Teacher_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!):

$$\begin{array}{rl}\text{Simple cubic: }& 8\times \left(\frac{1}{8}\right)=1\\ \text{BCC: }& 8\times \left(\frac{1}{8}\right)+1=2\\ \text{FCC: }& 8\times \left(\frac{1}{8}\right)+6\times \left(\frac{1}{2}\right)=1+3=4\end{array}$$

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

$$\begin{array}{rlrl}{n}_{\text{Si}}& =\frac{8}{a^3}& & \text{(8 atoms per unit cell of volume }{a}^3)\\ & =\frac{8}{(5.43\times 10^{-8}\text{ cm}{)}^3}& & \text{(substitute Si lattice parameter }a=5.43\text{ A˚})\\ & \approx 5\times 10^{22}{\text{ atoms/cm}}^3& & \text{(evaluate atomic density)}\end{array}$$

(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

$$\begin{array}{rlrl}{L}_{\text{diag}}& =\sqrt{a^2+a^2+a^2}=\sqrt{3}a& & \text{(Pythagoras in 3-D)}\end{array}$$

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

$$\begin{array}{rlrl}d& =\frac{1}{4}{L}_{\text{diag}}=\frac{\sqrt{3}}{4}a& & \text{(nearest neighbor is one-quarter of the body diagonal)}\\ & =\frac{\sqrt{3}}{4}(5.43\times 10^{-8}\text{ cm})& & \text{(substitute }a=5.43\times 10^{-8}\text{ cm)}\\ & =0.4330\times 5.43\times 10^{-8}\text{ cm}& & \text{(evaluate numerical coefficient }\sqrt{3}/4\approx 0.4330)\\ & \approx 2.35\times 10^{-8}\text{ cm}=2.35\text{ A˚}& & \text{(evaluate distance)}\end{array}$$ $$\boxed{d\approx 2.35\times 10^{-8}\text{ cm}=2.35\text{ A˚}}$$

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:

$$\begin{array}{rlrl}\text{Normalized intercepts: }& (-1,1,2)& & \text{(intercepts on }x,y,z\text{ axes in units of }a)\\ \text{Reciprocals: }& \left(\frac{1}{-1},\frac{1}{1},\frac{1}{2}\right)=\left(-1,1,\frac{1}{2}\right)& & \text{(take reciprocal of each intercept)}\\ \text{Multiply by LCM (2): }& (-2,2,1)& & \text{(clear fractions to get lowest integers)}\\ \text{Miller indices: }& (\bar{2}21)& & \text{(enclose in parentheses with overbar for negative index)}\end{array}$$ $$\boxed{(\bar{2}21)}$$

Direction:

$$\begin{array}{rlrl}\text{Projections on axes: }& (2a,a,0)& & \text{(vector components along }x,y,z)\\ \text{Divide by }a:& (2,1,0)& & \text{(already smallest whole numbers, no common factor)}\\ \text{Miller indices: }& [210]& & \text{(enclose in square brackets)}\end{array}$$ $$\boxed{[210]}$$

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

$$\begin{array}{rlrl}\text{Miller indices }(hkl):& (0,1,1)& & \\ \text{Reciprocals }(1/h,1/k,1/l):& \left(\frac{1}{0},\frac{1}{1},\frac{1}{1}\right)=(\infty ,1,1)& & \text{(intercepts on }x,y,z)\end{array}$$

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).

$$\boxed{\text{Plane intercepts: }(\infty ,1,1),\text{Direction vector: }[011]}$$

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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)

$$\begin{array}{rl}\text{(i) Elemental semiconductor: }& \boxed{\text{Silicon (Si) or Germanium (Ge)}}\\ \text{(ii) III-V compound semiconductor: }& \boxed{\text{GaAs (also InP, GaP, etc.)}}\end{array}$$

(b) A crystalline solid has atoms arranged in one orderly, periodic three-dimensional array throughout the entire volume; a polycrystalline solid is composed of many crystalline grains that are misoriented and disjointed relative to each other, so long-range order exists only within each individual grain.

(c) A unit cell is a small representative portion of a crystal lattice that, when translated and stacked in three dimensions, reproduces the entire crystal structure.

(d) Counting corner atoms ($1/8$ contribution), face-center atoms ($1/2$ contribution), and interior atoms ($1$ contribution):

$$\begin{array}{rl}\text{Simple cubic: }& 8\times \left(\frac{1}{8}\right)=\boxed{1\text{ atom/cell}}\\ \text{BCC: }& 8\times \left(\frac{1}{8}\right)+1=\boxed{2\text{ atoms/cell}}\\ \text{FCC: }& 8\times \left(\frac{1}{8}\right)+6\times \left(\frac{1}{2}\right)=1+3=\boxed{4\text{ atoms/cell}}\\ \text{Diamond: }& 4(\text{FCC sublattice})+4(\text{interior sublattice})=\boxed{8\text{ atoms/cell}}\end{array}$$

(e)

$$\boxed{1\text{ A˚}=10^{-8}\text{ cm}=10^{-10}\text{ m}}$$

(f) In a simple cubic lattice, nearest-neighbor atoms are separated by one cell edge:

$$\boxed{d=a}$$

(g)

$$\boxed{4\text{ nearest neighbors in both diamond and zincblende lattices}}$$

(h)

$$\begin{array}{rl}(hkl)& :\text{a specific crystal plane}\\ [hkl]& :\text{a specific crystal direction}\\ \{hkl\}& :\text{a set of symmetry-equivalent planes}\\ ⟨hkl⟩& :\text{a set of symmetry-equivalent directions}\end{array}$$

(i) In the Czochralski method:

1. Ultrapure polycrystalline silicon is placed in a quartz crucible inside an inert (argon) atmosphere and heated to form a melt.
2. A small, carefully oriented single-crystal seed crystal mounted on a rod is dipped into the molten silicon.
3. The melt temperature near the seed is lowered slightly so silicon freezes onto the seed as an exact structural extension of its lattice.
4. The seed is slowly rotated and withdrawn upward, pulling out a large cylindrical single-crystal ingot (up to ~200 mm diameter and 1–2 m long) that is subsequently sliced 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 $x=0.5$:

$$\begin{array}{rlrl}{N}_{\text{Group III sites}}& =4& & \text{(total Group III FCC sites per unit cell)}\\ N_{\mathrm{Al}}& =x\cdot 4=0.5\times 4=2& & \text{(Al atoms per unit cell)}\\ N_{\mathrm{Ga}}& =(1-x)\cdot 4=(1-0.5)\times 4=2& & \text{(Ga atoms per unit cell)}\\ N_{\mathrm{As}}& =4& & \text{(As atoms on the Group V sublattice)}\end{array}$$ $$\boxed{N_{\mathrm{Al}}=2,N_{\mathrm{Ga}}=2,N_{\mathrm{As}}=4}$$

So the unit cell is the same zincblende cube (lattice constant $a\approx 5.65\text{ A˚}$, 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 randomly distributed over those 4 sites. (For a sketch: replace two of the four Ga atoms in Fig. 1.4(b), for instance two face-center or corner Ga atoms, with Al atoms.) On average, every Group III site has equal (50%) probability of occupation by Al or 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 with 8 atoms per cubic unit cell:

$$\begin{array}{rlrl}n& =\frac{N}{V_{\text{cell}}}=\frac{8}{a^3}& & \text{(atomic density formula)}\\ a^3& =(5.65\times 10^{-8}\text{ cm}{)}^3& & \text{(substitute given lattice constant)}\\ & =180.36\times 10^{-24}{\text{ cm}}^3=1.8036\times 10^{-22}{\text{ cm}}^3& & \text{(calculate unit cell volume)}\\ n& =\frac{8}{1.8036\times 10^{-22}{\text{ cm}}^3}& & \text{(divide atom count by volume)}\\ & =4.4356\times 10^{22}{\text{ atoms/cm}}^3\approx 4.44\times 10^{22}{\text{ atoms/cm}}^3& & \text{(evaluate density)}\end{array}$$ $$\boxed{n\approx 4.44\times 10^{22}{\text{ atoms/cm}}^3}$$

Solution (b). Running the Exercise 1.3 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

Verification by direct computation:

$$\begin{array}{rl}n& =\frac{N\times 10^{24}}{a_{\text{A˚}}^3}=\frac{8\times 10^{24}}{5.65^3}=\frac{8\times 10^{24}}{180.36}=4.4356\times 10^{22}{\text{ cm}}^{-3}\end{array}$$

This agrees exactly 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. In a BCC lattice, the center atom touches all 8 corner atoms. The distance from the cube center to any corner is half of the body diagonal:

$$\begin{array}{rlrl}{L}_{\text{body}}& =\sqrt{a^2+a^2+a^2}=\sqrt{3}a& & \text{(length of the cube body diagonal)}\\ d_{\text{bcc}}& =\frac{1}{2}{L}_{\text{body}}=\frac{\sqrt{3}}{2}a& & \text{(half of the body diagonal)}\\ & \approx 0.866a& & \text{(evaluate numerical factor)}\end{array}$$

Since $0.866a<a$ (the corner-to-corner separation), this is indeed the nearest-neighbor distance.

$$\boxed{d_{\text{bcc}}=\frac{\sqrt{3}}{2}a\approx 0.866a}$$

Solution (b), fcc. In an FCC lattice, a corner atom and the atom at the center of an adjacent face touch each other, separated by half of the face diagonal:

$$\begin{array}{rlrl}{L}_{\text{face}}& =\sqrt{a^2+a^2}=\sqrt{2}a& & \text{(length of the face diagonal)}\\ d_{\text{fcc}}& =\frac{1}{2}{L}_{\text{face}}=\frac{\sqrt{2}}{2}a=\frac{a}{\sqrt{2}}& & \text{(half of the face diagonal)}\\ & \approx 0.707a& & \text{(evaluate numerical factor)}\end{array}$$

Since $0.707a<a$, this is the nearest-neighbor distance.

$$\boxed{d_{\text{fcc}}=\frac{a}{\sqrt{2}}\approx 0.707a}$$

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 centered on the $(100)$ face (corner atoms are each shared among 4 adjacent squares in the surface plane, while the face-center atom belongs wholly to this face):

$$\begin{array}{rlrl}{N}_{\text{face}}& =4\left(\frac{1}{4}\right)+1=2\text{ atoms}& & \text{(equivalent atoms centered on the face)}\\ A& =a^2=(5.43\times 10^{-8}\text{ cm}{)}^2& & \text{(area of the square unit face)}\\ & =2.9485\times 10^{-15}{\text{ cm}}^2& & \text{(evaluate area)}\\ & & & \\ sigm{a}_{(100)}& =\frac{N_{\text{face}}}{A}=\frac{2}{2.9485\times 10^{-15}{\text{ cm}}^2}& & \text{(surface density formula)}\\ & \approx 6.78\times 10^{14}{\text{ atoms/cm}}^2& & \text{(evaluate numerical value)}\end{array}$$ $$\boxed{{\sigma }_{(100)}\approx 6.78\times 10^{14}{\text{ atoms/cm}}^2}$$

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).

$$\begin{array}{rlrl}{N}_{(110)}& =1+1+2=4\text{ atoms}& & \text{(equivalent atoms centered on the diagonal rectangle)}\\ A& =\sqrt{2}{a}^2=\sqrt{2}(5.43\times 10^{-8}\text{ cm}{)}^2& & \text{(area of rectangle of dimensions }a\times \sqrt{2}a)\\ & =\sqrt{2}(2.9485\times 10^{-15}{\text{ cm}}^2)=4.1698\times 10^{-15}{\text{ cm}}^2& & \text{(evaluate rectangle area)}\\ {\sigma }_{(110)}& =\frac{N_{(110)}}{A}=\frac{4}{\sqrt{2}{a}^2}=\frac{2\sqrt{2}}{a^2}& & \text{(areal density formula)}\\ & =\frac{2.8284}{2.9485\times 10^{-15}{\text{ cm}}^2}\approx 9.59\times 10^{14}{\text{ atoms/cm}}^2& & \text{(evaluate numerical result)}\end{array}$$ $$\boxed{{\sigma }_{(110)}\approx 9.59\times 10^{14}{\text{ atoms/cm}}^2}$$

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).

$$\begin{array}{rlrl}\text{Normalized intercepts: }& (1,3,1)& & \text{(divide intercepts by }a)\\ \text{Reciprocals: }& \left(\frac{1}{1},\frac{1}{3},\frac{1}{1}\right)=\left(1,\frac{1}{3},1\right)& & \text{(invert each intercept)}\\ \text{Multiply by LCM (3): }& (3,1,3)& & \text{(multiply to obtain smallest whole numbers)}\\ \text{Miller indices: }& (313)& & \text{(enclose in parentheses)}\end{array}$$ $$\boxed{(313)}$$

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

$$\boxed{[313]}$$

Solution (b)(i).

$$\begin{array}{rlrl}\text{Normalized intercepts: }& (1,1,-2)& & \text{(divide intercepts by }a)\\ \text{Reciprocals: }& \left(\frac{1}{1},\frac{1}{1},\frac{1}{-2}\right)=\left(1,1,-\frac{1}{2}\right)& & \text{(invert each intercept)}\\ \text{Multiply by LCM (2): }& (2,2,-1)& & \text{(multiply to clear fractions)}\\ \text{Miller indices: }& (22\bar{1})& & \text{(enclose with overbar for negative index)}\end{array}$$ $$\boxed{(22\bar{1})}$$

Solution (b)(ii).

$$\begin{array}{rlrl}& & & \\ \text{Vector projections: }& \left(1a,1a,-\frac{1}{2}a\right)& & \text{(given vector projections on axes)}\\ \text{Divide by }a/2:& (2,2,-1)& & \text{(convert to smallest integers)}\\ \text{Miller indices: }& [22\bar{1}]& & \text{(enclose in square brackets)}\end{array}$$ $$\boxed{[22\bar{1}]}$$

(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). For $[h_1{k}_1{l}_1]=[100]$:

$$\begin{array}{r}1\cdot h_2+0\cdot k_2+0\cdot l_2=0⟹h_2=0\end{array}$$

Any direction $[0kl]$ is perpendicular to $[100]$. Two choices:

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

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

Solution (b). For $[h_1{k}_1{l}_1]=[111]$:

$$\begin{array}{r}1\cdot h_2+1\cdot k_2+1\cdot l_2=0⟹h_2+k_2+l_2=0\end{array}$$

Two simple solutions:

$$\begin{array}{rl}\text{For }[1\bar{1}0]:& 1-1+0=0✓\\ \text{For }[01\bar{1}]:& 0+1-1=0✓\end{array}$$ $$\boxed{[1\bar{1}0]\text{and}[01\bar{1}]}$$

(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:

$$h=0✓$$

Calculate the angle $\theta$ between the flat normal $[011]$ and the groove direction $[01\bar{1}]$:

$$\begin{array}{rl}\cos \theta & =\frac{(0)(0)+(1)(1)+(-1)(1)}{\sqrt{(0+1+1)(0+1+1)}}=\frac{1-1}{\sqrt{2}\sqrt{2}}=\frac{0}{2}=0⟹\theta =90^{\circ }\end{array}$$

Since $[011]$ is perpendicular to the major flat (normal to it), any in-plane direction perpendicular to $[011]$ must run parallel to the flat.

