left(e^{-\psi_s/V_t}-1\right)+e^{-2\phi_F/V_t}\,V_t\,e^{-V_c/V_t}\left(e^{\psi_s/V_t}-1\right)$$ - *Charge-sheet* (neglect the hole terms below the sheet; electrons treated as a sheet at the surface): $$F^2\approx\psi_s+V_t\,e^{(\psi_s-2\phi_F-V_c)/V_t}$$ **Step 2: channel charge.** - Charge-sheet: $Q_N=-A\left[F(\psi_s)-\sqrt{\psi_s}\right]$ (total charge minus the depletion-approximation bulk charge $-A\sqrt{\psi_s}$). - Exact: integrate the electron density through the semiconductor, using Gauss's law $\mathcal E(\psi)=\sqrt{2qN_A/K_S\epsilon_0}\,F(\psi)$ and $n(\psi)=n_{p0}e^{(\psi-V_c)/V_t}$, $n_{p0}=n_i^2/N_A$: $$Q_N=-q\int_0^{\psi_s}\frac{n(\psi)}{\mathcal E(\psi)}\,d\psi$$ (with $\psi=u^2$ to remove the $1/\sqrt\psi$ end-point singularity). **Step 3:** integrate over $V_c$ with `trapz`. Both models self-saturate (no artificial post-pinch-off construction) because at $V_c\approx V_G-V_T$ the sheet charge simply vanishes and the integrand becomes small. They also give a nonzero subthreshold current for $V_G<V_T$. ```matlab % Charge-sheet and exact-charge ID-VD (n-channel, cgs units) close; clear q=1.6e-19; e0=8.85e-14; Ks=11.8; Ko=3.9; Vt=0.0259; ni=1e10; NA=1e15; xo=0.05e-4; mu=550; Z=70e-4; L=7e-4; % Fig. 17.10 device Co=Ko*e0/xo; phiF=Vt*log(NA/ni); A=sqrt(2*q*Ks*e0*NA); npo=ni^2/NA; Fcs = @(p,Vc) sqrt(p + Vt*exp((p-2*phiF-Vc)/Vt)); Fex = @(p,Vc) sqrt(max(p + Vt*expm1(-p/Vt) + ... exp(-2*phiF/Vt)*Vt*exp(-Vc/Vt).*expm1(p/Vt), 1e-30)); VD = 0:0.25:9; figure; hold on for VG = 2:5 for model = 1:2 % 1 = charge sheet, 2 = exact ID = zeros(size(VD)); for k = 1:numel(VD) Vc = linspace(0,VD(k),40); integrand = zeros(size(Vc)); for j = 1:numel(Vc) F = Fcs; if model==2, F = Fex; end g = @(p) Co*(VG-p) - A*F(p,Vc(j)); ps = fzero(g,[1e-9 max(VG,1e-3)]); % surface potential if model==1 QN = -A*(F(ps,Vc(j)) - sqrt(ps)); else f = @(u) npo*exp((u.^2-Vc(j))/Vt)./ ... (sqrt(2*q*NA/(Ks*e0))*F(u.^2,Vc(j))).*2.*u; QN = -q*integral(f,0,sqrt(ps)); end integrand(j) = -QN; end if VD(k)>0, ID(k) = Z*mu/L*trapz(Vc,integrand); end end if model==1, plot(VD,ID*1e3,'--'); else, plot(VD,ID*1e3,'-'); end end end xlabel('V_D (volts)'); ylabel('I_D (mA)'); grid; hold off ``` Subthreshold check (Fig. 17.11): loop `VG = -0.6:0.05:0.5`, set `VD = 1`, and plot `semilogy(VG,ID)`. With the same code, $I_D$ falls exponentially at roughly $60\text{--}80$ mV per decade below $V_T$ rather than dropping abruptly to zero. Expected results (computed for the Fig. 17.10 device, $V_D=5$ V): $I_D\approx0.18$ mA ($V_G=2$ V), $0.60$–$0.68$ mA (3 V), $1.5$–$2.7$ mA (5 V depending on model details), in saturation by $V_D\approx V_G-V_T$ and with the two theories nearly overlapping, as in Fig. 17.10. Treat the numbers as approximate since exact agreement with the book's Appendix D equations cannot be guaranteed. ## Problem 17.17: Verify the bulk-charge entries of Table 17.1 **Solution.** Start from (17.28), $\xi=Z\bar\mu_nC_o/L$: $$I_D=\xi\left\{(V_G-V_T+V_W)V_D-\frac{V_D^2}{2}-\frac{4\phi_FV_W}{3}\left[\left(1+\frac{V_D}{2\phi_F}\right)^{3/2}-1\right]\right\}$$ **$g_d$ below pinch-off.** $$g_d=\left.\frac{\partial I_D}{\partial V_D}\right|_{V_G}=\xi\left\{V_G-V_T+V_W-V_D-\frac{4\phi_FV_W}{3}\cdot\frac32\cdot\frac{1}{2\phi_F}\left(1+\frac{V_D}{2\phi_F}\right)^{1/2}\right\}$$ $$\boxed{g_d=\xi\left[V_G-V_T-V_D+V_W\left(1-\sqrt{1+\frac{V_D}{2\phi_F}}\right)\right]}$$ (the factor: $\frac{4\phi_F}{3}\cdot\frac32\cdot\frac1{2\phi_F}=1$). **$g_m$ below pinch-off.** $V_G$ appears only in $(V_G-V_T+V_W)V_D$, so $$g_m=\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D}=\xi V_D$$ **Above pinch-off.** $I_D=I_{Dsat}(V_G)=I_D(V_D=V_{Dsat}(V_G),V_G)$ is independent of $V_D$, so $g_d=0$. By the chain rule: $$g_m=\frac{dI_{Dsat}}{dV_G}=\left.\frac{\partial I_D}{\partial V_G}\right|_{V_D=V_{Dsat}}+\underbrace{\left.\frac{\partial I_D}{\partial V_D}\right|_{V_D=V_{Dsat}}}_{=\,g_d(V_{Dsat})=0}\frac{dV_{Dsat}}{dV_G}=\xi V_{Dsat}$$ The second term vanishes because $g_d(V_{Dsat})=-Z\bar\mu_nQ_N(L)/L=0$ at pinch-off. $\blacksquare$ (Square-law entries follow the same way with $V_W=0$: $g_d=\xi(V_G-V_T-V_D)$, $g_m=\xi V_D$ below pinch-off, and $g_m=\xi(V_G-V_T)$ above.) ## Problem 17.18: Inversion charge and drain conductance **Problem.** $V_G=3$ V, $V_D=0$, $Z=70\ \mu$m, $L=7\ \mu$m, $\bar\mu_n=550$ cm$^2$/V-s, $x_o=0.05\ \mu$m, $V_T=1$ V. Square-law: (a) inversion charge per cm$^2$ at $y=L/2$; (b) $g_d$. **Solution.** $C_o=\dfrac{(3.9)(8.85\times10^{-14})}{5\times10^{-6}}=6.90\times10^{-8}$ F/cm$^2$. **(a)** With $V_D=0$ the channel potential is zero everywhere, $\phi(L/2)=0$: $$Q_N(L/2)=-C_o(V_G-V_T-\phi)=-(6.90\times10^{-8})(3-1-0)=-1.38\times10^{-7}\ \text{C/cm}^2$$ (about $8.6\times10^{11}$ electrons/cm$^2$). **(b)** $$g_d=\frac{Z\bar\mu_nC_o}{L}(V_G-V_T-V_D)=\frac{70}{7}(550)(6.90\times10^{-8})(2)=7.59\times10^{-4}\ \text{S}\approx0.76\ \text{mS}$$ ## Problem 17.19: Complementary pair **Problem.** n-channel: $Z=50\ \mu$m, $L=5\ \mu$m, $x_o=0.05\ \mu$m, $N_A=10^{16}$ cm$^{-3}$. The p-channel device has the same oxide thickness and doping (donors $N_D=10^{16}$) and must have the same $g_m$ at equivalent bias and $T=300$ K. Find $Z$ and $L$ of the p-channel device. Effective mobility is half the bulk mobility in both. **Solution.** Assumed bulk mobilities (lightly doped Si, 300 K): $\mu_n=1350$, $\mu_p=480$ cm$^2$/V-s (so effective values 675 and 240). Equal $x_o$ gives equal $C_o$; "equivalent bias" means equal overdrive magnitudes (same $|V_G-V_T|$ and $|V_D|$; the ideal $|V_T|$ is equal for equal doping). Since $g_m\propto(Z/L)\bar\mu\,C_o|V_D|$ (or $|V_G-V_T|$): $$\left(\frac{Z}{L}\right)_p\bar\mu_p=\left(\frac{Z}{L}\right)_n\bar\mu_n$$ $$\left(\frac{Z}{L}\right)_p=\left(\frac{Z}{L}\right)_n\frac{\mu_n^{bulk}/2}{\mu_p^{bulk}/2}=10\cdot\frac{1350}{480}=28.1$$ Choose the same (minimum) length $L_p=5\ \mu$m; then $Z_p=28.1\times5=140.6\ \mu$m, about **2.8 times wider** than the n-channel device. (Alternatively keep $Z_p=50\ \mu$m and shorten $L_p=1.78\ \mu$m, but then $f_{max}$ and other L-dependent properties no longer match, and $L$ is normally set to the technology minimum.) With 1350/480 the width ratio is $2.8$; if the bulk-mobility ratio used is $1500/600$ it becomes $2.5$. ## Problem 17.20: General MOSFET review **Problem.** Fig. P17.20 shows $I_D$–$V_D$ with $I_{Dsat}=10^{-3}$ A, $V_{Dsat}=5$ V, and three marked points. (The points' positions are not recoverable from the OCR. Assumption used here: point (1) lies in the rising region below pinch-off, point (2) also in the rising region at $V_D=2.5$ V, and point (3) lies in saturation at $V_D=10$ V, the right end of the axis. Formulas are given in general form so other positions can be substituted.) **Solution.** Square law: $V_{Dsat}=V_G-V_T=5$ V and $I_{Dsat}=\frac{\xi}{2}(5)^2=10^{-3}$ A: $$\xi=\frac{2I_{Dsat}}{(V_G-V_T)^2}=\frac{2\times10^{-3}}{25}=8\times10^{-5}\ \text{A/V}^2$$ **(a)** At point (1) (below pinch-off, $0<V_D<5$ V): inversion layer is a strip under the oxide connecting $n^+$ source and $n^+$ drain, **thick at the source and gradually thinner toward the drain but still nonzero at the drain end**. The depletion region (p-substrate) below it has constant minimum width at the source and widens toward the drain. Label the gate, oxide, $n^+$ source/drain, p-Si, inversion layer, depletion region, $L$, and the (absent) pinch-off point. **(b)** $V_G=V_T+V_{Dsat}=1+5=6$ V. **(c)** $x_o=0.1\ \mu$m: $C_o=\dfrac{(3.9)(8.85\times10^{-14})}{10^{-5}}=3.45\times10^{-8}$ F/cm$^2$. The charge at the drain end is $Q_N(L)=-C_o(V_G-V_T-V_D)$. At the assumed point (2), $V_D=2.5$ V: $$Q_N(L)=-(3.45\times10^{-8})(5-2.5)=-8.6\times10^{-8}\ \text{C/cm}^2$$ (If (2) were at $V_D\ge5$ V, i.e. at or beyond pinch-off, $Q_N(L)=0$.) **(d)** New overdrive $V_G-V_T=3$ V gives $V_{Dsat}=3$ V. $V_D=4$ V exceeds it, so the device is saturated: $$I_D=\frac{\xi}{2}(3)^2=\frac{8\times10^{-5}}{2}(9)=3.6\times10^{-4}\ \text{A}$$ **(e)** At the assumed saturated point (3), $g_d=0$. (If (3) were below pinch-off at $V_D$: $g_d=\xi(5-V_D)$.) **(f)** Saturated: $g_m=\xi V_{Dsat}=(8\times10^{-5})(5)=0.4$ mS (equivalently $2I_{Dsat}/(V_G-V_T)=0.4$ mS). (If below pinch-off: $g_m=\xi V_D$.) **(g)** $C_G$ versus $V_G$ with $V_D=0$: a **low-frequency MOS-C curve**. For $V_G\ll V_T$ (accumulation; for the ideal device, $V_G<0$) $C_G=C_oZL$ (maximum). In depletion it drops toward a minimum $C_{min}=\dfrac{C_oC_{dep}}{C_o+C_{dep}}ZL$ near $V_G\approx V_T$ (the minimum occurs slightly below $V_T$). In inversion ($V_G>V_T$) it rises back to $C_oZL$, because the source/drain islands supply electrons and the inversion layer screens the gate. So the shape is a flat high plateau, a dip around $0\le V_G\le V_T$, and a return to the same plateau. ## Problem 17.21: Small-signal characteristics **Problem.** Using the square-law entries of Table 17.1 and neglecting $\bar\mu_n(V_G)$, draw (one coordinate system per part): (a) $g_d/\xi$ versus $V_G$ ($0\le V_G\le5$ V) for $V_T=1$ V and $V_D=0,1,2$ V; (b) $g_d/\xi$ versus $V_D$ ($0\le V_D\le5$ V) for $V_G-V_T=1,2,3$ V; (c) $g_m/\xi$ versus $V_D$ ($0\le V_D\le5$ V) for $V_G-V_T=1,2,3$ V. **Solution.** **(a)** $g_d/\xi=V_G-V_T-V_D$ when $V_G-V_T\ge V_D$ (below pinch-off), and $0$ when $V_G-V_T<V_D$ (saturation or off). - $V_D=0$: zero for $V_G<1$ V, then a straight line of slope 1: $g_d/\xi=V_G-1$ (reaching 4 at $V_G=5$). - $V_D=1$ V: zero until $V_G=2$ V, then $V_G-2$ (reaching 3 at 5 V). - $V_D=2$ V: zero until $V_G=3$ V, then $V_G-3$ (reaching 2 at 5 V). Three parallel lines of slope 1, shifted right by $V_D$. Extrapolation of the $V_D=0$ line to the axis gives $V_T$ (as in Fig. 17.13a). **(b)** $g_d/\xi=(V_G-V_T)-V_D$ for $V_D\le V_G-V_T$, else 0. Each curve starts at $V_G-V_T$ at $V_D=0$ (1, 2, 3), decreases linearly with slope $-1$ and hits zero at $V_D=V_{Dsat}=V_G-V_T$ (at 1, 2, 3 V), then stays zero. **(c)** $g_m/\xi=V_D$ for $V_D\le V_G-V_T$ and $=V_G-V_T$ for $V_D>V_G-V_T$. Each curve is a straight line of slope 1 up to $V_D=V_G-V_T$ (1, 2, 3 V), then flat at the value $V_G-V_T$ (1, 2, 3). ```matlab VT=1; xi=1; VD=linspace(0,5,501); VG=linspace(0,5,501); % (a) figure; hold on for vd=[0 1 2] gd=xi*max(VG-VT-vd,0).*(VG-VT>=vd); plot(VG,gd); end xlabel('V_G (V)'); ylabel('g_d/(Z mu Co/L)'); grid; hold off % (b),(c) figure; hold on for vgt=[1 2 3] gd=xi*max(vgt-VD,0); plot(VD,gd); end xlabel('V_D (V)'); ylabel('g_d/(Z mu Co/L)'); grid; hold off figure; hold on for vgt=[1 2 3] gm=xi*min(VD,vgt); plot(VD,gm); end xlabel('V_D (V)'); ylabel('g_m/(Z mu Co/L)'); grid; hold off ``` ## Problem 17.22: Compare the MOSFET and the J-FET **Solution.** **Similarities.** - Both are majority-carrier, **voltage-controlled** unipolar devices with source, drain and gate terminals. A gate voltage modulates the conductance of a channel between source and drain. - Both have a **linear** region (channel behaves like a resistor, $I_D\propto V_D$ at small $V_D$) followed by sloping-over due to channel narrowing and **saturation after pinch-off**, where the pinched-off section absorbs the excess drain voltage. - Both have the gate input essentially open at low frequency (no d.c. gate current; reverse-biased junction in the J-FET, insulator in the MOSFET). Same small-signal circuit ($g_d$, $g_m$, $C_{gs}$, $C_{gd}$) and the same $f_{max}\sim g_m/2\pi C_G\propto1/L^2$. - Analysis follows the same gradual-channel method: integrate the local current equation along the channel from source to drain with a position-dependent channel charge, giving a square-law relation near saturation. $g_d$ and $g_m$ are defined identically. **Differences.** - *Structure and gate:* J-FET: the gate is a pn junction (or Schottky) and the channel is a **physical bulk n- or p-region** between depletion regions. MOSFET: the gate is an insulated MOS capacitor, and the channel is a **gate-induced surface inversion layer** that did not exist before the gate voltage was applied. - *How the gate controls the channel:* J-FET: the gate bias changes the **depletion width**, narrowing a built-in channel (normally-on, depletion mode). MOSFET: the gate bias changes the **inversion charge**; the standard device is **enhancement mode** (normally off, $V_T>0$ for n-channel). - *Gate polarity:* J-FET gates must be kept reverse biased (otherwise large gate current). MOSFET gate may be of either polarity, subject to oxide breakdown. - *Input impedance:* MOSFET is far higher (oxide leakage is tiny); J-FET limited by junction leakage. - *Transport:* J-FET carriers move in the bulk (bulk mobility); MOSFET carriers move in a surface layer and have an **effective mobility** reduced by surface scattering, which depends on $V_G$. - *Analysis:* J-FET: the channel charge is set by the depletion width $W(y)$ (the relation between $V_G$, $V_D$ and the channel opening $h(y)$, with pinch-off when the channels close), giving $I_D\propto(V_P-V_G)^2$-type results with the pinch-off voltage $V_P$. MOSFET: charge balance via $C_o$ (square law), with a second-order bulk-charge correction, a threshold voltage $V_T$, and the subthreshold current. - *Integration and application:* The MOSFET is easy to miniaturize and to build into dense complementary integrated circuits, so it dominates digital ICs and memories. J-FETs are used mostly as discrete low-noise amplifiers and analog switches. # Chapter 18: Nonideal MOS > **Where we are.** In Chapters 16 and 17 we met the *ideal* MOS structure: a perfect insulator, no stray charges, and a metal and semiconductor that "agree" on their work functions. It is a physicist's cartoon. Real devices are never that tidy. This chapter is the story of how reality departs from the cartoon, which villains cause the departures, how engineers hunted them down, and what is left over for the MOSFET threshold voltage. **Think of it like this.** The ideal MOS capacitor is a perfectly level billiard table. Real devices have a slightly tilted table (work-function difference), chalk dust on the felt (oxide charges), and a few sticky spots at the cushion (interface traps). The ball (the $C$-$V$ curve) still rolls, but its path is shifted, smeared, and sometimes unstable. This chapter teaches you to read those distortions as fingerprints. Because the MOS capacitor is simple to build, almost every nonideality is diagnosed from its $C$-$V$ curve. But the same imperfections hit the MOSFET, mostly through the threshold voltage $V_T$ (and, for interface traps, through $g_m$). The last section treats the transistor directly. **Constants used throughout** (Si, $T=300$ K): $K_s=11.8$, $K_O=3.9$, $\varepsilon_0=8.85\times10^{-14}$ F/cm, $n_i=10^{10}$ cm$^{-3}$, $kT=0.0259$ eV, $E_G=1.12$ eV, and $C_O=K_O\varepsilon_0/x_o$ (oxide capacitance per unit area). --- ## 18.1 Metal-Semiconductor Work-Function Difference ### The idea In the ideal MOS structure we assumed $\Phi_M=\Phi_S=\chi+(E_C-E_F)_{FB}$: the energy needed to lift an electron from the Fermi level to vacuum is the same in the metal and in the semiconductor. Then at $V_G=0$ nothing pulls on the semiconductor and the bands are flat. In a real Al-SiO$_2$-Si system the Fermi level sits at a different depth below vacuum in each isolated material: $$\Phi_M \neq \Phi_S=\chi+(E_C-E_F)_{FB}$$ (Fig. 18.1(a) draws the isolated components to scale.) ### A thought experiment: bringing the two materials together Imagine a wire connecting metal and semiconductor, and slide them toward each other in vacuum until they are a distance $x$ apart. 1. The wire lets charge flow, so the Fermi levels must line up (equilibrium). 2. If $\Phi_M\neq\Phi_S$, lining up the Fermi levels forces the **vacuum levels** of the two pieces to sit at different energies. 3. A vacuum-level difference means an electric field in the gap, and that field penetrates the semiconductor as band bending. The closer the pieces get, the stronger the field and bending. 4. Finally slip the insulator into the gap. It lowers the effective barriers ($\Phi_M\to\Phi_M'$, $\chi\to\chi'$, both by the same oxide electron affinity) and weakens the field by $K_O$. Result (Fig. 18.1(b)): at $V_G=0$ the semiconductor is **not** at flat band. There is a built-in potential, just as in a pn junction or metal-semiconductor diode. ### Deriving the built-in potential Equate the energy measured from the Fermi level up to the common reference along the two sides of the insulator: $$\underbrace{\Phi_M' + q\Delta\phi}_{\text{metal side}} \;=\; \underbrace{(E_C-E_F)_{FB}-q\phi_s+\chi'}_{\text{semiconductor side}}\qquad (18.1)$$ Here $\Delta\phi$ is the potential difference across the insulator and $\phi_s$ the surface potential. With the metal as zero-potential reference (the usual convention for built-in potentials), the potential difference across the whole structure is $$\phi_{bi}=-(\phi_s+\Delta\phi)=\phi_{MS}\qquad (18.2)$$ where $$\phi_{MS}=\frac{\Phi_M'-\chi'}{q}-\frac{(E_C-E_F)_{FB}}{q}\qquad (18.3)$$ Intuition: the built-in potential of an otherwise ideal structure is simply the work-function difference in volts. Note $\Phi_M'-\chi'=\Phi_M-\chi$ because the oxide lowers both by the same amount. ### Effect on the $C$-$V$ curve Take Fig. 18.1(b) as a p-type MOS-C ($\phi_{MS}<0$). To undo the built-in potential and reach $\phi_s=0$ we must apply $V_G=\phi_{MS}$, a negative voltage. Flat band means the same capacitance in real and ideal devices, so the real flat-band point is displaced $\phi_{MS}$ volts along the voltage axis (Fig. 18.2). The argument works at *any* point on the curve, since each $\phi_s$ maps to exactly one capacitance. So the whole real curve is translated rigidly: $$\boxed{\Delta V_G=(V_G-V_G')_{\text{same }C}=\phi_{MS}}\qquad (18.4)$$ where $V_G'$ is the ideal-device gate voltage. (The book uses a prime for ideal-device voltages only in this chapter, where real and ideal appear together.) ### Typical values $\phi_{MS}$ is computed from (18.3) using the appropriate $\Phi_M'-\chi'$ and the doping-dependent $(E_C-E_F)_{FB}$. At $T=300$ K: n$^+$ poly-Si gate has $\Phi_M-\chi=-0.18$ eV; Al gate has $\Phi_M-\chi=-0.03$ eV. Fig. 18.3 plots $\phi_{MS}$ against doping for both. Qualitatively: Al on p-Si gives about $-0.9$ V at $10^{15}$ cm$^{-3}$ and becomes more negative with doping; Al on n-Si is about $-0.3$ V at $10^{15}$; n$^+$ poly on p-Si is the most negative (about $-1$ V or more); n$^+$ poly on n-Si lies between Al(n-Si) and Al(p-Si). **Table 18.1: $\Phi_M-\chi'$ for selected metal-SiO$_2$-Si structures** | Gate metal | $\Phi_M-\chi'$ (eV) | |---|---| | Ag | 0.73 | | Au | 0.82 | | Cr | $-0.06$ | | Cu | 0.63 | | Mg | $-1.05$ | | Sn | $-0.83$ | Take-home: $\phi_{MS}$ is more often negative than not (especially for p-type) and is about a volt or less. It is small, perfectly predictable and cannot cause instabilities. ### Exercise 18.1 **Problem.** Can one choose gate material and doping to make $\phi_{MS}=0$, for $10^{14}\le N_A$ or $N_D\le10^{18}$ cm$^{-3}$ at 300 K? Use Table 18.1. **Solution.** We need $\Phi_M'-\chi'=(E_C-E_F)_{FB}$, so first find the range of $(E_C-E_F)_{FB}$. $$(E_C-E_F)_{FB}=(E_C-E_i)+(E_i-E_F)$$ $$\text{n-type: }\;(E_C-E_F)_{FB}\simeq\frac{E_G}{2}-kT\ln\frac{N_D}{n_i}\qquad\text{p-type: }\;(E_C-E_F)_{FB}\simeq\frac{E_G}{2}+kT\ln\frac{N_A}{n_i}$$ With $E_G/2=0.56$ eV: $$N_D=10^{14}:\;0.56-0.0259\ln(10^4)=0.56-0.24=0.32\text{ eV}$$ $$N_D=10^{18}:\;0.56-0.0259\ln(10^8)=0.56-0.48=0.08\text{ eV}$$ $$N_A=10^{14}:\;0.56+0.24=0.80\text{ eV}\qquad N_A=10^{18}:\;0.56+0.48=1.04\text{ eV}$$ So $0.08\le(E_C-E_F)_{FB}\le0.32$ eV for n-type and $0.80\le(E_C-E_F)_{FB}\le1.04$ eV for p-type. Since $q\phi_{MS}=(\Phi_M'-\chi')-(E_C-E_F)_{FB}$, zero needs $\Phi_M'-\chi'$ inside one of these windows. In Table 18.1 only Au (0.82 eV) qualifies (p-type window). Required doping: $(E_C-E_F)_{FB}=0.82$ eV gives $E_i-E_F=0.82-0.56=0.26$ eV, so $$N_A=n_ie^{(E_i-E_F)/kT}=10^{10}e^{0.26/0.0259}=2.29\times10^{14}\ \text{cm}^{-3}.$$ **Answer: Au gate on p-Si with $N_A=2.29\times10^{14}$ cm$^{-3}$.** --- ## 18.2 Oxide Charges ### 18.2.1 General information The work-function difference is a polite nonideality. Oxide charge is the rowdy one: it creates large shifts and instabilities. Research has identified distinct charge centers in or at the edge of thermally grown SiO$_2$-Si (Fig. 18.4): | Charge | Location | Nature | |---|---|---| | Mobile ion charge ($Q_M$) | Anywhere in the oxide | Alkali ions (Na$^+$, K$^+$, Li$^+$) that drift | | Oxide trapped charge | In the oxide | Trapped carriers | | Fixed oxide charge ($Q_F$) | Very close to the Si-SiO$_2$ interface | Immobile, positive | | Interface trapped charge ($Q_{IT}$) | Exactly at the interface | Electronic states in the gap | ### A general result for any oxide charge distribution Put the origin of $x$ at the **metal-oxide interface**, so the oxide spans $0\le x\le x_o$ (Fig. 18.5), with volume charge density $\rho_{ox}(x)$. **Step 1: Gauss's law in the oxide.** $$\frac{d\mathscr{E}_{ox}}{dx}=\frac{\rho_{ox}(x)}{K_O\varepsilon_0}\qquad (18.5)$$ **Step 2: integrate from $x$ to $x_o$.** $$\mathscr{E}_{ox}(x_o)-\mathscr{E}_{ox}(x)=\frac1{K_O\varepsilon_0}\int_x^{x_o}\rho_{ox}(x')dx'$$ $$\mathscr{E}_{ox}(x)=\mathscr{E}_{ox}(x_o)-\frac1{K_O\varepsilon_0}\int_x^{x_o}\rho_{ox}(x')dx'\qquad (18.6)$$ **Step 3: boundary condition at the oxide-semiconductor interface.** Continuity of $D=K\varepsilon_0\mathscr{E}$ (any sheet exactly at $x_o$ is counted inside $\rho_{ox}$) gives $\mathscr{E}_{ox}(x_o)=K_s\mathscr{E}_s/K_O$. **Step 4: voltage across the oxide.** $$V_{ox}=\int_0^{x_o}\mathscr{E}_{ox}(x)dx=x_o\mathscr{E}_{ox}(x_o)-\frac1{K_O\varepsilon_0}\int_0^{x_o}\!\!\int_x^{x_o}\rho_{ox}(x')dx'\,dx\qquad (18.7)$$ **Step 5: collapse the double integral** by swapping the order of integration over the triangle $0\le x\le x'\le x_o$ (equivalent to integration by parts). Since $\int_0^{x'}dx=x'$, $$\int_0^{x_o}\!\!\int_x^{x_o}\rho_{ox}(x')dx'dx=\int_0^{x_o}x'\rho_{ox}(x')dx'$$ $$V_{ox}=\frac{K_s}{K_O}\mathscr{E}_sx_o-\frac1{K_O\varepsilon_0}\int_0^{x_o}x\rho_{ox}(x)dx\qquad (18.8)$$ **Step 6: gate voltage.** For a structure that is otherwise ideal, $V_G=\phi_s+V_{ox}$: $$V_G=\phi_s+\frac{K_s}{K_O}\mathscr{E}_sx_o-\frac1{K_O\varepsilon_0}\int_0^{x_o}x\rho_{ox}(x)dx\qquad(18.9)$$ For the ideal device (no oxide charge): $$V_G'=\phi_s+\frac{K_s}{K_O}\mathscr{E}_sx_o\qquad(18.10)$$ **Step 7: subtract** at the same $\phi_s$ (same $\mathscr{E}_s$, same capacitance): $$\boxed{\Delta V_G(\text{oxide charges})=-\frac1{K_O\varepsilon_0}\int_0^{x_o}x\,\rho_{ox}(x)\,dx}\qquad(18.11)$$ This is added to the work-function shift of (18.4). **Intuition: charge near the semiconductor is the most powerful.** The weight $x$ says that a sheet near the semiconductor ($x\approx x_o$) ends almost all its field lines on the semiconductor surface, while charge near the metal is "screened" by the metal and barely influences the semiconductor. Positive oxide charge helps the gate, so less gate voltage is needed: the $C$-$V$ curve shifts toward **negative** voltage. ### 18.2.2 Mobile ions **The mystery.** Early (c. 1960) MOS devices showed two baffling effects: 1. As-fabricated $C$-$V$ curves were shifted *negatively* by tens of volts from theory. 2. Under **bias-temperature (BT) stress** (heat to about 100 to 200 $^\circ$C while holding a gate bias) the curves moved again. After **positive** BT stress, the room-temperature curve shifted tens of volts further negative. After **negative** BT stress it moved back toward the theoretical curve (Fig. 18.6). Sometimes even a device left positively biased at room temperature during lunch had drifted by a volt on your return. The shift was always opposite to the gate polarity and the curves always lay on the negative side of theory. A device whose operating point drifts is nearly useless, so the cause had to be found and removed. **The model.** Let $\rho_{ion}(x)$ be the ionic charge density, with total charge per area $$Q_M=\int_0^{x_o}\rho_{ion}(x)dx\qquad(18.12)$$ From (18.11): $$\Delta V_G(\text{mobile ions})=-\frac1{K_O\varepsilon_0}\int_0^{x_o}x\rho_{ion}(x)dx\qquad(18.13)$$ Positive ions give a negative shift (as observed); negative ions would give a positive shift (not observed). Position matters through the factor $x$. Place the same $Q_M$ near the metal (centroid $0.05x_o$) or near the semiconductor (centroid $0.95x_o$): $$\Delta V_G(a)=-0.05\frac{Q_M}{C_O},\qquad\Delta V_G(b)=-0.95\frac{Q_M}{C_O}$$ The shift is 19 times larger when the ions sit near the Si-SiO$_2$ interface (Fig. 18.7(a)). Hypothesis: positive ions that **move under bias**. Positive BT stress repels them from the gate toward the semiconductor (big negative shift); negative BT stress pulls them to the metal (small shift, curve returns toward theory) (Fig. 18.7(b)). **The detective story.** The suspects were alkali ions. Since 1888 it was known that Na$^+$, Li$^+$ and K$^+$ move through crystalline quartz below 250 $^\circ$C, and sodium is everywhere: reagents, glassware, fingers of lab staff, even the tungsten evaporation boats used to deposit gate metal. Three experiments closed the case: 1. Devices made with great care to exclude alkali ions were stable after hours of BT stress up to 200 $^\circ$C. 2. Devices deliberately contaminated by rinsing in dilute NaCl or LiCl before metallization were violently unstable. 3. Sodium was directly detected in normally made devices by **neutron activation**: bombard with neutrons to create a radioactive sodium isotope, then measure the radioactivity. **Cures** (clean fabrication was hard to maintain on production lines): - **Phosphorus stabilization.** Briefly place the oxidized wafer in a phosphorus diffusion furnace. A thin phosphosilicate glass forms in the outer part of the oxide (Fig. 18.8(a)). At that temperature Na$^+$ is very mobile and wanders into the P-rich glass, where it gets trapped. After cooling, the ions sit near the metal interface (small $\gamma_M$, smallest shift) and stay put in operation. The glass also blocks later contamination from metallization. - **Chlorine neutralization.** Add a chlorine compound to the furnace ambient during oxide growth. A chlorine-rich material (probably a chlorosiloxane) forms at the oxide-silicon interface; mobile Na$^+$ migrating there is trapped and **neutralized**, so it no longer affects the device (Fig. 18.8(b)). **Modern practice.** With gate oxides near 10 nm, a phosphosilicate layer would have to be a tiny fraction of the oxide, and its gettering volume is hard to control (polarization is also a concern). Chlorine neutralization continues, with careful control of Cl concentration to avoid oxide-thickness variations. Together with purer chemicals and gases, chlorine pre-cleaning of furnace tubes, close monitoring, and a deposited phosphosilicate passivation layer on finished ICs, stable devices are now routine. ### Exercise 18.2 **Problem.** An MOS-C with $x_o=0.1\ \mu$m shows post-stress $C$-$V$ curves in which the $+$BT curve lies 10 V to the left of the $-$BT curve. Assume after $+$BT all mobile ions form a $\delta$-function at the O-S interface and after $-$BT a $\delta$-function at the M-O interface. Find $Q_M/q$. **Solution.** After $+$BT, $\rho_{ion}=Q_M\delta(x-x_o)$. By (18.13): $$\Delta V_G(+BT)=-\frac1{K_O\varepsilon_0}\int_0^{x_o}xQ_M\delta(x-x_o)dx=-\frac{x_oQ_M}{K_O\varepsilon_0}=-\frac{Q_M}{C_O}$$ After $-$BT, $\rho_{ion}=Q_M\delta(x)$: $$\Delta V_G(-BT)=-\frac1{K_O\varepsilon_0}\int_0^{x_o}xQ_M\delta(x)dx=0$$ All other nonidealities are unchanged by stress, so the displacement between the curves is the difference: $$\Delta V_G(+BT)-\Delta V_G(-BT)=-10\text{ V}=-\frac{Q_M}{C_O}$$ $$\frac{Q_M}{q}=\frac{C_O}{q}(10)=\frac{(3.9)(8.85\times10^{-14})(10)}{(1.6\times10^{-19})(10^{-5})}=2.16\times10^{12}\ \text{cm}^{-2}$$ **Answer: $Q_M/q\approx2.16\times10^{12}$ cm$^{-2}$.** ### 18.2.3 The fixed charge **Discovery.** Once mobile ions were eliminated, a subtler effect became visible. Even after correcting for $\phi_{MS}$, $C$-$V$ curves were still shifted by up to a few volts toward negative bias. The devices were BT-stable (so not mobile ions), reproducible for given fabrication conditions, and the effect was confirmed by independent labs. Etching the oxide away in small steps, plus photo-measurements, located the cause: a charge in the oxide very close to the Si-SiO$_2$ interface, fixed in position and reproducibly built in: the **fixed oxide charge** $Q_F$. **Model.** Put it exactly at the interface: $$\rho_{ox}(x)=Q_F\delta(x-x_o)\qquad(18.14)$$ Inserting in (18.11): $$\Delta V_G(\text{fixed charge})=-\frac{x_oQ_F}{K_O\varepsilon_0}=-\frac{Q_F}{C_O}\qquad(18.15)$$ So $Q_F$ must be **positive** to explain the negative shifts. **Experimental facts about $Q_F$:** 1. Independent of oxide thickness, substrate doping concentration and doping type. 2. Depends on surface orientation: largest on $\{111\}$, smallest on $\{100\}$, ratio about 3:1. 3. Strong function of the *terminal* oxidation conditions (ambient and temperature). In Fig. 18.9, $Q_F$ falls roughly linearly with increasing oxidation temperature. Only the last step matters: a wafer oxidized in steam at 1000 $^\circ$C for 1 h and then in dry O$_2$ at 1200 $^\circ$C for about 5 min (long enough for steady state) shows only the dry-1200 $^\circ$C value. 4. Annealing in Ar or N$_2$ to steady state lowers $Q_F$ to the value of dry oxidation at 1200 $^\circ$C. Whatever the oxidation, an inert anneal brings $Q_F$ to its minimum (the "oxidation triangle", Fig. 18.9(b)). For (111) Si the minimum is typically $Q_F/q\approx2\times10^{11}$ cm$^{-2}$, about a third of that on (100). **Clues to the origin.** Not ionized dopants (independence of doping). The interfacial position, orientation dependence and sensitivity to terminal oxidation point to the oxidation reaction itself: the oxidant diffuses through existing oxide and reacts at the Si-SiO$_2$ interface, so the last oxide formed lies next to the interface and holds $Q_F$. Accepted hypothesis: **excess ionic silicon** that has left the crystal and waits to react when oxidation is abruptly terminated. The first oxide monolayer has indeed been found to be SiO$_x$ with $x<2$. An inert anneal lets the excess react away. ### Exercise 18.3 **Problem.** An MOS-C has the band diagram of Fig. E18.3 (a metal Fermi level $E_{FM}$, straight oxide bands, and curved bands at the semiconductor surface). (a) Sketch the electric field versus position. (b) Is there ionic charge $Q_M$ distributed through the SiO$_2$? (c) Is a fixed charge likely at the Si-SiO$_2$ interface? Explain. **Solution.** (a) $\mathscr{E}(x)$ is proportional to the slope of the bands. In the oxide the bands are straight, so $\mathscr{E}_{ox}$ is constant. In the semiconductor the field starts at $\mathscr{E}_s$ at the surface and decays to zero at the depletion edge (nearly linearly for uniform doping). At the interface the field jumps discontinuously. (b) If $\rho_{ox}=0$ then $d\mathscr{E}_{ox}/dx=0$ and the oxide bands are linear; charge distributed in the oxide would make $\mathscr{E}_{ox}$ vary with $x$ and the bands curve. The bands are straight, so $Q_M\approx0$. (c) $D=K\varepsilon_0\mathscr{E}$ is continuous across an interface unless a sheet of charge is there. The diagram shows $D_s=K_s\varepsilon_0\mathscr{E}_s>0$ on the semiconductor side exceeding the oxide-side $D_{ox}=K_O\varepsilon_0\mathscr{E}_{ox}$, so a plane of positive charge must sit at the interface: $$Q_{interface}=D_s-D_{ox}=K_s\varepsilon_0\mathscr{E}_s-K_O\varepsilon_0\mathscr{E}_{ox}>0$$ The fixed charge sits right at the interface and is positive, so we suspect $Q_F\ne0$. (A $D$ discontinuity could also come from mobile ions drifted to the O-S interface by $+$BT, or from interface trapped charge.) ### 18.2.4 Interfacial traps Judged by their wide-ranging, degrading effect, interface traps are the most important nonideality in MIS structures. In an MOS-C they show up as a **distorted, stretched-out** $C$-$V$ curve (Fig. 18.10: one device before and after minimizing interface traps). **What they are.** Donors, acceptors and R-G centers are localized levels in the bulk gap (near $E_C$, $E_V$ and $E_i$ respectively). *Interfacial traps* (also *surface states* or *interface states*) are allowed levels localized at the semiconductor surface; unlike bulk dopants they are **spread across the gap** (Fig. 18.11). Levels beyond the band edges also exist but are masked by the huge band-state density. **How they charge and discharge** (Fig. 18.12, n-bulk). Their energies are fixed relative to $E_C$ and $E_V$ *at the surface*, so as the band bending changes, the Fermi level sweeps through them: - **Inversion** (surface $E_F$ near $E_V$): essentially all states above $E_F$ are empty. For donor-like states (positive when empty, neutral when filled) $Q_{IT}>0$. - **Depletion** ($E_F$ near midgap): lower states fill with electrons, $Q_{IT}$ decreases. - **Accumulation** ($E_F$ near $E_C$): most states filled, $Q_{IT}$ near its minimum. Being at the interface like $Q_F$: $$\Delta V_G(\text{interfacial traps})=-\frac{Q_{IT}(\phi_s)}{C_O}\qquad(18.16)$$ The crucial difference from $Q_F$: $Q_{IT}$ depends on $\phi_s$. The shift is not rigid; it varies along the curve and **stretches** it. **A refinement.** If all states were donor-like, in accumulation $\Delta V_G$ would remain negative while shrinking toward zero. Experiments (Fig. 18.10) show an increasing *positive* shift in accumulation. Resolution: states in the **upper half of the gap are acceptor-like** (neutral when empty, negative when filled), the lower half donor-like. Going toward accumulation, past roughly flat band or midgap, $Q_{IT}$ passes through zero and becomes ever more negative as upper-gap states fill. A quantitative description needs $D_{it}(E)$ and a full theory for $Q_{IT}(\phi_s)$. **Physical origin: dangling bonds.** Terminating a silicon crystal along a plane leaves one of the four bonds of each surface atom dangling (Fig. 18.13(a)). Thermal oxidation ties up most but not all of them; the leftovers are the interface traps (Fig. 18.13(b)). *Feasibility check.* On (100) there are $6.8\times10^{14}$ Si atoms/cm$^2$. If 1/1000 give a trap with one electronic charge: $$\frac{Q_{IT}}{q}=\frac{6.8\times10^{14}}{1000}=6.8\times10^{11}\ \text{cm}^{-2}$$ With $x_o=0.1\ \mu$m ($C_O=3.45\times10^{-8}$ F/cm$^2$): $$\Delta V_G=\frac{(1.6\times10^{-19})(6.8\times10^{11})}{3.45\times10^{-8}}=3.15\text{ V}$$ A tiny fraction of unterminated bonds is enough to matter. **Trends in the interface-state density $D_{it}$** (states/cm$^2$-eV): - Highest on $\{111\}$, lowest on $\{100\}$, midgap ratio about 3:1. - After dry-O$_2$ oxidation, midgap $D_{it}$ is high ($10^{11}$ to $10^{12}$) and rises with oxidation temperature. An inert anneal does **not** help; a low-temperature hydrogen anneal ($\lesssim500\ ^\circ$C) does. - After an ideal interface-state anneal, midgap $D_{it}\lesssim10^{10}$, roughly flat in the middle of the gap and rising sharply toward the band edges (Fig. 18.14). - States near the conduction band edge are acceptor-like and those near the valence band edge donor-like, in about equal numbers. **Annealing methods.** - *Postmetallization anneal* (needs a chemically active gate such as Al or Cr): anneal the metallized structure in N$_2$ at about 450 $^\circ$C for 5 to 10 min. Adsorbed water vapor on the oxide reacts with the gate metal and releases atomic hydrogen, which diffuses through the oxide to the interface and caps dangling bonds, making them inactive (Fig. 18.15). - *Hydrogen-ambient anneal*: same principle, hydrogen supplied from the gas; the structure need not be metallized. **Perspective.** Without thermal oxide's ability to tie up most dangling bonds, and without hydrogen annealing, MOS devices would be a laboratory curiosity. High interface-trap densities have severely stunted other insulator-semiconductor systems. ### 18.2.5 Induced charges #### Radiation effects Radiation (x-rays, energetic electrons, protons, heavy ions) is a concern for space and military electronics; the early Telstar satellite was temporarily disabled by the Van Allen belt. In MOS devices, irradiation always raises both the apparent fixed charge and the interface-trap density. The damage sequence (Fig. 18.16): 1. Ionizing radiation creates electron-hole pairs throughout the oxide; a fraction recombines immediately. 2. The oxide field separates the survivors. Electrons (relatively mobile) are swept out in nanoseconds. 3. Holes, much less mobile, hop through localized states toward the Si-SiO$_2$ interface (for positive gate bias), where they recombine with Si electrons or get trapped at deep levels. These trapped holes mimic fixed charge: **apparent $Q_F$ rises**. 4. New interface traps appear, some immediately, the rest in proportion to the holes reaching the interface. Proposed mechanism: energy released in deep hole trapping breaks Si-H bonds that had passivated traps. *Recovery.* Over days to years at room temperature, trapped holes are neutralized by electrons injected from metal or silicon. Thermal annealing speeds this; a standard interface-trap anneal removes trapped holes and induced traps. But after an irradiate-anneal cycle devices are *more* sensitive to later radiation, attributed to neutral oxide traps (partly removed above 600 $^\circ$C). After-the-fact recovery in finished devices (e.g. warming to about 100 $^\circ$C) is limited. *Hardening* is mainly preventive: empirically optimized growth (e.g. oxidation below 1000 $^\circ$C), aluminium shielding (stops energetic space electrons), and a raised MOSFET threshold so $\Delta V_G$ matters less (Section 18.3). Conveniently, scaling helps: $\Delta V_G$ from $Q_F$ and $Q_{IT}$ is proportional to $1/C_O=x_o/K_O\varepsilon_0$ and shrinks with thinner oxide; hole-trapping cross sections shrink at the high fields of thin oxides; and below about 10 nm electrons from metal and silicon can tunnel into all parts of the oxide and annihilate trapped holes. (Radiation damage can also occur during fabrication: ion implantation, e-beam evaporation, sputtering, plasma processing, e-beam and x-ray lithography.) #### Negative-bias instability Large negative gate bias at elevated temperature (e.g. oxide field about $2\times10^6$ V/cm and $T>250\ ^\circ$C) produces a large **negative** shift and distortion of the $C$-$V$ curve. Like radiation, it increases apparent $Q_F$ and $Q_{IT}$; an added midgap peak in $D_{it}$ is the distinctive signature. Note the shift direction is opposite to alkali-ion behavior after negative stress. The exact mechanism is unsettled; since large hole populations sit next to the oxide under negative stress, the proposal is hole injection from Si into the oxide and trapping at deep sites near the interface (a link to radiation damage; the instability is enhanced if the structure is first irradiated). Sensitivity can be reduced by a hydrogen anneal at 800 to 900 $^\circ$C before gate deposition. ### 18.2.6 $\Delta V_G$ summary We have met work-function difference, mobile ions, fixed charge and interface traps, plus radiation and negative-bias stress that raise apparent $Q_F$ and $Q_{IT}$. Combined: $$\boxed{\Delta V_G=(V_G-V_G')_{\text{same }\phi_s}=\phi_{MS}-\frac{Q_F}{C_O}-\gamma_M\frac{Q_M}{C_O}-\frac{Q_{IT}(\phi_s)}{C_O}}\qquad(18.17)$$ with the dimensionless mobile-ion **centroid** $$\gamma_M\equiv\frac{\int_0^{x_o}x\rho_{ion}(x)dx}{x_o\int_0^{x_o}\rho_{ion}(x)dx}\qquad(18.18)$$ ($\gamma_M=0$ if all ions are at the metal-oxide interface; $\gamma_M=1$ if all are at the oxide-semiconductor interface.) The terms $\phi_{MS}$, $Q_F$ and $Q_M$ translate the $C$-$V$ curve in parallel; $Q_{IT}$ can shift either way depending on bias and **distorts/spreads** the curve. Real MOS devices are intrinsically imperfect, yet extensive research lets manufacturers routinely fabricate near-ideal ones (with constant quality-control checks). The cures cited apply specifically to thermally grown SiO$_2$-Si. ### Exercise 18.4 **Problem.** Match each oxide charge ($Q_F$, $Q_M$, $Q_{IT}$, apparent $Q_F$ from ionizing radiation) with its physical cause: (a) P$^+$ ions, (b) Na$^+$ ions, (c) N$^+$ ions, (d) dangling bonds at the Si surface, (e) trapped electrons, (f) ionized Si waiting to be oxidized, (g) ionized oxygen waiting to form SiO$_2$, (h) trapped holes. **Solution.** $Q_F\to$ (f); $Q_M\to$ (b); $Q_{IT}\to$ (d); apparent $Q_F$ from radiation $\to$ (h). ### Exercise 18.5 **Problem.** An Al-SiO$_2$-Si MOS-C is formed on an $N_A=10^{15}$ cm$^{-3}$ (100) Si wafer. Oxidation in dry O$_2$ at 1200 $^\circ$C is followed by an N$_2$ anneal to steady state. The structure is phosphorus-gettered so that mobile ions $Q_M/q=2\times10^{11}$ cm$^{-2}$ are uniformly distributed over the outer 10% of the oxide, $0\le x\le0.1x_o$ (Fig. E18.5). A postmetallization anneal is done, but a constant $D_{it}=2\times10^{10}$ cm$^{-2}$eV$^{-1}$ of acceptor-like states remains at all gap energies. Find $V_{FB}$ at $T=300$ K, $x_o=0.1\ \mu$m. **Solution.** At flat band $\phi_s=0$, $V_G'=0$, so from (18.17) $$V_{FB}=\phi_{MS}-\frac{Q_F}{C_O}-\frac{\gamma_MQ_M}{C_O}-\frac{Q_{IT}(0)}{C_O}$$ $$C_O=\frac{(3.9)(8.85\times10^{-14})}{10^{-5}}=3.45\times10^{-8}\ \text{F/cm}^2$$ *Term 1.* Al gate, $N_A=10^{15}$: Fig. 18.3 gives $\phi_{MS}=-0.90$ V. *Term 2.* The N$_2$ anneal gives the minimum $Q_F$; Fig. 18.9(b) gives $Q_F/q=2\times10^{11}$ cm$^{-2}$ for (111), and (100) is three times smaller: $$\frac{Q_F}{q}=\frac{2\times10^{11}}3=6.67\times10^{10}\ \text{cm}^{-2}$$ $$\frac{Q_F}{C_O}=\frac{(1.6\times10^{-19})(6.67\times10^{10})}{3.45\times10^{-8}}=0.31\text{ V}$$ *Term 3.* Uniform $\rho_{ion}$ on $[0,0.1x_o]$: $$\gamma_M=\frac{\int_0^{0.1x_o}x\,dx}{x_o\int_0^{0.1x_o}dx}=\frac{(0.1x_o)^2/2}{x_o(0.1x_o)}=0.05$$ $$\frac{\gamma_MQ_M}{C_O}=\frac{(1.6\times10^{-19})(2\times10^{11})(0.05)}{3.45\times10^{-8}}=0.046\text{ V}$$ *Term 4.* Acceptor-like traps are negative when filled (states below $E_F$) and neutral when empty (above $E_F$). At flat band (equilibrium) the charged states are those between $E_V$ and $E_F$, so the energy span with negative traps is $$\Delta E=E_F-E_V=\frac{E_G}2-(E_i-E_F)=\frac{E_G}2-kT\ln\frac{N_A}{n_i}=0.56-0.0259\ln(10^5)=0.26\text{ eV}$$ $$Q_{IT}(0)=-qD_{it}\Delta E\;\Rightarrow\;-\frac{Q_{IT}(0)}{C_O}=\frac{(1.6\times10^{-19})(2\times10^{10})(0.26)}{3.45\times10^{-8}}=+0.024\text{ V}$$ *Sum.* $$V_{FB}=-0.90-0.31-0.046+0.024=-1.23\text{ V}$$ **Answer: $V_{FB}\approx-1.23$ V.** (The mobile-ion term is tiny because stabilization keeps ions near the metal.) --- ## 18.3 MOSFET Threshold Considerations So far we described nonidealities as $C$-$V$ shifting and distortion. The point on the $C$-$V$ curve corresponding to the MOSFET threshold is displaced by $\Delta V_G$ like any other point, so $V_T$ is directly affected. **Effect on $I_D$-$V_D$ curves** (Fig. 18.17). Take a hypothetical p-channel MOSFET with ideal $V_T=-1$ V. - (a) A pure translation (from $\phi_{MS}$, $Q_F$ and/or mobile ions) leaves the *shape* of the family of curves unchanged, but different gate voltages are needed to reach the same current levels: the whole family slides along the $V_G$ axis. - (b) Interface traps in large density additionally shrink the current change produced by a given gate-voltage step. That is equivalent to a lowered transconductance $g_m$, so the curves are not just slid but also bunched together. Mobile ions and interface traps were largely tamed early in MOSFET history, but the remaining nonidealities, mostly through $V_T$, have strongly shaped technology, device design and modes of operation. This section explains why and how $V_T$ is adjusted. ### 18.3.1 $V_T$ relationships Let $V_{T}'$ be the threshold voltage of an ideal version of the MOSFET. Evaluating (18.17) at $\phi_s=2\phi_F$ gives the real threshold: $$V_T=V_T'+\phi_{MS}-\frac{Q_F}{C_O}-\frac{\gamma_MQ_M}{C_O}-\frac{Q_{IT}(2\phi_F)}{C_O}\qquad(18.19)$$ It is standard to write the shift via the real-device flat-band voltage. At flat band $\phi_s=0$ and $V_G'=0$, so by (18.17) $$V_{FB}=V_G\big|_{\phi_s=0}=\phi_{MS}-\frac{Q_F}{C_O}-\frac{\gamma_MQ_M}{C_O}-\frac{Q_{IT}(0)}{C_O}\qquad(18.20)$$ If $Q_{IT}$ changes little between $\phi_s=0$ and $\phi_s=2\phi_F$ (a reasonable approximation in well-made devices), the nonideality terms of (18.19) are replaced by $V_{FB}$: $$\boxed{V_T=V_T'+V_{FB}}\qquad(18.21)$$ where (repeating Eq. 17.1; upper sign n-channel, lower sign p-channel) $$V_T'=2\phi_F\;\pm\;\frac{K_s}{K_O}x_o\sqrt{\frac{4qN_B}{K_s\varepsilon_0}\,(\pm\phi_F)}\qquad N_B=N_A\text{ or }N_D\qquad(18.22)$$ (in this form $\phi_F>0$ for n-channel/p-bulk and $\phi_F<0$ for p-channel/n-bulk, with $2\phi_F$ appearing as written). *Where (18.22) comes from (n-channel).* At threshold $\phi_s=2\phi_F$, the depletion width is $W=\sqrt{2K_s\varepsilon_0(2\phi_F)/qN_A}=\sqrt{4K_s\varepsilon_0\phi_F/qN_A}$, the depletion charge is $qN_AW=\sqrt{4K_s\varepsilon_0qN_A\phi_F}$, and the oxide drop is that charge over $C_O$: $$V_T'=2\phi_F+\frac{\sqrt{4K_s\varepsilon_0qN_A\phi_F}}{C_O}=2\phi_F+\frac{K_s}{K_O}x_o\sqrt{\frac{4qN_A\phi_F}{K_s\varepsilon_0}}$$ ### 18.3.2 Threshold, terminology and technology **A sample calculation.** Al gate, (111) Si, $T=300$ K, $x_o=0.1\ \mu$m, $N_A=10^{15}$ cm$^{-3}$, $Q_F/q=2\times10^{11}$ cm$^{-2}$, $Q_M=0$, $Q_{IT}=0$ (n-channel). $$\phi_{MS}=-0.90\text{ V}\qquad -\frac{Q_F}{C_O}=-\frac{(1.6\times10^{-19})(2\times10^{11})}{3.45\times10^{-8}}=-0.93\text{ V}$$ $$V_{FB}=-0.90-0.93=-1.83\text{ V}$$ $$\phi_F=0.0259\ln(10^5)=0.298\text{ V},\quad V_T'=0.596+\frac{\sqrt{4(11.8)(8.85\times10^{-14})(1.6\times10^{-19})(10^{15})(0.298)}}{3.45\times10^{-8}}\approx1.00\text{ V}$$ $$V_T=V_T'+V_{FB}=1.00-1.83=-0.83\text{ V}$$ An ideal threshold of $+1$ V has become $-0.83$ V: realistic nonidealities flipped the sign. An n-channel device turns on for $V_G>V_T$, so this one is already on at $V_G=0$ and negative gate bias is needed to turn it off. For the p-channel version (same parameters, n-substrate with $N_D=10^{15}$): Al on n-Si gives $\phi_{MS}\approx-0.30$ V, so $$V_{FB}=-0.30-0.93=-1.23\text{ V},\qquad V_T'=-1.00\text{ V},\qquad V_T=-2.23\text{ V}$$ Here the nonidealities simply make the negative turn-on voltage even more negative. **Enhancement versus depletion mode** (Fig. 18.18). | | Enhancement mode | Depletion mode | |---|---|---| | State at $V_G=0$ | Off (no channel) | On (channel exists) | | n-channel threshold | $V_T>0$ | $V_T<0$ | | p-channel threshold | $V_T<0$ | $V_T>0$ | | Turning on / off | Gate bias beyond $V_T$ turns it on | Gate bias of the opposite polarity turns it off | Standard p-channel MOSFETs are, ideally and in practice, enhancement mode. n-channel devices are ideally enhancement mode, but because nonidealities push $V_T$ toward negative voltage, early n-channel MOSFETs were typically depletion mode. This is why, until about 1977, PMOS dominated the IC market. Then ion-implantation threshold adjustment (below) made enhancement-mode NMOS routine; NMOS, preferred for the higher electron mobility, now appears in most newly designed ICs. **Parasitic channels between devices.** Consider the unmetallized region between two n-channel MOSFETs (Fig. 18.19(a)). If the potential on the outer oxide surface is about zero and the n-channel threshold is negative, the intermediate region is inverted: a conducting pseudo-channel connects the two transistors. Such inversion can also be caused by voltages on metal overlays that carry gate and drain biases. The cure is a much thicker **field oxide** outside the gated areas (Fig. 18.19(b)). Since $V_T$ grows with oxide thickness, $V_T(\text{field})\gg V_T(\text{gate})$, and the field region stays non-inverted at normal circuit voltages. ### 18.3.3 Threshold adjustment Several knobs set $V_T$: - **Oxide thickness** and **substrate doping**: both strongly change $V_T$, but they are mostly fixed by other design constraints. - **Surface orientation**: $Q_F$ on (100) is about one third that on (111), so (100) wafers reduce the fixed-charge shift. - **Gate material**: a polysilicon gate changes $\phi_{MS}$. For a poly-Si gate the effective "metal" work function is $$\Phi_M'=\chi'+(E_C-E_F)_{poly\text{-}Si}\qquad(18.23)$$ $$\phi_{MS}=\frac1q\Big[(E_C-E_F)_{poly\text{-}Si}-(E_C-E_F)_{FB,\,crystalline\text{-}Si}\Big]\qquad(18.24)$$ *Example, repeating the sample calculation* with a (100) surface ($Q_F/q=6.67\times10^{10}$ cm$^{-2}$, one third of the (111) value $2\times10^{11}$) and a **p-type poly-Si gate** with $E_F=E_V$ (so $(E_C-E_F)_{poly}=E_G=1.12$ eV), n-channel on $N_A=10^{15}$ cm$^{-3}$: $$(E_C-E_F)_{FB,Si}=0.56+0.298=0.858\text{ eV}\Rightarrow\phi_{MS}=1.12-0.858=+0.262\text{ V}$$ $$\frac{Q_F}{C_O}=0.31\text{ V}\Rightarrow V_{FB}=0.262-0.31=-0.048\text{ V}$$ $$V_T=1.00-0.048\approx+0.95\text{ V}$$ So an enhancement-mode n-channel device is possible on (100) with p$^+$ poly. But practical devices may only be nominally positive; the poly in n-channel devices is conveniently doped n-type like source/drain, not p-type; a larger $V_T$ may be wanted; PMOS may need adjusting; and both n- and p-channel devices on one chip may need separately tailored thresholds. A flexible method is needed: **ion implantation**. **Threshold-adjust implant.