Sketch: Draw the circular wafer with the major flat at the bottom edge; draw parallel grooves as horizontal straight lines running parallel to the flat (left-to-right). (If the problem had instead asked for grooves along $[011]$ itself, they would run perpendicular to the flat: top-to-bottom vertical lines.)

$$\boxed{\text{The etched grooves run parallel to the major flat}}$$

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 Cartesian axis places the atom at the cube corners and gives atoms at all 8 corners: this generates the simple cubic lattice (atoms are separated by distance $a$ in each Cartesian direction).

$$\boxed{\text{Simple cubic lattice}}$$

Solution (b). Exactly one atom per unit cell of volume $a^3$:

$$\begin{array}{rl}n& =\frac{N}{V}=\frac{1}{a^3}\end{array}$$ $$\boxed{n=\frac{1}{a^3}}$$

Solution (c). The $(110)$ plane is a diagonal rectangle of side lengths $a$ and $\sqrt{2}a$ whose four corners are lattice points. Each corner atom is shared among 4 coplanar rectangles, and no other lattice points lie on the plane:

$$\begin{array}{rlrl}N& =4\left(\frac{1}{4}\right)=1\text{ atom}& & \text{(equivalent atom count on the plane)}\\ A& =a\times \sqrt{2}a=\sqrt{2}{a}^2& & \text{(area of diagonal rectangle)}\\ {\sigma }_{(110)}& =\frac{N}{A}=\frac{1}{\sqrt{2}{a}^2}\approx \frac{0.707}{a^2}& & \text{(areal density of atoms)}\end{array}$$ $$\boxed{{\sigma }_{(110)}=\frac{1}{\sqrt{2}{a}^2}\approx \frac{0.707}{a^2}}$$

Solution (d). Vector drawn from the corner at the origin $(0,0,0)$ to the center at $(a/2,a/2,a/2)$:

$$\begin{array}{rlrl}\text{Projections: }& \left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)& & \text{(components along }x,y,z)\\ \text{Divide by }a/2:& (1,1,1)& & \text{(clear common factor)}\end{array}$$ $$\boxed{[111]}$$

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 choices) times 2 signs ($\pm$):

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

(b) $\{110\}$: two indices are $\pm 1$ and one is $0$. Choose which index is zero (3 choices) times sign combinations for the two nonzero indices ($2^2=4$):

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

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

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

(d) $⟨100⟩$: same counting as (a):

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

(e) $⟨110⟩$: same counting as (b):

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

(f) $⟨111⟩$: same counting as (c):

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

(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 atomic sphere radius is $r=d/2$, the volume of one spherical atom is $\frac{4}{3}\pi r^3$, and the packing fraction is

$$\begin{array}{rl}f& =\frac{N\cdot \frac{4}{3}\pi r^3}{a^3}\end{array}$$

with $N$ atoms per cubic unit cell of volume $a^3$.

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

$$\begin{array}{rl}f& =\frac{1\cdot \frac{4}{3}\pi {\left(\frac{a}{2}\right)}^3}{a^3}=\frac{\frac{4}{3}\pi \cdot \frac{a^3}{8}}{a^3}=\frac{\pi }{6}\approx 0.5236⟹\boxed{52\%}\end{array}$$

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

$$\begin{array}{rl}{r}^3& ={\left(\frac{\sqrt{3}}{4}a\right)}^3=\frac{3\sqrt{3}}{64}{a}^3\\ f& =\frac{2\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{64}{a}^3}{a^3}=2\cdot \frac{4\cdot 3\sqrt{3}\pi }{3\cdot 64}=2\cdot \frac{\sqrt{3}\pi }{16}=\frac{\sqrt{3}\pi }{8}\approx 0.6802⟹\boxed{68\%}\end{array}$$

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

$$\begin{array}{rl}{r}^3& ={\left(\frac{\sqrt{2}}{4}a\right)}^3=\frac{2\sqrt{2}}{64}{a}^3=\frac{\sqrt{2}}{32}{a}^3\\ f& =\frac{4\cdot \frac{4}{3}\pi \cdot \frac{\sqrt{2}}{32}{a}^3}{a^3}=4\cdot \frac{4\sqrt{2}\pi }{3\cdot 32}=\frac{\sqrt{2}\pi }{6}\approx 0.7405⟹\boxed{74\%}\end{array}$$

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

$$\begin{array}{rl}{r}^3& ={\left(\frac{\sqrt{3}}{8}a\right)}^3=\frac{3\sqrt{3}}{512}{a}^3\\ f& =\frac{8\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{512}{a}^3}{a^3}=8\cdot \frac{4\sqrt{3}\pi }{3\cdot 512}=\frac{\sqrt{3}\pi }{16}\approx 0.3401⟹\boxed{34\%}\end{array}$$

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:

$$1\text{ eV}=1.60\times 10^{-19}\text{ J}$$

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:

$$\sum _{\text{filled band}}\mathbf{p}=0⟹\text{no net current}$$

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?

$$\begin{array}{rlrl}1\text{ eV}& =(1.602\times 10^{-19}\text{ C})(1\text{ V})& & \text{(energy gained by charge }q\text{ across 1 V)}\\ & =1.602\times 10^{-19}\text{ J}& & (\text{C}\cdot \text{V}=\text{J})\end{array}$$ $$\boxed{1\text{ eV}=1.602\times 10^{-19}\text{ J}}$$

(b) $kT$ in eV at $300\text{ K}$:

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5}\text{ eV/K})(300\text{ K})& & \text{(Boltzmann constant times }T)\\ & =2.5851\times 10^{-2}\text{ eV}\approx 0.0259\text{ eV}& & \text{(evaluate thermal energy)}\end{array}$$ $$\boxed{kT\approx 0.0259\text{ eV}=25.9\text{ meV}}$$

This number, about $26\text{ meV}$, 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 $0.1\text{ eV}$ (measured $0.04$ to $0.07\text{ eV}$ for common dopants).

$$\boxed{|E_B|\sim 0.04\text{ to }0.07\text{ eV}\approx 0.1\text{ eV}}$$

(d) Band gap of Si at $300\text{ K}$:

$$\boxed{E_G(\text{Si})=1.12\text{ eV}}$$

(e) Band gap of ${\text{SiO}}_2$:

$$\boxed{E_G({\text{SiO}}_2)\approx 8\text{ eV}}$$

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

$$\begin{array}{rl}|E_H(n=1)|& =\frac{13.6}{1^2}\text{ eV}=13.6\text{ eV}\end{array}$$ $$\boxed{|E_H(n=1)|=13.6\text{ eV}}$$

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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 problem condition states:

$$\begin{array}{rlrl}f(E_c)& =1-f(E_v)& & \text{(given condition)}\\ \frac{1}{1+e^{(E_c-E_F)/kT}}& =1-\frac{1}{1+e^{(E_v-E_F)/kT}}& & \text{(definition of Fermi function)}\\ & =\frac{e^{(E_v-E_F)/kT}}{1+e^{(E_v-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}& & \text{(algebraic simplification of RHS)}\\ \frac{1}{1+e^{(E_c-E_F)/kT}}& =\frac{1}{1+e^{(E_F-E_v)/kT}}& & \text{(set LHS equal to RHS)}\\ \frac{E_c-E_F}{kT}& =\frac{E_F-E_v}{kT}& & \text{(equate denominators and exponents)}\\ E_c-E_F& =E_F-E_v& & \text{(multiply through by }kT)\\ 2E_F& =E_c+E_v& & \text{(rearrange terms)}\\ E_F& =\frac{E_c+E_v}{2}& & \text{(solve for }{E}_F)\end{array}$$ $$\boxed{E_F=\frac{E_c+E_v}{2}\text{(midgap)}}$$

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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:

$$\begin{array}{rlrl}{E}_G(300)& =1.17-\frac{(4.730\times 10^{-4})(300{)}^2}{300+636}=1.17-\frac{42.57}{936}=1.1245\text{ eV}& & \text{(band gap fit)}\\ E_G-E_{ex}& =1.1245-0.0074=1.1171\text{ eV}& & \text{(exciton correction)}\\ \frac{m_n^{\ast }}{m_0}& =1.028+0.1833-0.0278=1.1835& & \text{(electron mass ratio)}\\ \frac{m_p^{\ast }}{m_0}& =0.610+0.2349-0.0401=0.8048& & \text{(hole mass ratio)}\\ {\left(\frac{m_n^{\ast }}{m_0}\frac{m_p^{\ast }}{m_0}\right)}^{3/4}& =(1.1835\times 0.8048{)}^{0.75}=(0.9524{)}^{0.75}=0.9641& & \text{(mass factor)}\\ kT& =(8.617\times 10^{-5})(300)=0.025851\text{ eV}& & \text{(thermal energy)}\\ \frac{E_G-E_{ex}}{2kT}& =\frac{1.1171}{0.051702}=21.607& & \text{(exponent factor)}\\ e^{-21.607}& =4.13\times 10^{-10}& & \text{(exponential)}\\ n_i& =(2.510\times 10^{19})(0.9641)(1)(4.13\times 10^{-10})\approx 1.00\times 10^{10}{\text{ cm}}^{-3}& & \text{(evaluate }{n}_i)\end{array}$$ $$\boxed{n_i\approx 1.00\times 10^{10}{\text{ cm}}^{-3}}$$

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:

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

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.

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

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

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

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 ($N_A=10^{14}{\text{ cm}}^{-3}$).

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}=\frac{3}{4}(0.0259)\ln \frac{0.81}{1.18}& & \text{(Eq. 2.36, Table 2.1)}\\ & =0.01943\times (-0.3761)=-0.0073\text{ eV}& & \text{(evaluate intrinsic level shift)}\end{array}$$ $$\boxed{E_i-E_{\text{mid}}=-0.0073\text{ eV}(0.0073\text{ eV below midgap})}$$

Next, since $N_A\gg n_i$, apply Eq. (2.38b):

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

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

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

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

Because $N_A\sim n_i$, apply Eq. (2.37) using the exact hole concentration $p$:

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

$E_F$ 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 $0\le T\le 600\text{ K}$ and note $E_G(300)$:

$$\begin{array}{rlrl}{E}_G(0)& =1.170\text{ eV}& & (\alpha T^2=0)\\ E_G(100)& =1.170-\frac{(4.73\times 10^{-4})(10^4)}{736}=1.170-0.0064=1.1636\text{ eV}& & \\ E_G(200)& =1.170-\frac{(4.73\times 10^{-4})(4\times 10^4)}{836}=1.170-0.0226=1.1474\text{ eV}& & \\ E_G(300)& =1.170-\frac{(4.73\times 10^{-4})(9\times 10^4)}{936}=1.170-0.0455=1.1245\text{ eV}& & \\ E_G(400)& =1.170-\frac{(4.73\times 10^{-4})(1.6\times 10^5)}{1036}=1.170-0.0731=1.0969\text{ eV}& & \\ E_G(500)& =1.170-\frac{(4.73\times 10^{-4})(2.5\times 10^5)}{1136}=1.170-0.1041=1.0659\text{ eV}& & \\ E_G(600)& =1.170-\frac{(4.73\times 10^{-4})(3.6\times 10^5)}{1236}=1.170-0.1378=1.0322\text{ eV}& & \end{array}$$ $$\boxed{E_G(300\text{ K})=1.1245\text{ eV}\approx 1.12\text{ eV}}$$

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.

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

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

$$\boxed{\text{Si on Ga site: Donor (n-type)}}$$

(d) Si on an As site.

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

A bond is left incomplete, so a neighboring valence electron is accepted, leaving a mobile hole: Si on As is an acceptor, so the GaAs is p-type. Si in GaAs is therefore amphoteric.

$$\boxed{\text{Si on As site: Acceptor (p-type)}}$$

(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:

$$\begin{array}{rlrl}N& ={\int }_{E_c}^{E_c+\gamma kT}{g}_c(E)dE& & \text{(number of states in energy range)}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{E_c}^{E_c+\gamma kT}(E-E_c{)}^{1/2}dE& & \text{(insert density of states formula Eq. 2.6a)}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\left[\frac{2}{3}(E-E_c{)}^{3/2}\right]}_{E_c}^{E_c+\gamma kT}& & \left(\text{using }\int u^{1/2}du=\frac{2}{3}{u}^{3/2}\right)\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}\cdot \frac{2}{3}(\gamma kT{)}^{3/2}& & \text{(substitute limits of integration)}\\ & =\frac{2\sqrt{2}(m_n^{\ast }\gamma kT{)}^{3/2}}{3{\pi }^2{\hslash }^3}{\text{ cm}}^{-3}& & (\text{combine mass terms: }{m}_n^{\ast }\sqrt{m_n^{\ast }}=(m_n^{\ast }{)}^{3/2})\end{array}$$ $$\boxed{N=\frac{2\sqrt{2}(m_n^{\ast }\gamma kT{)}^{3/2}}{3{\pi }^2{\hslash }^3}{\text{ cm}}^{-3}}$$

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

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Problem 2.6: Fermi function questions

(a) At $E=E_F$:

$$\begin{array}{rl}f(E_F)& =\frac{1}{1+e^0}=\frac{1}{1+1}=\frac{1}{2}\end{array}$$ $$\boxed{f(E_F)=\frac{1}{2}}$$

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

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

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

$$\begin{array}{rlrl}f(E_c-kT)& =1-f(E_v+kT)& & \text{(given equality of probabilities)}\\ \frac{1}{1+e^{(E_c-kT-E_F)/kT}}& =\frac{1}{1+e^{(E_F-E_v-kT)/kT}}& & \left(\text{using }1-f(E)=\frac{1}{1+e^{(E_F-E)/kT}}\right)\\ \frac{E_c-kT-E_F}{kT}& =\frac{E_F-E_v-kT}{kT}& & \text{(equate exponents)}\\ E_c-kT-E_F& =E_F-E_v-kT& & \text{(multiply through by }kT)\\ E_c-E_F& =E_F-E_v& & \text{(add }kT\text{ to both sides)}\\ 2E_F& =E_c+E_v⟹E_F=\frac{E_c+E_v}{2}& & \text{(solve for }{E}_F)\end{array}$$ $$\boxed{E_F=\frac{E_c+E_v}{2}\text{(midgap)}}$$

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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

$$\begin{array}{rlrl}{g}_c(E)f(E)& =C\sqrt{x}{e}^{-(E_c-E_F)/kT}{e}^{-x/kT}\propto \sqrt{x}{e}^{-x/kT}& & (C\text{ is constant})\\ \frac{d}{dx}\left(\sqrt{x}{e}^{-x/kT}\right)& =e^{-x/kT}\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{kT}\right)=0& & \text{(peak condition: first derivative zero)}\\ \frac{1}{2\sqrt{x}}& =\frac{\sqrt{x}}{kT}& & \text{(set expression in brackets to zero)}\\ x& =\frac{kT}{2}& & \text{(solve for }x)\\ E_{\text{peak}}& =E_c+\frac{kT}{2}& & \text{(electron distribution peak position)}\end{array}$$ $$\boxed{E_{\text{peak}}=E_c+\frac{kT}{2}}$$

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 $g_v(E)\propto \sqrt{y}$, so $g_v(1-f)\propto \sqrt{y}{e}^{-y/kT}$, whose peak occurs at $y=kT/2$ by the identical differentiation:

$$\begin{array}{rlrl}{E}_{\text{peak}}& =E_v-\frac{kT}{2}& & \text{(hole distribution peak position)}\end{array}$$ $$\boxed{E_{\text{peak}}=E_v-\frac{kT}{2}}$$

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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).

$$\begin{array}{rlrl}\frac{h(5kT)}{h(kT/2)}& =\frac{\sqrt{5kT}{e}^{-5}}{\sqrt{kT/2}{e}^{-1/2}}& & \text{(ratio of distribution values)}\\ & =\sqrt{\frac{5}{0.5}}{e}^{-5+0.5}=\sqrt{10}{e}^{-4.5}& & (\sqrt{5/0.5}=\sqrt{10})\\ & =(3.1623)(0.01111)\approx 0.0351& & (e^{-4.5}\approx 0.01111)\\ & =3.51\%& & \text{(expressed as a percentage)}\end{array}$$ $$\boxed{\frac{h(5kT)}{h(kT/2)}=\sqrt{10}{e}^{-4.5}\approx 3.51\%}$$

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.

$$\begin{array}{rlrl}{N}_C& =(2.510\times 10^{19})(1.18{)}^{3/2}=3.22\times 10^{19}{\text{ cm}}^{-3}& & \text{(Table 2.1)}\\ N_V& =(2.510\times 10^{19})(0.81{)}^{3/2}=1.83\times 10^{19}{\text{ cm}}^{-3}& & \\ n& =N_C{e}^{-(E_c-E_F)/kT}=3.22\times 10^{19}{e}^{-10.83}\approx 6.5\times 10^{14}{\text{ cm}}^{-3}& & \\ p& =N_V{e}^{-(E_F-E_v)/kT}=1.83\times 10^{19}{e}^{-32.5}\approx 1.5\times 10^5{\text{ cm}}^{-3}& & \left(\text{check: }\frac{n_i^2}{n}=\frac{10^{20}}{6.5\times 10^{14}}\approx 1.5\times 10^5\right)\end{array}$$ $$\boxed{n\approx 6.5\times 10^{14}{\text{ cm}}^{-3},p\approx 1.5\times 10^5{\text{ cm}}^{-3}}$$

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

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

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

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

which yields the stated result:

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

(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 $-\infty \to {\eta }^{\prime }=\infty$:

$$\begin{array}{rlrl}p& =\frac{\sqrt{2}(m_p^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\frac{\sqrt{{\eta }^{\prime }}d{\eta }^{\prime }}{1+e^{{\eta }^{\prime }-{\eta }_v}}=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)& & \text{(Eq. 2.14b)}\end{array}$$ $$\boxed{p=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v),{\eta }_v=\frac{E_v-E_F}{kT}}$$

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:

$$\begin{array}{rlrl}\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)& \approx \frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}{\int }_0^{\infty }\sqrt{{\eta }^{\prime }}{e}^{-{\eta }^{\prime }}d{\eta }^{\prime }=\frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}\left(\frac{\sqrt{\pi }}{2}\right)=e^{(E_v-E_F)/kT}& & \\ p& =N_V{e}^{(E_v-E_F)/kT}& & \text{(Eq. 2.16b)}\end{array}$$ $$\boxed{p=N_V{e}^{(E_v-E_F)/kT}}$$

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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:

$$\begin{array}{rlrl}n& ={\int }_{E_c}^{\infty }\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}{e}^{-(E-E_c)/kT}dE& & \text{(integrate the distribution over band)}\\ & =\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}[-kT{e}^{-(E-E_c)/kT}{]}_{E_c}^{\infty }& & \text{(evaluate antiderivative)}\\ & =\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}[0-(-kT)]=N_C{e}^{(E_F-E_c)/kT}& & \text{(bracket yields }kT)\end{array}$$ $$\boxed{n=N_C{e}^{(E_F-E_c)/kT}}$$

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.

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5}\text{ eV/K})(1.602\times 10^{-19}\text{ J/eV})(300\text{ K})=4.141\times 10^{-21}\text{ J}& & \text{(thermal energy in Joules)}\\ 2\pi m_{0}kT& =(6.2832)(9.109\times 10^{-31}\text{ kg})(4.141\times 10^{-21}\text{ J})=2.370\times 10^{-50}\text{ kg J}& & \text{(numerator product)}\\ h^2& =(6.626\times 10^{-34}\text{ J}\cdot \text{s}{)}^2=4.390\times 10^{-67}{\text{ J}}^2{\text{s}}^2& & \text{(Planck constant squared)}\\ \frac{2\pi m_{0}kT}{h^2}& =\frac{2.370\times 10^{-50}}{4.390\times 10^{-67}}=5.40\times 10^{16}{\text{ m}}^{-2}& & \text{(ratio)}\\ {\left(5.40\times 10^{16}\right)}^{3/2}& =1.255\times 10^{25}{\text{ m}}^{-3}& & \text{(raise to power }3/2)\\ N& =2\times 1.255\times 10^{25}{\text{ m}}^{-3}=2.510\times 10^{25}{\text{ m}}^{-3}=2.510\times 10^{19}\mathbf{c}{\mathbf{m}}^{-3}& & ({\text{convert to cm}}^{-3})\end{array}$$

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

$$\boxed{N_{C,V}=(2.510\times 10^{19}{\text{ cm}}^{-3}){\left(\frac{m^{\ast }}{m_0}\right)}^{3/2}}$$

(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 $2\times 10^{13}$.

$$\boxed{T\approx 429\text{ K}(\approx 156^{\circ }\text{C})}$$

(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.)

$$\boxed{T\approx 595\text{ K}(\approx 322^{\circ }\text{C})}$$

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

$$\begin{array}{rlrl}\frac{n_{iA}}{n_{iB}}& =\frac{\sqrt{N_C{N}_V}{e}^{-E_{GA}/2kT}}{\sqrt{N_C{N}_V}{e}^{-E_{GB}/2kT}}=e^{(E_{GB}-E_{GA})/2kT}& & \text{(prefactors cancel)}\\ & =e^{(2-1)/(2\times 0.025851)}=e^{1/0.051702}=e^{19.34}\approx 2.5\times 10^8& & \text{(insert numbers)}\end{array}$$ $$\boxed{\frac{n_{iA}}{n_{iB}}\approx 2.5\times 10^8}$$

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:

$$\begin{array}{rlrl}{T}^{3/2}& =300^{1.5}=5196.15& & \text{(temperature power)}\\ \frac{0.392}{kT}& =\frac{0.392}{0.025851}=15.1637& & \text{(exponent argument)}\\ e^{-15.1637}& =2.60\times 10^{-7}& & \text{(exponential)}\\ n_i& =(1.76\times 10^{16})(5196.15)(2.60\times 10^{-7})\approx 2.4\times 10^{13}{\text{cm}}^{-3}& & \text{(product)}\end{array}$$ $$\boxed{n_i(300\text{ K})\approx 2.4\times 10^{13}{\text{cm}}^{-3}}$$

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:

$$\boxed{p=0,n=0}$$

(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).

$$\boxed{\text{Donor impurity}}$$

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

$$\begin{array}{rl}p& =\frac{n_i^2}{n}=\frac{(10^{10}{)}^2}{10^5}=10^{15}{\text{cm}}^{-3}\end{array}$$ $$\boxed{p=10^{15}{\text{cm}}^{-3}}$$

(The sample is strongly p-type.)

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

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

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

$$\begin{array}{rlrl}np& =n_i^2& & \text{(np product)}\\ (2p)(p)& =2p^2=(10^{13}{)}^2=10^{26}& & \text{(insert }n=2p)\\ p& =\sqrt{5\times 10^{25}}\approx 7.07\times 10^{12}{\text{cm}}^{-3}& & \text{(solve for }p)\\ n& =2p\approx 1.41\times 10^{13}{\text{cm}}^{-3}& & \text{(solve for }n)\\ p-n+N_D-N_A& =0⟹N_D=n-p+N_A& & \text{(charge neutrality)}\\ N_D& =1.41\times 10^{13}-7.07\times 10^{12}+0\approx 7.07\times 10^{12}{\text{cm}}^{-3}& & \text{(solve for }{N}_D)\end{array}$$ $$\boxed{N_D\approx 7.07\times 10^{12}{\text{cm}}^{-3}}$$

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Problem 2.17: Carrier concentrations in Si

For all parts: nondegenerate, totally ionized dopants.

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

$$\begin{array}{rl}n& \approx N_D=10^{15}{\text{cm}}^{-3}\\ p& =\frac{n_i^2}{n}=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}\end{array}$$ $$\boxed{n\approx 10^{15}{\text{cm}}^{-3},p=10^5{\text{cm}}^{-3}}$$

(b) $T=300\text{ K}$, $N_A=10^{16}{\text{ cm}}^{-3}$, $N_D=0$:

$$\begin{array}{rl}p& \approx N_A=10^{16}{\text{cm}}^{-3}\\ n& =\frac{10^{20}}{10^{16}}=10^4{\text{cm}}^{-3}\end{array}$$ $$\boxed{p\approx 10^{16}{\text{cm}}^{-3},n=10^4{\text{cm}}^{-3}}$$

(c) $T=300\text{ K}$, $N_A=9\times 10^{15}{\text{ cm}}^{-3}$, $N_D=10^{16}{\text{ cm}}^{-3}$ (compensated). Net doping:

$$\begin{array}{rl}{N}_D-N_A& =10^{16}-9\times 10^{15}=10^{15}{\text{cm}}^{-3}\gg n_i\\ n& \approx N_D-N_A=10^{15}{\text{cm}}^{-3}\\ p& =\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}\end{array}$$ $$\boxed{n\approx 10^{15}{\text{cm}}^{-3},p=10^5{\text{cm}}^{-3}}$$

(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):

$$\begin{array}{rl}n& =\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}=5\times 10^{13}+\sqrt{(5\times 10^{13}{)}^2+(4.7\times 10^{13}{)}^2}\\ & =5\times 10^{13}+\sqrt{2.5\times 10^{27}+2.22\times 10^{27}}\\ & =5\times 10^{13}+6.87\times 10^{13}\approx 1.19\times 10^{14}{\text{cm}}^{-3}\\ p& =\frac{n_i^2}{n}=\frac{2.22\times 10^{27}}{1.19\times 10^{14}}\approx 1.87\times 10^{13}{\text{cm}}^{-3}\approx 1.9\times 10^{13}{\text{cm}}^{-3}\end{array}$$

(Check: $n-p=1.19\times 10^{14}-1.87\times 10^{13}=1.0\times 10^{14}=N_D$.)

$$\boxed{n\approx 1.19\times 10^{14}{\text{cm}}^{-3},p\approx 1.9\times 10^{13}{\text{cm}}^{-3}}$$

(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:

$$\begin{array}{rl}n& =\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}\approx 5\times 10^{13}+1.146\times 10^{16}\approx 1.15\times 10^{16}{\text{cm}}^{-3}\\ p& =n-N_D=1.15\times 10^{16}-10^{14}\approx 1.14\times 10^{16}{\text{cm}}^{-3}\end{array}$$ $$\boxed{n\approx 1.15\times 10^{16}{\text{cm}}^{-3},p\approx 1.14\times 10^{16}{\text{cm}}^{-3}}$$

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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).

Solution (a). 300 K, $N_D=10^{15}{\text{ cm}}^{-3}$:

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

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, $N_A=10^{16}{\text{ cm}}^{-3}$: $E_i$ again $0.0073\text{ eV}$ below midgap;

$$\begin{array}{rl}{E}_i-E_F& =kT\ln \frac{N_A}{n_i}=0.0259\ln (10^6)=0.0259(13.816)=0.358\text{eV}\end{array}$$ $$\boxed{E_i-E_F=0.358\text{ eV}}$$

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 $n=10^{15}{\text{ cm}}^{-3}$: same as (a): $E_F-E_i=0.298\text{ eV}$ (the carrier numbers, not the individual dopings, set $E_F$). $E_i$ is $0.0073\text{ eV}$ below midgap.

$$\boxed{E_F-E_i=0.298\text{ eV}}$$

(d) 450 K, $E_G=1.08\text{ eV}$, $kT=0.038777\text{ eV}$, $n=1.19\times 10^{14}$, $n_i=4.7\times 10^{13}{\text{ cm}}^{-3}$:

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

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, $E_G=1.015\text{ eV}$, $kT=0.056011\text{ eV}$, $n=1.15\times 10^{16}$, $n_i=1.146\times 10^{16}{\text{ cm}}^{-3}$:

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

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).

---

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

$$\begin{array}{rlrl}{n}_{\max }& =N_C{e}^{-3}=(3.22\times 10^{19}{\text{ cm}}^{-3})(0.049787)\approx 1.60\times 10^{18}{\text{cm}}^{-3}& & \text{(Eq. 2.16a)}\\ p_{\max }& =N_V{e}^{-3}=(1.83\times 10^{19}{\text{ cm}}^{-3})(0.049787)\approx 9.11\times 10^{17}{\text{cm}}^{-3}& & \text{(Eq. 2.16b)}\end{array}$$

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

$$\boxed{N_{D,\max }\approx 1.6\times 10^{18}{\text{cm}}^{-3},N_{A,\max }\approx 9.1\times 10^{17}{\text{cm}}^{-3}}$$

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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.