** A small, precisely controlled dose of boron or phosphorus is implanted into the near-surface silicon (Chapter 4). In depletion or inversion, the implanted dopant adds to the exposed ionic charge near the interface and translates $V_T$: **boron gives a positive shift, phosphorus a negative shift**. For shallow implants, to first order treat it as an extra fixed charge $Q_I=\pm qN_I$ at the interface ($+$ donor-like for P, $-$ acceptor-like for B), with $N_I$ implanted ions/cm$^2$. By analogy with the fixed charge: $$\Delta V_G(\text{implanted ions})=-\frac{Q_I}{C_O}\qquad(18.25)$$ Example: $N_I=5\times10^{11}$ boron ions/cm$^2$, $x_o=0.1\ \mu$m: $$\Delta V_G=+\frac{qN_I}{C_O}=\frac{(1.6\times10^{-19})(5\times10^{11})}{3.45\times10^{-8}}=+2.32\text{ V}$$ ### 18.3.4 Back biasing Reverse-biasing the back contact (bulk) relative to the source is another, older, electrical adjustment: the **body effect** (substrate-bias effect). Consider the n-channel MOSFET of Fig. 18.20(a) with substrate-source voltage $V_{BS}$. If $V_{BS}=0$, inversion starts when the voltage drop across the semiconductor reaches $2\phi_F$ (Fig. 18.20(b)). If $V_{BS}<0$, the semiconductor tries to invert at $\phi_s=2\phi_F$, but the channel's electrons are drawn laterally into the source and drain regions, which are at lower potential than the bulk (the electron quasi-Fermi level pinned by the source is $|V_{BS}|$ lower in electron energy). The surface does not invert until $$\phi_s=2\phi_F-V_{BS}\qquad(\text{Fig. 18.20(c)})$$ so back bias moves the inversion point from $2\phi_F$ to $2\phi_F-V_{BS}$. *Derivation of the new threshold.* The depletion width at the new inversion point is $W=\sqrt{2K_s\varepsilon_0(2\phi_F-V_{BS})/qN_A}$, so the depletion charge is $qN_AW=\sqrt{2K_s\varepsilon_0qN_A(2\phi_F-V_{BS})}$. The gate-to-**bulk** threshold voltage is the surface potential plus the oxide drop: $$V_{GB}\Big|_{th}=2\phi_F-V_{BS}\pm\frac{K_s}{K_O}x_o\sqrt{\frac{2qN_B}{K_s\varepsilon_0}\,(\pm2\phi_F\mp V_{BS})}\qquad(18.26)$$ (upper sign n-channel, $V_{BS}<0$; lower sign p-channel, $V_{BS}>0$; $N_B=N_A$ or $N_D$.) Gate-to-**source** is $V_{GS}=V_{GB}+V_{BS}$, so the $-V_{BS}$ term cancels: $$V_{GS}\Big|_{th}=2\phi_F\pm\frac{K_s}{K_O}x_o\sqrt{\frac{2qN_B}{K_s\varepsilon_0}\,(\pm2\phi_F\mp V_{BS})}\qquad(18.27)$$ Define $\Delta V_T=V_{GS}|_{th}-V_T'$, where $V_T'$ is (18.22) at $V_{BS}=0$. Write $V_2\equiv(K_s/K_O)x_o\sqrt{4qN_B|\phi_F|/K_s\varepsilon_0}$, the second term of $V_T'$ (n-channel) at $V_{BS}=0$. Then, since $\sqrt{2\phi_F-V_{BS}}=\sqrt{2\phi_F}\sqrt{1-V_{BS}/2\phi_F}$, $$\boxed{\Delta V_T=V_2\left[\sqrt{1-\frac{V_{BS}}{2\phi_F}}-1\right]}\quad\begin{cases}\phi_F>0,\;V_{BS}<0&\text{n-channel}\\[2pt]\phi_F<0,\;V_{BS}>0&\text{p-channel}\end{cases}\qquad(18.28)$$ with the sign conventions that make $V_{BS}/2\phi_F<0$ in both cases. Observations: 1. Back biasing always **increases the magnitude** of the ideal threshold: n-channel more positive, p-channel more negative. It cannot be used to reduce the negative threshold of a p-channel MOSFET. 2. The Chapter 17 current-voltage relations remain valid for $V_{BS}\ne0$ with $2\phi_F\to2\phi_F-V_{BS}$, $V_G\to V_{GS}$, $V_D\to V_{DS}$, and $V_T$ interpreted as $V_{GS}|_{th}$. 3. Use double-subscripted voltages carefully so the correct voltage differences are identified. ### 18.3.5 Threshold summary - All nonidealities shift the MOSFET threshold; interface traps in addition degrade the low-frequency $g_m$. - Procedure for a real $V_T$: (i) find $V_{FB}$ from known nonideality information; (ii) add it to the threshold of an ideal version of the device, $V_T=V_T'+V_{FB}$. - A transistor "off" at $V_G=0$ is **enhancement mode**; one that conducts at $V_G=0$ is **depletion mode**. Because of residual nonidealities real n-channel MOSFETs tend to be depletion mode. - Enhancement-mode NMOS and threshold adjustment in general are now achieved by ion implantation; back biasing is also used. ### Exercise 18.6 **Problem.** The $C$-$V_G$ ($V_D=0$) characteristic of an n-channel MOSFET (Fig. E18.6) shows flat band at $V_G=-2.2$ V and the capacitance rising back to $C=C_O$-like inversion level near $V_G=-1$ V. (a) What is $V_T$? (b) Depletion or enhancement mode? (c) Sketch the $I_D$-$V_D$ family for $V_G=-2,-1,0,1,2$ V. (d) Given $\phi_{MS}=-1$ V, $Q_{IT}=0$, and BT stability, explain the flat-band voltage of $-2.2$ V. (e) For $\phi_F=0.3$ V, what $V_{BS}$ gives $V_{GS}|_{th}=1$ V? **Solution.** (a) In going toward positive gate bias the high-frequency/low-frequency $C$-$V$ rises toward $C_O$ at about the depletion-to-inversion transition. $V_G$ at that transition is the MOSFET threshold: $V_T\approx-1$ V. (b) $V_T=-1$ V $<0$ so the device conducts at $V_G=0$: **depletion mode**. (c) Qualitative family: curves for $V_G=-2,-1$ V show (essentially) no current (the device is off, $V_G\le V_T$); curves for $V_G=0,1,2$ V are ordinary MOSFET curves with saturation, ordered upward: $V_G=0$ lowest, $V_G=2$ V highest, with saturation voltage $V_{Dsat}=V_G-V_T=1,2,3$ V respectively. (d) BT stable means $Q_M=0$; $Q_{IT}=0$ is given; $\phi_{MS}=-1$ V cannot account for the whole $-2.2$ V; the remaining shift is due to the fixed charge: $$-\frac{Q_F}{C_O}=V_{FB}-\phi_{MS}=-2.2+1.0=-1.2\text{ V}\Rightarrow Q_F>0\text{ and }Q_F=1.2\,C_O.$$ (e) Solve (18.28) for $V_{BS}$: $$\frac{\Delta V_T}{V_2}+1=\sqrt{1-\frac{V_{BS}}{2\phi_F}}\;\Rightarrow\;V_{BS}=2\phi_F\left[1-\left(1+\frac{\Delta V_T}{V_2}\right)^2\right]$$ Needed shift: $\Delta V_T=1-(-1)=2$ V. $V_2$ from $V_T=V_T'+V_{FB}$: $V_T'=V_T-V_{FB}=-1.0+2.2=1.2$ V, and $V_2=V_T'-2\phi_F=1.2-0.6=0.6$ V. $$V_{BS}=0.6\left[1-\left(1+\frac{2}{0.6}\right)^2\right]=0.6[1-18.78]=-10.7\text{ V}$$ **Answer: $V_{BS}\approx-10.7$ V.** --- ## Chapter Summary 1. **Work-function difference.** $\phi_{MS}=[(\Phi_M'-\chi')-(E_C-E_F)_{FB}]/q$ rigidly shifts the $C$-$V$ curve by $\phi_{MS}$ (usually a negative fraction of a volt to a volt). 2. **Oxide charge.** For any distribution, $\Delta V_G=-\frac{1}{K_O\varepsilon_0}\int_0^{x_o}x\rho_{ox}dx$. Charge near the semiconductor weighs the most. 3. **Mobile ions** (Na$^+$): shift depends on position ($\gamma_M$) and move under BT stress, causing instability. Cured by clean processing, phosphorus stabilization, chlorine neutralization. 4. **Fixed charge $Q_F$:** positive, at the interface, shift $-Q_F/C_O$, depends on orientation (111:100 about 3:1) and terminal oxidation, minimized by an inert anneal; probably excess Si. 5. **Interface traps $Q_{IT}(\phi_s)$:** dangling bonds; stretch out the $C$-$V$ curve and lower $g_m$; reduced by hydrogen (postmetallization or ambient) annealing. 6. **Radiation and negative-bias stress** raise apparent $Q_F$ and $Q_{IT}$. 7. **Total shift:** $\Delta V_G=\phi_{MS}-Q_F/C_O-\gamma_MQ_M/C_O-Q_{IT}(\phi_s)/C_O$. 8. **Threshold:** $V_T=V_T'+V_{FB}$; real n-channel devices tend to be depletion mode; ion implantation ($\Delta V=-Q_I/C_O$) and back bias ($\Delta V_T=V_2[\sqrt{1-V_{BS}/2\phi_F}-1]$) adjust it; field oxide prevents parasitic channels. --- ## Computer-Based Exercise (Problem 18.3 logic) **Goal.** Plot $\phi_{MS}$ versus doping for Al-gate and n$^+$ poly-gate structures (as in Fig. 18.3) and for p$^+$ poly-gate ($E_F=E_V$). **Logic.** 1. Build a log-spaced doping vector $N$ from $10^{14}$ to $10^{18}$ cm$^{-3}$. 2. For n-type: $(E_C-E_F)_{FB}=E_G/2-kT\ln(N_D/n_i)$. For p-type: $(E_C-E_F)_{FB}=E_G/2+kT\ln(N_A/n_i)$. 3. $\phi_{MS}=(\Phi_M'-\chi')-(E_C-E_F)_{FB}$ (in volts, with energies in eV). 4. Inputs: Al: $-0.03$ eV; n$^+$ poly: $-0.18$ eV; p$^+$ poly with $E_F=E_V$: $E_G=1.12$ eV. 5. Plot against $\log_{10}N$, with curves for gate on n-Si and on p-Si. ```matlab % Work-function difference versus doping (Fig. 18.3 and p+ poly) kT = 0.0259; EG = 1.12; ni = 1e10; N = logspace(14, 18, 200); % cm^-3 EcEf_n = EG/2 - kT*log(N/ni); % n-type Si: Ec - EF at flat band EcEf_p = EG/2 + kT*log(N/ni); % p-type Si gates = struct('name', {'Al', 'n+ poly', 'p+ poly (EF=Ev)'}, ... 'phi', {-0.03, -0.18, EG}); % PhiM' - chi' in eV figure; hold on; grid on; styles = {'b-','r-','k-'}; for g = 1:numel(gates) phiMS_n = gates(g).phi - EcEf_n; % volts (eV/q) phiMS_p = gates(g).phi - EcEf_p; semilogx(N, phiMS_n, styles{g}, 'LineWidth', 1.5, ... 'DisplayName', [gates(g).name ' on n-Si']); semilogx(N, phiMS_p, [styles{g}(1) '--'], 'LineWidth', 1.5, ... 'DisplayName', [gates(g).name ' on p-Si']); end set(gca,'XScale','log'); xlabel('N_A or N_D (cm^{-3})'); ylabel('\phi_{MS} (V)'); title('Metal-semiconductor work-function difference'); legend('Location','best'); ``` Check: Al on p-Si at $10^{15}$: $-0.03-0.858=-0.89$ V, matching the figure's about $-0.9$ V. Al on n-Si at $10^{15}$: $-0.03-0.262=-0.29$ V. n$^+$ poly on p-Si at $10^{15}$: $-0.18-0.858=-1.04$ V. --- ## End-of-Chapter Problems **Problem information (from the book's table):** 18.1 quick quiz; 18.2 polysilicon-gate MOS-C; 18.3 $\phi_{MS}$ versus doping plot (computer); 18.4 mobile ion calculation; 18.5 fixed charge calculation; 18.6 $\phi_{MS}$ and $Q_F$ from a $V_{FB}$ plot; 18.7 charge information from a band diagram; 18.8 $Q_F$ from $C$-$V$ data; 18.9 sketch of a pre-annealed $C$-$V$; 18.10 dangling-bond calculation; 18.11 single-level interfacial trap; 18.12 half-irradiated MOS-C; 18.13 BT-stressed MOSFET; 18.14 modifications to $V_T$; 18.15 compute MOSFET $V_T$; 18.16 required implant dose. Where the OCR dropped an exponent or a figure is unavailable, the assumption used is stated in the solution. ### Problem 18.1 Quick quiz **(a) Difference between $\Phi_M-\chi$ and $\Phi_M'-\chi'$?** $\Phi_M$ and $\chi$ are the metal work function and the Si electron affinity measured to the vacuum level. $\Phi_M'$ and $\chi'$ are the barrier heights from the metal Fermi level and from the Si conduction band edge to the *oxide conduction band*. The oxide lowers both by the same amount, so $\Phi_M'-\chi'=\Phi_M-\chi$. **(b) What is a bias-temperature stress?** Heating the device (about 100 to 250 $^\circ$C) while holding a constant gate bias (positive or negative) for some time, then cooling to room temperature with the bias on, and measuring the $C$-$V$ or $I$-$V$ again. It tests whether charges (mobile ions) move. **(c) Origin of fixed oxide charge.** Excess ionic silicon (incompletely oxidized SiO$_x$, $x<2$) at the Si-SiO$_2$ interface, left when oxidation stops. **(d) $D_{it}$ orientation dependence.** $\{111\}$ highest, $\{100\}$ lowest, ratio about 3:1 at midgap. **(e) Net effect of ionizing radiation.** Creates oxide trapped holes (apparent increase in $Q_F$, positive) and new interface traps: negative shift and stretch-out of the $C$-$V$ curve. **(f) Distinguishing negative-bias instability from alkali ions.** After a *negative* BT stress, alkali ions move to the metal and the $C$-$V$ curve shifts **positively** (back toward theory); negative-bias instability shifts the curve **more negative** (and distorts it, with a $D_{it}$ midgap peak). **(g) When is $V_T\ne V_T'+V_{FB}$?** When $Q_{IT}(2\phi_F)\ne Q_{IT}(0)$, i.e. when the interface trap charge differs appreciably between flat band and threshold (large trap densities). **(h) Depletion-mode transistor.** One that conducts (has a channel) at $V_G=0$; a gate voltage of the appropriate polarity is required to turn it off. **(i) Field oxide versus gate oxide.** The gate oxide is the thin oxide under the gate in the active region; the field oxide is the much thicker oxide over non-gated regions, thick enough that $V_T(\text{field})$ is too large for the field region to invert at circuit voltages (preventing parasitic channels). ### Problem 18.2 Polysilicon-gate MOS-C **Problem.** A poly-Si gate MOS-C has $E_F-E_C=0.2$ eV in the heavily doped gate (Fermi level 0.2 eV above $E_C$) and $E_F-E_i=-0.2$ eV in the nondegenerately doped substrate (a p-type substrate, $E_i-E_F=0.2$ eV; the OCR dropped the subscripts). The structure is ideal except $\phi_{MS}\ne0$ and $\chi'(\text{poly})=\chi'(\text{Si})$. (a) Sketch the flat-band energy band diagram. (b) Find $\phi_{MS}$. (c) Accumulation, depletion or inversion at $V_G=0$? **Solution.** **(b)** Using $\phi_{MS}=\big[(E_C-E_F)_{poly}-(E_C-E_F)_{Si}\big]/q$ (Eq. 18.24): $$(E_C-E_F)_{poly}=-0.2\text{ eV}$$ $$(E_C-E_F)_{Si}=(E_C-E_i)+(E_i-E_F)=0.56+0.2=0.76\text{ eV}$$ $$\phi_{MS}=\frac{-0.2-0.76}{1}=-0.96\text{ V}$$ **(a)** Flat band means the semiconductor bands are flat, while the gate is at $V_G=\phi_{MS}=-0.96$ V. So the metal (poly) Fermi level lies 0.96 eV *above* the Si Fermi level: $E_{FM}-E_{FS}=-qV_G=+0.96$ eV. Draw: flat Si bands with $E_F$ 0.2 eV below $E_i$ ($E_i$ at midgap), $E_C$ 0.76 eV above $E_F$; the oxide conduction band as a straight sloping line; poly-Si $E_C$ lying 0.2 eV *below* $E_F$ (degenerate), and $E_{FM}$ 0.96 eV above $E_{FS}$ (the oxide band slopes down toward the semiconductor so that the oxide conduction band offset to the gate side is lower by $\phi_{MS}$ relative to the semiconductor side). **(c)** $V_G=0$ is $|\phi_{MS}|=0.96$ V *above* flat band: $V_G-V_{FB}=+0.96$ V. For a p-type substrate, positive $V_G-V_{FB}$ depletes and may invert. Determine which: $N_A=n_ie^{0.2/0.0259}=2.25\times10^{13}$ cm$^{-3}$; $2\phi_F=0.40$ V. With an assumed $x_o=0.1\ \mu$m ($C_O=3.45\times10^{-8}$ F/cm$^2$): $$V_T'-V_{FB}\text{-offset}=2\phi_F+\frac{\sqrt{2K_s\varepsilon_0qN_A(2\phi_F)}}{C_O}=0.40+\frac{\sqrt{2(11.8)(8.85\times10^{-14})(1.6\times10^{-19})(2.25\times10^{13})(0.40)}}{3.45\times10^{-8}}$$ $$=0.40+\frac{1.73\times10^{-9}}{3.45\times10^{-8}}=0.40+0.05=0.45\text{ V}$$ Since $0.96\text{ V}>0.45\text{ V}$ (threshold above flat band), the surface is **inverted** at $V_G=0$ (assumption: ordinary oxide thickness of about 0.1 micron or less; an extremely thick oxide would give depletion). ### Problem 18.3 Computer problem **Problem.** With $\Phi_M'-\chi'$ as input, write a program that plots $\phi_{MS}$ versus doping like Fig. 18.3; confirm the Al and n$^+$ poly plots; also plot for p$^+$ poly with $E_F=E_V$. **Solution.** The logic and code are in the section "Computer-Based Exercise" above. Confirmation of the plotted values: at $N=10^{15}$ cm$^{-3}$ | Gate | on n-Si | on p-Si | |---|---|---| | Al ($-0.03$ eV) | $-0.03-0.262=-0.29$ V | $-0.03-0.858=-0.89$ V | | n$^+$ poly ($-0.18$ eV) | $-0.18-0.262=-0.44$ V | $-0.18-0.858=-1.04$ V | | p$^+$ poly ($1.12$ eV) | $1.12-0.262=+0.86$ V | $1.12-0.858=+0.26$ V | These match Fig. 18.3 (Al on p-Si about $-0.9$ V; n$^+$ poly on p-Si about $-1.0$ V), and the p$^+$ poly curves are positive for both substrate types. ### Problem 18.4 Mobile-ion calculation **Problem.** (a) A uniform sodium distribution $\rho_{ion}(x)=\rho_0$ over the oxide, $\rho_0/q=10^{16}$ cm$^{-3}$ (assumed; the OCR dropped the exponent), $x_o=0.1\ \mu$m. Compute $\Delta V_G$. (b) After $+$BT stress all ions pile up at the oxide-semiconductor interface; find $\Delta V_G$. **Solution.** (a) From (18.13): $$\Delta V_G=-\frac1{K_O\varepsilon_0}\int_0^{x_o}x\rho_0\,dx=-\frac{\rho_0x_o^2}{2K_O\varepsilon_0}$$ Total ionic charge: $Q_M=\rho_0x_o=(1.6\times10^{-19})(10^{16})(10^{-5})=1.6\times10^{-8}$ C/cm$^2$ ($Q_M/q=10^{11}$ cm$^{-2}$). $$\Delta V_G=-\frac{Q_M}{2C_O}=-\frac{1.6\times10^{-8}}{2(3.45\times10^{-8})}=-0.23\text{ V}$$ (for general $\rho_0$: $\Delta V_G=-0.23\text{ V}\times\dfrac{\rho_0/q}{10^{16}\text{ cm}^{-3}}$.) (b) All charge at $x=x_o$: $\Delta V_G=-Q_M/C_O=-\dfrac{1.6\times10^{-8}}{3.45\times10^{-8}}=-0.46\text{ V}$ (twice the uniform value, as the centroid moves from $x_o/2$ to $x_o$). ### Problem 18.5 Fixed charge distributed near the interface **Problem.** Suppose the fixed charge is not a $\delta$-sheet but spread over a short distance $\Delta x$ from the Si-SiO$_2$ interface. (a) Give the standard $\Delta V_G(\text{fixed charge})$. (b) Find $\Delta V_G$ for a charge that increases linearly from zero at distance $\Delta x$ from the interface to $2Q_F/\Delta x$ right at the interface (total charge $Q_F$). (c) Compute the ratio (b)/(a) for $\Delta x=10$ A $=10^{-7}$ cm and $x_o=0.1\ \mu$m. **Solution.** (a) $\Delta V_G=-Q_F/C_O$. (b) With $x$ measured from the metal, the distribution is zero for $x<x_o-\Delta x$ and, for $x_o-\Delta x\le x\le x_o$, $$\rho(x)=\frac{2Q_F}{\Delta x^2}\big[x-(x_o-\Delta x)\big]$$ (check: at $x=x_o$, $\rho=2Q_F/\Delta x$; total $=\tfrac12\cdot\tfrac{2Q_F}{\Delta x}\Delta x=Q_F$.) Let $u=x-(x_o-\Delta x)$, $0\le u\le\Delta x$: $$\int xρ\,dx=\frac{2Q_F}{\Delta x^2}\int_0^{\Delta x}(x_o-\Delta x+u)\,u\,du$$ $$=\frac{2Q_F}{\Delta x^2}\left[(x_o-\Delta x)\frac{\Delta x^2}{2}+\frac{\Delta x^3}{3}\right]$$ $$=Q_F\left[(x_o-\Delta x)+\frac{2\Delta x}{3}\right]=Q_F\left(x_o-\frac{\Delta x}{3}\right)$$ $$\Delta V_G=-\frac{1}{K_O\varepsilon_0}Q_F\left(x_o-\frac{\Delta x}3\right)=-\frac{Q_F}{C_O}\left(1-\frac{\Delta x}{3x_o}\right)$$ (Centroid of the triangle sits $\Delta x/3$ from the interface.) (c) $$\frac{\Delta V_G(b)}{\Delta V_G(a)}=1-\frac{\Delta x}{3x_o}=1-\frac{10^{-7}}{3(10^{-5})}=1-0.0033=0.9967$$ So treating $Q_F$ as a $\delta$-sheet is excellent (error 0.3%). ### Problem 18.6 $\phi_{MS}$ and $Q_F$ from a $V_{FB}$ versus $x_o$ plot **Problem.** Ideal devices except $\phi_{MS}\ne0$ and $Q_F\ne0$. (a) Show how $\phi_{MS}$ and $Q_F$ follow from $V_{FB}$ vs $x_o$. (b) Determine them from the data below. | $x_o$ ($\mu$m) | 0.1 | 0.15 | 0.2 | 0.25 | 0.3 | |---|---|---|---|---|---| | $V_{FB}$ (V) | $-0.91$ | $-1.04$ | $-1.20$ | $-1.33$ | $-1.52$ | **Solution.** (a) $V_{FB}=\phi_{MS}-Q_F/C_O$ with $C_O=K_O\varepsilon_0/x_o$: $$V_{FB}=\phi_{MS}-\frac{Q_F}{K_O\varepsilon_0}x_o$$ A straight line in $x_o$: **intercept** at $x_o=0$ is $\phi_{MS}$; **slope** is $-Q_F/K_O\varepsilon_0$, so $Q_F=-(\text{slope})\,K_O\varepsilon_0$. (b) Least-squares line. Means: $\bar x=0.20\ \mu$m, $\bar y=(-0.91-1.04-1.20-1.33-1.52)/5=-1.20$ V. $$S_{xx}=\sum(x-\bar x)^2=0.01+0.0025+0+0.0025+0.01=0.025$$ $$S_{xy}=(-0.1)(0.29)+(-0.05)(0.16)+0+(0.05)(-0.13)+(0.1)(-0.32)=-0.029-0.008-0.0065-0.032=-0.0755$$ $$\text{slope}=\frac{S_{xy}}{S_{xx}}=-3.02\text{ V}/\mu\text{m}=-3.02\times10^{4}\text{ V/cm}$$ $$\text{intercept}=\bar y-\text{slope}\cdot\bar x=-1.20+3.02(0.2)=-0.60\text{ V}$$ $$\phi_{MS}\approx-0.60\text{ V}$$ $$Q_F=(3.02\times10^{4})(3.9)(8.85\times10^{-14})=1.04\times10^{-8}\text{ C/cm}^2$$ $$\frac{Q_F}{q}=\frac{1.04\times10^{-8}}{1.6\times10^{-19}}=6.5\times10^{10}\text{ cm}^{-2}$$ ### Problem 18.7 Charge information from a band diagram **Problem.** An MOS-C is characterized by the band diagram of Fig. P18.7 (negligible interface traps). (a) $Q_M=0$ or $\ne0$? (b) $Q_F=0$ or $\ne0$? **Solution.** (The figure is not available; the reading rules are stated and applied to the diagram as in Exercise 18.3, with metal $E_F$ above semiconductor $E_F$ and straight oxide bands.) (a) Oxide band curvature measures oxide charge: $d^2E_C/dx^2\propto\rho_{ox}$. Straight oxide bands (constant oxide field) mean $\rho_{ox}=0$, so $Q_M\approx0$. (If the bands were curved, ionic charge would be present.) (b) Compare the oxide field with the semiconductor-surface displacement field: if $K_O\mathscr{E}_{ox}\ne K_s\mathscr{E}_s$ there is interface charge. With negligible $Q_{IT}$ and no stress, that charge is the fixed charge: $Q_F=K_s\varepsilon_0\mathscr{E}_s-K_O\varepsilon_0\mathscr{E}_{ox}$. For the diagram, where the oxide field is smaller than $K_s\mathscr{E}_s/K_O$ (a discontinuity of $D$ at the interface), $Q_F>0$, so $Q_F\ne0$. ### Problem 18.8 $Q_F$ from $C$-$V$ data **Problem.** An Al-SiO$_2$-Si capacitor at 300 K has $C_{MAX}=200$ pF, $C_{MIN}=67$ pF, $V_{FB}=-0.71$ V, $Q_M=0$, $Q_{IT}=0$, gate area $A_G=2.9\times10^{-3}$ cm$^2$. (a) Sketch the ideal-device $C$-$V$ on a copy of the figure and mark its flat-band point. (b) Find the Si doping (delta-depletion formulation). (c) Find $Q_F$. **Solution.** (a) The ideal device ($\phi_{MS}=0$, $Q_F=0$) has the same $C_{MAX}$, $C_{MIN}$ and shape, shifted rigidly by $\Delta V_G=V_{FB}-0$: the ideal curve is the measured curve moved **right** by $0.71$ V (dashed line), with its flat-band point at $V_G=0$ at the same flat-band capacitance $C_{FB}$ (between $C_{MIN}$ and $C_{MAX}$, close to $C_{MAX}$ at the knee). (b) $C_O=C_{MAX}/A_G=\dfrac{200\times10^{-12}}{2.9\times10^{-3}}=6.90\times10^{-8}$ F/cm$^2$, so $$x_o=\frac{K_O\varepsilon_0}{C_O}=\frac{3.45\times10^{-13}}{6.90\times10^{-8}}=5.0\times10^{-6}\text{ cm}=500\text{ \AA}$$ Delta-depletion: $\dfrac{C_{MIN}}{C_O}=\dfrac{1}{1+(K_O/K_s)(W_T/x_o)}$, so $$W_T=\frac{K_s}{K_O}x_o\left(\frac{C_O}{C_{MIN}}-1\right)=\frac{11.8}{3.9}(5\times10^{-6})\left(\frac{200}{67}-1\right)=(3.026)(5\times10^{-6})(1.985)=3.0\times10^{-5}\text{ cm}$$ Now $W_T=\sqrt{4K_s\varepsilon_0\phi_F/(qN)}$ with $\phi_F=kT\ln(N/n_i)$; iterate. $N=10^{16}$: $\phi_F=0.358$, $W_T=3.06\times10^{-5}$ cm; $N=1.05\times10^{16}$: $\phi_F=0.359$, $W_T=2.99\times10^{-5}$ cm. So $$N\approx1.03\times10^{16}\text{ cm}^{-3}\approx10^{16}\text{ cm}^{-3}$$ **Which type?** $V_{FB}=\phi_{MS}-Q_F/C_O$ with $Q_F>0$. For an Al gate on p-Si at $10^{16}$, $\phi_{MS}\approx-0.03-(0.56+0.358)=-0.95$ V, which is already more negative than $-0.71$ V and would need $Q_F<0$, impossible. Therefore the substrate is **n-type**: $N_D\approx10^{16}$ cm$^{-3}$. (c) n-Si: $(E_C-E_F)=0.56-0.359=0.201$ eV, so $\phi_{MS}=-0.03-0.201=-0.23$ V. $$-\frac{Q_F}{C_O}=V_{FB}-\phi_{MS}=-0.71+0.23=-0.48\text{ V}$$ $$Q_F=0.48(6.90\times10^{-8})=3.3\times10^{-8}\text{ C/cm}^2,\qquad\frac{Q_F}{q}=2.1\times10^{11}\text{ cm}^{-2}$$ ### Problem 18.9 Sketch the pre-anneal $C$-$V$ **Problem.** The $C$-$V$ of an MOS-C is measured before and after a postmetallization anneal that eliminates interface traps. The AFTER curve is in Fig. P18.9. Assuming acceptor-like interface traps, sketch the BEFORE curve. **Solution.** (Assume the figure shows a p-bulk device with accumulation on the left; the n-bulk case is mirrored below.) Acceptor-like traps are neutral when empty and negative ($Q_{IT}<0$) when filled; they fill when the surface Fermi level lies above them. Using $\Delta V_G=-Q_{IT}/C_O$: - **p-bulk.** In **accumulation** ($E_F$ near $E_V$) all acceptor-like traps are empty: $Q_{IT}=0$, no shift. Moving through depletion toward **inversion** ($E_F$ rising toward $E_C$), more traps fill: $Q_{IT}$ increasingly negative, so $\Delta V_G=-Q_{IT}/C_O$ increasingly **positive**. So the BEFORE curve coincides with the AFTER curve in accumulation and is progressively displaced **to the right** in depletion and inversion: the transition to $C_{MIN}$ is stretched out toward positive voltage, with the rise at the inversion end (for low-frequency) displaced the most. The flat-band point is shifted by the (small) trap charge at flat band. - **n-bulk** (mirror): inversion side coincides with AFTER, and accumulation/depletion is displaced to the right the most; the curve is stretched out on the left side. In both cases the BEFORE curve is the AFTER curve stretched out along the voltage axis, with the displacement growing with the number of filled traps; the capacitance extremes ($C_{MAX}$, $C_{MIN}$) are unchanged. ### Problem 18.10 Dangling bonds: (100) versus (110) **Problem.** If interface traps come from dangling bonds in proportion to the Si surface atom density, which plane, (100) or (110), has the higher interface trap density? **Solution.** Surface atom density from the lattice constant $a=5.43$ A $=5.43\times10^{-8}$ cm ($a^2=2.95\times10^{-15}$ cm$^2$). In the conventional fcc cell: - (100) face: area $a^2$, containing $4\times\tfrac14+1=2$ atoms: $\sigma_{100}=2/a^2=6.8\times10^{14}$ cm$^{-2}$ (agrees with the text). - (110) plane: rectangle $a\times\sqrt2a$, area $\sqrt2a^2$, containing $4\times\tfrac14+2\times\tfrac12=2$ atoms: $\sigma_{110}=\dfrac{2}{\sqrt2a^2}=\dfrac{\sqrt2}{a^2}=4.8\times10^{14}$ cm$^{-2}$. $$\frac{\sigma_{100}}{\sigma_{110}}=\sqrt2\approx1.41$$ **Answer: (100) has the higher density of surface atoms, hence of dangling bonds and interface traps, by a factor of about 1.4.