---

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

$$\begin{array}{rl}\frac{g_v}{g_c}& ={\left(\frac{m_p^{\ast }}{m_n^{\ast }}\right)}^{3/2}={\left(\frac{0.52}{0.066}\right)}^{3/2}=(7.8788{)}^{3/2}\approx 22.12\approx 22\end{array}$$ $$\boxed{\frac{g_v}{g_c}\approx 22}$$

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:

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}& & \text{(Eq. 2.36)}\\ & =\frac{3}{4}(0.025851)\ln (7.8788)=0.019388\times 2.0642\approx 0.040\text{eV}& & \end{array}$$ $$\boxed{E_i-E_{\text{mid}}\approx 0.040\text{eV}(0.040\text{ eV above midgap})}$$

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

(d) Maximum nondegenerate doping at 300 K:

$$\begin{array}{rl}{N}_C& =(2.510\times 10^{19})(0.066{)}^{3/2}\approx 4.26\times 10^{17}{\text{cm}}^{-3}\\ N_V& =(2.510\times 10^{19})(0.52{)}^{3/2}\approx 9.41\times 10^{18}{\text{cm}}^{-3}\\ N_{D,\max }& =N_C{e}^{-3}=(4.26\times 10^{17})(0.049787)\approx 2.1\times 10^{16}{\text{cm}}^{-3}\\ N_{A,\max }& =N_V{e}^{-3}=(9.41\times 10^{18})(0.049787)\approx 4.7\times 10^{17}{\text{cm}}^{-3}\end{array}$$ $$\boxed{N_{D,\max }\approx 2.1\times 10^{16}{\text{cm}}^{-3},N_{A,\max }\approx 4.7\times 10^{17}{\text{cm}}^{-3}}$$

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,

$$\begin{array}{rl}{E}_F& \to \frac{E_v+E_A}{2}=E_v+\frac{|E_B|}{2}\approx E_v+0.02\text{eV}(\text{B in Si})\end{array}$$ $$\boxed{E_F\to \frac{E_v+E_A}{2}=E_v+\frac{|E_B|}{2}\approx E_v+0.02\text{ eV}}$$

(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 $v_d$.

$$\text{Distance upstream: }\Delta x=v_{d}t$$

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

$$\text{Volume enclosing carriers: }\Delta V=v_{d}tA$$

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

$$\text{Number of holes: }\Delta N_p=p{v}_{d}tA$$

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

$$\text{Charge crossing in time }t:\Delta Q=qp{v}_{d}tA$$

Multiply by the charge per hole $+q$.

$$\text{Charge per unit time: }I=\frac{\Delta Q}{t}=qp{v}_{d}A$$

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

$$\boxed{I_{P|\text{drift}}=qp{v}_{d}A}\text{\hspace{1em}(3.1)}$$

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 $J=I/A$:

$$\boxed{J_{P|\text{drift}}=qp{v}_d}\text{\hspace{1em}(3.2)}$$

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

$$v_d=\frac{{\mu }_0\mathcal{E}}{{\left[1+{\left(\frac{{\mu }_0\mathcal{E}}{v_{\text{sat}}}\right)}^{\beta }\right]}^{1/\beta }}\text{\hspace{1em}(3.3)}$$

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

$$\begin{array}{rlrl}\mathcal{E}\to 0:& v_d\to {\mu }_0\mathcal{E}& & \text{(linear regime; the proportionality constant is the mobility }\mu )\\ \mathcal{E}\to \infty :& v_d\to \frac{{\mu }_0\mathcal{E}}{{\mu }_0\mathcal{E}/v_{\text{sat}}}=v_{\text{sat}}& & \text{(saturation velocity regime; independent of }\mathcal{E})\end{array}$$

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

$$\boxed{J_{P|\text{drift}}=q{\mu }_{p}p\mathcal{E}}\text{\hspace{1em}(3.4a)}$$

and similarly for electrons

$$\boxed{J_{N|\text{drift}}=q{\mu }_{n}n\mathcal{E}}\text{\hspace{1em}(3.4b)}$$

${\mu }_n$ 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:

$$R_{\text{total}}=R_L+R_I$$

($R_L$ = 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:

\begin{aligned} \mathcal{E} &= \rho\,J \tag{3.5a} \\ J &= \sigma\mathcal{E} = \frac{1}{\rho}\mathcal{E} \tag{3.5b} \end{aligned}

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

$$\begin{array}{rl}{J}_{\text{drift}}& =J_{N|\text{drift}}+J_{P|\text{drift}}=q({\mu }_{n}n+{\mu }_{p}p)\mathcal{E}\end{array}\text{\hspace{1em}(3.6)}$$

Comparing with (3.5b):

$$\boxed{\rho =\frac{1}{q({\mu }_{n}n+{\mu }_{p}p)}}\text{\hspace{1em}(3.7)}$$

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

$$\boxed{\rho \approx \frac{1}{q{\mu }_n{N}_D}\text{(n-type)}}\text{\hspace{1em}(3.8a)}$$ $$\boxed{\rho \approx \frac{1}{q{\mu }_p{N}_A}\text{(p-type)}}\text{\hspace{1em}(3.8b)}$$

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:

$$\rho =2\pi s\frac{V}{I}\Gamma \text{\hspace{1em}(3.9)}$$

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

$$\begin{array}{rl}\text{K.E. of electrons}& =E-E_C\\ \text{K.E. of holes}& =E_V-E\end{array}$$

(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),

\begin{aligned} \text{P.E.} &= -qV \tag{3.10} \\ \text{P.E.} &= E_C - E_{\text{ref}} \tag{3.11} \end{aligned}

Equate them and solve for $V$:

$$\begin{array}{rl}V& =-\frac{1}{q}(E_C-E_{\text{ref}})\end{array}\text{\hspace{1em}(3.12)}$$

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

$$\begin{array}{rl}\mathcal{E}& =-\frac{dV}{dx}\end{array}\text{\hspace{1em}(3.14)}$$

Differentiate (3.12):

$$\begin{array}{rl}\mathcal{E}& =-\frac{dV}{dx}=\frac{1}{q}\frac{d{E}_C}{dx}\end{array}$$

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

$$\boxed{\mathcal{E}=\frac{1}{q}\frac{d{E}_C}{dx}=\frac{1}{q}\frac{d{E}_V}{dx}=\frac{1}{q}\frac{d{E}_i}{dx}}\text{\hspace{1em}(3.15)}$$

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 $E_i-E_F=E_G/4=0.28\text{ eV}$, so the material is p-type with:

$$\begin{array}{rl}{N}_A& \approx p=n_i{e}^{(E_i-E_F)/kT}=10^{10}{e}^{0.28/0.0259}\approx 4.96\times 10^{14}{\text{cm}}^{-3}\end{array}$$

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

$$\begin{array}{rl}\rho & =\frac{1}{q{\mu }_p{N}_A}=\frac{1}{(1.602\times 10^{-19}\text{ C})(459{\text{ cm}}^2/\text{V}\cdot \text{s})(4.96\times 10^{14}{\text{ cm}}^{-3})}\\ & \approx 27.5\Omega \cdot \text{cm}\end{array}$$ $$\boxed{\rho \approx 27.5\Omega \cdot \text{cm}}$$

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

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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:

$$\mathbf{F}=-D\nabla \eta \text{\hspace{1em}(3.16)}$$

where $\mathbf{F}$ is the flux (particles per ${\text{cm}}^2$ 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:

$$\boxed{{\mathbf{J}}_{P|\text{diff}}=-q{D}_P\nabla p}\text{\hspace{1em}(3.17a)}$$ $$\boxed{{\mathbf{J}}_{N|\text{diff}}=+q{D}_N\nabla n}\text{\hspace{1em}(3.17b)}$$

$D_P$ and $D_N$ (units ${\text{cm}}^2/\text{s}$) 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):

$$\boxed{{\mathbf{J}}_P={\mathbf{J}}_{P|\text{drift}}+{\mathbf{J}}_{P|\text{diff}}=q{\mu }_{p}p\mathcal{E}-q{D}_P\nabla p}\text{\hspace{1em}(3.18a)}$$ $$\boxed{{\mathbf{J}}_N={\mathbf{J}}_{N|\text{drift}}+{\mathbf{J}}_{N|\text{diff}}=q{\mu }_{n}n\mathcal{E}+q{D}_N\nabla n}\text{\hspace{1em}(3.18b)}$$ $$\boxed{\mathbf{J}={\mathbf{J}}_N+{\mathbf{J}}_P}\text{\hspace{1em}(3.19)}$$

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:

$$\begin{array}{rl}{J}_N& =J_{N|\text{drift}}+J_{N|\text{diff}}=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}=0\end{array}\text{\hspace{1em}(3.20)}$$

Use the band-bending relation for the field,

$$\begin{array}{rl}\mathcal{E}& =\frac{1}{q}\frac{d{E}_i}{dx}\end{array}\text{\hspace{1em}(3.21)}$$

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

$$\begin{array}{rl}n& =n_i{e}^{(E_F-E_i)/kT}\end{array}\text{\hspace{1em}(3.22)}$$

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

$$\begin{array}{rl}\frac{dn}{dx}& =n_i{e}^{(E_F-E_i)/kT}\cdot \left(-\frac{1}{kT}\frac{d{E}_i}{dx}\right)\\ \frac{dn}{dx}& =-\frac{n}{kT}\frac{d{E}_i}{dx}\end{array}\text{\hspace{1em}(3.23)}$$

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

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

Collect terms:

$$\begin{array}{rl}\left(n\frac{d{E}_i}{dx}\right)\left[{\mu }_n-\frac{q}{kT}{D}_N\right]& =0\end{array}\text{\hspace{1em}(3.24)}$$

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

$$\boxed{\frac{D_N}{{\mu }_n}=\frac{kT}{q}}\text{\hspace{1em}(3.25a)}$$

The same argument for holes gives

$$\boxed{\frac{D_P}{{\mu }_p}=\frac{kT}{q}}\text{\hspace{1em}(3.25b)}$$

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 strictly invariant with position (a horizontal line). By Fact 1 this is the signature of thermodynamic equilibrium:

$$\boxed{\frac{d{E}_F}{dx}=0⟹\text{Thermodynamic equilibrium}}$$

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

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

(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 $N_A\approx 5\times 10^{14}{\text{cm}}^{-3}$ and ${\mu }_p\approx 459{\text{cm}}^2/\text{V}\cdot \text{s}$. By the Einstein relationship:

$$\begin{array}{rl}{D}_P& =\frac{kT}{q}{\mu }_p=(0.0259\text{ V})(459{\text{ cm}}^2/\text{V}\cdot \text{s})\approx 11.9{\text{cm}}^2/\text{s}\end{array}$$ $$\boxed{D_P\approx 11.9{\text{cm}}^2/\text{s}}$$

---

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

$$I=I_0{e}^{-\alpha x}\text{\hspace{1em}(3.26)}$$

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:

$${\frac{\partial n}{\partial t}|}_{\text{light}}={\frac{\partial p}{\partial t}|}_{\text{light}}=G_L(x,\lambda )\text{\hspace{1em}(3.27)}$$ $$G_L(x,\lambda )=G_{L0}{e}^{-\alpha x}\text{\hspace{1em}(3.28)}$$

$G_L$ 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 ${\text{cm}}^3$

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: $N_D=10^{15}{\text{ cm}}^{-3}$ Si at room temperature, perturbation $\Delta p=\Delta n=10^{10}{\text{ cm}}^{-3}$. Then $n_0=N_D=10^{15}{\text{ cm}}^{-3}$ and $p_0\approx n_i^2/N_D=10^5{\text{ cm}}^{-3}$. So $n=n_0+\Delta n\approx n_0$ and $\Delta p=10^{10}{\text{ cm}}^{-3}\ll n_0$: low-level injection. But look at the minority carriers: $p=p_0+\Delta p\approx 10^{10}{\text{ cm}}^{-3}\gg p_0=10^5{\text{ cm}}^{-3}$! 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 $⟹$ 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):

$${\frac{\partial p}{\partial t}|}_R=-c_p{N}_{T}p\text{\hspace{1em}(3.29)}$$

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):

$${\frac{\partial p}{\partial t}|}_G=+c_p{N}_T{p}_0\text{\hspace{1em}(3.30)}$$

Net rate:

$${\frac{\partial p}{\partial t}|}_{R-G}={\frac{\partial p}{\partial t}|}_R+{\frac{\partial p}{\partial t}|}_G=-c_p{N}_T(p-p_0)\text{\hspace{1em}(3.31)}$$

or

$${\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}=-c_p{N}_T\Delta p\text{(holes in n-type)}\text{\hspace{1em}(3.32a)}$$

and analogously

$${\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=-c_n{N}_T\Delta n\text{(electrons in p-type)}\text{\hspace{1em}(3.32b)}$$

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 $1/\text{time}$ (since $\partial p/\partial t$ is $\text{concentration}/\text{time}$ and $\Delta p$ is $\text{concentration}$). Define

$${\tau }_p\equiv \frac{1}{c_p{N}_T}\text{\hspace{1em}(3.33a)}$$ $${\tau }_n\equiv \frac{1}{c_n{N}_T}\text{\hspace{1em}(3.33b)}$$

so

$$\boxed{{\frac{\partial p_n}{\partial t}|}_{\text{thermal R-G}}=-\frac{\Delta p_n}{{\tau }_p}\text{(holes in n-type)}}\text{\hspace{1em}(3.34a)}$$ $$\boxed{{\frac{\partial n_p}{\partial t}|}_{\text{thermal R-G}}=-\frac{\Delta n_p}{{\tau }_n}\text{(electrons in p-type)}}\text{\hspace{1em}(3.34b)}$$

(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):

$${\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}={\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=-\frac{np-n_i^2}{{\tau }_p(n+n_1)+{\tau }_n(p+p_1)}\text{\hspace{1em}(3.35)}$$

with

$$n_1\equiv n_i{e}^{(E_T-E_i)/kT}\text{\hspace{1em}(3.36a)}$$ $$p_1\equiv n_i{e}^{(E_i-E_T)/kT}\text{\hspace{1em}(3.36b)}$$

(Problem 3.15 asks you to show that (3.35) reduces to (3.34); a complete derivation is given there.)

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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 mobilities $\mu$ and diffusion coefficients $D$, there are no handy plots of $\tau$ versus doping, because $\tau$ depends on the poorly controlled R-G center concentration $N_T$ (Eq. 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 ${\tau }_{n(p)}\sim 1\text{ ms}$ in Si.
• Intentional gold doping raises $N_T$ and gives ${\tau }_{n(p)}\sim 1\text{ ns}$.
• 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)

$$\Delta p(t)=\Delta p_0{e}^{-t/{\tau }_p}\text{\hspace{1em}(3.37)}$$

($\Delta p_0$ = excess at $t=0$ when light is switched off). The conductivity is

\begin{aligned} \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} \end{aligned}

With $\Delta n=\Delta p$ and, in n-type material, $n_0\approx N_D\gg p_0$:

$$\sigma =\underset{{\sigma }_{\text{dark}}}{\underbrace{q{\mu }_n{N}_D}}+\underset{\Delta \sigma (t)}{\underbrace{q({\mu }_n+{\mu }_p)\Delta p}}\text{\hspace{1em}(3.39)}$$

where ${\sigma }_{\text{dark}}$ 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:

$$v_L=i{R}_L=V_A\frac{R_L}{R_s+R_L}\text{\hspace{1em}(3.40)}$$

with the sample resistance (length $\ell$, area $A$)

$$R_s=\frac{\rho \ell }{A}=\frac{\ell }{\sigma A}=\frac{\ell }{({\sigma }_{\text{dark}}+\Delta \sigma )A}=\frac{R_{sd}}{1+\Delta \sigma /{\sigma }_{\text{dark}}}\text{\hspace{1em}(3.41)}$$

where $R_{sd}=\ell /({\sigma }_{\text{dark}}A)$ is the dark resistance. Substituting (3.41) into (3.40) and rearranging:

$$v_L=\frac{V_A}{1+\frac{R_{sd}}{R_L}\cdot \frac{1}{1+\Delta \sigma /{\sigma }_{\text{dark}}}}\text{\hspace{1em}(3.42)}$$

If the load is matched, $R_L=R_{sd}$, and the light is weak so $\Delta \sigma /{\sigma }_{\text{dark}}\ll 1$ (equivalent to low-level injection), expand:

$$v_L\approx \frac{V_A}{2}+\frac{V_A}{4}\frac{\Delta \sigma }{{\sigma }_{\text{dark}}}=\underset{\text{d.c.}}{\underbrace{\frac{V_A}{2}}}+\underset{\text{decay transient}}{\underbrace{v_0{e}^{-t/{\tau }_p}}}\text{\hspace{1em}(3.43)}$$

(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 $x=\Delta \sigma /{\sigma }_{\text{dark}}$.) 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 $\sim 1\mu \text{s}$, high-intensity flashes spaced from $2.5\text{ ms}$ to $0.5\text{ s}$ 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) versus 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, ${\tau }_p\approx 150\mu \text{s}$.

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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:

$$\frac{\partial n}{\partial t}={\frac{\partial n}{\partial t}|}_{\text{drift}}+{\frac{\partial n}{\partial t}|}_{\text{diff}}+{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial n}{\partial t}|}_{\text{other (light, etc.)}}\text{\hspace{1em}(3.44a)}$$ $$\frac{\partial p}{\partial t}={\frac{\partial p}{\partial t}|}_{\text{drift}}+{\frac{\partial p}{\partial t}|}_{\text{diff}}+{\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial p}{\partial t}|}_{\text{other (light, etc.)}}\text{\hspace{1em}(3.44b)}$$

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:

$${\frac{\partial n}{\partial t}|}_{\text{drift}}+{\frac{\partial n}{\partial t}|}_{\text{diff}}=\frac{1}{q}\left(\frac{\partial J_{Nx}}{\partial x}+\frac{\partial J_{Ny}}{\partial y}+\frac{\partial J_{Nz}}{\partial z}\right)=\frac{1}{q}\nabla \cdot {\mathbf{J}}_N\text{\hspace{1em}(3.45a)}$$ $${\frac{\partial p}{\partial t}|}_{\text{drift}}+{\frac{\partial p}{\partial t}|}_{\text{diff}}=-\frac{1}{q}\nabla \cdot {\mathbf{J}}_P\text{\hspace{1em}(3.45b)}$$

(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):

$$\boxed{\frac{\partial n}{\partial t}=\frac{1}{q}\nabla \cdot {\mathbf{J}}_N+{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial n}{\partial t}|}_{\text{other processes}}}\text{\hspace{1em}(3.46a)}$$ $$\boxed{\frac{\partial p}{\partial t}=-\frac{1}{q}\nabla \cdot {\mathbf{J}}_P+{\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial p}{\partial t}|}_{\text{other processes}}}\text{\hspace{1em}(3.46b)}$$

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. Eq. 3.34 or the general Eq. 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.

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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

$$\frac{1}{q}\nabla \cdot {\mathbf{J}}_N\to \frac{1}{q}\frac{\partial J_N}{\partial x}\text{\hspace{1em}(3.47)}$$

The electron current is

$$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{\partial n}{\partial x}\to q{D}_N\frac{\partial n}{\partial x}\text{\hspace{1em}(3.48)}$$

(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 $n_0\ne n_0(x)$:

$$\frac{\partial n}{\partial x}=\frac{\partial n_0}{\partial x}+\frac{\partial \Delta n}{\partial x}=\frac{\partial \Delta n}{\partial x}\text{\hspace{1em}(3.49)}$$

Combine (3.47)ΓÇô(3.49):

$$\frac{1}{q}\nabla \cdot {\mathbf{J}}_N\to D_N\frac{{\partial }^2\Delta n}{\partial x^2}\text{\hspace{1em}(3.50)}$$

R-G term. Indirect thermal R-G dominance plus low-level injection and minority carriers allow use of (3.34):

$${\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=-\frac{\Delta n}{{\tau }_n}\text{\hspace{1em}(3.51)}$$

Other processes. Only light is allowed:

$${\frac{\partial n}{\partial t}|}_{\text{other processes}}=G_L\text{\hspace{1em}(3.52)}$$

($G_L=0$ if the sample is dark.)

Left-hand side. The equilibrium concentration is constant in time, $\partial n_0/\partial t=0$, so

$$\frac{\partial n}{\partial t}=\frac{\partial n_0}{\partial t}+\frac{\partial \Delta n}{\partial t}=\frac{\partial \Delta n}{\partial t}\text{\hspace{1em}(3.53)}$$

Putting (3.50)ΓÇô(3.53) into (3.46a) and treating holes analogously:

$$\boxed{\frac{\partial \Delta n_p}{\partial t}=D_N\frac{{\partial }^2\Delta n_p}{\partial x^2}-\frac{\Delta n_p}{{\tau }_n}+G_L}\text{\hspace{1em}(3.54a)}$$ $$\boxed{\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L}\text{\hspace{1em}(3.54b)}$$

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 ($G_L$).

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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.)

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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 $N_D=10^{15}{\text{ cm}}^{-3}$, ${\tau }_p=10^{-6}\text{ s}$, and a light-induced creation of $10^{17}$ electron-hole pairs per ${\text{cm}}^3$ per second throughout the semiconductor. Determine $\Delta p_n(t)$ for $t>0$.

Step 1. Silicon, $300\text{ K}$, uniform $N_D=10^{15}{\text{ cm}}^{-3}$, $G_L=10^{17}{\text{ cm}}^{-3}\cdot {\text{s}}^{-1}$ everywhere. Equilibrium holds for $t<0$.

Step 2 (equilibrium). With $n_i=10^{10}{\text{ cm}}^{-3}$ and $N_D\gg n_i$:

$$\begin{array}{rl}{n}_0& =N_D=10^{15}{\text{ cm}}^{-3}\\ p_0& =\frac{n_i^2}{N_D}=\frac{(10^{10}{)}^2}{10^{15}}=\frac{10^{20}}{10^{15}}=10^5{\text{ cm}}^{-3}\end{array}$$

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:

$$G_L=\frac{\Delta p_n(\infty )}{{\tau }_p}⟹\Delta p_{n,\max }=G_L{\tau }_p$$

(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

$$\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L\text{\hspace{1em}(3.55)}$$

with initial condition

$${\Delta p_n(t)|}_{t=0}=0\text{\hspace{1em}(3.56)}$$

Since $\Delta p_n$ does not depend on $x$, the curvature term vanishes and (3.55) becomes an ordinary differential equation:

$$\frac{d\Delta p_n}{dt}+\frac{\Delta p_n}{{\tau }_p}=G_L\text{\hspace{1em}(3.57)}$$

Solve it. Particular solution (steady value): $\Delta p_n=G_L{\tau }_p$. Homogeneous solution: $A{e}^{-t/{\tau }_p}$. So

$$\Delta p_n(t)=G_L{\tau }_p+A{e}^{-t/{\tau }_p}\text{\hspace{1em}(3.58)}$$

Apply $\Delta p_n(0)=0$:

$$0=G_L{\tau }_p+A⟹A=-G_L{\tau }_p\text{\hspace{1em}(3.59)}$$ $$\boxed{\Delta p_n(t)=G_L{\tau }_p\left(1-e^{-t/{\tau }_p}\right)}\text{\hspace{1em}(3.60)}$$

Step 5 (examine). $G_L{\tau }_p$ has the dimensions $({\text{cm}}^{-3}\cdot {\text{s}}^{-1})(\text{s})={\text{cm}}^{-3}$, 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:

$$\Delta p_{n,\max }=(10^{17}{\text{ cm}}^{-3}\cdot {\text{s}}^{-1})(10^{-6}\text{ s})=10^{11}{\text{ cm}}^{-3}\ll n_0=10^{15}{\text{ cm}}^{-3}\text{(low-level injection confirmed)}$$

so

$$p_{\max }=p_0+10^{11}\approx 10^{11}{\text{ cm}}^{-3}$$

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 $\sim 1\mu \text{s}$ while ${\tau }_p\lesssim 150\mu \text{s}$, so $t_{\text{on}}/{\tau }_p\ll 1$ 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:

$$\Delta p_n(t)=\Delta p_n(0)e^{-t/{\tau }_p}\text{\hspace{1em}(3.61)}$$

(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).

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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 $N_D=10^{15}{\text{ cm}}^{-3}$ is illuminated so as to create $\Delta p_{n0}=10^{10}{\text{ cm}}^{-3}$ excess holes at $x=0$ (Fig. 3.27a). The wavelength is such that no light penetrates into the interior ($x>0$). Determine $\Delta p_n(x)$.

Steps 1 and 2. Silicon, steady state (we are asked for $\Delta p_n(x)$ not $\Delta p_n(x,t)$), $N_D=10^{15}{\text{ cm}}^{-3}$, so $n_0=10^{15}{\text{ cm}}^{-3}$, $p_0=10^5{\text{ cm}}^{-3}$ uniform. Boundary conditions: $\Delta p_n(0)=\Delta p_{n0}=10^{10}{\text{ cm}}^{-3}$ 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 $300\text{ K}$.

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 $x>0$:

$$D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}=0,x>0\text{\hspace{1em}(3.62)}$$

Boundary conditions:

$${\Delta p_n|}_{x=0^{+}}={\Delta p_n|}_{x=0}=\Delta p_{n0}\text{\hspace{1em}(3.63)}$$ $${\Delta p_n|}_{x\to \infty }=0\text{\hspace{1em}(3.64)}$$

This is Solution No. 1 of Table 3.2:

$$\Delta p_n(x)=A{e}^{-x/L_P}+B{e}^{x/L_P}\text{\hspace{1em}(3.65)}$$ $$L_P=\sqrt{D_P{\tau }_p}\text{\hspace{1em}(3.66)}$$

Since $e^{x/L_P}\to \infty$ as $x\to \infty$, (3.64) can be satisfied only if $B=0$. Then (3.63) gives

$$A=\Delta p_{n0}\text{\hspace{1em}(3.67)}$$ $$\boxed{\Delta p_n(x)=\Delta p_{n0}{e}^{-x/L_P}}\text{\hspace{1em}(3.68)}$$

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:

$$\boxed{L_P\equiv \sqrt{D_P{\tau }_p}\text{(holes in n-type)}}\text{\hspace{1em}(3.69a)}$$ $$\boxed{L_N\equiv \sqrt{D_N{\tau }_n}\text{(electrons in p-type)}}\text{\hspace{1em}(3.69b)}$$

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

$$⟨x⟩=\frac{{\int }_0^{\infty }x\Delta p_n(x)dx}{{\int }_0^{\infty }\Delta p_n(x)dx}=\frac{{\int }_0^{\infty }x{e}^{-x/L_P}dx}{{\int }_0^{\infty }{e}^{-x/L_P}dx}=\frac{L_P^2}{L_P}=L_P\text{\hspace{1em}(3.70)}$$

(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 $T=300\text{ K}$, $N_D=10^{15}{\text{ cm}}^{-3}$ Si (${\mu }_p=458{\text{ cm}}^2/\text{V}\cdot \text{s}$) and ${\tau }_p=10^{-6}\text{ s}$:

$$L_P=\sqrt{D_P{\tau }_p}=\sqrt{\frac{kT}{q}{\mu }_p{\tau }_p}={\left[(0.0259\text{ V})(458{\text{ cm}}^2/\text{V}\cdot \text{s})(10^{-6}\text{ s})\right]}^{1/2}=3.44\times 10^{-3}\text{ cm}$$

(about $34\mu \text{m}$). Because lifetimes vary over orders of magnitude, diffusion lengths do too.

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3.5.2 Quasi-Fermi Levels

Why we need them. In equilibrium the Fermi level $E_F$ nails down both carrier concentrations:

$$n_0=n_i{e}^{(E_F-E_i)/kT}\text{\hspace{1em}(3.71a)}$$ $$p_0=n_i{e}^{(E_i-E_F)/kT}\text{\hspace{1em}(3.71b)}$$

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: $n_0=N_D=10^{15}{\text{ cm}}^{-3}$, $p_0=10^5{\text{ cm}}^{-3}$ (equilibrium diagram, Fig. 3.28a). After the steady state is reached: $\Delta n=\Delta p=G_L{\tau }_p=10^{11}{\text{ cm}}^{-3}$, $p=p_0+\Delta p\approx 10^{11}{\text{ cm}}^{-3}$ 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):

$$n\equiv n_i{e}^{(F_N-E_i)/kT}\text{or}{F}_N=E_i+kT\ln \left(\frac{n}{n_i}\right)\text{\hspace{1em}(3.72a)}$$ $$p\equiv n_i{e}^{(E_i-F_P)/kT}\text{or}{F}_P=E_i-kT\ln \left(\frac{p}{n_i}\right)\text{\hspace{1em}(3.72b)}$$

$F_N$ 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 $p=10^{11}{\text{ cm}}^{-3}$, $n_i=10^{10}{\text{ cm}}^{-3}$, $kT=0.0259\text{ eV}$:

$$F_P=E_i-kT\ln \frac{p}{n_i}=E_i-0.0259\ln (10)=E_i-0.060\text{ eV}$$

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 $0.30\text{ eV}$ below $E_F$… i.e., $E_F-E_i=0.298\text{ eV}$, so $E_F$ is $0.30\text{ eV}$ above $E_i$ and $F_P$ is $0.06\text{ eV}$ below $E_i$) up toward the mid-gap.