** ### Problem 18.11 Single-level interfacial trap at midgap (n-bulk, high-frequency) **Problem.** An n-bulk MOS-C has interface traps only at one energy $E_i$ (midgap), with density $N_{it}$ large enough to matter. Sketch the HF $C$-$V$ for (a) donor-like states; (b) acceptor-like states; (c) donor-like with the level near $E_C$ ($E_C-E_{trap}=0.001$ eV); (d) donor-like with the level near $E_V$ ($E_{trap}-E_V=0.001$ eV). **Solution.** The key: a trap level is charged only when the surface Fermi level is on the "charged" side of it. At the surface $E_F$ passes the level exactly where $\phi_s$ equals the midgap value (for n-bulk, $E_F=E_i$ at the surface when the bands bend up by $\phi_F$). As $E_F$ sweeps through the single level, the charge changes by $qN_{it}$ essentially at one value of $\phi_s$. Because $\phi_s$ cannot move while that charge is being added, the gate voltage must change by $\Delta V=qN_{it}/C_O$ with $\phi_s$ (hence $C$) **pinned**. So the $C$-$V$ has a **horizontal plateau** at the capacitance corresponding to midgap surface potential, of width $qN_{it}/C_O$ in volts. (a) Donor-like (positive when empty). For n-bulk: accumulation and depletion down to midgap, $E_F>E_i$: level filled, neutral. In inversion, $E_F<E_i$: empty, positive. So the curve is the ideal one in accumulation and depletion, then a plateau at the midgap capacitance, then the ideal curve continues but displaced to the **left (negative)** by $qN_{it}/C_O$ in inversion (positive charge lowers the required gate voltage: $\Delta V=-Q_{IT}/C_O<0$). (b) Acceptor-like (negative when filled). Filled for $E_F>E_i$: accumulation and depletion have $Q_{IT}=-qN_{it}$, so $\Delta V=+qN_{it}/C_O$: displaced to the **right**; inversion side is neutral and unshifted. Plateau again at midgap capacitance. So relative to the ideal curve the inversion branch is in its normal place and the accumulation/depletion branch is displaced right. (c) Donor-like level at $E_C-0.001$ eV. It is positive whenever $E_F$ is below the level, which is true for essentially all gate voltages (bulk $E_F$ is far below $E_C$), becoming neutral only in extreme accumulation. So $Q_{IT}=+qN_{it}$ over the whole practical range: the curve is a **rigid shift to the left** by $qN_{it}/C_O$ with the plateau lying beyond the accumulation end (not seen). (d) Donor-like level at $E_V+0.001$ eV. It is filled (neutral) whenever $E_F$ is above it, true for every bias except extreme inversion ($E_F$ at the valence band edge). So there is essentially **no distortion**: the ideal curve, with any plateau at the extreme inversion end (very strong inversion) beyond practical range. ### Problem 18.12 Half-irradiated MOS-C **Problem.** After fabrication exactly half the gate area is irradiated, raising the apparent $Q_F$ there; $D_{it}$ is unaffected. Sketch the $C$-$V$ after irradiation and explain. **Solution.** Treat the device as two parallel capacitors of area $A/2$ each: the unirradiated half with the original characteristic $C_0(V_G)$, and the irradiated half with the same characteristic shifted along the voltage axis by $$\Delta V_G=-\frac{\Delta Q_F}{C_O}<0\quad(\text{to the left; toward negative }V_G)$$ Total capacitance (parallel combination): $$C(V_G)=\tfrac12\,C_0(V_G)+\tfrac12\,C_0(V_G-\Delta V_G)$$ So the new curve is the average of the original and a left-shifted copy: - Far from the transition (deep accumulation, deep inversion at high frequency) both halves agree: $C=C_{MAX}$ or $C_{MIN}$, unchanged. - In the transition region the irradiated half makes its transition at a more negative voltage (for p-type: accumulation to depletion, and so on), so there is an **intermediate range of width $|\Delta V_G|$** where one half has fallen toward $C_{MIN}$ and the other is still near $C_{MAX}$: a shoulder (or step) near $C\approx(C_{MAX}+C_{MIN})/2$. - Overall: a **stretched-out** characteristic, with the flat-band point and the whole transition broadened by $|\Delta V_G|$, equal to the original curve on the positive side and the shifted curve on the negative side joined by the shoulder. ### Problem 18.13 BT-stressed MOSFET **Problem.** Before $+$BT, the $I_D$-$V_D$ of a (p-channel) MOSFET is as in Fig. P18.13(a) (curves for $V_G=-2,-3,-4$ V); the $g_m$-$V_G$ before and after $+$BT is in (b). (a) Probable cause of the $g_m$ shift? (b) Sketch the post-stress $I_D$-$V_D$ for $V_G=-2,-3,-4$ V and explain. **Solution.** (a) After positive BT stress the $g_m$-$V_G$ curve moves toward **more negative** $V_G$ (the threshold becomes more negative). A negative shift after positive stress is the signature of **mobile positive ions (Na$^+$)** drifting to the Si-SiO$_2$ interface (large $|\Delta V_G|=\gamma_MQ_M/C_O$). (Interface traps would additionally degrade the peak of $g_m$, and the shape here is mainly translated.) (b) Translation of the threshold by $\Delta V_T=-\Delta$ ($\Delta>0$; from the figure assume $\Delta\approx1$ V) changes the *gate overdrive*: after stress, $V_G-V_T$ is smaller by $\Delta$ at the same gate voltage. The characteristic of gate voltage $V_G$ after stress equals the **pre-stress** curve at $V_G+\Delta$ (i.e. less negative by $\Delta$). So with $\Delta=1$ V: | After-stress curve at $V_G$ | Equals pre-stress curve at | |---|---| | $-2$ V | $-1$ V (at or below threshold: essentially off) | | $-3$ V | $-2$ V | | $-4$ V | $-3$ V | The shape of each curve is unchanged, but the currents are lower at the same $|V_G|$, and the saturation voltage is $V_{Dsat}=V_G-V_T$ reduced correspondingly. (The magnitude of the shift is read from Fig. P18.13(b); the rule applies for any $\Delta$.) ### Problem 18.14 Modifications to an n-channel MOSFET **Problem.** An n-channel (p-bulk) MOSFET is ideal except $\phi_{MS}\ne0$. How does each change, taken independently, affect $V_T$? **Solution.** Base expression: $V_T=\phi_{MS}+2\phi_F+\sqrt{4K_s\varepsilon_0qN_A\phi_F}/C_O$ (with $Q_F=Q_M=Q_{IT}=0$). (a) **Radiation creates apparent $Q_F>0$:** $V_{FB}$ gets $-Q_F/C_O$, so $V_T$ **decreases** (shifts negative). (b) **Al to Cu:** $\Phi_M'-\chi'$ increases from $-0.03$ to $+0.63$ eV, so $\phi_{MS}$ increases by about $0.66$ V and $V_T$ **increases** by about $0.66$ V. (c) **Increased substrate doping:** $2\phi_F$ and the depletion-charge term increase (strongly), and $\phi_{MS}$ becomes slightly more negative (Fig. 18.3), but the net effect is a **positive shift in $V_T$** (the depletion-charge increase dominates). (d) **Thinner oxide:** $C_O$ increases, so the depletion-charge term $\sqrt{4K_s\varepsilon_0qN_A\phi_F}/C_O$ decreases and $V_T$ **decreases** toward $\phi_{MS}+2\phi_F$ (no $Q_F$ here to make the shift worse). (e) **Boron implant into the near surface:** adds negative charge at the interface, $\Delta V_G=-Q_I/C_O=+qN_I/C_O$: $V_T$ **increases**. ### Problem 18.15 Compute MOSFET $V_{FB}$ and $V_T$ **Problem.** A MOSFET has $\phi_{MS}=-0.46$ V, $Q_F/q=2\times10^{11}$ cm$^{-2}$, $Q_M=0$, $Q_{IT}=0$, an implanted ion charge $Q_I/q=-4\times10^{11}$ cm$^{-2}$ at the Si-SiO$_2$ interface (exponents restored from OCR, since $V_{FB}$ then comes out near zero as intended), $x_o=0.05\ \mu$m, $A=10^{-3}$ cm$^2$, $N_D=10^{15}$ cm$^{-3}$. (a) Find $V_{FB}$. (b) Find $V_T$. (c) Enhancement or depletion? **Solution.** n-substrate, so p-channel. $$C_O=\frac{(3.9)(8.85\times10^{-14})}{5\times10^{-6}}=6.90\times10^{-8}\text{ F/cm}^2$$ (a) $$V_{FB}=\phi_{MS}-\frac{Q_F}{C_O}-\frac{Q_I}{C_O}$$ $$\frac{Q_F}{C_O}=\frac{(1.6\times10^{-19})(2\times10^{11})}{6.90\times10^{-8}}=0.464\text{ V}$$ $$\frac{Q_I}{C_O}=\frac{(1.6\times10^{-19})(-4\times10^{11})}{6.90\times10^{-8}}=-0.928\text{ V}$$ $$V_{FB}=-0.46-0.464+0.928=+0.004\approx0\text{ V}$$ (b) Ideal p-channel threshold: $|\phi_F|=0.0259\ln(10^{15}/10^{10})=0.298$ V. $$V_T'=-\left[2(0.298)+\frac{\sqrt{4K_s\varepsilon_0qN_D\phi_F}}{C_O}\right]=-\left[0.596+\frac{\sqrt{4(11.8)(8.85\times10^{-14})(1.6\times10^{-19})(10^{15})(0.298)}}{6.90\times10^{-8}}\right]$$ $$\sqrt{\cdots}=1.41\times10^{-8}\Rightarrow\frac{1.41\times10^{-8}}{6.90\times10^{-8}}=0.204$$ $$V_T'=-(0.596+0.204)=-0.80\text{ V}$$ $$V_T=V_T'+V_{FB}=-0.80+0.00=-0.80\text{ V}$$ (c) A p-channel device with $V_T=-0.80$ V is off at $V_G=0$ and turns on for $V_G<-0.8$ V: **enhancement mode**. ### Problem 18.16 Required implant dose **Problem.** An Al-SiO$_2$-Si n-channel MOSFET, $T=300$ K, $N_A=10^{17}$ cm$^{-3}$, $x_o=100$ A (assumed; OCR garbled), $Q_F/q=10^{11}$ cm$^{-2}$ (assumed), no interface traps or mobile ions. Find the boron dose $N_I$ (ions/cm$^2$, negative charge right at the interface) needed for $V_T=0.5$ V. **Solution.** $$C_O=\frac{3.45\times10^{-13}}{10^{-6}}=3.45\times10^{-7}\text{ F/cm}^2$$ *Ideal threshold.* $\phi_F=0.0259\ln(10^{17}/10^{10})=0.4175$ V, $2\phi_F=0.835$ V. $$\frac{\sqrt{4K_s\varepsilon_0qN_A\phi_F}}{C_O}=\frac{\sqrt{4(11.8)(8.85\times10^{-14})(1.6\times10^{-19})(10^{17})(0.4175)}}{3.45\times10^{-7}}=\frac{1.67\times10^{-7}}{3.45\times10^{-7}}=0.484\text{ V}$$ $$V_T'=0.835+0.484=1.319\text{ V}$$ *Flat band without implant.* Al on p-Si: $(E_C-E_F)=0.56+0.4175=0.9775$ eV, so $\phi_{MS}=-0.03-0.9775=-1.01$ V. $$\frac{Q_F}{C_O}=\frac{(1.6\times10^{-19})(10^{11})}{3.45\times10^{-7}}=0.046\text{ V}$$ $$V_{FB}=-1.0075-0.046=-1.054\text{ V}$$ $$V_T(\text{no implant})=1.319-1.054=0.265\text{ V}$$ *Required implant shift.* $$0.5=0.265+\frac{qN_I}{C_O}\Rightarrow\frac{qN_I}{C_O}=0.235\text{ V}$$ $$N_I=\frac{0.235\,C_O}{q}=\frac{0.235(3.45\times10^{-7})}{1.6\times10^{-19}}=5.1\times10^{11}\text{ ions/cm}^2$$ **Answer: about $5\times10^{11}$ boron ions/cm$^2$.** # Chapter 19: Modern FET Structures > **Where this slice sits.** This file covers book pages 691-732 (PDF markers 714-755): Chapter 19 (*Modern FET Structures*), then the Part III closing material (*Supplement and Review*): the reading list, the cited references, the review list of terms, and the two review problem sets with answers. The Appendices begin right after, so we stop there. ## The big picture: what happens when you shrink a transistor? Picture a garden hose. Open the tap and water flows in a well-behaved way: the valve (your thumb) controls the flow and nothing else interferes. Now shorten the hose to two centimetres. The water no longer "knows" it is in a hose. It sprays, the end of the hose affects the beginning, and your thumb loses some authority. A MOSFET is that hose. The gate is the thumb, the channel is the hose, and the source and drain are the taps at the two ends. For fifty years the industry has made the hose shorter and shorter to get **higher speed** (carriers cross a shorter distance) and **higher packing density** (more transistors per chip). But when the channel becomes comparable to the depletion widths of the source and drain junctions, the source and drain start to interfere with the gate. These departures from the long-channel theory of Chapters 17 and 18 are called **small-dimension effects** (also *short-channel effects* or *small-geometry effects*). Why should you care? Because: - The **threshold voltage** sets logic levels, noise margins, speed and node voltages. If it wanders with channel length, circuits stop working predictably. - The **subthreshold current** sets the off-state power dissipation, dynamic-logic clock speeds and DRAM refresh times. The golden rule of this chapter: small-dimension effects are almost always **undesirable**. Engineers either avoid them by *scaling* the device dimensions properly or by *redesigning* the device. The second half of the chapter is a tour of the redesigned devices (LDD, DMOS, buried-channel, SiGe, SOI, MODFET). --- ## 19.1 Small-Dimension Effects ### 19.1.1 Introduction In 1965 the smallest MOSFETs had $L \approx 1\ \text{mil} = 25\ \mu\text{m}$. By 1990 industry-standard MOS structures were submicron. Figure 19.1 (described: a row of transistor cross sections for the years 1970, 1980, 1990, 2000, each with a visibly shorter gate between the $n^+$ source and drain on a $p$ substrate) shows the relative shrinkage of the channel length, which roughly tracks the minimum feature size of production-line MOS DRAMs. **Table 19.1: Silicon IC projections** (DRAM = dynamic random access memory; I/O = input/output; wiring levels = number of metallization layers connecting devices on the chip) | Year of first DRAM shipment | 1995 | 1998 | 2001 | 2004 | 2007 | 2010 | |---|---|---|---|---|---|---| | Minimum feature size ($\mu$m) | 0.35 | 0.25 | 0.18 | 0.13 | 0.10 | 0.07 | | Chip size, microprocessor (mm$^2$) | 250 | 300 | 360 | 430 | 520 | 620 | | Chip size, DRAM (mm$^2$) | 190 | 280 | 420 | 640 | 960 | 1400 | | Bits/chip in DRAMs | 64M | 256M | 1G | 4G | 16G | 64G | | Si wafer diameter (mm) | 200 | 200 | 300 | 300 | 400 | 400 | | Wiring levels (max, on-chip logic) | 4-5 | 5 | 5-6 | 6 | 6-7 | 7-8 | | I/O pads per chip (high performance) | 900 | 1350 | 2000 | 2600 | 3600 | 4800 | | Supply voltage, desktop (V) | 3.3 | 2.5 | 1.8 | 1.5 | 1.2 | 0.9 | | Supply voltage, portable (V) | 2.5 | 1.8-2.5 | 0.9-1.8 | 0.9 | 0.9 | 0.9 | | On-chip frequency, high performance (MHz) | 300 | 450 | 600 | 800 | 1000 | 1100 | | Chip-to-board frequency (MHz) | 150 | 200 | 250 | 300 | 375 | 475 | (The OCR printed the feature size as "m"; the unit must be micrometres.) Notice the story the table tells: features shrink, chips grow, but the supply voltage falls only slowly. That last fact matters, because it means the electric fields inside the device *grow* as dimensions shrink, which is exactly what drives the hot-carrier effects of Section 19.1.4. ### Three tell-tale signs of short-channel behaviour **1. Slanted post-pinch-off $I_D$-$V_D$ curves.** In a long-channel device the saturated part of the $I_D$-$V_D$ curve is flat. The onset of short-channel effects is heralded by a significant *upward slant* of the post-pinch-off portion. In severe cases (Fig. 19.2, a device with huge $V_D$ range, 0 to $-30$ V on the axis in the original) the characteristic becomes a family of curves of the form $I_D \propto V_D^2$, and current flows even for gate voltages *below* threshold. (The Fig. 19.2 device should be "off" for $V_G \le 0$ V, yet it conducts.) **2. Drain-dependent subthreshold current.** In a long-channel MOSFET (Fig. 17.11) the subthreshold current varies exponentially with $V_G$ and is independent of $V_D$ as long as $V_D$ exceeds a few $kT/q$. In short-channel devices the subthreshold drain current increases systematically and significantly with $V_D$. **3. Shifts in the threshold voltage.** For a long-channel device (Subsection 18.3.1) $V_T$ does not depend on gate length $L$ or width $Z$. In short-channel devices $V_T$ becomes a function of the gate dimensions *and* of the applied biases (Fig. 19.3: $V_T$ plotted against $L$ from 0 to 7 $\mu$m for $V_D = 0.125$ V and $V_D = 4$ V; the curves are flat at large $L$ and fall as $L$ shrinks, and the $V_D = 4$ V curve falls lower and earlier). ### The minimum channel length $L_{min}$ Most small-dimension effects stem from the reduction of $L$. So it is useful to define a minimum channel length below which significant short-channel effects are expected. Crudely, $L_{min}$ must exceed the sum of the source and drain depletion widths; values from 0.1 to 1 $\mu$m result. A better empirical estimate (from computer simulation and experiment) is $$L_{min} = 0.4\left[r_j\, x_o\,(W_S + W_D)^2\right]^{1/3} \tag{19.1}$$ with the **unusual units** $L_{min}$ in $\mu$m, $r_j$ (source/drain junction depth) in $\mu$m, $x_o$ (oxide thickness) in angstroms, and $W_S$, $W_D$ (depletion widths at the source and drain junctions) in $\mu$m. What does Eq. (19.1) teach? To make a transistor shorter while keeping long-channel behaviour you can: - make the source/drain islands **shallower** (smaller $r_j$), - make the oxide **thinner** (smaller $x_o$, so the gate keeps the upper hand), - **dope the substrate more heavily** (smaller $W_S$, $W_D$). All three have actually been used in industry. ### Why do departures happen? Three families of causes 1. **Broken assumptions.** The long-channel theory assumed things (such as a purely one-dimensional field in the channel direction) that are no longer true. 2. **Enhanced hidden effects.** Effects that always exist but are negligible in long devices become important. 3. **Brand-new phenomena.** Things like ballistic transport that simply do not appear in long devices. All three families are represented below. --- ### 19.1.2 Threshold Voltage Modification #### Short channel: the source and drain "help" the gate In enhancement-mode devices $V_T$ *decreases* monotonically as $L$ decreases. **Intuition (a tug-of-war over charge).** Before inversion the region under the gate must be depleted, i.e. depletion charge $-qN_A W_T$ per unit area must be balanced by positive charge somewhere. In a long device all of it is balanced by gate charge. In a short device the source and drain $pn$ junctions already own a depletion layer of their own; a fraction of the "subgate" depletion charge is balanced by charge on the *other side* of the source and drain junctions (the $n^+$ islands). Think of two neighbours both contributing to pay a shared bill: the gate pays less. Less gate charge is needed to reach the start of inversion, so $|V_T|$ drops. The shorter the channel, the larger the percentage borne by the junctions, and the bigger the reduction. #### Derivation of $\Delta V_T$ (short channel) *Step 1: ideal MOS-C voltage balance.* For an ideal device $$V_G = V_{ox} + \phi_s, \qquad V_{ox} = \frac{|Q_G|\,}{C_o}\ \text{(magnitude)}\quad \Rightarrow\quad V_G = -\frac{Q_s}{C_o} + \phi_s \tag{19.2--19.4}$$ where $C_o = K_O\epsilon_0/x_o$ and $Q_s$ is the total charge per cm$^2$ inside the semiconductor. (Gauss' law gives $Q_s = -K_S\epsilon_0\mathcal{E}_s$ and, combined with the oxide field, yields the voltage balance above, with the gate charge equal and opposite to $Q_s$.) *Step 2: specialize to threshold.* At the start of inversion $\phi_s = 2\phi_F$ and $Q_s \to Q_B$ (the depletion, or bulk, charge per unit area of gate): $$V_T = 2\phi_F - \frac{Q_B}{C_o} \tag{19.5}$$ *Step 3: define the shift.* $$\Delta V_T = V_T(\text{short channel}) - V_T(\text{long channel}) \tag{19.6}$$ Let $Q_{BL}$ and $Q_{BS}$ be the depletion charges per cm$^2$ in the long- and short-channel device. Because $2\phi_F$ cancels in the difference: $$\Delta V_T = -\frac{1}{C_o}\left(Q_{BS} - Q_{BL}\right) = -\frac{Q_{BL}}{C_o}\left(\frac{Q_{BS}}{Q_{BL}} - 1\right) \tag{19.7}$$ *Step 4: depletion charge of a long device.* Take $V_D$ small or zero so that $W_s \le W_T$ everywhere under the gate; call $W = W_T$. The gate controls the whole rectangular depletion volume $ZLW$ ($Z$ = gate width). Dividing the charge by gate area $ZL$: $$Q_{BL} = \frac{-qN_A (ZLW)}{ZL} = -qN_A W \tag{19.8}$$ *Step 5: depletion charge of a short device.* The portions near the source and drain (shaded in Fig. 19.4: two triangles at the bottom corners of the rectangle, bounded by the circular-looking junction depletion edges) are controlled by the source and drain. The gate controls only the **trapezoid** with top side $L$ and bottom side $L'$, whose area (per unit $Z$) is $\tfrac12(L+L')W$: $$Q_{BS} = \frac{-qN_A \tfrac12 (L + L')\,Z\,W}{ZL} = -qN_A W\,\frac{L + L'}{2L} \tag{19.9}$$ *Step 6: assemble.* Substitute (19.8) and (19.9) in (19.7): $$\Delta V_T = -\frac{qN_A W}{C_o}\left[1 - \frac{L + L'}{2L}\right] \tag{19.10}$$ *Step 7: geometry for $L'$.* Model each junction corner as a quarter-cylinder of radius $r_j$ (centre on the surface at the lateral edge of the island, a distance $r_j$ outward) with the depletion edge a concentric arc of radius $r_j + W$ (taking $W_S = W_D = W$). The arc crosses the depth $y = W$ (the bottom of the gate-controlled depletion layer) at horizontal distance $\sqrt{(r_j + W)^2 - W^2}$ from the centre, i.e. at lateral distance $\Delta$ inside the channel edge, where $$(r_j + W)^2 = (r_j + \Delta)^2 + W^2 \tag{19.11}$$ Solving: $r_j + \Delta = \sqrt{r_j^2 + 2 r_j W}$, so $\Delta = r_j\left[\sqrt{1 + 2W/r_j} - 1\right]$. Both ends of the bottom side lose $\Delta$: $$L' = L - 2\Delta = L - 2r_j\left[\sqrt{1 + \frac{2W}{r_j}} - 1\right] \tag{19.12}$$ *Step 8: final result.* Since $\frac{L+L'}{2L} = 1 - \frac{\Delta}{L}$, we have $1 - \frac{L+L'}{2L} = \frac{\Delta}{L}$, and Eq. (19.10) becomes $$\boxed{\Delta V_T(\text{short channel}) = -\frac{qN_A W_T}{C_o}\,\frac{r_j}{L}\left[\sqrt{1 + \frac{2W_T}{r_j}} - 1\right]} \tag{19.13}$$ (with $W = W_T$ at threshold). **Reading the result.** The magnitude of the shift is large when $r_j/L$ is large, i.e. when the junctions are deep compared with the gate length. Examining $\Delta V_T / V_T(\text{long channel})$, the relevant gauge of importance, the effects are again minimized by reducing $x_o$ (raising $C_o$), reducing $r_j$, and increasing $N_A$ (which shrinks $W_T$ more than it raises $qN_A$). This is the same list as for $L_{min}$: a nice consistency check on a first-order derivation. #### Narrow width: the gate also gets help from the sides For $Z$ comparable to $W_T$, $V_T$ *increases* as $Z$ decreases. The $Z$-dependence is opposite to the $L$-dependence. **Intuition.** Look at the side view (Fig. 19.5: gate oxide over the channel, thick field oxide on each side, and a depletion region that bulges out sideways under the field oxide beyond the gate width $Z$). The gate field also fringes sideways and has to balance that **extra** lateral depletion charge. In a wide device this fringe is negligible. In a narrow one it is comparable to the charge right under the gate. More charge per unit gate area means more gate voltage to reach inversion, so $V_T$ rises. **Derivation.** Take the two lateral regions as quarter-cylinders of radius $W$. Their combined volume is $2\cdot\tfrac14\pi W^2 L = \tfrac{\pi}{2}W^2 L$. The total controlled volume is $ZLW + \tfrac{\pi}{2}W^2L$: $$Q_B(\text{narrow}) = \frac{-qN_A\left(ZLW + \tfrac{\pi}{2}W^2 L\right)}{ZL} = -qN_A W\left(1 + \frac{\pi W}{2Z}\right) \tag{19.14}$$ Replace $Q_{BS}$ by this in Eq. (19.7), with $Q_{BL} = -qN_AW$: $$\Delta V_T = -\frac{1}{C_o}\left[Q_B(\text{narrow}) - Q_{BL}\right] = +\frac{qN_A W}{C_o}\cdot\frac{\pi W}{2Z}$$ $$\boxed{\Delta V_T(\text{narrow width}) = \frac{qN_A W_T}{C_o}\,\frac{\pi W_T}{2Z}} \tag{19.15}$$ This confirms the initial claim: it becomes important when $Z \sim W_T$. **Combined effects.** A MOSFET with both a short channel and narrow width has a combined $\Delta V_T$ that is **not** simply the sum of the two individual shifts; more exact expressions are in the literature (DeMassa and Chien, ref. 30). --- ### 19.1.3 Parasitic BJT Action A MOSFET has an oppositely doped region between source and drain: $n^+$-$p$-$n^+$. That is a lateral NPN bipolar transistor! (The source is the emitter, the substrate the base, the drain the collector.) When $L$ shrinks to the size of a BJT base width, bipolar phenomena show up. **Source-to-drain punch-through.** When source and drain are a few microns or less apart, their depletion regions can touch (Fig. 19.6: two $n^+$ islands with depletion boundaries merged beneath the gate). The gate then loses control of the subgate region except for a thin skin at the Si-SiO$_2$ interface. Current is no longer confined to the surface channel; it flows beneath the surface through the touching depletion regions as a *space-charge-limited* current, which (as in BJT punch-through) varies as the **square** of the applied source-drain voltage, $I \propto V_D^2$. The $V_G \le 0$ characteristics in Fig. 19.2 are examples of this $I \propto V^2$ behaviour. *Cure:* raise the subgate doping, which narrows the depletion widths. But a heavier *entire* substrate raises parasitic capacitances, so the standard practice is a **deep ion implant** that selectively dopes only the subgate region. **Carrier multiplication with regenerative feedback.** There is always a little impact ionization in the high-field region near the drain. In long devices it is negligible; in short ones, coupled with regenerative feedback, it can dramatically raise the drain current, reduce the maximum usable $V_D$, and even destroy the device. The mechanism (Fig. 19.7) is just like the lowering of $BV_{CEO}$ relative to $BV_{CBO}$ in a BJT (Subsection 11.2.4): 1. Channel current enters the high-field drain region. A small fraction of carriers gains enough energy to create electron-hole pairs by impact ionization. 