Compact current expressions. Start from the hole current:

$$J_P=q{\mu }_{p}p\mathcal{E}-q{D}_P\nabla p\text{\hspace{1em}(3.73)}$$

(same as 3.18a). Differentiate (3.72b):

\begin{aligned} \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} \end{aligned}

where we used $\mathcal{E}=\nabla E_i/q$ (3-D version of Eq. 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 $\nabla F_P$:

$$J_P=qp\left({\mu }_p-\frac{q{D}_P}{kT}\right)\mathcal{E}+{\mu }_p\left(\frac{q{D}_P}{kT{\mu }_p}\right)p\nabla F_P\text{\hspace{1em}(3.75)}$$

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:

$$\boxed{{\mathbf{J}}_P={\mu }_{p}p\nabla F_P}\text{\hspace{1em}(3.76a)}$$

and similarly

$$\boxed{{\mathbf{J}}_N={\mu }_{n}n\nabla F_N}\text{\hspace{1em}(3.76b)}$$

(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.

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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? ($N_D=10^{15}{\text{ cm}}^{-3}$, $\Delta p_{n0}=10^{10}{\text{ cm}}^{-3}$, Si, $300\text{ K}$.)

Solution.

(a) Since $n\approx n_0$, from (3.72a):

$$\boxed{F_N\approx E_F}$$

For holes, substitute $p(x)=p_0+\Delta p_{n0}{e}^{-x/L_P}$ into (3.72b):

$$\begin{array}{rl}{F}_P& =E_i-kT\ln \frac{p}{n_i}\\ & =\boxed{E_i-kT\ln \left[\frac{p_0}{n_i}+\frac{\Delta p_{n0}}{n_i}{e}^{-x/L_P}\right]}\end{array}$$

(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:

$$\begin{array}{rl}{F}_P& =E_i-kT\ln \frac{\Delta p_{n0}}{n_i}-kT\ln \left(e^{-x/L_P}\right)\\ & =\boxed{E_i-kT\ln \frac{\Delta p_{n0}}{n_i}+kT\frac{x}{L_P}}\end{array}$$

so $F_P$ is a linear function of $x$ with slope $+kT/L_P$.

(c) Numbers: $\Delta p_{n0}=10^{10}{\text{ cm}}^{-3}$, $n_i=10^{10}{\text{ cm}}^{-3}$, $p_0=n_i^2/N_D=10^5{\text{ cm}}^{-3}$.

• (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 $\ln (\Delta p_{n0}/n_i)=\ln (1)=0$ and $F_P(0)=E_i$.
• (iii) For large $x$, $F_P\to E_F$ (back to equilibrium).
• (iv) $F_N-E_i\approx E_F-E_i=kT\ln (N_D/n_i)=0.0259\ln (10^5)=0.0259(11.51)\approx 0.30\text{ eV}$.

Sketch: $E_C,E_V,E_i$ are flat lines across the bar (since $\mathcal{E}=0$). $F_N$ is a horizontal line $0.30\text{ eV}$ 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: $J_{N|\text{drift}}=q{\mu }_{n}n\mathcal{E}$, $J_{P|\text{drift}}=q{\mu }_{p}p\mathcal{E}$ (Eqs. 3.4).
• Diffusion: migration from regions of high concentration to regions of low concentration due to random thermal motion. Current: Eqs. 3.17ΓÇô3.19.
• 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

$${\mathbf{J}}_N={\mathbf{J}}_{N|\text{drift}}+{\mathbf{J}}_{N|\text{diff}}=q{\mu }_{n}n\mathcal{E}+q{D}_N\nabla n,{\frac{\partial n}{\partial t}|}_{\text{i-thermal R-G}}=-\frac{\Delta n}{{\tau }_n}$$ $${\mathbf{J}}_P={\mathbf{J}}_{P|\text{drift}}+{\mathbf{J}}_{P|\text{diff}}=q{\mu }_{p}p\mathcal{E}-q{D}_P\nabla p,{\frac{\partial p}{\partial t}|}_{\text{i-thermal R-G}}=-\frac{\Delta p}{{\tau }_p}$$ $$\mathbf{J}={\mathbf{J}}_N+{\mathbf{J}}_P$$

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

$$\rho =\frac{1}{q({\mu }_{n}n+{\mu }_{p}p)},\rho \approx \frac{1}{q{\mu }_n{N}_D}\text{ (n-type)},\rho \approx \frac{1}{q{\mu }_p{N}_A}\text{ (p-type)}$$ $$\mathcal{E}=\frac{1}{q}\frac{d{E}_C}{dx}=\frac{1}{q}\frac{d{E}_V}{dx}=\frac{1}{q}\frac{d{E}_i}{dx},V=-\frac{1}{q}(E_C-E_{\text{ref}})$$

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 1ΓÇô3 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 (${\mathrm{O}}_2$ or ${\mathrm{H}}_2\mathrm{O}$) diffuses through the oxide already present and reacts at the Si-$\mathrm{Si}{\mathrm{O}}_2$ 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

$$\mathrm{Si}+{\mathrm{O}}_2\to \mathrm{Si}{\mathrm{O}}_2\text{(4.1a, dry oxidation)}$$ $$\mathrm{Si}+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Si}{\mathrm{O}}_2+2{\mathrm{H}}_2\text{(4.1b, wet oxidation)}$$

Property	Dry Oxidation	Wet Oxidation
Oxidant	${\mathrm{O}}_2$ gas	Water vapour
Growth rate	Slower	Faster
Interface quality	Superior $\mathrm{Si}$-$\mathrm{Si}{\mathrm{O}}_2$ 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 ($\mathrm{SiC}$ 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 ${\mathrm{N}}_2$) 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, $800$ΓÇô$1200{}^{\circ }\mathrm{C}$, and afterwards the wafer wears a layer of $\approx 0.01$ΓÇô$1.0\mu \mathrm{m}$ 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 $100\text{ hr}$) 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 $10^{21}{\text{ cm}}^{-3}$), 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 ($\approx 900$ΓÇô$1200{}^{\circ }\mathrm{C}$). 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 $\frac{d}{d\eta }\mathrm{erfc}\eta =-\frac{2}{\sqrt{\pi }}{e}^{-{\eta }^2}$.

$$\begin{array}{rl}\frac{\partial \eta }{\partial x}& =\frac{1}{2\sqrt{Dt}}\\ \frac{\partial \eta }{\partial t}& =-\frac{\eta }{2t}\\ \frac{\partial N}{\partial t}& =N_0\left(-\frac{2}{\sqrt{\pi }}{e}^{-{\eta }^2}\right)\left(-\frac{\eta }{2t}\right)=\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }t}\\ \frac{\partial N}{\partial x}& =N_0\left(-\frac{2}{\sqrt{\pi }}{e}^{-{\eta }^2}\right)\left(\frac{1}{2\sqrt{Dt}}\right)=-\frac{N_0}{\sqrt{\pi Dt}}{e}^{-{\eta }^2}\\ \frac{{\partial }^{2}N}{\partial x^2}& =-\frac{N_0}{\sqrt{\pi Dt}}\frac{\partial }{\partial x}\left(e^{-{\eta }^2}\right)=-\frac{N_0}{\sqrt{\pi Dt}}(-2\eta e^{-{\eta }^2})\left(\frac{1}{2\sqrt{Dt}}\right)=\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }Dt}\\ D\frac{{\partial }^{2}N}{\partial x^2}& =\frac{N_0\eta e^{-{\eta }^2}}{\sqrt{\pi }t}=\frac{\partial N}{\partial t}✓\end{array}$$

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.

$$\begin{array}{rl}Q& ={\int }_0^{\infty }{N}_{1}dx=N_{0}2\sqrt{D_1{t}_1}{\int }_0^{\infty }\mathrm{erfc}(\eta )d\eta \\ & =N_{0}2\sqrt{D_1{t}_1}\left(\frac{1}{\sqrt{\pi }}\right)\\ & =\frac{2N_0\sqrt{D_1{t}_1}}{\sqrt{\pi }}\end{array}$$

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 $10\text{ min}$, then drive-in at $1200{}^{\circ }\mathrm{C}$ for $30\text{ min}$, with $N_0\sim 10^{21}{\text{ cm}}^{-3}$ and $D\approx 2.58\times 10^{-14}{\text{ cm}}^2/\text{s}$ for the predeposition (the remaining parameters are cut off in the OCR). The figure is a semilog plot of phosphorus concentration ($10^{15}$ to $10^{21}{\text{ cm}}^{-3}$ 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: $D_1{t}_1=(2.58\times 10^{-14}{\text{ cm}}^2/\text{s})(600\text{ s})=1.55\times 10^{-11}{\text{ cm}}^2$, so $2\sqrt{D_1{t}_1}=2\sqrt{1.55\times 10^{-11}}=7.9\times 10^{-6}\text{ cm}=0.079\mu \mathrm{m}.$ Dose (with $N_0=10^{21}{\text{ cm}}^{-3}$): $Q=\frac{2N_0\sqrt{D_1{t}_1}}{\sqrt{\pi }}=2\times 10^{21}\times \frac{3.94\times 10^{-6}}{1.772}\approx 4.4\times 10^{15}{\text{ cm}}^{-2}.$ 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 $5\text{ keV}$ to $1\text{ MeV}$, 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 $\Phi$ (${\text{ions/cm}}^2$). 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:

$$N(x)=\frac{\Phi }{\sqrt{2\pi }\Delta R_p}\exp \left\{-\frac{1}{2}{\left[\frac{x-R_p}{\Delta R_p}\right]}^2\right\}\text{\hspace{1em}(4.3)}$$

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 $10^{14}{\text{ cm}}^{-2}$ for energies $25,50,75,100\text{ keV}$: 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 $10^{20}{\text{ cm}}^{-3}$; higher energy means deeper $R_p$, broader curve and lower peak (same area).

Worked illustration (mine; approximate range statistics). A $50\text{ keV}$ phosphorus implant into Si has $R_p\approx 0.06\mu \mathrm{m}$ and $\Delta R_p\approx 0.024\mu \mathrm{m}$. For $\Phi =10^{14}{\text{ cm}}^{-2}$: $N_{\max }=\frac{10^{14}}{\sqrt{2\pi }(2.4\times 10^{-6}\text{ cm})}=\frac{10^{14}}{6.02\times 10^{-6}}\approx 1.7\times 10^{19}{\text{ cm}}^{-3}.$

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 $80$ΓÇô$100{}^{\circ }\mathrm{C}$ 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:

Property	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, $30\text{ min}$ at $120$ΓÇô$180{}^{\circ }\mathrm{C}$: 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.

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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 aΓÇôl), with a one-line meaning.

1. Starting silicon and well definition (4.17a). $150\text{ mm}$ (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). $5000$ΓÇô$10,000\text{ A˚}$ 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 $1.5\text{ cm}$ square containing $3.1\text{ million}$ 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	$N_1=N_0\mathrm{erfc}\left(x/2\sqrt{D_1{t}_1}\right)$
Drive-in profile	Gaussian $\propto \exp (-x^2/4D_2{t}_2)$, with $D_2{t}_2\gg D_1{t}_1$
Implant profile	$N=\frac{\Phi }{\sqrt{2\pi }\Delta R_p}\exp \left\{-\frac{1}{2}{\left[\frac{x-R_p}{\Delta R_p}\right]}^2\right\}$

End-of-chapter problems / computer exercises: none (read-only chapter).

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Part I Supplement and Review

Part I covered Chapters 1ΓÇô3 (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	1ΓÇô6
Neamen	A	Undergraduate	1ΓÇô6
Pierret	S	Introductory graduate	1ΓÇô6
Streetman	A	Undergraduate	1ΓÇô4
Tyagi	A/S	Advanced undergrad / intro grad	1ΓÇô5
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.)

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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.

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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	$n_i\gg \text{doping}$, 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 $\frac{1}{2}{m}^{\ast }{v}_{\text{th}}^2\sim \frac{3}{2}kT$
36	drift velocity	average velocity from an applied field
37	saturation velocity	limiting drift velocity at high field ($\sim 10^7\text{ cm/s}$ 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 ${\text{cm}}^2$, $\Phi$
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

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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 $T=300\text{ K}$, $n_i=10^{10}{\text{ cm}}^{-3}$, $kT=0.0259\text{ eV}$.

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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) Determine the Miller indices of the plane through A, B, C:

1. Intercepts with coordinate axes in units of $a$: $$(x,y,z)=(1,1,1)$$
2. Reciprocals of intercepts: $$\left(\frac{1}{1},\frac{1}{1},\frac{1}{1}\right)=(1,1,1)$$
3. No fractions to clear, so the plane is: $$\boxed{(111)}$$

(b) Determine the direction vector from $(0,0,0)$ to D at $(a,a/2,a/2)$:

1. Components of the vector: $$\Delta x=a,\Delta y=\frac{a}{2},\Delta z=\frac{a}{2}$$
2. Divide by $a/2$ to clear fractions and find smallest integers: $$\left(\frac{a}{a/2},\frac{a/2}{a/2},\frac{a/2}{a/2}\right)=(2,1,1)$$
3. Enclose in square brackets: $$\boxed{[211]}$$

(c) In the diamond lattice, each atom is tetrahedrally bonded to four nearest neighbours located at one-quarter of the body diagonal:

1. The cube body diagonal has length: $$D_{\text{body}}=\sqrt{a^2+a^2+a^2}=\sqrt{3}a$$
2. The nearest-neighbour distance is one-fourth of the body diagonal: $$\boxed{d_{nn}=\frac{\sqrt{3}}{4}a}$$

(d) A donor is a group-V atom (e.g., P, As, Sb) having five valence electrons that substitutes for a host Si atom on a lattice site. Four of its valence electrons participate in the covalent $s{p}^3$ bonds with neighboring Si atoms; the fifth electron is weakly bound electrostatically to the positive donor core in a hydrogen-like orbit and is easily ionized into the conduction band at room temperature.