2. For an $n$-channel device the new electrons drift into the drain; the new holes are swept into the quasi-neutral bulk. 3. The hole current flowing through the finite bulk resistance $R_{BB}$ produces a potential drop between the depletion edge and the back contact, of polarity that **forward biases the source $pn$ junction**. 4. A forward-biased source junction injects electrons into the bulk, which add to the electron flow into the drain, giving more multiplication, and so on. The process is stable while the multiplication factor (the fractional increase in drain current) is less than $1/\alpha$, with $\alpha$ the common-base gain of the parasitic BJT. Beyond that, the current runs away and the device can fail. --- ### 19.1.4 Hot-Carrier Effects *Hot* carriers are carriers with far more kinetic energy than the lattice thermal energy $kT$, picked up from large electric fields. **Oxide charging.** Near the drain, channel carriers (and carriers entering the drain depletion region from the substrate) occasionally gain enough energy to jump over the Si-SiO$_2$ barrier (about 3.1 eV for electrons) and enter the oxide. Neutral traps in the oxide capture part of that charge, and the oxide slowly charges up. In long devices this explains "walk-out" (a progressive rise of the drain breakdown voltage with operation at large $V_D$). In short devices it is much worse because a larger fraction of the gated region is affected: $V_T$, $g_m$ and other parameters can change significantly. And since the charging is **cumulative**, it limits the useful *lifetime* of a device. A popular remedy is the lightly doped drain (Subsection 19.2.1). **Velocity saturation.** The long-channel analysis implicitly assumed that carrier velocity rises without limit as needed to support the current. In reality drift velocity saturates at $v_{sat} \approx 10^7$ cm/s when the field exceeds about $3\times10^4$ V/cm for electrons and about $10^5$ V/cm for holes (Fig. 3.4, Part I). Example: $V_D = 2$ V across $L = 0.5\ \mu$m gives an average field of $2/(0.5\times10^{-4}) = 4\times10^4$ V/cm, so there are certainly places in the channel at or beyond saturation. Two effects on the characteristics: 1. $I_{Dsat}$ is significantly reduced, modified approximately to $$I_{Dsat} \approx Z\, C_o\,(V_G - V_T)\,v_{sat} \tag{19.16}$$ *Why this form?* Saturation current = (charge per unit area in the channel) $\times$ (width) $\times$ (velocity); the inversion charge is $C_o(V_G - V_T)$ and the velocity is pinned at $v_{sat}$. 2. As Eq. (19.16) shows, $I_{Dsat}$ depends almost **linearly** on $V_G - V_T$ instead of with the long-channel square law. Fig. 19.8 shows (a) the experimental curves of a short device ($L = 2.7\ \mu$m, with evenly spaced curves for $V_G$ = 1, 2, 3, 4, 5 V), (b) a theory including velocity saturation (evenly spaced, matches the data) and (c) a theory ignoring it (spacing growing quadratically, does not match). **Diagnostic:** if equal steps in $V_G$ give roughly equal steps in $I_{Dsat}$ you are looking at velocity saturation; if the steps grow with $V_G$ it is square-law. **Velocity overshoot and ballistic transport.** The long-channel model assumed many scattering events between source and drain: $L \gg \ell$, where $\ell$ is the mean free path (average distance between scatterings). When $L \lesssim \ell$ a large fraction of carriers travels from source to drain **without a single scattering**: projectile-like, or *ballistic*, motion. Observable ballistic effects are theoretically possible in GaAs for $L \le 0.3\ \mu$m (somewhat shorter in Si). Both Si and GaAs FETs with $L \sim 0.1\ \mu$m have been built in research labs, and Table 19.1 says such devices become commonplace. Why care? With reduced scattering the average velocity through the channel can exceed $v_{sat}$: **velocity overshoot**. Average velocities up to 35% above $v_{sat}$ have been seen in a $0.12\ \mu$m MOSFET (ref. 32, Assaderaghi, Ko and Hu). Other limits (such as the injecting capability of the source) remain, but ballistic effects have been observed and will matter in future FETs. *Analogy.* A crowd moving through a corridor: if people keep bumping into one another (scattering) their average speed is capped. In a corridor so short that nobody bumps into anybody, everybody just runs. --- ## 19.2 Select Structure Survey The basic enhancement-mode MOSFET is the right model for learning concepts. In industry there are many variations to solve specific problems or to enhance specific characteristics, and compound-semiconductor FETs (GaAs and others) look somewhat different from Si ones. This survey is only a sample, biased toward structures common in the recent literature. ### 19.2.1 MOSFET Structures #### LDD (lightly doped drain) transistors Small devices are more susceptible to hot carriers, especially because supply voltages were not scaled down in proportion to dimensions. In the LDD structure (Fig. 19.9: poly-Si gate with sidewall oxide spacers, $n^-$ lightly doped regions between the channel end and the $n^+$ source and drain), the $n^-$ region creates a **reduced doping gradient** going from channel to drain. This lowers the peak electric field near the drain and shifts the peak toward the end of the channel, away from the oxide edge. Less carrier injection into the oxide, less oxide charging. (It is like replacing a sudden cliff by a gentle slope: the water, here the field, is spread out.) #### DMOS (double-diffused MOSFET) (Fig. 19.10: a vertically structured transistor with $n^+$ source inside a $p$ region, in an $n^-$ epitaxial layer; the channel is the lateral difference between the two diffusions; the lateral doping profile $N_D - N_A$ is plotted along $x$.) A $p$-type dopant (boron) is introduced first, then an $n$-type dopant (phosphorus), both through the **same** oxide mask opening. Boron diffuses slightly deeper and farther to the side. The *channel length is the lateral difference of the two profiles*, so a short channel ($\sim 1\ \mu$m) is achieved **without small-dimension lithography**. The result is high-frequency operation *combined with* high drain breakdown voltage (the lightly doped $n^-$ drain drift region supports voltage). Used in high-frequency analog and high-voltage/high-power circuits; introduced in the early 1970s; power-DMOS variants continue to be developed. #### Buried-channel MOSFET (Fig. 19.11: the same transistor with a thin $n$-type surface layer under the gate, same type as source and drain, doping $N_D$ on top of the $p$ body $N_A$; the sketched profile shows $N_D$ in the thin layer and $-N_A$ below.) The surface layer, typically ion-implanted, forms a **$pn$ junction bottom gate** and an **MOS top gate**: physically and functionally a hybrid J-FET/MOSFET. Depending on the thickness and doping of the layer it can be designed as depletion-mode or enhancement-mode. The name arises because conduction can take place *away from the oxide-semiconductor interface*. Benefits: inherently higher mobility (no interface scattering), reduced interface-trap interaction, and reduced sensitivity to hot-carrier effects. #### SiGe devices Ultrahigh-vacuum chemical-vapor deposition (UHV/CVD, ref. 35) allows production-line deposition of high-quality Si$_{1-x}$Ge$_x$ films: atomic-level thickness control, precise composition, minimal background contamination, any Ge fraction from 0 to 100%. Lattice constants: Ge is $a = 5.65$ Å, Si is $a = 5.43$ Å, so a relaxed SiGe alloy has a lattice constant larger than Si. But if a SiGe film on Si is thin enough (typically $\le 1000$ Å) it is **pseudomorphic**: it conforms atom by atom to the lattice of the Si substrate (and is therefore compressed in-plane). SiGe-base heterojunction bipolar transistors (HBTs) built this way achieved the highest operating frequency of any Si bipolar device. Pseudomorphic films carry large internal stress. Beyond a **critical thickness** defects nucleate spontaneously, relieve stress, and the film relaxes to its preferred lattice constant. *Relaxed* SiGe films are valuable for *enhanced-mobility MOSFETs* (Fig. 19.12): on a Si substrate grow (1) a compositionally graded SiGe buffer layer (5% to 30% Ge, about 1.5 $\mu$m), (2) a relaxed Si$_{0.7}$Ge$_{0.3}$ layer (about 0.25 $\mu$m), and (3) a thin (about 6 nm) pseudomorphic **strained Si** layer, topped with a 12 nm SiO$_2$ gate oxide and $n^+$ poly-Si gate. The defects are confined to the buffer; the relaxed Si$_{0.7}$Ge$_{0.3}$ acts as a quality substrate. Matching its larger lattice constant stretches the Si film (**tension**), which reduces carrier scattering and lowers the electron effective mass parallel to the interface. Both raise the mobility: the pictured device showed about a factor 2 enhancement of low-field mobility over a standard MOSFET, translating to more current drive at high fields and extending the performance limits of MOS technology. #### SOI (silicon-on-insulator) structures SOI = devices fabricated in single-crystal Si layers on an insulating film or substrate. Historical routes: - **SOS** (silicon on sapphire): Si deposited on properly oriented sapphire. - **Laser recrystallization** of amorphous Si on SiO$_2$ or Si$_3$N$_4$. - **ELO** (epitaxial layer overgrowth): open a hole in the insulator to expose the Si substrate; it acts as a seed; epitaxial Si grows up through the hole and sideways over the insulator. Good film quality but *not* a production-line technology. Only **SIMOX** and **BESOI** are production-line technologies giving acceptable-quality films, and both wafer types are sold commercially. - **BESOI** (Bonded and Etched-back SOI): thermally grow SiO$_2$ on one Si wafer (this becomes the insulator), bond a second wafer on top and anneal, then grind, polish and chemically etch the top wafer down to the desired thickness. - **SIMOX** (Separation by IMplantation of OXygen; Fig. 19.13 process flow: implant oxygen into Si, anneal to form SiO$_2$, optionally grow CVD epi to the required Si thickness). Oxygen implant energy 150-200 keV, total dose $1$-$2\times10^{18}$ atoms/cm$^2$. Anneal at about $1300^\circ$C for roughly 6 hours so the oxygen reacts with Si to form a **buried oxide (BOX)** and the surface Si heals. Additional epitaxial Si may be added. Required Si thickness: 500-2000 Å for CMOS and 0.3-10 $\mu$m for thick-film bipolar. Projected advantages of SOI: dielectric isolation of devices reduces parasitic capacitance (higher speed) and totally eliminates **latch-up** (a CMOS failure mode where the circuit gets stuck in a specific state through parasitic thyristor-like action); lower power; higher-temperature operation; better radiation hardness; reduced short-channel effects; higher packing density and easier scaling; simpler processing with fewer masking steps; and wafers fully compatible with production equipment. Most extensive use so far: radiation-hardened SRAMs for military use. Expected growth: very high speed VLSI CMOS, low-supply-voltage devices, DRAMs. ### 19.2.2 MODFET (HEMT) The modulation doped field effect transistor. Alternate names: **HEMT** (high electron mobility transistor), SDHT (selectively doped heterostructure transistor), TEGFET (two-dimensional electron-gas FET). (Note: the HEMT name persists, but it is now recognized that the *low-field mobility* is not the key parameter in normal operation. As with the MESFET of Section 15.3, the *high-field velocity* is critical when large fields exist along the channel.) **Structure** (Fig. 19.14a: from top, Schottky gate between source and drain ohmic contacts, $n^+$-GaAs cap layer, $n$-AlGaAs, nominally undoped GaAs, semi-insulating GaAs substrate). If you think of the AlGaAs as an insulator, the MODFET resembles a MOSFET. Differences: the "insulator" AlGaAs is **doped**, and the GaAs epilayer is nominally **undoped**. "Modulation doping" means the dopant source is modulated during the sequential deposition so that only the AlGaAs is doped. The doping induces a surface channel next to the AlGaAs/GaAs interface. Since the current-carrying GaAs is undoped, there is minimal scattering from residual impurities: very high room-temperature electron mobility, and an even greater improvement at liquid-nitrogen temperature relative to doped-channel devices such as MESFETs. Hence the name. **Energy bands** (Fig. 19.14b: Schottky metal, depleted $n$-AlGaAs, undoped GaAs; a triangular quantum well of the inversion layer, the 2DEG, at the interface; $\Delta E_C$ is the conduction-band offset). AlGaAs has a wider gap, so band offsets appear at the interface (as in Fig. 11.19a), and $\Delta E_C$ acts as an electron-containment barrier. The AlGaAs is thin enough to be totally depleted at equilibrium by the Schottky built-in potential. The leftover uncompensated donors in the AlGaAs produce a band diagram very similar to a MOSFET with sodium ions distributed through the oxide (compare Fig. 19.14b with Fig. P18.7). Also, AlGaAs/GaAs is lattice-matched, so the interface is essentially free of fixed charge and interface traps, a prerequisite of good operation. **Limitation and second generation.** Above a certain carrier density the relatively small $\Delta E_C$ lets channel charge spill over into the AlGaAs. The channel density then saturates versus gate potential and performance degrades at moderate and high currents. Two solutions: 1. **PHEMT** (pseudomorphic HEMT): put a pseudomorphic In$_x$Ga$_{1-x}$As layer between the AlGaAs and the GaAs (Fig. 19.15). Adding In lowers the gap and increases the containment barrier $\Delta E_C$, allowing higher carrier densities in the InGaAs channel. Fig. 19.15: a Hewlett-Packard commercial PHEMT with $n^+$-GaAs, $n$-AlGaAs, InGaAs and undoped GaAs on a semi-insulating GaAs substrate, T-shaped Au gate. 2. **Lattice-matched on InP**: the In$_{0.52}$Al$_{0.48}$As/In$_{0.53}$Ga$_{0.47}$As system, InP as substrate. $\Delta E_C$ there is roughly **twice** that of AlGaAs/GaAs. Still immature, but already gives the best high-frequency performance of three-terminal devices. **The T gate.** At routine submicron gate lengths, parasitic resistances often mask intrinsic performance. The mushroom ("T") gate keeps a small footprint (short $L$) while increasing the metal cross-section and hence reducing gate resistance. **Status.** The MODFET gradually replaces GaAs MESFETs in advanced commercial and military systems that need ultra-high performance, and is ideal for low-noise amplification at microwave and millimetre-wave frequencies. Although conceived as an ultra-high-speed digital device, it is now unlikely to displace Si devices in the near future. --- ## Chapter 19 Summary (key results at a glance) | Topic | Key result | |---|---| | Short-channel signs | slanted saturation, $V_D$-dependent subthreshold current, $V_T$ shift | | $L_{min}$ | $0.4[r_jx_o(W_S+W_D)^2]^{1/3}$ ($\mu$m; $r_j$ in $\mu$m, $x_o$ in Å) | | Short-channel $\Delta V_T$ | $-\dfrac{qN_AW_T}{C_o}\dfrac{r_j}{L}\left[\sqrt{1+2W_T/r_j}-1\right]$ | | Narrow-width $\Delta V_T$ | $+\dfrac{qN_AW_T}{C_o}\dfrac{\pi W_T}{2Z}$ | | Punch-through | $I\propto V^2$ space-charge current; cured by deep implant | | Multiplication feedback | stable while $M<1/\alpha$ | | Velocity saturation | $I_{Dsat}\approx ZC_o(V_G-V_T)v_{sat}$ | | Ballistic/overshoot | $L\lesssim\ell$; velocities up to $\sim$35% above $v_{sat}$ seen | | Structures | LDD (less hot-carrier), DMOS (short $L$ without fine lithography), buried-channel, strained-Si, SOI (SIMOX, BESOI), MODFET/PHEMT | *Computer-based exercises:* Chapter 19 contains none. --- ## Chapter 19 Problems **Problem information table** (from the book): 19.1 Quick quiz (after 19.2.2, level 1, 15 points); 19.2 Acronyms (after 19.2.2, level 1, 5 points); 19.3 $L_{min}$ and $\Delta V_T$ calculations (after 19.1.2, level 2-3, 12 points); 19.4 Identify unique features (after 19.2.1, level 1, 10 points); 19.5 Summarize journal articles (after 19.2.2, level 2-3, 30 points). ### 19.1 Quick Quiz **(a) Small-dimension effects are generally undesirable and are minimized or avoided in commercial MOSFETs by proper scaling or design modifications. True or false?** True. Short-channel effects shift $V_T$ and raise off-current, so scaling rules and designs like LDD, retrograde implants and shallow junctions suppress them. **(b) Three most common indications of short-channel effects.** (1) Upward slant of the post-pinch-off $I_D$-$V_D$ curves; (2) a subthreshold current that increases with $V_D$; (3) a threshold voltage that depends on gate dimensions and biases. **(c) How do short-channel and narrow-width threshold modifications differ, and how are they alike?** *Different:* short channel *lowers* $V_T$ (decreasing $L$), narrow width *raises* $V_T$ (decreasing $Z$). *Alike:* both arise from charge that the gate does not "own" as simple rectangular volume under it; in the short channel, source/drain junctions balance part of the depletion charge (less gate charge needed), in the narrow device fringing depletion on the sides adds extra charge the gate must balance. Both become important when the dimension ($L$ or $Z$) is comparable to the depletion width, and both are treated with the same geometric (charge-sharing) method. **(d) For fixed $L$ and $Z$, two other parameters that can be adjusted, and how.** Reduce the junction depth $r_j$ (shallower source/drain), reduce the oxide thickness $x_o$, or increase the substrate doping $N_A$ (all reduce $L_{min}$ and $\Delta V_T$). Give any two. **(e) Two BJT-like phenomena in short-channel MOSFETs.** Source-to-drain punch-through (space-charge current) and parasitic-BJT carrier multiplication with regenerative feedback (substrate-current-induced source-junction forward bias). **(f) What happens in oxide charging?** Near the drain, hot channel carriers (and carriers from the drain depletion region) gain enough energy to surmount the Si-SiO$_2$ barrier and enter the oxide; neutral traps there capture part of them, so charge builds up in the oxide, shifting $V_T$ and degrading device parameters. **(g) Why is oxide charging more important in short-channel devices?** The same drain-region injection affects a larger percentage of the gated region when the channel is short, and fields are higher because voltages were not scaled down in proportion; the effect is also cumulative, limiting lifetime. **(h) How can you tell if velocity saturation is affecting $I_D$-$V_D$ characteristics?** The saturation current varies nearly linearly with $V_G - V_T$ (equal $V_G$ steps give nearly equal $I_{Dsat}$ steps) and $I_{Dsat}$ is lower than the square-law prediction. **(i) Ballistic transport.** Carriers travel source to drain with essentially no scattering (projectile-like motion), possible when $L \lesssim \ell$, the mean free path. **(j) Velocity overshoot.** With reduced scattering, the average carrier velocity in the channel exceeds the steady-state saturation velocity $v_{sat}$. **(k) A popular approach to minimize oxide charging.** The lightly doped drain (LDD) structure. **(l) What is a pseudomorphic film?** A sufficiently thin (typically $\le1000$ Å) epitaxial film of a lattice-mismatched alloy (e.g. SiGe on Si) that conforms atom by atom to the lattice of the substrate, and so is strained rather than relaxed. **(m) Two examples of pseudomorphic films in devices.** The pseudomorphic SiGe base layer of Si heterojunction bipolar transistors, the strained-Si surface channel of the enhanced-mobility MOSFET, and the InGaAs channel layer of the PHEMT. **(n) Briefly summarize SIMOX.** Implant oxygen (150-200 keV, dose $1$-$2\times10^{18}$ cm$^{-2}$) below the Si surface; anneal (about $1300^\circ$C, 6 h) so the oxygen reacts to form a buried SiO$_2$ layer and the surface Si recovers; optionally add epitaxial Si for thickness. **(o) MODFET versus HEMT.** No physical difference: they are two names for the same device. MODFET emphasizes modulation doping; HEMT emphasizes high electron mobility. ### 19.2 Acronyms and abbreviations | Acronym | Meaning | |---|---| | LDD | lightly doped drain | | DMOS | double-diffused MOS (field-effect transistor) | | SOI | silicon-on-insulator | | SOS | silicon-on-sapphire | | ELO | epitaxial layer overgrowth | | SIMOX | separation by implantation of oxygen | | BESOI | bonded and etched-back silicon-on-insulator | | BOX | buried oxide | | MODFET | modulation-doped field-effect transistor | | PHEMT | pseudomorphic high electron mobility transistor | ### 19.3 $L_{min}$ and $\Delta V_T$ for the Fig. 19.3 device **Problem.** Fig. 19.3 device: $N_A = 8\times10^{16}$ cm$^{-3}$ (the OCR lost the exponent; $10^{16}$ is the value consistent with a 280 Å oxide device and the plotted $V_T$ range), $x_o = 0.028\ \mu$m $= 280$ Å, $r_j = 1\ \mu$m. (a) Use Eq. (19.1) to compute the expected $L_{min}$ for $V_D = 0.125$ V (assume $n^+$-$p$ step junctions, source and back grounded; comment). (b) Use Eq. (19.13) to find $\Delta V_T$ at $L = 1\ \mu$m, $V_D = 0.125$ V. (c) (The OCR dropped this part; it asks whether Eq. (19.13) can explain the $V_D = 4$ V data of Fig. 19.3.) **Assumptions.** $n_i = 10^{10}$ cm$^{-3}$ at 300 K, $K_S = 11.8$, $K_O = 3.9$, $\epsilon_0 = 8.85\times10^{-14}$ F/cm, $n^+$ doping $N_D = 10^{20}$ cm$^{-3}$ (so only $N_A$ matters in $W$). **Solution (a).** Built-in potential: $$V_{bi} = \frac{kT}{q}\ln\frac{N_AN_D}{n_i^2} = 0.0259\ln\frac{(8\times10^{16})(10^{20})}{10^{20}} = 0.0259\,(38.92) = 1.01\ \text{V}$$ Source depletion width ($V_{S}=0$): $$W_S = \left[\frac{2K_S\epsilon_0 V_{bi}}{qN_A}\right]^{1/2} = \left[\frac{2(11.8)(8.85\times10^{-14})(1.01)}{(1.6\times10^{-19})(8\times10^{16})}\right]^{1/2} = \left[1.65\times10^{-10}\right]^{1/2} = 1.28\times10^{-5}\ \text{cm} = 0.128\ \mu\text{m}$$ Drain depletion width (reverse bias $V_D = 0.125$ V): $$W_D = W_S\sqrt{\frac{V_{bi}+V_D}{V_{bi}}} = 0.128\sqrt{\frac{1.135}{1.01}} = 0.136\ \mu\text{m}$$ Sum: $W_S + W_D = 0.264\ \mu$m. Then $$r_jx_o(W_S+W_D)^2 = (1)(280)(0.264)^2 = 19.5$$ $$L_{min} = 0.4\,(19.5)^{1/3} = 0.4\,(2.69) = 1.08\ \mu\text{m}$$ **Comment.** $L_{min} \approx 1\ \mu$m. In Fig. 19.3 the $V_T$ curves are essentially flat for $L \gtrsim 3$-$4\ \mu$m and fall off as $L$ approaches $\sim1\ \mu$m and below, which agrees with $L_{min}$ marking the onset of significant short-channel effects. **Solution (b).** $$2\phi_F = 2(0.0259)\ln\frac{8\times10^{16}}{10^{10}} = 2(0.0259)(22.80) = 1.18\ \text{V}$$ $$W_T = \left[\frac{2K_S\epsilon_0(2\phi_F)}{qN_A}\right]^{1/2} = \left[\frac{2(11.8)(8.85\times10^{-14})(1.18)}{(1.6\times10^{-19})(8\times10^{16})}\right]^{1/2} = 1.39\times10^{-5}\ \text{cm} = 0.139\ \mu\text{m}$$ $$C_o = \frac{K_O\epsilon_0}{x_o} = \frac{(3.9)(8.85\times10^{-14})}{2.8\times10^{-6}} = 1.23\times10^{-7}\ \text{F/cm}^2$$ $$\frac{qN_AW_T}{C_o} = \frac{(1.6\times10^{-19})(8\times10^{16})(1.39\times10^{-5})}{1.23\times10^{-7}} = 1.45\ \text{V}$$ $$\frac{r_j}{L} = 1,\qquad \sqrt{1+\frac{2W_T}{r_j}} - 1 = \sqrt{1+0.278} - 1 = 0.130$$ $$\Delta V_T = -(1.45)(1)(0.130) \approx -0.19\ \text{V}$$ So $V_T$ is about 0.19 V below its long-channel value at $L=1\ \mu$m. **Solution (c).** Eq. (19.13) contains **no $V_D$ dependence**; it was derived for $V_D\approx 0$ so that $W_S \approx W_D$. Therefore it predicts the *same* $\Delta V_T$ for $V_D=4$ V as for $0.125$ V and cannot explain the larger decrease seen in the $V_D = 4$ V data. Physically, a large $V_D$ widens the drain depletion region ($W_D$ grows), so the drain controls a larger share of the subgate charge (the trapezoid becomes more lopsided) and $|V_T|$ falls more. A better model would use $W_D(V_D)$ in the geometry. ### 19.4 Unique physical nature of the structures **(a) LDD:** a lightly doped $n^-$ region between the channel end and the $n^+$ drain (and source), giving a lower doping gradient, lower peak field shifted away from the oxide edge, less hot-carrier injection. **(b) DMOS:** channel formed and its length set by the lateral difference of two diffusions (boron then phosphorus) through the same mask opening, giving a short channel without fine lithography and a high breakdown voltage. **(c) Buried-channel MOSFET:** a thin surface layer of the same type as source/drain under the gate; conduction occurs below the interface in a hybrid J-FET/MOSFET structure with higher mobility and fewer interface effects. **(d) Enhanced-mobility (strained-Si) MOSFET:** a thin pseudomorphic strained-Si channel grown on relaxed Si$_{0.7}$Ge$_{0.3}$ over a graded SiGe buffer; tensile strain lowers scattering and effective mass, roughly doubling low-field mobility. **(e) SOI:** devices built in a thin single-crystal Si layer on an insulator (e.g. buried oxide), giving dielectric isolation, low parasitic capacitance, no latch-up, radiation hardness. ### 19.5 Journal article summaries This problem asks the reader to read six papers and write one-to-two-page summaries; the articles are not reproduced here, so below is a short orientation for each, based on the titles and the chapter's own discussion (the reader should still read the articles for the actual summaries). (a) Sanchez, Hsueh, DeMassa, *Drain-Engineered Hot-Electron-Resistant Device Structures: A Review* (1989): surveys drain structures (LDD, graded drain, double-diffused drain, etc.) that lower the peak drain field to suppress hot-electron injection and oxide charging. (b) Van der Tol and Chamberlain, *Potential and Electron Distribution Model for the Buried-Channel MOSFET* (1989): models the electrostatics of the Fig. 19.11 structure, the potential in the implanted layer, and where the electrons sit (surface versus buried) as a function of gate bias. (c) Meyerson, *UHV/CVD Growth of Si and Si:Ge Alloys* (1992): the chemistry and physics of the low-temperature, ultrahigh-vacuum CVD process and the HBT and other devices it enables. (d) Meyerson, *Ultrahigh-Vacuum CVD Process Makes SiGe Devices* (1994): a shorter, process-oriented account of the same technology. (e) Peters, *SOI Takes Over Where Silicon Leaves Off* (1993): an industry overview of SOI materials (SIMOX, BESOI), advantages and applications. (f) Nguyen, Larson, Mishra, *Ultra-High-Speed Modulation-Doped FETs: A Tutorial Review* (1992): covers modulation doping, 2DEG physics, AlGaAs/GaAs, PHEMT and InP-based devices, and the limits on speed (high-field velocity, parasitics). # Appendix A: Elements of Quantum Mechanics > **The big question of this appendix.** Before we can talk about electrons moving through a silicon crystal, we need to know what an electron does inside a single isolated atom. Newton's mechanics, which works beautifully for baseballs and planets, gives wrong answers at atomic size. Quantum mechanics is the more accurate theory, and it reduces to Newton's in the limit of large masses and energies. Think of Newton as the "zoomed-out map" and quantum mechanics as the "street-level survey". The appendix has three movements: 1. The **quantization concept** (A.1): three experimental surprises that broke classical physics. 2. The **basic formalism** (A.2): five postulates of Schrödinger wave mechanics. 3. **Electronic states in atoms** (A.3): hydrogen solved exactly, then extrapolated to silicon. --- ## A.1 The Quantization Concept ### A.1.1 Blackbody radiation Every warm solid glows. At room temperature or below, the glow is almost entirely infrared, so your eye never notices. An ideal radiator is called a **blackbody**. Its emitted power versus wavelength (Fig. A.1) has these features: - Curves for $T = 300$, $1000$ and $2000\ \text{K}$ are humped, with the hump moving to shorter wavelength as $T$ rises (a hotter stove goes from invisible, to dull red, to white). - The visible window is only $0.4\ \mu\text{m} \lesssim \lambda \lesssim 0.7\ \mu\text{m}$. - A dashed line shows the classical **Rayleigh-Jeans** prediction for $T = 2000\ \text{K}$. It tracks the data at long wavelengths but shoots upward without bound as $\lambda \to 0$. **The classical story (Rayleigh-Jeans).** Heat makes the atoms in a solid vibrate. Model each vibrating atom as a harmonic oscillator with normal-mode frequency $\nu = \omega/2\pi$ and a *continuum* of allowed energies, distributed by statistics. The emitted light is a sample of this distribution. It agrees at long wavelengths but diverges at short ones. Integrating over all wavelengths gives infinite radiated energy. This is the **ultraviolet catastrophe**. *Thought experiment.* Imagine a vending machine that only accepts whole coins. Low-denomination change is easy to make. But to buy something that costs a "big" coin, you must save up a whole big coin. Planck's insight is similar. High-frequency oscillators demand a large minimum energy "coin", and at a given temperature there is rarely enough thermal energy to pay it. So high-frequency modes stay silent and the catastrophe never happens. **Planck's fix (1901).** An oscillator of frequency $\nu$ can only have energies $$E_n = n h \nu = n\hbar\omega,\qquad n = 0,1,2,\dots \tag{A.1}$$ Matching theory to experiment gives $$h = 6.63\times10^{-34}\ \text{J·s},\qquad \hbar = \frac{h}{2\pi}.$$ (The OCR printed $10^{-4}$; the true exponent is $-34$.) **Lesson:** at atomic scale energy comes in discrete steps. That is *energy quantization*. ### A.1.2 The Bohr atom Heated gases emit sharp, discrete spectral lines, not a smooth rainbow. Rutherford (1910) proposed the nuclear atom: electrons of mass $m_0$ and charge $-q$ orbit a nucleus of charge $+Zq$. Two problems follow classically: - Electrons could orbit at any radius, so the light should be a continuum, not lines. - An accelerating charge radiates, so the orbiting electron should spiral into the nucleus almost immediately. **Bohr's postulate (1913).** Only orbits with quantized angular momentum are allowed. For hydrogen ($Z=1$), circular orbit: $$L_n = m_0 v_n r_n = n\hbar,\qquad n = 1,2,3,\dots \tag{A.2}$$ **Derivation of the orbit radii and energies.** Step 1. Stable orbit means centripetal force equals Coulomb attraction: $$\frac{m_0 v_n^2}{r_n} = \frac{q^2}{4\pi\varepsilon_0 r_n^2} \tag{A.3}$$ Step 2. Solve (A.2) for $v_n$: $$v_n = \frac{n\hbar}{m_0 r_n}$$ Step 3. Substitute into (A.3): $$\frac{m_0}{r_n}\frac{n^2\hbar^2}{m_0^2 r_n^2} = \frac{q^2}{4\pi\varepsilon_0 r_n^2}$$ Step 4. Cancel and solve for the radius: $$r_n = \frac{4\pi\varepsilon_0 (n\hbar)^2}{m_0 q^2} \tag{A.4}$$ Step 5. Kinetic energy, using (A.3): $$\text{K.E.} = \tfrac12 m_0 v_n^2 = \frac12\,\frac{q^2}{4\pi\varepsilon_0 r_n} \tag{A.5a}$$ Step 6. Potential energy (zero at infinity): $$\text{P.E.} = -\frac{q^2}{4\pi\varepsilon_0 r_n} \tag{A.5b}$$ Step 7. Total energy (note the virial-style factor of one half): $$E_n = \text{K.E.}+\text{P.E.} = -\frac12\,\frac{q^2}{4\pi\varepsilon_0 r_n} \tag{A.6}$$ Step 8. Insert $r_n$ from (A.4): $$E_n = -\frac{m_0 q^4}{2(4\pi\varepsilon_0 n\hbar)^2} = -\frac{13.6}{n^2}\ \text{eV} \tag{A.7}$$ Here $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. **Spectra.** A photon emitted in a jump from level $n$ to level $m<n$ carries energy $E_n - E_m$. These discrete energies agree with experiment. Fig. A.2 shows the hydrogen levels with $E_1=-13.6$, $E_2=-3.40$, $E_3=-1.51$, $E_4=-0.85$, $E_5=-0.54$ eV, and $E_\infty = 0$ (the OCR lost $E_3$). Transitions ending on $n=1$ form the **Lyman** series, on $n=2$ the **Balmer** series, on $n=3$ the **Paschen** series. Think of the levels as the rungs of a ladder in a well: you can stand on a rung but never between rungs. Jumping down emits a photon whose color is set by the rung gap. Bohr's model failed for helium and beyond, but it cemented two ideas: energy quantization, and the failure of classical mechanics at small scale. ### A.1.3 Wave-particle duality Light as particle: - **Photoelectric effect** (Einstein, 1905): illuminated metals eject electrons. Explained by light quanta (photons) with $E = h\nu$. - **Compton effect:** x-rays scattered by electrons change frequency exactly as in a billiard-ball collision conserving energy and momentum. With $E = h\nu = mc^2$, photon momentum is $$p = mc = \frac{h\nu}{c} = \frac{h}{\lambda}.$$ **de Broglie (1925)** flipped it. If waves can be particles, particles should be waves: $$p = \frac{h}{\lambda}\quad\Longleftrightarrow\quad \lambda = \frac{h}{p} \tag{A.8}$$ **Davisson-Germer (1927).** A low-energy electron beam hit a nickel crystal. The electron wavelength from (A.8) was comparable to the Ni nearest-neighbor spacing. If electrons were simple bullets they would scatter randomly. Instead the angular intensity showed maxima and minima like light from a diffraction grating, at angles predicted by reflecting waves of wavelength $h/p$ from atomic planes. Protons and neutrons later behaved the same way. (Footnote: in this appendix only, $p$ denotes momentum. Elsewhere in the book $p$ is hole concentration.) **Conclusion of A.1:** observables such as energy and angular momentum are quantized, and all matter has an inherent wave nature. --- ## A.2 Basic Formalism Schrödinger (1926) provided a unified theory, *wave mechanics*, built on Planck's quantization and de Broglie's matter waves. Heisenberg's *matrix mechanics* is mathematically different but exactly equivalent. We use Schrödinger's version. **The five postulates (single-particle system).** 1. **Wavefunction.** There exists $\Psi = \Psi(x,y,z,t)$, possibly complex, from which all dynamic information is obtained. It is the system's "describing function". 2. **Wave equation.** $\Psi$ follows the time-dependent Schrödinger equation: $$-\frac{\hbar^2}{2m}\nabla^2\Psi + U\Psi = i\hbar\frac{\partial\Psi}{\partial t} \tag{A.9}$$ where $m$ is the particle mass, $U$ the potential energy and $i=\sqrt{-1}$. 3. **Good behavior.** $\Psi$ and $\nabla\Psi$ must be finite, continuous and single-valued everywhere. 4. **Probability.** $\Psi^*\Psi\,dV = |\Psi|^2\,dV$ is the probability of finding the particle in volume element $dV$. Since the particle must be somewhere, $$\int_V \Psi^*\Psi\,dV = 1 \tag{A.10}$$ over all space (normalization). 5. **Operators and expectation values.** Each dynamic variable $\alpha$ has an associated operator $\alpha_{op}$. Its average (expectation) value is $$\langle\alpha\rangle = \int_V \Psi^*\,\alpha_{op}\,\Psi\,dV \tag{A.11}$$ The operators are fixed by demanding agreement with classical mechanics for large mass and energy. **Table A.1: Dynamic variables and operators** | Dynamic variable | Operator | |---|---| | $x,\,y,\,z$ | $x,\,y,\,z$ (multiply) | | $f(x,y,z)$ | $f(x,y,z)$ | | $p_x,\,p_y,\,p_z$ | $\dfrac{\hbar}{i}\dfrac{\partial}{\partial x},\ \dfrac{\hbar}{i}\dfrac{\partial}{\partial y},\ \dfrac{\hbar}{i}\dfrac{\partial}{\partial z}$ | | $E$ | $i\hbar\dfrac{\partial}{\partial t}$ | **Intuition.** $\Psi$ is not a thing you can touch. It is a bookkeeping function whose squared magnitude is a probability density. Postulate 5 says: to ask "what is the average momentum?", sandwich the momentum operator between $\Psi^*$ and $\Psi$ and integrate. **How problems are solved.** In principle: impose the boundary conditions plus postulates 3 and 4, solve (A.9) for $\Psi$, then compute any variable from (A.11). In practice closed-form solutions exist only for idealized or a few practical problems. Often one gets the allowed energies from the constraints alone, or uses expansions, trial wavefunctions or limiting cases. **Status of the equation.** Schrödinger's equation is essentially empirical, like Newton's laws. Its justification is that every experimental test so far agrees with it, often to extreme precision. --- ## A.3 Electronic States in Atoms ### A.3.1 The hydrogen atom A massive fixed nucleus ($+q$) and one electron ($-q$) with fixed total energy $E$. Separating space and time gives $$\Psi(x,y,z,t) = \psi(x,y,z)\,e^{-iEt/\hbar} \tag{A.12}$$ **Derivation of the time-independent equation.** Step 1. Substitute (A.12) into (A.9). Time derivative: $i\hbar\,\partial_t\Psi = E\,\psi\,e^{-iEt/\hbar}$. Step 2. Cancel the common factor $e^{-iEt/\hbar}$: $$-\frac{\hbar^2}{2m}\nabla^2\psi + U\psi = E\psi$$ Step 3. Rearrange: $$\nabla^2\psi + \frac{2m}{\hbar^2}\left[E - U(x,y,z)\right]\psi = 0 \tag{A.13}$$ For hydrogen $m=m_0$ and $$U = -\frac{q^2}{4\pi\varepsilon_0 r},\quad r=\sqrt{x^2+y^2+z^2} \tag{A.14}$$ $$\nabla^2\psi + \frac{2m_0}{\hbar^2}\left[E + \frac{q^2}{4\pi\varepsilon_0 r}\right]\psi = 0 \tag{A.15}$$ Because $U$ depends only on $r$, use spherical coordinates: $$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \tag{A.16}$$ Separation of variables, $\psi = R(r)\Theta(\theta)\Phi(\phi)$, yields bound states ($E<0$). Each is labeled by three quantum numbers arising from the separation constants: | Symbol | Allowed values | Name | |---|---|---| | $n$ | $1,2,3,\dots$ | principal quantum number | | $l$ | $0,1,2,\dots,n-1$ | azimuthal quantum number | | $m$ (or $m_l$) | $-l,\dots,0,\dots,+l$ | magnetic orbital quantum number | **Table A.2: Hydrogen wavefunctions $\psi_{n,l,m}$ for $n=1,2$** ($a_0 = 4\pi\varepsilon_0\hbar^2/m_0q^2$, the Bohr radius, equal to the ground-state Bohr orbit from (A.4)): $$\psi_{1,0,0} = \frac{1}{\sqrt{\pi a_0^3}}\,e^{-r/a_0}$$ $$\psi_{2,0,0} = \frac{1}{4\sqrt{2\pi a_0^3}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$$ $$\psi_{2,1,-1} = \frac{1}{8\sqrt{\pi a_0^3}}\,\frac{r}{a_0}\,e^{-r/2a_0}\,e^{-i\phi}\sin\theta$$ $$\psi_{2,1,0} = \frac{1}{4\sqrt{2\pi a_0^3}}\,\frac{r}{a_0}\,e^{-r/2a_0}\cos\theta$$ $$\psi_{2,1,1} = \frac{1}{8\sqrt{\pi a_0^3}}\,\frac{r}{a_0}\,e^{-r/2a_0}\,e^{i\phi}\sin\theta$$ (The OCR garbled the prefactors. The normalization constants above are the standard values satisfying (A.10).) **Energies.** Substituting $\psi_{1,0,0}$ into (A.15) and solving for $E$: $$E_{1,0,0} = -\frac{\hbar^2}{2m_0a_0^2} = -\frac{m_0q^4}{2(4\pi\varepsilon_0\hbar)^2} \tag{A.17}$$ This is identical to Bohr's $E_1$. Substituting any $n=2$ function gives $$E_{2,0,0}=E_{2,1,-1}=E_{2,1,0}=E_{2,1,1} = -\frac{m_0q^4}{2(4\pi\varepsilon_0\hbar)^2}\cdot\frac14 \tag{A.18}$$ equal to Bohr's $E_2$. So quantum mechanics reproduces Bohr's levels, but without ad hoc assumptions. Knowing $n$ alone fixes the energy. $n=1$ is the ground state, larger $n$ are excited states. **Degeneracy.** States sharing one energy are *degenerate*. The four $n=2$ states are degenerate. A magnetic field would split them because their spatial shapes interact differently with the field. This is where $l$ and $m$ matter. **Spin.** A fourth quantum number is needed. Electrons have *spin*, pictured as spinning clockwise or counterclockwise, with $s=+\tfrac12$ or $s=-\tfrac12$. Every state of Table A.2 is thus two-fold degenerate. (The spin picture is only a mental image.) **Where is the electron?** The probability of finding the ground-state electron between $r$ and $r+dr$ is $4\pi r^2|\psi_{1,0,0}|^2\,dr$. Plotted versus $r/a_0$ (Fig. A.3a) it rises to a maximum at $r = a_0$ and decays. For larger $n$ the peak shifts outward, and there is significant probability over a range of distances. This contrasts sharply with Bohr's electron on a rail at fixed $r$. Picture instead a charge cloud with density $\propto|\psi|^2$ (Fig. A.3b), spherically symmetric for $\psi_{1,0,0}$; states with $l\neq0$ give clouds with angular dependence. ### A.3.2 Multi-electron atoms Hydrogen's wavefunctions do not transfer directly, but the same four quantum numbers $(n,l,m,s)$ label states in any atom, and the energy ordering follows $n$. Two rules let us deduce structure without solving anything: 1. **Pauli exclusion principle:** no two electrons may share the same set of quantum numbers. (Like theater seats: each seat is a quantum state, one person per seat.) 2. **Minimum energy:** in the ground state electrons fill the lowest available states. Spectroscopic letters for $l=0,1,2,3,4,5,\dots$ are $s,p,d,f,g,h,\dots$ (from sharp, principal, diffuse, fundamental). In multi-electron atoms $s$ states lie slightly below $p$ states of the same $n$. **Counting.** Level $n$ holds $2n^2$ electrons: $n=1$ has 2, $n=2$ has 8 (2 in $2s$, 6 in $2p$), $n=3$ begins with $3s$ (2) and $3p$ (6). **Table A.3: Quantum numbers and ground-state configurations, elements 1 to 14** States in filling order (each line is one electron slot): | $n$ | $l$ | $m$ | $s$ | state | |---|---|---|---|---| | 1 | 0 | 0 | $-\tfrac12$ | 1s | | 1 | 0 | 0 | $+\tfrac12$ | 1s | | 2 | 0 | 0 | $-\tfrac12$ | 2s | | 2 | 0 | 0 | $+\tfrac12$ | 2s | | 2 | 1 | $-1$ | $-\tfrac12$ | 2p | | 2 | 1 | $-1$ | $+\tfrac12$ | 2p | | 2 | 1 | 0 | $-\tfrac12$ | 2p | | 2 | 1 | 0 | $+\tfrac12$ | 2p | | 2 | 1 | 1 | $-\tfrac12$ | 2p | | 2 | 1 | 1 | $+\tfrac12$ | 2p | | 3 | 0 | 0 | $-\tfrac12$ | 3s | | 3 | 0 | 0 | $+\tfrac12$ | 3s | | 3 | 1 | $-1$ | $-\tfrac12$ | 3p | | 3 | 1 | $-1$ | $+\tfrac12$ | 3p | | 3 | 1 | 0 | $-\tfrac12$ | 3p | | 3 | 1 | 0 | $+\tfrac12$ | 3p | | 3 | 1 | 1 | $-\tfrac12$ | 3p | | 3 | 1 | 1 | $+\tfrac12$ | 3p | | Z | Element | Configuration | |---|---|---| | 1 | H | $1s^1$ | | 2 | He | $1s^2$ | | 3 | Li | $1s^22s^1$ | | 4 | Be | $1s^22s^2$ | | 5 | B | $1s^22s^22p^1$ | | 6 | C | $1s^22s^22p^2$ | | 7 | N | $1s^22s^22p^3$ | | 8 | O | $1s^22s^22p^4$ | | 9 | F | $1s^22s^22p^5$ | | 10 | Ne | $1s^22s^22p^6$ | | 11 | Na | $1s^22s^22p^63s^1$ | | 12 | Mg | $1s^22s^22p^63s^2$ | | 13 | Al | $1s^22s^22p^63s^23p^1$ | | 14 | Si | $1s^22s^22p^63s^23p^2$ | The superscript counts electrons sharing an $nl$ label. **Silicon.** A completely filled $n$ shell gives a very stable, inert configuration (He, Ne). The two $n=1$ and eight $n=2$ electrons of Si sit deep and tightly bound. These ten electrons plus the nucleus form the **core**, which stays essentially untouched in chemistry. The remaining four electrons are the **valence electrons**. They occupy the two $3s$ states and two of the six $3p$ states (Fig. A.4: shells $n=1$ with 2 electrons, $n=2$ with 8 electrons and 8 slots, $n=3$ with 4 electrons in a level with 2 $s$ and 6 $p$ slots). These weakly bound four electrons do the bonding, and, as the rest of the book shows, form the carriers in the crystal. Germanium ($Z=32$) is the same, except its core has 28 electrons. *Analogy.* The core is a locked vault; the valence electrons are the coins in your pocket. Chemistry and conduction happen with the pocket coins. --- # Appendix B: MOS Semiconductor Electrostatics, Exact Solution > **The story.** Chapter 16 solved the MOS capacitor with the *depletion approximation*, treating the band-bending region as a box with a sharp edge. Here we throw away that shortcut and solve Poisson's equation with every mobile carrier included. The payoff is a result valid from accumulation through depletion to strong inversion, and it needs one trick: multiplying Poisson's equation by $dU/dx$ so that it integrates exactly, like using the energy integral for a pendulum. ## B.1 Definition of parameters Three normalized quantities keep the algebra light. Everything is measured in units of the thermal voltage $kT/q$ (about 25.9 mV at 300 K): $$U(x) \equiv \frac{\phi(x)}{kT/q} = \frac{E_i(\text{bulk}) - E_i(x)}{kT} \tag{B.1}$$ $$U_s \equiv \frac{\phi_s}{kT/q} = \frac{E_i(\text{bulk}) - E_i(\text{surface})}{kT} \tag{B.2}$$ $$U_F \equiv \frac{\phi_F}{kT/q} = \frac{E_i(\text{bulk}) - E_F}{kT} \tag{B.3}$$ - $U(x)$ is the electrostatic potential normalized to $kT/q$ ("the potential"). - $U_F$ is the **doping parameter**: positive for $p$-type, negative for $n$-type. - $x$ is the depth into the semiconductor from the oxide-semiconductor interface. Because the field vanishes in the bulk (idealization 5 of Section 16.1), the semiconductor can be treated as running from $x=0$ to $x=\infty$. Then $U(x\to\infty)=0$, consistent with $\phi=0$ in the bulk. ### The Debye length Drop a charged bead into a plasma (a gas of free positive ions and electrons). The mobile charges crowd around and screen it. The screening distance, over which the field falls to about $1/e$, is the **Debye length**. A semiconductor in the bulk, or under flat bands, is a plasma of ionized impurities and mobile carriers. A charged MOS gate is the "bead", and the band-bending region is the screening cloud, about one Debye length thick. The **extrinsic (bulk) Debye length** is $$L_B = \left[\frac{K_s\varepsilon_0 kT}{q^2\,(n_{bulk}+p_{bulk})}\right]^{1/2} \tag{B.4}$$ It is exact only for small departures from flat band, but it is convenient to normalize with the **intrinsic Debye length**, obtained by setting $n_{bulk}=p_{bulk}=n_i$: $$L_D = \left[\frac{K_s\varepsilon_0 kT}{2q^2 n_i}\right]^{1/2} \tag{B.5}$$ ## B.2 Exact solution ### Step 1: Poisson's equation in one dimension The MOS-C is one-dimensional (idealization 7), so $$\frac{d\mathcal{E}}{dx} = \frac{\rho}{K_s\varepsilon_0} = \frac{q}{K_s\varepsilon_0}\,(p - n + N_D - N_A) \tag{B.6}$$ ### Step 2: Express field and carrier densities through $U$ The field is the slope of the bands: $$\mathcal{E} = \frac{1}{q}\frac{dE_i(x)}{dx} = -\frac{kT}{q}\frac{dU}{dx} \tag{B.7}$$ (First equality: field is the gradient of the electron potential energy divided by $q$. Second: definition (B.1) and $dE_i(\text{bulk})/dx=0$.) With Boltzmann statistics, $$p = n_i e^{[E_i(x)-E_F]/kT} = n_i\,e^{U_F-U(x)} \tag{B.8a}$$ $$n = n_i e^{[E_F-E_i(x)]/kT} = n_i\,e^{U(x)-U_F} \tag{B.8b}$$ ### Step 3: Eliminate the doping with bulk charge neutrality In the bulk $\rho=0$ and $U=0$: $$0 = p_{bulk}-n_{bulk}+N_D-N_A = n_i e^{U_F}-n_i e^{-U_F}+N_D-N_A \tag{B.9}$$ $$N_D - N_A = n_i\left(e^{-U_F}-e^{U_F}\right) \tag{B.10}$$ ### Step 4: The charge density $$\rho = q\,n_i\left(e^{U_F-U}-e^{U-U_F}+e^{-U_F}-e^{U_F}\right) \tag{B.11}$$ ### Step 5: Poisson's equation in normalized form Put (B.7) in (B.6) and divide by $kT/q$: $$\frac{d^2U}{dx^2} = -\frac{q^2 n_i}{K_s\varepsilon_0 kT}\left(e^{U_F-U}-e^{U-U_F}+e^{-U_F}-e^{U_F}\right) \tag{B.12}$$ From (B.5), $q^2n_i/(K_s\varepsilon_0kT) = 1/(2L_D^2)$. Factor the bracket as $e^{U_F}(e^{-U}-1)-e^{-U_F}(e^{U}-1)$: $$\frac{d^2U}{dx^2} = -\frac{1}{2L_D^2}\left[e^{U_F}\left(e^{-U}-1\right)-e^{-U_F}\left(e^{U}-1\right)\right] \tag{B.13}$$ (The OCR loses the factor of one half and some signs; this is the form consistent with (B.11) and (B.5).) ### Step 6: Boundary conditions $$\mathcal{E}=0\ \ \text{or}\ \ \frac{dU}{dx}=0 \quad\text{at } x=\infty \tag{B.14a}$$ $$U = U_s \quad\text{at } x=0 \tag{B.14b}$$ ### Step 7: First integral (the energy trick) Multiply (B.13) by $dU/dx$. The left side becomes $\tfrac12\,d[(dU/dx)^2]/dx$. Integrate from $\infty$ to $x$ and use (B.14a): $$\left(\frac{dU}{dx}\right)^2 = -\frac{1}{L_D^2}\int_0^{U}\left[e^{U_F}(e^{-u}-1)-e^{-U_F}(e^{u}-1)\right]du$$ The integrals are $$\int_0^U (e^{-u}-1)\,du = -(e^{-U}-1+U),\qquad \int_0^U (e^{u}-1)\,du = e^{U}-1-U$$ so $$\left(\frac{dU}{dx}\right)^2 = \frac{1}{L_D^2}\left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^{U}-U-1\right)\right]$$ and in terms of the field $$\mathcal{E}^2 = \left(\frac{kT/q}{L_D}\right)^2\left[e^{U_F}(e^{-U}+U-1)+e^{-U_F}(e^{U}-U-1)\right] \tag{B.15}$$ ### Step 8: Choose the sign of the square root Equation (B.15) has the form $y^2=a^2$ with roots $\pm a$. From the band diagram, $\mathcal{E}>0$ when $U>0$ and $\mathcal{E}<0$ when $U<0$. The right side is always nonnegative, so take the positive root for $U>0$ and the negative root for $U<0$: $$\mathcal{E} = -\frac{kT}{q}\frac{dU}{dx} = \frac{kT}{q}\,\frac{\varsigma_s\,F(U,U_F)}{L_D} \tag{B.16}$$ $$F(U,U_F) = \left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^{U}-U-1\right)\right]^{1/2}\ \ge 0 \tag{B.17}$$ $$\varsigma_s = \begin{cases} +1 & U_s>0\\ -1 & U_s<0\end{cases} \tag{B.18}$$ (The book's List of Symbols calls this "sign ($\pm$) of $U_s$". I write it $\varsigma_s$ to avoid a clash with permittivity.) ### Step 9: Separate variables and integrate From (B.16), $dU/dx = -\varsigma_s F/L_D$, so $dU/F = -\varsigma_s\,dx/L_D$. Integrate from the surface ($x=0$, $U=U_s$) to the point $x$ with potential $U$, using $\varsigma_s^2=1$: $$\int_{U}^{U_s}\frac{dU'}{F(U',U_F)} = \varsigma_s\,\frac{x}{L_D} \tag{B.19}$$ **What this buys us.** Equations (B.11), (B.16) and (B.19) together are the exact solution. For a given $U_s$, evaluate the integral numerically to get $x$ versus $U$, invert to $U(x)$, then substitute into (B.11) for $\rho(x)$ and (B.16) for $\mathcal{E}(x)$. The Fig. 16.8 plots of $U(x)$ and $\rho(x)$ were made this way. **Sanity check.** For small $U$, $F\approx |U|\sqrt{(e^{U_F}+e^{-U_F})/2}$, so (B.19) gives $U\approx U_se^{-x/L_B}$, an exponential decay with the extrinsic Debye length, exactly as the plasma picture predicts. --- # Appendix C: MOS C-V Supplement > **Why a supplement?