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Problem A2: n-Type Wafer Table

Problem. A Si wafer is uniformly doped $N_D=10^{17}{\text{ cm}}^{-3}$ under equilibrium at $300\text{ K}$, with ${\tau }_p=10^{-6}\text{ s}$. 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 / Formulation
$E_G$	$1.12\text{ eV}$	Silicon band gap at $300\text{ K}$
$n_i$	$10^{10}{\text{ cm}}^{-3}$	Standard value at $300\text{ K}$
$n_0$	$10^{17}{\text{ cm}}^{-3}$	$n_0=N_D$ since $N_D\gg n_i$ and complete ionization
$p_0$	$10^3{\text{ cm}}^{-3}$	$p_0=n_i^2/n_0=10^{20}/10^{17}$
$E_F-E_i$	$0.417\text{ eV}$	$E_F-E_i=kT\ln (n_0/n_i)=0.0259\ln (10^7)$
Charge density	$0{\text{ C/cm}}^3$	Exact charge neutrality in uniform equilibrium semiconductor
${\mu }_n$	$801{\text{ cm}}^2/\text{V}\cdot \text{s}$	Empirical mobility curve (Fig. 3.5) at $N_D=10^{17}{\text{ cm}}^{-3}$
$\rho$	$0.078\Omega \cdot \text{cm}$	$\rho \approx 1/(q{\mu }_n{N}_D)$
$D_p$	$8.57{\text{ cm}}^2/\text{s}$	Einstein relation $D_p=(kT/q){\mu }_p$ with ${\mu }_p\approx 331{\text{ cm}}^2/\text{V}\cdot \text{s}$
$L_p$	$2.93\times 10^{-3}\text{ cm}$	$L_p=\sqrt{D_p{\tau }_p}$

Step-by-step quantitative evaluations:

1. Equilibrium minority hole concentration: $$\begin{array}{rl}{p}_0& =\frac{n_i^2}{n_0}=\frac{(10^{10}{\text{ cm}}^{-3}{)}^2}{10^{17}{\text{ cm}}^{-3}}=\frac{10^{20}}{10^{17}}\\ & =\boxed{10^3{\text{ cm}}^{-3}}\end{array}$$
2. Fermi level position relative to intrinsic Fermi level: $$\begin{array}{rl}{E}_F-E_i& =kT\ln \left(\frac{n_0}{n_i}\right)=(0.0259\text{ eV})\ln \left(\frac{10^{17}}{10^{10}}\right)\\ & =(0.0259\text{ eV})\ln (10^7)=(0.0259)(16.118)\\ & =\boxed{0.417\text{ eV}}\end{array}$$
3. Resistivity: $$\begin{array}{rl}\rho & =\frac{1}{q{\mu }_n{N}_D}=\frac{1}{(1.602\times 10^{-19}\text{ C})(801{\text{ cm}}^2/\text{V}\cdot \text{s})(10^{17}{\text{ cm}}^{-3})}\\ & =\frac{1}{12.83{\Omega }^{-1}\cdot {\text{cm}}^{-1}}\\ & =\boxed{0.078\Omega \cdot \text{cm}}\end{array}$$
4. Minority hole diffusion coefficient (using ${\mu }_p\approx 331{\text{ cm}}^2/\text{V}\cdot \text{s}$ from Fig. 3.5 at $N_D=10^{17}{\text{ cm}}^{-3}$): $$\begin{array}{rl}{D}_p& =\frac{kT}{q}{\mu }_p=(0.0259\text{ V})(331{\text{ cm}}^2/\text{V}\cdot \text{s})\\ & =\boxed{8.57{\text{ cm}}^2/\text{s}}\end{array}$$
5. Minority hole diffusion length: $$\begin{array}{rl}{L}_p& =\sqrt{D_p{\tau }_p}=\sqrt{(8.57{\text{ cm}}^2/\text{s})(10^{-6}\text{ s})}=\sqrt{8.57\times 10^{-6}{\text{ cm}}^2}\\ & =\boxed{2.93\times 10^{-3}\text{ cm}}(29.3\mu \text{m})\end{array}$$

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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 $300\text{ K}$ 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 $N_A=10^{16}{\text{ cm}}^{-3}$ on the left and $N_D=10^{16}{\text{ cm}}^{-3}$ 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) Relative energy band positions:

• $E_i$ (intrinsic level): located very near the middle of the band gap ($E_G/2$).
• $E_D$ (donor level): a discrete level located a few hundredths of an $\text{eV}$ below $E_C$.
• $E_A$ (acceptor level): a discrete level located a few hundredths of an $\text{eV}$ above $E_V$.
• $E_T$ (R-G trap center): a localized deep level typically residing near midgap ($E_T\approx E_i$).
• $E_F$ for degenerate $p$-type material: $E_F$ lies within $3kT$ of the valence band edge, or penetrates into the valence band ($E_F\le E_V+3kT$).

(b) Band diagram construction:

1. In equilibrium, the Fermi energy $E_F$ is strictly invariant with position: $$\frac{d{E}_F}{dx}=0$$
2. The relation between band slope and electric field is: $$\mathcal{E}=\frac{1}{q}\frac{d{E}_C}{dx}=\frac{1}{q}\frac{d{E}_i}{dx}=\frac{1}{q}\frac{d{E}_V}{dx}$$
   • In region $0\le x\le x_a$: $\mathcal{E}=0⟹$ all band edges are horizontal.
   • In region $x_a\le x\le x_b$: $\mathcal{E}=-{\mathcal{E}}_0<0⟹$ the band edges slope downward with a constant negative slope: $$\frac{d{E}_i}{dx}=-q{\mathcal{E}}_0<0$$
   • In region $x_b\le x\le x_c$: $\mathcal{E}=0⟹$ all band edges are horizontal.
3. Quantify band positions relative to $E_F$:
   • On the neutral p-side ($0\le x\le x_a$): $$\begin{array}{rl}{E}_i-E_F& =kT\ln \left(\frac{N_A}{n_i}\right)=(0.0259\text{ eV})\ln \left(\frac{10^{16}}{10^{10}}\right)\\ & =(0.0259)(13.816)=\boxed{0.358\text{ eV}}\end{array}$$ Thus $E_i$ is $0.358\text{ eV}$ above $E_F$.
   • On the neutral n-side ($x_b\le x\le x_c$): $$\begin{array}{rl}{E}_F-E_i& =kT\ln \left(\frac{N_D}{n_i}\right)=(0.0259\text{ eV})\ln \left(\frac{10^{16}}{10^{10}}\right)\\ & =\boxed{0.358\text{ eV}}\end{array}$$ Thus $E_i$ is $0.358\text{ eV}$ below $E_F$.
   • $E_C$ and $E_V$ remain parallel to $E_i$ at separations of $+E_G/2=+0.56\text{ eV}$ and $-E_G/2=-0.56\text{ eV}$, respectively.

(c) Find the potential difference $V(x_c)-V(0)$:

1. Relate electrostatic potential to intrinsic level: $$V(x)=-\frac{E_i(x)}{q}+\text{constant}$$
2. Compute the total potential drop: $$\begin{array}{rl}V(x_c)-V(0)& =-\frac{1}{q}\left[E_i(x_c)-E_i(0)\right]\\ & =-\frac{1}{q}\left[(E_F-0.358\text{ eV})-(E_F+0.358\text{ eV})\right]\\ & =-\frac{1}{q}\left[-0.716\text{ eV}\right]\\ & =\boxed{+0.716\text{ V}}\end{array}$$

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Problem A4: Two Light Flashes

Problem. An $N_D=10^{15}{\text{ cm}}^{-3}$ Si sample at room temperature receives two flashes of light. Each creates $\Delta p_f=10^{10}{\text{ cm}}^{-3}$ 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 ${\mu }_n=1350{\text{ cm}}^2/\text{V}\cdot \text{s}$, ${\mu }_p=458{\text{ cm}}^2/\text{V}\cdot \text{s}$ at this doping; the numbers are recovered from the printed answer).

Solution.

(a) Sketch of $\Delta p(t)$:

• At $t=0$, $\Delta p$ instantaneously jumps from $0$ to $\Delta p_f=10^{10}{\text{ cm}}^{-3}$.
• For $0<t<t_f$, $\Delta p(t)$ decays exponentially as $\Delta p_f{e}^{-t/{\tau }_p}$, reaching $\Delta p_f{e}^{-t_f/{\tau }_p}$ just before $t=t_f$.
• At $t=t_f$, the second impulse adds another $\Delta p_f$, causing an instantaneous step to $\Delta p_f(1+e^{-t_f/{\tau }_p})$.
• For $t>t_f$, $\Delta p(t)$ decays exponentially toward zero as $\Delta p_f(1+e^{-t_f/{\tau }_p})e^{-(t-t_f)/{\tau }_p}$.

(b) Verify low-level injection:

1. Maximum possible excess carrier concentration occurs at $t=t_f^{+}$: $$\Delta p_{\max }=\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)<2\Delta p_f=2\times 10^{10}{\text{ cm}}^{-3}$$
2. Compare with majority carrier concentration $n_0=N_D=10^{15}{\text{ cm}}^{-3}$: $$\frac{\Delta p_{\max }}{n_0}<\frac{2\times 10^{10}{\text{ cm}}^{-3}}{10^{15}{\text{ cm}}^{-3}}=2\times 10^{-5}\ll 1$$ $$\boxed{\text{Low-level injection is strictly valid at all times.}}$$

(c) Solve the minority-carrier diffusion equation for $\Delta p(t)$: For a uniformly illuminated sample without gradients (${\nabla }^2\Delta p=0$) and zero electric field ($\mathcal{E}=0$), the equation between flashes ($G_L=0$) is:

$$\frac{d\Delta p}{dt}+\frac{\Delta p}{{\tau }_p}=0$$

1. Interval $0\le t<t_f$: $$\begin{array}{rl}\Delta p(t)& =A_1{e}^{-t/{\tau }_p}\\ \Delta p(0^{+})& =\Delta p_f⟹A_1=\Delta p_f\\ \Delta p(t)& =\boxed{\Delta p_f{e}^{-t/{\tau }_p}}\end{array}$$
2. Interval $t\ge t_f$: $$\begin{array}{rl}\Delta p(t_f^{-})& =\Delta p_f{e}^{-t_f/{\tau }_p}\\ \Delta p(t_f^{+})& =\Delta p(t_f^{-})+\Delta p_f=\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)\\ \Delta p(t)& =\Delta p(t_f^{+})e^{-(t-t_f)/{\tau }_p}\\ & =\boxed{\Delta p_f\left(1+e^{-t_f/{\tau }_p}\right)e^{-(t-t_f)/{\tau }_p}}\end{array}$$

(d) Compute the fractional conductivity change right after the first flash:

1. Dark equilibrium conductivity ($t=0^{-}$): $$\sigma (0^{-})=q({\mu }_n{n}_0+{\mu }_p{p}_0)\approx q{\mu }_n{N}_D(\text{since }{N}_D\gg p_0)$$
2. Conductivity immediately after the flash ($t=0^{+}$): $$\begin{array}{rl}\sigma (0^{+})& =q\left[{\mu }_n(n_0+\Delta n)+{\mu }_p(p_0+\Delta p)\right]\\ & =q{\mu }_n{N}_D+q({\mu }_n+{\mu }_p)\Delta p_f\end{array}$$
3. Conductivity perturbation: $$\Delta \sigma =\sigma (0^{+})-\sigma (0^{-})=q({\mu }_n+{\mu }_p)\Delta p_f$$
4. Fractional change symbolically: $$\begin{array}{rl}\frac{\Delta \sigma }{\sigma (0^{-})}& =\frac{q({\mu }_n+{\mu }_p)\Delta p_f}{q{\mu }_n{N}_D}\\ & =\boxed{\left(1+\frac{{\mu }_p}{{\mu }_n}\right)\frac{\Delta p_f}{N_D}}\end{array}$$
5. Numerical evaluation: $$\begin{array}{rl}\frac{\Delta \sigma }{\sigma (0^{-})}& =\left(1+\frac{458}{1350}\right)\frac{10^{10}{\text{ cm}}^{-3}}{10^{15}{\text{ cm}}^{-3}}\\ & =(1+0.33926)(10^{-5})\\ & =\boxed{1.34\times 10^{-5}}\end{array}$$

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Problem Set B (Closed Book)

Use $n_i=10^{10}{\text{ cm}}^{-3}$, $kT=0.0259\text{ eV}$.

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.

Step-by-step counting:

1. There are 4 vertical edges, each shared by 4 adjacent unit cells: $$N_{\text{edges}}=4\times \frac{1}{4}=1$$
2. There are 2 face-centered positions (top and bottom), each shared by 2 unit cells: $$N_{\text{faces}}=2\times \frac{1}{2}=1$$
3. Total atoms per cell: $$N=1+1=2$$

$$\boxed{\text{Answer: (b)}}$$

(2) Lattice name? Options: simple cubic, bcc, fcc, diamond.

Shift the origin by $(a/2,a/2,0)$ to map atoms: every atom is at a cube corner or the exact body center of displaced cubes.

$$\boxed{\text{Answer: (b) bcc}}$$

(3) Miller indices of the plane with intercepts $x=a$, $y=-2a$, $z=3a$.

1. Intercepts in units of lattice parameter $a$: $$(1,-2,3)$$
2. Reciprocals of intercepts: $$\left(1,-\frac{1}{2},\frac{1}{3}\right)$$
3. Multiply by lowest common denominator ($6$): $$6\times \left(1,-\frac{1}{2},\frac{1}{3}\right)=(6,-3,2)$$
4. Represent negative index with an overbar: $$(6\bar{3}2)$$

$$\boxed{\text{Answer: (e)}}$$

(4) Which direction is not perpendicular to $[100]$? Options: $[0\bar{1}1]$, $[032]$, $[001]$, $[\bar{1}00]$.

Two directions $[u_1{v}_1{w}_1]$ and $[u_2{v}_2{w}_2]$ are perpendicular if their dot product vanishes:

$$[100]\cdot [uvw]=1\cdot u+0\cdot v+0\cdot w=u$$

• For $[0\bar{1}1]$: $u=0⟹$ perpendicular.
• For $[032]$: $u=0⟹$ perpendicular.
• For $[001]$: $u=0⟹$ perpendicular.
• For $[\bar{1}00]$: $u=-1\ne 0⟹$ parallel (antiparallel), not perpendicular.

$$\boxed{\text{Answer: (d)}}$$

(5) The occupation probability at $E_C+kT$ is $e^{-10}$. Find $E_F$.

1. Write the Fermi-Dirac distribution function: $$f(E)=\frac{1}{1+\exp \left(\frac{E-E_F}{kT}\right)}=e^{-10}$$
2. Invert: $$1+\exp \left(\frac{E_C+kT-E_F}{kT}\right)=e^{10}$$
3. Since $e^{10}\approx 22026\gg 1$: $$\exp \left(\frac{E_C+kT-E_F}{kT}\right)\approx e^{10}$$
4. Take natural logarithms: $$\frac{E_C+kT-E_F}{kT}\approx 10⟹E_C+kT-E_F\approx 10kT$$ $$\boxed{E_C-E_F\approx 9kT}$$

$$\boxed{\text{Answer: (b)}}$$

(6) Dominant scattering in lightly doped Si at room temperature: Lattice vibrations (phonons) dominate over ionized impurity scattering when doping is low and temperature is moderate.

$$\boxed{\text{Answer: (b) lattice scattering}}$$

(7) $D_N$ versus $N_D$ in Si at $300\text{ K}$. For a nondegenerate semiconductor, the Einstein relation gives $D_N=(kT/q){\mu }_n$. Thus, $D_N(N_D)$ mimics the mobility curve ${\mu }_n(N_D)$: flat at low doping where lattice scattering dominates, then monotonically falling as ionized impurity scattering increases.

$$\boxed{\text{Answer: sketch (c)}}$$

(8) Donor in the bonding model: A group-V impurity atom (with 5 valence electrons) substitutes for a tetravalently bonded Si atom. Four electrons complete covalent bonds with adjacent Si atoms; the fifth electron is weakly bound to the donor core and is thermally released into the conduction band at room temperature.