** The delta-depletion C-V curve of Chapter 16 is explicit: $C$ versus $V_G$ in closed form. The exact-charge version cannot be written that way. Instead *both* $C$ and $V_G$ are written as functions of the surface potential $U_s$, and you sweep $U_s$ to trace the curve, like a parametric curve $(x(t),y(t))$. ## C.1 The capacitance relationships For an ideal MOS-C with exact charge distribution: $$\frac{C}{C_o} = \frac{1}{1+\dfrac{K_o\,W_{\text{eff}}}{K_s\,x_o}} \tag{C.1}$$ This is two capacitors in series, $C=\left(C_o^{-1}+C_s^{-1}\right)^{-1}$, with $C_o=K_o\varepsilon_0/x_o$ and $C_s \equiv K_s\varepsilon_0/W_{\text{eff}}$. The "effective width" depends on the bias region. **Accumulation** ($p$-type, $U_s<0$): $$W_{\text{eff}} = \frac{U_s}{|U_s|}\,L_D\,\frac{2F(U_s,U_F)}{e^{U_F}\left(1-e^{-U_s}\right)+e^{-U_F}\left(e^{U_s}-1\right)} \tag{C.2a}$$ **Flat band** ($U_s=0$): $$W_{\text{eff}}=W_{fb} = L_D\left[\frac{2}{e^{U_F}+e^{-U_F}}\right]^{1/2} \tag{C.2b}$$ **Depletion/inversion** ($p$-type, $U_s>0$): $$W_{\text{eff}} = \frac{U_s}{|U_s|}\,L_D\,\frac{2F(U_s,U_F)}{e^{U_F}\left(1-e^{-U_s}\right)+e^{-U_F}\left(e^{U_s}-1\right)\left(1+\Delta\right)} \tag{C.2c}$$ where $\Delta$ distinguishes the measurement frequency: $$\Delta = 0\qquad\text{(low-frequency limit)} \tag{C.3a}$$ $$\Delta = \frac{e^{U_s}-U_s-1}{F(U_s,U_F)}\times\big[\text{prefactor, Eq. (C.3b)}\big]\times\int_{0}^{U_s}\frac{\left(1-e^{-U}\right)\left(e^{U}-U-1\right)}{2F^3(U,U_F)}\,dU\qquad\text{(high-frequency limit)} \tag{C.3b}$$ The integrand is the function $$\frac{\left(1-e^{-U}\right)\left(e^{U}-U-1\right)}{2F^3(U,U_F)} \tag{C.4}$$ **Reading note on (C.3b).** In the scan the factor between $(e^{U_s}-U_s-1)/F$ and the integral is garbled beyond recovery (fragments of $e^{U_F}$, $(1-e^{-U_s})$ and $(e^{-U}-1)$ survive). The structure shown and the integrand (C.4) are legible. This correction is Brews's "rearrangement capacitance", the contribution of inversion-layer carriers shifting position (not number) under the small-signal probe. **Why $\Delta=0$ at low frequency?** Minority carriers can be generated and recombined fast enough to follow the signal, so the semiconductor responds fully and the charge change is just $dQ_s/d\phi_s$. At high frequency the minority carriers cannot be exchanged with the bulk, so the capacitance stays near its minimum in inversion. Footnotes: all relations except (C.2c) and (C.3b) hold for both $n$- and $p$-type devices. The low-frequency relation traces to Grove, Deal, Snow and Sah (Solid-State Electronics, 8, 145, 1965); the high-frequency formula is adapted from J. R. Brews, J. Appl. Phys. 45, 1276 (1974). ## C.2 Definitions used $$F(U,U_F) = \left[e^{U_F}(e^{-U}+U-1)+e^{-U_F}(e^{U}-U-1)\right]^{1/2} \tag{C.5}$$ $$L_D = \left[\frac{K_s\varepsilon_0 kT}{2q^2 n_i}\right]^{1/2} \tag{C.6}$$ $$U_F = \frac{\phi_F}{kT/q} \tag{C.7}$$ $$U_s = \frac{\phi_s}{kT/q} \tag{C.8}$$ $$\varsigma_s = \begin{cases}+1 & U_s>0\\-1 & U_s<0\end{cases} \tag{C.9}$$ ## C.3 The gate voltage $$V_G = \frac{kT}{q}\left[U_s + \varsigma_s\,\frac{K_s\,x_o}{K_o\,L_D}\,F(U_s,U_F)\right] \tag{C.10}$$ **Derivation.** 1. For an ideal device $V_G=\phi_s+V_{ox}$. 2. Gauss: $Q_s=-K_s\varepsilon_0\mathcal{E}_s$, with $\mathcal{E}_s=\varsigma_s(kT/q)F(U_s,U_F)/L_D$ from (B.16). 3. Oxide drop: $V_{ox} = -Q_s/C_o = \varsigma_s\dfrac{kT}{q}\dfrac{K_sx_o}{K_oL_D}F$. 4. Add $\phi_s=(kT/q)U_s$ to get (C.10). (The OCR drops the $x_o$ in the printed fraction.) ## C.4 How to compute a C-V curve $C$ cannot be written explicitly in $V_G$, but both are known functions of $U_s$. The efficient recipe: 1. Choose $U_s$ values, stepped by whole numbers ($\dots,-5,-4,\dots$) or half units, over the normal range $U_F-21\le U_s\le U_F+21$ at room temperature (the MATLAB program uses steps of 0.5). 2. For each $U_s$ compute $F$, then $V_G$ from (C.10), then $W_{\text{eff}}$ from (C.2), then $C/C_o$ from (C.1). 3. Plot the $(V_G,C/C_o)$ pairs. Cautions: - At $U_s=0$ the accumulation and depletion/inversion expressions become $0/0$. Use the flat-band expression (C.2b) there. - For $n$-type devices the text gives modified forms of (C.2c) and (C.3b) (the exponentials are mirrored). The simplest alternative: compute the equivalently doped $p$-type device and **change the sign of every computed $V_G$**. This works because ideal $n$- and $p$-type devices are voltage mirror images. - The low-frequency calculation is simple enough for a hand calculator. --- # Appendix D: MOSFET I-V Supplement > **The picture.** The square-law MOSFET equation treats channel charge crudely. The exact-charge analysis (Pao and Sah) uses the true carrier distribution in the semiconductor. The charge-sheet analysis (Brews) squeezes the inversion layer into a zero-thickness sheet, which is nearly as accurate and far easier. Both rest on the electrostatics of Appendix B. ## D.1 The two current relationships For an ideal $n$-channel (p-bulk) MOSFET the exact-charge relationship is Eq. (D.1) and the charge-sheet relationship is Eq. (D.3). The exact-charge integrand uses the generalized field function $$F(U,U_F,\xi) = \left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^{U-\xi}-U-e^{-\xi}\right)\right]^{1/2} \tag{D.2}$$ where $\xi$ is the normalized channel quasi-Fermi potential, running from 0 at the source to $U_D$ at the drain. With $\xi=0$, (D.2) collapses to the equilibrium $F(U,U_F)$ of (B.17), as it must. **Reading note.** In the scan the typeset bodies of (D.1), (D.3) and the definition (D.4) are damaged beyond reliable recovery. What is legible: an overall prefactor $Z\mu_n/L$ (width, mobility, length); in (D.1) a second term proportional to $K_s/(K_oL_D)$; integrals between the source-end surface potential $U_{s0}$ and drain-end surface potential $U_{sL}$; and in the charge-sheet case a voltage $V_B$ defined in (D.4) from $K_s$, $N_A$, $K_o$, $L_D$ and $n_i$. The physics common to both is the drift-diffusion channel current $$I_D=\frac{Z\mu_n}{L}\int_{\text{source}}^{\text{drain}}\left|Q_N\right|\,dV_c ,$$ where $Q_N$ is the electron charge per unit area in the channel and $V_c$ the channel quasi-Fermi potential. $Q_N$ comes from the exact electrostatics (exact charge) or from gate-charge balance with the bulk depletion charge subtracted (charge sheet). Sources: R. F. Pierret and J. A. Shields, "Simplified Long-Channel MOSFET Theory," Solid-State Electronics, 26, 143 (1983); H. C. Pao and C. T. Sah, Solid-State Electronics, 9, 927 (1966) for the exact-charge analysis; J. R. Brews, Solid-State Electronics, 21, 345 (1978) for charge-sheet. ## D.2 Normalized voltages (both theories) $$U_G = \frac{V_G}{kT/q} \tag{D.5}$$ $$U_{s0} = \frac{V_{s0}}{kT/q} \tag{D.6}$$ $$U_{sL} = \frac{V_{sL}}{kT/q} \tag{D.7}$$ $$U_D = \frac{V_D}{kT/q} \tag{D.8}$$ $V_{s0}$ and $V_{sL}$ are the surface potentials at the source end ($y=0$) and the drain end ($y=L$). (Only the first symbol of (D.5) is clear in the scan; $U_G=V_G/(kT/q)$ is the quantity that fits.) ## D.3 Surface potentials at the channel ends From the gate-voltage relationship of Appendix C, now with the channel quasi-Fermi level: $$V_G = \frac{kT}{q}\left[U_{s0} + \frac{K_s x_o}{K_oL_D}\,F(U_{s0},U_F,0)\right]\quad (U_{s0}>0) \tag{D.9a}$$ $$V_G = \frac{kT}{q}\left[U_{sL} + \frac{K_s x_o}{K_oL_D}\,F(U_{sL},U_F,U_D)\right]\quad (U_{sL}>0) \tag{D.9b}$$ At the source the channel quasi-Fermi potential is 0, at the drain it is $U_D$. ## D.4 Procedure to generate $I_D$-$V_D$ characteristics 1. Step $V_G$ and $V_D$ systematically over the desired range. 2. For each $(V_G,V_D)$ pair, iterate (D.9a) and (D.9b) to find $U_{s0}$ and $U_{sL}$. (Since $V_G$ rises monotonically with $U_s$ for $U_s>0$, bisection or Newton iteration works.) 3. With $U_{s0}$ and $U_{sL}$ known, compute $I_D$ from (D.1) or (D.3). 4. Repeat for each $V_G$-$V_D$ combination. 5. For a $p$-channel device, run the calculation for an equivalently doped and biased $n$-channel device and reverse the bias polarities when plotting. --- # Appendix E: List of Symbols > A field guide to the notation. Many symbols are reused: a letter that means "hole concentration" in the main text means "momentum" in Appendix A, and $U$ is potential energy in Appendix A but normalized electrostatic potential in Appendices B to D. Subscripts carry the meaning (B, E, C for BJT regions; n, p for carrier type; 0 for equilibrium or oxide). Entries below are grouped alphabetically (Latin first, then Greek), repairing OCR damage from context. ## E.1 Latin symbols | Symbol | Meaning | |---|---| | $A$ | anode; area; arbitrary constant | | $a$ | lattice constant; grading constant; half-width of the channel region in a J-FET; width of the channel region in a MESFET | | $a_0$ | Bohr radius | | $A^{**}$ | Richardson's constant (120 A/cm$^2$-K$^2$) | | $A^{*}$ | modified Richardson's constant (see Eq. 14.19) | | $A_G$ | gate area | | $B$ | base | | $C$ | capacitance; collector | | $c$ | speed of light | | $C_D$ | diffusion capacitance | | $C_{\pi}$ | emitter-to-base capacitance in the high-frequency hybrid-pi model | | $C_G$ | MOSFET gate capacitance | | $C_{gd},\,C_{gs}$ | capacitances in the small-signal circuit for the J-FET and MOSFET (gate-drain, gate-source) | | $C_J$ | junction or depletion region capacitance | | $c_n$ | electron capture coefficient | | $C_o$ | oxide capacitance (pF) | | $C_o'$ | oxide capacitance per unit area (pF/cm$^2$) | | $c_p$ | hole capture coefficient | | $C_s$ | semiconductor capacitance | | $D$ | drain | | $D_B,\,D_C,\,D_E$ | minority-carrier diffusion coefficient in the BJT base, collector, emitter | | $D_T$ | density of interfacial traps (states/cm$^2$-eV) | | $D_N,\,D_P$ | electron, hole diffusion coefficient (cm$^2$/s) | | $D_{ox}$ | dielectric displacement in the oxide | | $D_{semi}$ | dielectric displacement in the semiconductor | | $E$ | emitter; energy | | $\mathcal{E}$ | electric field | | $\mathcal{E}_{ox}$ | electric field in the oxide | | $\mathcal{E}_s$ | surface electric field, field in the semiconductor at the oxide-semiconductor interface | | $\mathcal{E}_y$ | $y$-direction component of the electric field | | $E_0$ | vacuum level: minimum energy an electron needs to free itself completely from a material | | $E_A$ | acceptor energy level | | $E_B$ | binding energy at dopant (donor, acceptor) sites | | $E_C$ | minimum conduction band energy | | $E_D$ | donor energy level | | $E_F$ | Fermi energy or Fermi level | | $E_{FM}$ | Fermi level in the metal | | $E_{Fn},\,E_{Fp}$ | Fermi level on the $n$-side, $p$-side of a pn junction | | $E_{FS}$ | Fermi level in the semiconductor | | $E_G$ | band gap or forbidden-gap energy | | $E_H$ | electron binding energy in the hydrogen atom | | $E_i$ | intrinsic Fermi level | | $E_n$ | energy of the $n$th level (Appendix A) | | $E_{ph}$ | photon energy ($h\nu$) | | $E_t$ | trap or R-G center energy level | | $E_V$ | maximum valence band energy | | $F$ | force | | $f$ | frequency (Hz) | | $f(E)$ | Fermi function | | $F(U,U_F)$ | field function (see Eq. B.17) | | $F_{1/2}$ | Fermi-Dirac integral of order 1/2 | | $FF$ | fill factor | | $F_N,\,F_P$ | quasi-Fermi level (or energy) for electrons, holes | | $f_T$ | unity-beta (unity-gain) frequency of a BJT | | $G$ | conductance; gate | | $G_0$ | low-frequency conductance of a pn junction diode; channel conductance in a J-FET if there were no depletion regions | | $g_C(E)$ | density of conduction band states | | $G_D$ | diffusion conductance | | $g_d$ | drain or channel conductance | | $G_L$ | photogeneration rate (number of electron-hole pairs created per cm$^3$-s) | | $g_m$ | transconductance | | $g_V(E)$ | density of valence band states | | $h$ | Planck's constant | | $\hbar$ | $h/2\pi$ | | $I$ | current; light intensity | | $i$ | a.c. current; $\sqrt{-1}$ in Appendix A | | $I_0$ | saturation current in an ideal diode; light intensity at $x=0$ | | $I_{AK}$ | anode-to-cathode current | | $I_B$ | d.c. base current | | $i_B$ | total (a.c. + d.c.) base current | | $i_b$ | a.c. base current | | $I_C$ | d.c. collector current | | $i_C$ | total (a.c. + d.c.) collector current | | $i_c$ | a.c. collector current | | $I_{CBO}$ | collector-to-base current when $I_E=0$ | | $I_{CEO}$ | collector-to-emitter current when $I_B=0$ | | $I_{Cn},\,I_{Cp}$ | d.c. collector current due to electrons, holes | | $I_D$ | d.c. drain current in a field-effect transistor | | $i_d$ | small-signal drain current | | $I_{DO}$ | $V_G=0$ saturation drain current in a J-FET | | $I_{dark}$ | dark current | | $I_{DIFF}$ | diffusion current (same as the ideal diode current) | | $i_{diff}$ | a.c. component of the diffusion current | | $I_{Dsat}$ | saturation drain current | | $I_E$ | d.c. emitter current | | $I_{En},\,I_{Ep}$ | d.c. emitter current due to electrons, holes | | $I_F$ | steady-state forward-bias current | | $I_{F0}$ | effective diode forward saturation current (Ebers-Moll model) | | $I_G$ | gate current in an SCR | | $I_L$ | current due to light | | $I_{M\to S}$ | current due to electrons drifting from the metal to the semiconductor in an MS ($n$-type) diode | | $I_P$ | hole current | | $I_{P|drift}$ | hole current due to drift | | $I_R$ | steady-state reverse-bias current | | $I_{R\text{-}G}$ | recombination-generation current | | $I_{R0}$ | effective diode reverse saturation current (Ebers-Moll model) | | $I_S$ | reverse-bias saturation current in an MS diode | | $I_{SC}$ | short-circuit current in a solar cell | | $I_{S\to M}$ | current due to electrons drifting from the semiconductor to the metal in an MS ($n$-type) diode | | $J,\,\mathbf{J}$ | current density (A/cm$^2$) | | $J_{drift}$ | total current density due to drift | | $J_N,\,\mathbf{J}_N$ | electron current density | | $J_{Nx},J_{Ny},J_{Nz}$ | $x,y,z$ components of the electron current density | | $J_{N|diff},\,J_{N|drift}$ | electron current density due to diffusion, drift | | $J_P,\,\mathbf{J}_P$ | hole current density | | $J_{Px},J_{Py},J_{Pz}$ | $x,y,z$ components of the hole current density | | $J_{P|diff},\,J_{P|drift}$ | hole current density due to diffusion, drift | | $K$ | cathode | | $k$ | Boltzmann constant ($8.617\times10^{-5}$ eV/K) | | $\mathbf{K}$ | wavenumber (parameter proportional to the electron crystal momentum) | | K.E. | kinetic energy | | $K_O$ | oxide dielectric constant | | $K_S$ | semiconductor (usually Si) dielectric constant | | $L$ | length of the J-FET or MOSFET channel | | $l$ | azimuthal quantum number | | $L'$ | reduced channel length defined in Fig. 19.4 | | $L_B$ | minority-carrier diffusion length in the BJT base; extrinsic Debye length | | $L_C$ | minority-carrier diffusion length in the BJT collector | | $L_D$ | intrinsic Debye length | | $L_E$ | minority-carrier diffusion length in the BJT emitter | | $L_{min}$ | minimum MOSFET channel length yielding long-channel behavior | | $L_n$ | electron minority-carrier diffusion length; angular momentum for the $n$ quantum number (Appendix A) | | $L_p$ | hole minority-carrier diffusion length | | $M$ | carrier multiplication factor | | $m$ | particle mass; magnetic orbital quantum number | | $m_0$ | electron rest mass | | $m_n^*,\,m_p^*$ | electron, hole effective mass | | $n$ | energy quantum number; electron carrier concentration (electrons/cm$^3$) | | $n^+$ | heavily doped $n$-type material | | $n_0$ | equilibrium electron concentration | | $n_P,\,n_N$ | defined electron concentration (see Eq. 3.36a) | | $N_A$ | total number of acceptor atoms/cm$^3$ | | $N_A^-$ | number of ionized (negatively charged) acceptors/cm$^3$ | | $N_B$ | bulk semiconductor doping ($N_A$ or $N_D$ as appropriate); doping concentration in the BJT base | | $n_{bulk}$ | electron concentration in the semiconductor bulk | | $N_C$ | effective density of conduction band states; doping concentration in the BJT collector | | $n_{C0}$ | equilibrium electron concentration in the collector of a pnp BJT | | $N_D$ | total number of donor atoms/cm$^3$ | | $N_D^+$ | number of ionized (positively charged) donors/cm$^3$ | | $N_E$ | doping concentration in the BJT emitter | | $n_{E0}$ | equilibrium electron concentration in the emitter of a pnp BJT | | $N_I$ | number of implanted ions/cm$^2$ | | $n_i$ | intrinsic carrier concentration | | $N_T$ | number of R-G centers/cm$^3$ | | $N_V$ | effective density of valence band states | | $p$ | hole concentration (holes/cm$^3$); momentum in Appendix A | | $p^+$ | heavily doped $p$-type material | | P.E. | potential energy | | $p_0$ | equilibrium hole concentration | | $p_N,\,p_P$ | defined hole concentration (see Eq. 3.36b) | | $p_{B0}$ | equilibrium hole concentration in the base of a pnp BJT | | $p_{bulk}$ | hole concentration in the semiconductor bulk | | $p_s$ | hole concentration at the semiconductor surface (number/cm$^3$) | | $Q$ | general designation for a charge | | $q$ | magnitude of the electronic charge ($1.60\times10^{-19}$ C) | | $Q_B$ | excess minority-carrier charge in the quasineutral base; bulk or depletion-region charge per unit area of the MOSFET gate | | $Q_{BL}$ | $Q_B$ in a long-channel MOSFET | | $Q_{BS}$ | $Q_B$ in a short-channel MOSFET | | $Q_F$ | fixed oxide charge per unit area at the oxide-semiconductor interface | | $Q_I$ | implant-related charge/cm$^2$ located at the oxide-semiconductor interface | | $Q_M$ | total mobile ion charge within the oxide per unit area of the MOS gate | | $Q_N$ | total electronic charge/cm$^2$ in the MOSFET channel ($n$-channel device) | | $Q_{o\text{-}s}$ | charge per unit area located at the oxide-semiconductor interface | | $Q_P$ | excess hole charge | | $Q_s$ | total charge in the semiconductor per unit area of the gate | | $R$ | ramp rate (see Fig. 16.17) | | $r_B,\,r_b$ | base resistance | | $r_C,\,r_c$ | collector resistance | | $R_D$ | channel-to-drain resistance in a J-FET or MOSFET | | $r_E,\,r_e$ | emitter resistance | | $r_j$ | depth of the source and drain islands in a MOSFET | | $R_L$ | load resistor | | $r_n$ | radius of the Bohr orbit corresponding to the $n$ quantum number | | $r_o$ | output resistance in the BJT hybrid-pi model | | $R_P$ | projected range in ion implantation | | $R_S$ | series resistance; sample resistance; source-to-channel resistance in a J-FET or MOSFET | | $r_\mu$ | feedthrough resistance in the BJT hybrid-pi model | | $r_\pi$ | input resistance in the BJT hybrid-pi model | | $S$ | source; probe-to-probe spacing in a four-point probe | | $s$ | spin quantum number | | $T$ | temperature | | $t$ | time | | $t_{ON}$ | triggering time in an SCR | | $TR$ | tuning ratio | | $t_r$ | recovery time (pn diode); rise time (BJT) | | $t_{rr}$ | reverse recovery time (pn diode) | | $t_s$ | storage delay time (pn diode) | | $t_{sd}$ | storage delay time (BJT) | | $U$ | potential energy in Appendix A; electrostatic potential normalized to $kT/q$ in Appendices B to D | | $U_D$ | drain voltage normalized to $kT/q$ | | $U_F$ | semiconductor doping parameter | | $U_s$ | normalized surface potential, $U$ evaluated at the oxide-semiconductor interface | | $\varsigma_s$ | sign ($\pm$) of $U_s$ | | $U_{s0}$ | normalized surface potential at $x=0$ (source end) in a MOSFET | | $U_{sL}$ | normalized surface potential at $x=L$ (drain end) in a MOSFET | | $V$ | voltage, electrostatic potential; volume; velocity | | $V_A$ | applied d.c. voltage | | $v_a$ | applied a.c. voltage | | $V_{AK}$ | anode-to-cathode voltage | | $V_B$ | defined voltage (see Eq. D.4) | | $v_{be}$ | a.c. base-to-emitter voltage | | $V_{BBF}$ | forward-bias blocking voltage in PNPN devices | | $V_{bi}$ | "built-in" junction voltage | | $V_{BR}$ | reverse-bias pn junction breakdown voltage; reverse-bias blocking voltage in PNPN devices | | $V_{BS}$ | back-to-source voltage | | $V_{CB}$ | d.c. collector-to-base voltage | | $V_{CBO}$ | collector-to-base breakdown voltage when $I_E=0$ | | $v_{ce}$ | a.c. collector-to-emitter voltage | | $V_{CEO}$ | collector-to-emitter breakdown voltage when $I_B=0$ | | $V_D$ | d.c. drain voltage | | $\mathbf{v}_d$ | drift velocity vector | | $v_d$ | drift velocity; a.c. drain voltage | | $V_{DS}$ | drain-to-source voltage | | $V_{Dsat}$ | saturation drain voltage | | $v_{sat}$ | saturation drift velocity | | $V_{EB}$ | d.c. emitter-to-base voltage | | $V_{EC}$ | d.c. emitter-to-collector voltage | | $V_{FB}$ | flat-band voltage | | $V_G$ | d.c. gate voltage | | $v_g$ | a.c. gate voltage | | $V_G'$ | d.c. gate voltage applied to an ideal device | | $V_{GS}$ | gate-to-source voltage | | $V_{GS}'$ | gate-to-source voltage applied to an ideal device | | $V_J$ | junction voltage | | $V_{OC}$ | open-circuit voltage of a solar cell | | $V_P$ | pinch-off gate voltage in a J-FET | | $V_S$ | d.c. source voltage | | $v_s$ | pulsed source voltage | | $V_{S0}$ | surface potential at $x=0$ in a MOSFET | | $V_{SL}$ | surface potential at $x=L$ in a MOSFET | | $V_T$ | turn-on (threshold) voltage | | $V_T'$ | ideal-device inversion-depletion transition point gate voltage | | $V_x$ | defined voltage (see Eq. 17.24) | | $V_y$ | defined voltage (see Eq. 16.36) | | $W$ | depletion width; quasineutral width of the BJT base | | $W_{N2}'$ | quasineutral width of the N2 base in an SCR | | $W_{P3}'$ | quasineutral width of the P3 base in an SCR | | $W_{BB}$ | total width of the base in a BJT | | $W_D$ | drain pn junction depletion width in a MOSFET | | $W_{eff}$ | effective depletion width in a MOS capacitor (see Appendix C) | | $W_{N2}$ | width of the N2 base in an SCR | | $W_{P3}$ | width of the P3 base in an SCR | | $W_S$ | source pn junction depletion width in a MOSFET | | $W_T$ | MOS depletion width when the semiconductor is biased at the inversion-depletion transition point | | $x_c$ | width of the base in a narrow-base pn junction diode; depth of the MOSFET channel | | $x_n$ | $n$-side width of the pn junction depletion region | | $x_o$ | oxide thickness | | $x_p$ | $p$-side width of the pn junction depletion region | | $Y$ | admittance | | $Y_D$ | diffusion admittance | | $Z$ | width of the J-FET or MOSFET channel | ## E.2 Greek symbols | Symbol | Meaning | |---|---| | $\alpha$ | absorption coefficient | | $\alpha_{dc}$ | common-base d.c. current gain | | $\alpha_F$ | $\alpha_F=\alpha_{dc}$, forward gain (Ebers-Moll model) | | $\alpha_R$ | reverse gain (Ebers-Moll model) | | $\alpha_T$ | base transport factor | | $\beta_{dc}$ | common-emitter d.c. current gain | | $\chi$ | semiconductor electron affinity | | $\chi'$ | effective semiconductor electron affinity in an MOS structure | | $\chi_i$ | insulator (oxide) electron affinity | | $\chi_{Si}$ | silicon electron affinity | | $\Delta$ | frequency parameter in the exact-charge C-V theory (see Appendix C) | | $\Delta E_C$ | conduction band offset energy in a heterojunction | | $\Delta E_V$ | valence band offset energy in a heterojunction | | $\Delta_{ox}$ | voltage drop across the oxide | | $\Delta_{semi}$ | voltage drop across the semiconductor | | $\Delta L$ | decrease in the channel length under above-pinch-off conditions | | $\Delta n$ | $\Delta n=n-n_0$, deviation in electron concentration from its equilibrium value | | $\Delta n_C$ | excess electron concentration in the collector of a pnp BJT | | $\Delta n_E$ | excess electron concentration in the emitter of a pnp BJT | | $\Delta n_p$ | $\Delta n$ in $p$-type material | | $\Delta p$ | $\Delta p=p-p_0$, deviation in hole concentration from its equilibrium value | | $\Delta p_B$ | excess hole concentration in the base of a pnp BJT | | $\Delta p_n$ | $\Delta p$ in $n$-type material | | $\Delta Q$ | general designation for a change in charge | | $\Delta Q_G$ | change in the gate charge/cm$^2$ | | $\Delta Q_{semi}$ | change in the charge/cm$^2$ inside the semiconductor | | $\Delta R_P$ | straggle in ion implantation | | $\Delta V_G$ | difference between the actual device and ideal device gate voltage required to achieve a given semiconductor surface potential | | $\Delta V_T$ | change in the threshold voltage due to small-dimension effects | | $\Delta V_{T}'$ | change in the threshold voltage due to back biasing (specifically applied to ideal devices) | | $\Delta\Phi_B$ | change in the MS barrier height caused by Schottky barrier lowering | | $\varepsilon$ | permittivity | | $\varepsilon_0$ | permittivity of free space ($8.85\times10^{-14}$ F/cm) | | $\phi$ | electrostatic potential inside the semiconductor component of an MOS device | | $\Phi_B$ | surface potential-energy barrier height in an MS diode | | $\Phi_{B0}$ | $\Phi_B$ when $\mathcal{E}=0$ at the MS interface | | $\phi_F$ | reference voltage related to the semiconductor doping concentration | | $\Phi_M$ | metal work function | | $\Phi_M'$ | $\Phi_M'=\Phi_M-\chi_i$ (effective metal work function in an MOS structure) | | $\phi_{MS}$ | metal-semiconductor work-function difference expressed in volts | | $\phi_{ox}$ | voltage inside the oxide | | $\Phi_S$ | semiconductor work function | | $\phi_s$ | semiconductor surface potential | | $\Gamma$ | four-point probe correction factor | | $\gamma$ | emitter efficiency | | $\gamma_M$ | normalized centroid of mobile ion charge in the oxide | | $\eta$ | $\eta=(E-E_F)/kT$; power conversion efficiency of a solar cell; external efficiency of an LED | | $\eta_c$ | $\eta_c=(E_F-E_C)/kT$ | | $\eta_v$ | $\eta_v=(E_V-E_F)/kT$ | | $\lambda$ | wavelength of light | | $\lambda_G$ | wavelength of light corresponding to the semiconductor band gap | | $\mu_0$ | low-field electron or hole mobility; mobility fit parameter | | $\mu_{bulk}$ | carrier mobility in the bulk of a semiconductor | | $\mu_n$ | electron mobility | | $\mu_p$ | hole mobility | | $\mu_{n\,eff}$ | effective electron mobility | | $\mu_{p\,eff}$ | effective hole mobility | | $\nu$ | frequency of light | | $\rho$ | resistivity ($\Omega$-cm); charge density (C/cm$^3$) | | $\rho_{ion}$ | ionic charge density inside the oxide | | $\rho_{ox}$ | charge density in the oxide | | $\sigma$ | conductivity | | $\tau_0$ | defined carrier lifetime (see Eq. 6.44) | | $\tau_B$ | minority-carrier lifetime in the BJT base | | $\tau_C$ | minority-carrier lifetime in the BJT collector | | $\tau_E$ | minority-carrier lifetime in the BJT emitter | | $\tau_n$ | electron minority-carrier lifetime | | $\tau_p$ | hole minority-carrier lifetime | | $\tau_t$ | base transit time | | $\Psi$ | time-dependent wavefunction | | $\psi$ | time-independent wavefunction | | $\omega$ | angular frequency (radians) | *Where the OCR corrupted a subscript in the source (for example several entries for $C_d$, $C_s$, $g_{d}$, $V_s$), the symbol was repaired by its description and the surrounding alphabetical order. Entries whose scan lost both symbol and subscript (for example a repeated "$E_n$" row and an "$a$" row after "emitter-to-base capacitance") are folded into the nearest legible entry.* --- # Appendix M: MATLAB Program Scripts > **What these are.** Four programs, each tied to an end-of-chapter exercise, that turn the book's equations into pictures: the equilibrium band diagram of a BJT (Exercise 10.2), the BJT input/output characteristics with and without non-ideal effects (Exercises 11.7 and 11.10), and the exact-charge MOS C-V curve (Exercise 16.5). Treat each one as a lab bench: change a doping, a width, a thickness, and watch the physics respond. General notes on the cleaned code: - The scan lost some characters (decimal exponents such as `1.5e16`, the digit 0 in names like `IF0`, `IR0`, `VCB0`, `VCE0`, `nE0`, `pB0`, `nC0`, parentheses, quotes, and Greek letters in names). They are restored below. Greek-letter variable names (mu) are written in ASCII (`mun_min`, `mup_min`, `mun0`, `mup0`) because MATLAB identifiers cannot contain Greek letters. - A few lines in the scan were printed upside-down or reversed; they are rebuilt from the corresponding physics equation, and noted where this happened. - Units are CGS-flavored: cm, cm$^{-3}$, cm$^2$/V-s, F/cm, with energies in eV. --- ## M.1 Exercise 10.2: `BJT_Eband` (equilibrium energy band diagram of a BJT) **Purpose.** Plots $E_C$, $E_V$, $E_i$ and $E_F$ versus position through an emitter-base-collector structure at equilibrium, for any combination of $n$ and $p$ doping (npn, pnp, or even odd combinations). The user supplies a vector `DOPING = [NE NB NC]` where the **sign encodes the type** (positive = $p$-type, negative = $n$-type) and the magnitude is the concentration in cm$^{-3}$. Authored originally by Aaron Luft as a course project for Prof. Gerry Neudeck; major revisions by R. F. Pierret. **The physics inside.** 1. Built-in potentials of each junction from the Fermi-level offsets: $$V_{bi} = \left(E_i-E_F\right)_{\text{side 1}} - \left(E_i-E_F\right)_{\text{side 2}},\qquad (E_i-E_F)=\pm kT\ln\frac{N}{n_i}$$ 2. Depletion widths of an abrupt junction between regions with concentrations $N_1,N_2$: $$x_1 = \sqrt{\frac{2K_S\varepsilon_0}{q}\,\frac{N_2\,V_{bi}}{N_1\,(N_1+N_2)}}$$ 3. Quasineutral base width $W = W_B - x_{B,EB} - x_{B,CB}$ (error if the base is totally depleted). 4. Parabolic band bending in each depletion region, plotted as $E_C=-V(x)+E_G/2+V_{max}/2$ and $E_V=E_C-E_G$ with $E_i$ at midgap. 5. The two junction plots are stitched together with an offset so the bands are continuous across the base. ```matlab % BJT Equilibrium Energy Band Diagram Generator % This program plots out the BJT equilibrium energy band diagram. % Original version authored by Aaron Luft as a course project for Prof. Gerry Neudeck. % Major revisions by R.F. Pierret. % DOPING = [NE NB NC] must be defined before running (sign: +p-type, -n-type). WB = 1.0e-4; % Total base width in cm; 1.0e-4 cm = 1 micrometer close % Constants T = 300; % Temperature in Kelvin k = 8.617e-5; % Boltzmann constant, eV/K e0 = 8.85e-14; % permittivity of free space (F/cm) q = 1.602e-19; % charge on an electron (coul) KS = 11.8; % Dielectric constant of Si at 300 K ni = 1.0e10; % intrinsic conc. of silicon at 300 K EG = 1.12; % Silicon band gap (eV) % end constants % General computations and manipulations NE = DOPING(1); % Emitter doping and type NB = DOPING(2); % Base doping and type NC = DOPING(3); % Collector doping and type sE = sign(NE); sB = sign(NB); sC = sign(NC); NE = abs(NE); % Emitter doping NB = abs(NB); % Base doping NC = abs(NC); % Collector doping Ei_emitter = [(sE*k*T*log(NE/ni)) (-sB*k*T*log(NB/ni))]; Ei_collector = [(sB*k*T*log(NB/ni)) (-sC*k*T*log(NC/ni))]; Vbi = [sum(Ei_emitter) sum(Ei_collector)]; svbi = sign(Vbi); Vbi = abs(Vbi); % Depletion width on emitter side of EB junction xE = sqrt(2*KS*e0/q*NB*Vbi(1)/(NE*(NE+NB))); % Depletion width on base side of EB junction xBeb = sqrt(2*KS*e0/q*NE*Vbi(1)/(NB*(NE+NB))); % Depletion width on base side of CB junction xBcb = sqrt(2*KS*e0/q*NC*Vbi(2)/(NB*(NC+NB))); % Depletion width on collector side of CB junction xC = sqrt(2*KS*e0/q*NB*Vbi(2)/(NC*(NB+NC))); W = WB - xBeb - xBcb; if W < 0 error('For the given DOPING and WB, the base is totally depleted.') end if (xC > xE) % Adjust the x-axis for optimum looking plot HIGH_X = 1.5; LOW_X = xC/xE; else HIGH_X = xE/xC; LOW_X = 1.5; end VMAX = 3; % Maximum plot voltage plot([-LOW_X*xE HIGH_X*xC+WB], [0 VMAX], 'w'); hold on; % EB JUNCTION xlft = -LOW_X*xE; % Leftmost x position xrght = xBeb + W/2; % Rightmost x position x = linspace(xlft, xrght, 200); sVx = -svbi(1)*sE*sB; Vx1 = sVx*(Vbi(1) - q*NB.*(xBeb-x).^2/(2*KS*e0).*(x<=xBeb)).*(x>=0); Vx2 = sVx*(q*NE.*(x+xE).^2/(2*KS*e0).*(x<0).*(x>=-xE)); Vx = Vx1 + Vx2; % V as a function of x EF = Vx(1) + VMAX/2 - sE*k*T*log(NE/ni); % Fermi level Ec = -Vx + EG/2 + VMAX/2; Ev = -Vx - EG/2 + VMAX/2; Ei = -Vx + VMAX/2; LEc = Ec(1); LEv = Ev(1); LEi = Ei(1); % Plot V vs x plot(x, Ec); % Ec plot(x, Ev); % Ev plot(x, Ei, 'w:'); % Ei plot([xlft 0], [EF EF], 'w'); % EF on left plot([0 0], [0.15 VMAX-0.15], 'w-'); % Junction center % CB JUNCTION xlft = -xBcb - W/2; % Leftmost x position xrght = HIGH_X*xC; % Rightmost x position x = linspace(xlft, xrght, 200); sVx = -svbi(2)*sC*sB; Vx1 = sVx*(Vbi(2) - q*NC.*(xC-x).^2/(2*KS*e0).*(x<=xC)).*(x>=0); Vx2 = sVx*(q*NB.*(x+xBcb).^2/(2*KS*e0).*(x<0).*(x>=-xBcb)); Vx = Vx1 + Vx2; % V as a function of x OFFSET = (Ec(200)) - (-Vx(1) + EG/2 + VMAX/2); Ec = (-Vx + EG/2 + VMAX/2) + OFFSET; Ev = (-Vx - EG/2 + VMAX/2) + OFFSET; Ei = (-Vx + VMAX/2) + OFFSET; x = x + WB; % Plot V vs x plot(x, Ec); % Ec plot(x, Ev); % Ev plot(x, Ei, 'w:'); % Ei plot([0 xrght+WB], [EF EF], 'w'); % EF on right plot([WB WB], [0.15 VMAX-0.15], 'w-'); % Junction center if (sC == -1) RIGHT = 'N'; else RIGHT = 'P'; end if (sB == -1) MIDL = 'N'; else MIDL = 'P'; end if (sE == -1) LEFT = 'N'; else LEFT = 'P'; end A = -LOW_X*xE/2; B = WB + ((HIGH_X*xC+WB) - WB)/2; text(A, 2.5, LEFT); text(WB/2, 2.5, MIDL); text(B, 2.5, RIGHT); text(x(200), Ec(200), 'Ec'); text(x(200), Ei(200), 'Ei'); text(x(200), Ev(200), 'Ev'); text(x(200), EF, 'EF'); REG = [LEFT(1) MIDL(1) RIGHT(1)]; TITLE = ['Energy band diagram for the ', REG, ' device']; title(TITLE); ``` *Reconstruction notes.* In the scan, the lines defining `xE`, `Vx1` and `Vx2` (EB and CB) were printed reversed or garbled. They were rebuilt from the abrupt-junction depletion formulas above: on the lightly doped side the potential rises as $V_{bi}-qN(x_{edge}-x)^2/2K_S\varepsilon_0$ and on the heavily doped side as $qN(x+x_{edge})^2/2K_S\varepsilon_0$, which match each other at $x=0$ and give total drop $V_{bi}$. The statement `x = x + WB` shifts the CB junction plot to the right by the base width. --- ## M.2 Exercises 11.7 and 11.10: `BJT` and `BJTplus` (with `BJT0` and `BJTmod`) **Purpose.** Computes and plots the four classic BJT characteristics from the Ebers-Moll model: common-base input ($I_E$ vs $V_{EB}$), common-base output ($I_C$ vs $V_{CB}$), common-emitter input ($I_B$ vs $V_{EB}$) and common-emitter output ($I_C$ vs $V_{EC}$). `BJT` (Exercise 11.7) is the ideal version. `BJTplus` (Exercise 11.10) adds **base-width modulation** (Early effect) and **carrier multiplication** (avalanche in the collector junction). The book's note: the italicized lines in the `BJT` listing are the additions that form `BJTplus`. Subprogram `BJT0` is a run-time requirement of both; subprogram `BJTmod` must also be present for `BJTplus`. **Division of labor.** - `BJT0`: constants, device geometry, doping, mobility model, lifetimes, diffusion coefficients and lengths, equilibrium minority concentrations, and the Ebers-Moll parameters at $W=W_B$. - `BJTmod`: recomputes the Ebers-Moll parameters with the voltage-dependent quasineutral base width $W=W_B-x_{nEB}-x_{nCB}$. - `BJT`/`BJTplus`: menu-driven main program. **Equations implemented.** Ebers-Moll in terms of $\alpha_F,\alpha_R$ and the saturation currents $I_{F0}$, $I_{R0}$: $$I_E = I_{F0}\left(e^{V_{EB}/kT}-1\right)-\alpha_R I_{R0}\left(e^{V_{CB}/kT}-1\right)$$ $$I_C = \alpha_F I_E - \left(1-\alpha_F\alpha_R\right)I_{R0}\left(e^{V_{CB}/kT}-1\right)$$ (the second form is used for the common-base output). With a finite base, $$f_B = \frac{D_B}{L_B}\,p_{B0}\coth\frac{W}{L_B},\qquad I_{F0}=qA\left[\frac{D_E}{L_E}n_{E0}+f_B\right],\qquad I_{R0}=qA\left[\frac{D_C}{L_C}n_{C0}+f_B\right]$$ $$\alpha_F=\frac{qA\,(D_B/L_B)\,p_{B0}/\sinh(W/L_B)}{I_{F0}},\qquad \alpha_R=\frac{qA\,(D_B/L_B)\,p_{B0}/\sinh(W/L_B)}{I_{R0}}$$ Breakdown models (empirical): $V_{CB0}=60\,(N_C/10^{16})^{-3/4}$ volts, $V_{CE0}=V_{CB0}(1-\alpha_F)^{1/m}$ with $m=6$. Multiplication factor $M=1/[1-(|V_{CB}|/V_{CB0})^m]$, applied as $\alpha_F\to M\alpha_F$. ### BJT / BJTplus (main program) ```matlab % BJT Common Base/Emitter Input/Output Characteristics % Modified version of BJT including Base-Width Modulation and % Carrier Multiplication (BJTplus) % Input Ebers-Moll Parameters BJT0 % Limiting Voltages used in Calculation VbiE = kT*log(NE*NB/ni^2); VbiC = kT*log(NC*NB/ni^2); VCB0 = 50; VCE0 = 50; VCB0 = 60*(NC/1.0e16)^(-3/4); m = 6; VCE0 = VCB0*(1-aF)^(1/m); % Choice of Characteristic and Special Calculations format compact echo on % THIS PROGRAM COMPUTES BJT INPUT AND OUTPUT CHARACTERISTICS % Subprograms BJT0 and BJTmod are run-time requirements. % Modify entries in BJT0 to change device/material parameters. % Modify axis commands to change plot min/max values. echo off close c = menu('Specify the desired characteristic', 'Common Base Input', ... 'Common Base Output', 'Common Emitter Input', 'Common Emitter Output'); j = input('Specify number of curves per plot.'); if c ~= 2, bw = input('Include base-width modulation? 1-Yes, 2-No.'); else end ii = 2; if c == 4 & bw == 1, ii = input('Include impact ionization? 1-Yes, 2-No.'); else end % Calculation Proper for i = 1:j, % Common-Base Input Characteristics if c == 1, VCB = -(i-1)*10; VEB = 0:0.005:VbiE; jj = length(VEB); if bw == 1, BJTmod % Base-Width Modulation subprogram else end IE = (IF0.*(exp(VEB/kT)-1) - aR.*IR0.*(exp(VCB/kT)-1))*1.0e3; % 1.0e3 in the preceding equation changes IE units to mA if i == 1, plot(VEB, IE); axis([0.35 0.85 0 5]); grid; xlabel('VEB (volts)'); ylabel('IE (mA)'); else plot(VEB, IE); end % Common-Base Output Characteristics elseif c == 2, IE = (j-i)*1.0e-3; VCB1 = 2:-0.01:0; VCB2 = 0:-VCB0/200:-VCB0; VCB = [VCB1, VCB2]; jj = length(VCB); IC = (aF*IE - (1-aF*aR)*IR0*(exp(VCB/kT)-1))*(1.0e3); if i == 1, plot(-VCB, IC); axis([-VCB0/10 VCB0 0 1.3e3*IE]); grid; xlabel('-VCB (volts)'); ylabel('IC (mA)'); text(5, 1.1e3*IE, 'IE step = 1 mA'); else plot(-VCB, IC); end else end % Common-Emitter Input Characteristics if c == 3, VEC = (i-1)*5; VEB = 0:0.005:VbiE; jj = length(VEB); if bw == 1, VCB = VEB - VEC; BJTmod else end IB0 = (1-aF).*IF0 + (1-aR).*IR0; IB1 = (1-aF).*IF0 + (1-aR).*IR0.*exp(-VEC/kT); IB = (IB1.*exp(VEB/kT) - IB0)*(1.0e6); if i == 1, plot(VEB, IB); axis([.35 .85 -5 20]); grid; xlabel('VEB (volts)'); ylabel('IB (uA)'); else plot(VEB, IB); end % Common-Emitter Output Characteristics elseif c == 4, IB = (j-i)*2.5e-6; VECA = 0:0.01:VCE0/50; VECB = VCE0/50:VCE0/200:VCE0; VEC = [VECA, VECB]; jj = length(VEC); if bw == 1, VEB = 0; % Neglect xnEB variation with bias VCB = VEB - VEC; BJTmod else end if ii == 1, M = 1.0./(1 - (-VCB/VCB0).^m); aF = M.*aF; else end IB0 = (1-aF).*IF0 + (1-aR).*IR0; IB1 = (1-aF).*IF0 + (1-aR).*IR0.*exp(-VEC/kT); IC = ((aF.*IF0 - IR0.*exp(-VEC/kT)).*(IB+IB0)./IB1 + IR0 - aF.*IR0)*(1.0e3); if i == 1, jA = length(VECA); plot(VEC, IC); axis([0 VCE0 0 2.5*IC(jA)]); grid; xlabel('VEC (volts)'); ylabel('IC (mA)'); text(5, 2*IC(jA), 'IB step = 2.5 uA'); else plot(VEC, IC); end else end hold on end hold off ``` **What each branch does.** | Menu choice | Sweeps | Curve parameter | Output | |---|---|---|---| | 1. Common-base input | $V_{EB}$ from 0 to $V_{biE}$ in 5 mV steps | $V_{CB}=0,-10,-20,\dots$ V | $I_E$ (mA) vs $V_{EB}$; shows how base-width modulation shifts the curves | | 2. Common-base output | $V_{CB}$ from +2 V down to $-V_{CB0}$ | $I_E$ stepped in 1 mA | $I_C$ (mA) vs $-V_{CB}$; nearly flat lines (high $\alpha_F$) and slightly negative-$V_{CB}$ saturation region | | 3. Common-emitter input | $V_{EB}$ | $V_{EC}=0,5,10,\dots$ V | $I_B$ ($\mu$A) vs $V_{EB}$ | | 4. Common-emitter output | $V_{EC}$ from 0 to $V_{CE0}$ | $I_B$ stepped in 2.5 $\mu$A | $I_C$ (mA) vs $V_{EC}$; with the options on, shows Early-effect slope and avalanche upturn | *Reconstruction notes.* In the scan the comparison `ii == 1` shows as `i == 1` in the multiplication step; physics and the earlier prompt show it must test `ii` (the impact-ionization flag). The digit zero and letter O were confused in `IF0`, `IR0`, `VCB0`, `VCE0`, `IB0`; the zeros are restored. Axis command arguments, run together in the scan, were separated by position (for example `[.35 .85 -5 20]`). ### BJT0 (parameters and Ebers-Moll constants) ```matlab % BJT Constants and Ebers-Moll Parameters (subprogram BJT0) % Universal Constants q = 1.602e-19; k = 8.617e-5; e0 = 8.85e-14; % Device/Miscellaneous Parameters A = 1.0e-4; % A in cm^2 WB = 2.5e-4; % WB in cm T = 300; kT = k*T; % Material Parameters ni = 1.0e10; KS = 11.8; NE = 1.0e18; NB = 1.5e16; NC = 1.5e15; % Mobility Fit Parameters NDref = 1.3e17; NAref = 2.35e17; mun_min = 92; mup_min = 54.3; mun0 = 1268; mup0 = 406.9; an = 0.91; ap = 0.88; muE = mun_min + mun0./(1 + (NE/NDref).^an); muB = mup_min + mup0./(1 + (NB/NAref).^ap); muC = mun_min + mun0./(1 + (NC/NDref).^an); TauE = 1.0e-7; TauB = 1.0e-6; TauC = 1.0e-6; DE = kT*muE; DB = kT*muB; DC = kT*muC; LE = sqrt(DE*TauE); LB = sqrt(DB*TauB); LC = sqrt(DC*TauC); nE0 = ni^2/NE; pB0 = ni^2/NB; nC0 = ni^2/NC; % Ebers-Moll Parameter Computation (W = WB) W = WB; fB = (DB/LB)*pB0*(cosh(W/LB)/sinh(W/LB)); IF0 = q*A*((DE/LE)*nE0 + fB); IR0 = q*A*((DC/LC)*nC0 + fB); aF = q*A*(DB/LB)*(pB0/sinh(W/LB))/IF0; aR = q*A*(DB/LB)*(pB0/sinh(W/LB))/IR0; ``` **Explanation.** With $N_E=10^{18}$, $N_B=1.5\times10^{16}$, $N_C=1.5\times10^{15}$ (cm$^{-3}$), the default is a heavily doped emitter, moderately doped base and lightly doped collector (minority concentrations $n_{E0}$, $p_{B0}$, $n_{C0}$ and the quantities named in the List of Symbols). Mobility follows the Caughey-Thomas style fit $$\mu = \mu_{min}+\frac{\mu_0}{1+(N/N_{ref})^{\alpha}}$$ with the electron and hole constants listed. $D=\mu kT/q$ (with $kT$ in eV, so $D=kT\mu$), $L=\sqrt{D\tau}$. ### BJTmod (base-width-modulation subprogram) ```matlab % Base-width modulation: recalculation of Ebers-Moll parameters % Subprogram BJTmod xnEB = sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-VEB)); xnCB = sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB)); W = WB - xnEB - xnCB; fB = (DB/LB)*pB0*(cosh(W/LB)./sinh(W/LB)); IF0 = q*A.*((DE/LE)*nE0 + fB); IR0 = q*A.*((DC/LC)*nC0 + fB); aF = q*A*(DB/LB)*(pB0./sinh(W/LB))./IF0; aR = q*A*(DB/LB)*(pB0./sinh(W/LB))./IR0; ``` **Explanation.** `xnEB` and `xnCB` are the depletion widths penetrating into the base from the emitter and collector junctions at the present bias, $x=\sqrt{\dfrac{2K_S\varepsilon_0}{q}\dfrac{N_{\text{other}}}{N_B(N_{\text{other}}+N_B)}(V_{bi}-V)}$. Subtracting them from $W_B$ gives the **voltage-dependent quasineutral base width** $W$, and the element-wise operators (`./`, `.*`) let all voltage points be processed as arrays. Increasing reverse collector bias shrinks $W$, raises $\alpha_F$ and gives the upward slope of the output curves (Early effect). --- ## M.3 Exercise 16.5: `MOS_CV` (exact-charge MOS capacitor C-V curves) **Purpose.** Computes and plots the low-frequency and/or high-frequency C-V characteristics of a $p$-type MOS-C using the exact-charge relationships of Appendix C. The user chooses the plot type (low-f, high-f or both), enters $N_A$, $x_o$, and the gate-voltage window. Subprogram `CVintgrd` provides the integrand of Eq. (C.4) and is a run-time requirement. **Algorithm.** 1. $U_F=\ln(N_A/n_i)$, $L_D=\sqrt{kTK_S\varepsilon_0/(2q\,n_i)}$ (with $kT$ in eV). 2. Create the vector $U_s$ from $U_F-21$ to $U_F+21$ in steps of 0.5. 3. Evaluate $F(U_s,U_F)$ and the gate voltage $V_G=kT\left[U_s+\operatorname{sgn}(U_s)\,\dfrac{K_Sx_o}{K_OL_D}F\right]$ (C.10). 4. Low-frequency: $W_L=\operatorname{sgn}(U_s)L_D\,2F/\text{DENOML}$ with DENOML $=e^{U_F}(1-e^{-U_s})+e^{-U_F}(e^{U_s}-1)$, then $C_L/C_o=1/[1+K_OW_L/(K_Sx_o)]$. 5. High-frequency (only for $U_s\ge3$): accumulate the integral $\int_0^{U_s}$(C.4) with numerical quadrature, form $\Delta$ (C.3b), $W_H$ from (C.2c), then $C_H/C_o$. Below $U_s=3$ the two curves coincide. 6. Plot, label, and annotate with $N_A$ and $x_o$. ```matlab % LOW and/or HIGH-frequency p-type MOS-C C-V CHARACTERISTICS % Subprogram CVintgrd is a run-time requirement. % Initialization and Input format compact close clear s = menu('Choose the desired plot', 'Low-f C-V', 'High-f C-V', 'Both'); NA = input('Please input the bulk doping in /cm3, NA='); xo = input('Please input the oxide thickness in cm, xo='); xmin = input('Specify VGmin (volts), VGmin='); xmax = input('Specify VGmax (volts), VGmax='); global UF % Constants and Parameters e0 = 8.85e-14; q = 1.6e-19; k = 8.617e-5; KS = 11.8; KO = 3.9; ni = 1.0e10; T = 300; kT = k*T; % Computed Constants UF = log(NA/ni); LD = sqrt((kT*KS*e0)/(2*q*ni)); % Gate Voltage Computation US = UF-21:0.5:UF+21; F = sqrt(exp(UF).*(exp(-US)+US-1) + exp(-UF).*(exp(US)-US-1)); VG = kT*(US + (US./abs(US)).*(KS*xo)/(KO*LD).*F); % Low-frequency Capacitance Computation DENOML = exp(UF).*(1-exp(-US)) + exp(-UF).*(exp(US)-1); WL = (US./abs(US)).*LD.*(2*F)./DENOML; cL = 1./(1 + (KO*WL)/(KS*xo)); % High-frequency Capacitance Computation if s ~= 1, jj = length(US); nn = 0; for ii = 1:jj, if US(ii) < 3, elseif nn == 0, INTG = quad('CVintgrd', 0.001, US(ii), 0.001); nn = 1; else INTG = INTG + quad('CVintgrd', US(ii-1), US(ii), 0.001); end if US(ii) < 3, cH(ii) = cL(ii); else % d implements Delta of Eq. (C.3b): (exp(Us)-Us-1)/F times the % prefactor of (C.3b) times the integral INTG d = (exp(US(ii))-US(ii)-1)/F(ii) * INTG; % prefactor omitted (illegible in scan) DENOMH = exp(UF).*(1-exp(-US(ii))) + exp(-UF).*((exp(US(ii))-1)*(1+d)); WH = LD.*(2*F(ii))./DENOMH; cH(ii) = 1/(1 + (KO*WH)/(KS*xo)); end end else end % Plotting the Result if s == 1, plot(VG, cL); elseif s == 2, plot(VG, cH); else plot(VG, cL, '--', VG, cH); text(0.8*xmin, .17, '--- Low-f', 'color', [0 0 1]); text(0.8*xmin, .12, '___ High-f', 'color', [1 0 0]); end axis([xmin, xmax, 0, 1]); text(0.8*xmin, .27, ['NA = ', num2str(NA), ' /cm3']); text(0.8*xmin, .22, ['xo = ', num2str(xo), ' cm']); xlabel('VG (volts)'); ylabel('C/CO'); grid ``` ```matlab function [y] = CVintgrd(U) % Integrand of Eq. (C.4): (1-exp(-U))(exp(U)-U-1)/(2 F^3) global UF F = sqrt(exp(UF).*(exp(-U)+U-1) + exp(-UF).*(exp(U)-U-1)); y = (1-exp(-U)).*(exp(U)-U-1)./(2*F.^3); ``` *Reconstruction notes.* In the scan the lines for `VG`, `cL`, `d` and `cH(ii)` were printed reversed and are rebuilt from (C.10), (C.1) and (C.3b). The decisive caveat is the line for `d`: the scan's version is unreadable in its prefactor, so the line above shows the legible factors of (C.3b) only and is annotated accordingly. Everything else (flow control, quadrature accumulation, plot annotations) is intact. The first quadrature call in the scan has swapped limits; the version above integrates from $0.001$ (avoiding the $0/0$ at $U=0$) up to `US(ii)` and then accumulates segment by segment. If $U_s$ ever equals exactly 0 in the vector `US`, the sign factor `US./abs(US)` yields NaN; use the flat-band formula (C.2b) or shift the grid slightly. --- # End Matter: Index and Document Notes ## Index (book pages 781 to 792, scan pages 804 to 815) The subject index is an alphabetical locator rather than new technical content. It is organized from **A** (abrupt junction, absorption coefficient, accumulation, acceptors, admittance, aluminum, amorphous, angular momentum, annealing, avalanche) to **Z** (Zener process, zincblende unit cell). Its largest entries are the book's main device topics, each with sub-entries and page ranges: | Heavily indexed topic | Where it lives (book pages) | |---|---| | BJT (biasing modes, gain, Ebers-Moll, hybrid-pi, switching) | 371 to 459 | | pn junction and pn diode (electrostatics, ideal diode, breakdown, C-V, small signal, transients) | 195 to 340 | | MOS-C (biasing, C-V, deep depletion, exact solution) | 563 to 600, 749 to 754 | | MOSFET (square law, bulk-charge, charge-sheet, small-signal, short-channel) | 611 to 637, 674 to 707, 755 to 756 | | J-FET, MESFET | 525 to 556 | | Schottky diode and MS contacts | 477 to 500 | | Photodiode, solar cell, LED | 347 to 368 | | SCR and other PNPN devices | 463 to 476 | | Carrier properties (mobility, effective mass, Fermi function, concentrations) | 32 to 67, 75 to 137 | | Quantum mechanics and Bohr atom | 733 to 748 | | Debye length | 749 to 751 | Useful navigation tips: C-V topics include exact-charge relationships at pages 753 to 754; hydrogen atom quantum states at 741 to 745; Debye length at 750 to 751; the exact-charge and charge-sheet MOSFET relationships at 755 to 756. ## The last page The final scan page (816) is not textbook content. It is a provenance notice from the digitization source (a DuXiu-collection scan archived around 2023 to 2024, with embedded page images and file metadata such as 792 main pages and 815 total pages). It carries no technical information and can be ignored. --- # Study Notes for This Part - **Appendix A** is where the core ideas begin: Planck's quantized energies, Bohr's $E_n=-13.6/n^2$ eV, de Broglie's $\lambda=h/p$, Schrödinger's equation and its five postulates, hydrogen quantum numbers $(n,l,m,s)$, the Pauli principle, and the Si configuration $1s^22s^22p^63s^23p^2$ (ten-electron core plus four valence electrons). - **Appendix B** is one beautiful trick: $d^2U/dx^2=f(U)$ is integrated once by multiplying by $dU/dx$, giving $\mathcal{E}=\varsigma_s(kT/q)F/L_D$ and a quadrature for $x(U)$. - **Appendix C** turns B into a parametric C-V curve through $W_{\text{eff}}$ and $V_G(U_s)$, with the low- and high-frequency difference carried by $\Delta$. - **Appendix D** uses the same $F$ with a channel quasi-Fermi shift $\xi$ to obtain the drain current, evaluated between $U_{s0}$ and $U_{sL}$. - **Appendix E** is a reference table; remember that symbols are overloaded between appendices. - **Appendix M** converts the equations into four MATLAB experiments; changing a single parameter in `BJT0` or the inputs of `MOS_CV` is the best way to build intuition.