(9) Freeze-out in band-diagram language: As $T\to 0\text{ K}$, thermal energy $kT\to 0$. Mobile electrons in the conduction band lose thermal energy and fall back into the localized donor energy levels $E_D$ located slightly below $E_C$. Conduction band states become completely empty ($n\to 0$).

(10) Resistivities:

$${\rho }_n=\frac{1}{q{\mu }_n{N}_D},{\rho }_p=\frac{1}{q{\mu }_p{N}_A}$$

In silicon at $300\text{ K}$, electron mobility is substantially higher than hole mobility (${\mu }_n>{\mu }_p$). For equal dopings $N_D=N_A$:

$$q{\mu }_n{N}_D>q{\mu }_p{N}_A⟹{\rho }_p>{\rho }_n$$

The $n$-type semiconductor has the smaller resistivity.

(11) Recombination via R-G centres: A two-step transition:

1. A conduction band electron is trapped by a localized R-G level $E_T$ located near midgap.
2. The trapped electron transitions down to the valence band to annihilate a hole (or the center captures a hole).

(12) Drift velocity is the average net carrier velocity superimposed on random thermal motion as a result of an applied electric field $\mathcal{E}$. At low fields:

$$v_d=\mu \mathcal{E}$$

where $\mu$ is carrier mobility.

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II. Interpretation of an Energy Band Diagram

The diagram is for a Si device at $300\text{ K}$ with a single flat $E_F$. $E_C$ 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 $x=L$.

(13) Electrostatic potential $V(x)$: Since $V(x)=-E_i(x)/q$, the potential profile is the exact vertical mirror image (upside down) of the energy band edges.

$$\boxed{\text{Answer: (d)}}$$

(14) Electric field $\mathcal{E}(x)$: $\mathcal{E}=\frac{1}{q}\frac{d{E}_i}{dx}$. The slope of $E_i$ is zero for $x<L/3$, constant negative for $L/3\le x\le 2L/3$, and zero for $x>2L/3$. Thus the field is: zero, constant negative, zero.

$$\boxed{\text{Answer: (e)}}$$

(15) State of the system: The Fermi level $E_F$ is strictly horizontal (constant) throughout the entire device:

$$\frac{d{E}_F}{dx}=0⟹\text{Thermodynamic equilibrium}$$ $$\boxed{\text{Answer: (a)}}$$

(16) Degeneracy location: A semiconductor is degenerate where $E_F$ is within $3kT$ of or inside a band edge. This occurs near $x=L$.

$$\boxed{\text{Answer: (c)}}$$

(17) Concentration at $x=x_2$: At this point, $E_F$ is $E_G/3$ from the band edge, so $|E_i-E_F|=E_G/2-E_G/3=E_G/6$:

$$\begin{array}{rl}\text{Concentration}& =n_i\exp \left(\frac{E_G}{6kT}\right)=10^{10}\exp \left(\frac{1.12}{6\times 0.0259}\right)\\ & =10^{10}\exp (7.207)=10^{10}(1349)\\ & =\boxed{1.35\times 10^{13}{\text{ cm}}^{-3}}\end{array}$$ $$\boxed{\text{Answer: (b)}}$$

(18) Net current: Thermodynamic equilibrium implies zero net current of either carrier type: $J_N=0,J_P=0$.

$$\boxed{\text{Answer: (a)}}$$

(19) Hole drift current density where $E_i=E_F$:

1. At $E_i=E_F$, the carrier concentration is intrinsic: $p=n_i$.
2. In the sloped region, the electric field is: $$\mathcal{E}=\frac{1}{q}\frac{\Delta E_i}{\Delta x}=\frac{1}{q}\frac{E_G/3}{L/3}=\frac{E_G}{qL}$$
3. Hole drift current density: $$\begin{array}{rl}{J}_{P|\text{drift}}& =q{\mu }_{p}p\mathcal{E}=q{\mu }_p{n}_i\left(\frac{E_G}{qL}\right)\\ & =\boxed{\frac{{\mu }_p{n}_i{E}_G}{L}}\end{array}$$

$$\boxed{\text{Answer: (b)}}$$

(20) Hole kinetic energy: Hole kinetic energy is measured downward from the valence-band edge to the state energy, which is $E_G/3$.

$$\boxed{\text{Answer: (c)}}$$

---

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, $\Delta p_{n0}\le n_0$; $N_D=10^{15}{\text{ cm}}^{-3}$.

(21) Equilibrium carrier concentrations:

$$n_0\approx N_D=10^{15}{\text{ cm}}^{-3}$$ $$p_0=\frac{n_i^2}{n_0}=\frac{10^{20}}{10^{15}}=10^5{\text{ cm}}^{-3}$$ $$\boxed{\text{Answer: (c)}}$$

(22) Injection level: Since $\Delta p_{n,\max }=\Delta p_{n0}\ll n_0=10^{15}{\text{ cm}}^{-3}$, low-level injection is satisfied.

$$\boxed{\text{Answer: (a)}}$$

(23) Governing differential equation:

1. Start with minority carrier diffusion equation: $$\frac{\partial \Delta p_n}{\partial t}=D_P\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L$$
2. Steady state implies $\partial \Delta p_n/\partial t=0$.
3. No light implies $G_L=0$.
4. Resulting ODE: $$0=D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}⟺0=\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{L_P^2}$$

$$\boxed{\text{Answer: (e)}}$$

(24) Form of the general solution: Characteristic equation $r^2-1/L_P^2=0⟹r=\pm 1/L_P$:

$$\Delta p_n(x)=A{e}^{-x/L_P}+B{e}^{x/L_P},L_P=\sqrt{D_P{\tau }_p}$$ $$\boxed{\text{Answer: (a)}}$$

(25) Boundary conditions: At $x=0$, $\Delta p_n(0)=\Delta p_{n0}$; at $x=L$, $\Delta p_n(L)=0$.

$$\boxed{\text{Answer: (c)}}$$

Completing the problem (step-by-step solution):

1. Apply boundary conditions to the general solution: $$A+B=\Delta p_{n0}⟹B=\Delta p_{n0}-A$$ $$A{e}^{-L/L_P}+B{e}^{L/L_P}=0⟹A{e}^{-L/L_P}+(\Delta p_{n0}-A)e^{L/L_P}=0$$
2. Solve for $A$ and $B$: $$A\left(e^{L/L_P}-e^{-L/L_P}\right)=\Delta p_{n0}{e}^{L/L_P}⟹A=\Delta p_{n0}\frac{e^{L/L_P}}{e^{L/L_P}-e^{-L/L_P}}$$ $$B=\Delta p_{n0}-A=-\Delta p_{n0}\frac{e^{-L/L_P}}{e^{L/L_P}-e^{-L/L_P}}$$
3. Combine using hyperbolic functions: $$\begin{array}{rl}\Delta p_n(x)& =\Delta p_{n0}\frac{e^{(L-x)/L_P}-e^{-(L-x)/L_P}}{e^{L/L_P}-e^{-L/L_P}}\\ & =\boxed{\Delta p_{n0}\frac{\sinh \left(\frac{L-x}{L_P}\right)}{\sinh \left(\frac{L}{L_P}\right)}}\end{array}$$

---

Part II: pn Junction Diodes and Chapter 5: pn Junction Electrostatics

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.

Family Member	Where It Appears	Description
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 Four-Step Recipe of Every Device Analysis

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.

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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. 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

Idealization	Profile Definition	When Appropriate
Step junction	$N_D-N_A=\{\begin{array}{ll}-N_A,& x<0\\ +N_D,& x>0\end{array}$	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

---

5.1.2 Poisson’s Equation

From electromagnetism, the 3-D form is

$$\nabla \cdot \mathcal{E}=\frac{\rho }{K_S{\epsilon }_0}\text{\hspace{1em}(5.1)}$$

In one dimension, $\mathcal{E}={\mathcal{E}}_x$:

$$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}\text{\hspace{1em}(5.2)}$$

• $K_S$ is the semiconductor dielectric constant ($11.8$ for Si) and ${\epsilon }_0=8.85\times 10^{-14}\text{ F/cm}$.
• In the electrostatics context $\rho$ is the charge density in ${\text{C/cm}}^3$.

With totally ionized dopants:

$$\rho =q(p-n+N_D-N_A)\text{\hspace{1em}(5.3)}$$

---

5.1.3 Qualitative Solution

Building the equilibrium band diagram in three moves (Fig. 5.3)

1. Far from the metallurgical junction the material is uniform p-type on the left and uniform n-type on the right. On the p-side $E_F$ sits near $E_V$; on the n-side $E_F$ sits near $E_C$.
2. 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. 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.

Reading the diagram for $V$, $\mathcal{E}$, $\rho$ (Fig. 5.4)

• Potential: $V(x)$ has the same shape as upside-down $E_C$ (or $E_i$ or $E_V$), because electron energy $E=-qV$. Setting $V=0$ on the far p-side, $V$ rises to $V_{bi}$ on the n-side.
• Field: $\mathcal{E}=-dV/dx$, i.e. proportional to the slope of $E_C$. 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, positive on the n-side.

---

5.1.4 The Built-in Potential ($V_{bi}$)

The voltage drop across the depletion region in equilibrium is the built-in potential $V_{bi}$.

Derivation (Eqs. 5.4ΓÇô5.10)

1. Relate electric field and potential: $$\mathcal{E}=-\frac{dV}{dx}\text{\hspace{1em}(5.4)}$$
2. Integrate across the depletion region: $$-{\int }_{-x_p}^{x_n}\mathcal{E}dx={\int }_{V(-x_p)}^{V(x_n)}dV=V(x_n)-V(-x_p)=V_{bi}\text{\hspace{1em}(5.5)}$$
3. In equilibrium, the electron current density must identically vanish: $$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}=0\text{\hspace{1em}(5.6)}$$
4. Solve for $\mathcal{E}$ using the Einstein relation $D_N/{\mu }_n=kT/q$: $$\mathcal{E}=-\frac{D_N}{{\mu }_n}\frac{1}{n}\frac{dn}{dx}=-\frac{kT}{q}\frac{1}{n}\frac{dn}{dx}\text{\hspace{1em}(5.7)}$$
5. Substitute (5.7) into (5.5) and integrate: \begin{aligned} V_{bi} &= -\int_{-x_p}^{x_n}\left(-\frac{kT}{q}\frac{1}{n}\frac{dn}{dx}\right)dx = \frac{kT}{q}\int_{n(-x_p)}^{n(x_n)}\frac{dn}{n} \end{aligned}$$$$ \mathcal{E}= \frac{kT}{q}\ln\left[\frac{n(x_n)}{n(-x_p)}\right] \tag{5.8}
6. Specialize to a nondegenerate step junction: $$n(x_n)=N_D\text{\hspace{1em}(5.9a)}$$ $$n(-x_p)=\frac{n_i^2}{p(-x_p)}=\frac{n_i^2}{N_A}\text{\hspace{1em}(5.9b)}$$
7. Substitute boundary values: $$\boxed{V_{bi}=\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)}\text{\hspace{1em}(5.10)}$$

Alternative derivation from the band diagram

$$V_{bi}=V(x_n)-V(-x_p)\text{\hspace{1em}(5.11a)}$$ $$q{V}_{bi}=(E_i{)}_{p\text{-side}}-(E_i{)}_{n\text{-side}}\text{\hspace{1em}(5.11b)}$$

Add and subtract $E_F$:

$$V_{bi}=\frac{1}{q}\left[(E_i-E_F{)}_{p\text{-side}}+(E_F-E_i{)}_{n\text{-side}}\right]\text{\hspace{1em}(5.12)}$$

For a nondegenerate step junction:

$$(E_i-E_F{)}_{p\text{-side}}=kT\ln \left(\frac{N_A}{n_i}\right)\text{\hspace{1em}(5.13a)}$$ $$(E_F-E_i{)}_{n\text{-side}}=kT\ln \left(\frac{N_D}{n_i}\right)\text{\hspace{1em}(5.13b)}$$

Substituting gives Eq. (5.10).

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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 $300\text{ K}$, for $10^{14}{\text{ cm}}^{-3}\le N_A\text{ or }{N}_D\le 10^{17}{\text{ cm}}^{-3}$.

Solution.

1. On the $p^{+}$ side, $E_F=E_V⟹(E_i-E_F{)}_{p\text{-side}}=E_i-E_V=E_G/2$.
2. On the lightly doped n-side: $$(E_F-E_i{)}_{n\text{-side}}=kT\ln \left(\frac{N_D}{n_i}\right)$$
3. Built-in potential: $$\begin{array}{rl}{V}_{bi}& =\frac{1}{q}\left[\frac{E_G}{2}+kT\ln \left(\frac{N_D}{n_i}\right)\right]\\ & =\boxed{\frac{E_G}{2q}+\frac{kT}{q}\ln \left(\frac{N_D}{n_i}\right)}\end{array}$$

Hand checks ($E_G/2q=0.56\text{ V}$):

• At $N_D=10^{14}{\text{ cm}}^{-3}$: $$V_{bi}=0.56+0.0259\ln (10^4)=0.56+0.0259(9.210)=0.56+0.2385=\boxed{0.799\text{ V}}$$
• At $N_D=10^{17}{\text{ cm}}^{-3}$: $$V_{bi}=0.56+0.0259\ln (10^7)=0.56+0.0259(16.118)=0.56+0.4175=\boxed{0.978\text{ V}}$$

% 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')

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5.1.5 The Depletion Approximation

Poisson’s equation:

$$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}=\frac{q}{K_S{\epsilon }_0}(p-n+N_D-N_A)\text{\hspace{1em}(5.14)}$$

Statement of the approximation (Fig. 5.6)

1. Within the depletion region $-x_p\le x\le x_n$: $$p\ll |N_D-N_A|,n\ll |N_D-N_A|$$
2. Outside the depletion region ($x<-x_p$ and $x>x_n$): $$\rho \equiv 0$$

Poisson’s equation simplifies to:

$$\frac{d\mathcal{E}}{dx}=\frac{q(N_D-N_A)}{K_S{\epsilon }_0},-x_p\le x\le x_n\text{\hspace{1em}(5.15a)}$$ $$\frac{d\mathcal{E}}{dx}=0,x\le -x_p\text{ and }x\ge x_n\text{\hspace{1em}(5.15b)}$$

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5.2 Quantitative Electrostatic Relationships

5.2.1 Assumptions and Definitions

• One-dimensional device geometry along $x$.
• $x=0$ is the metallurgical junction.
• Ohmic non-rectifying contacts at terminal boundaries.
• $V_A$ is the applied voltage ($V_A>0$ forward bias, $V_A<0$ reverse bias).

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5.2.2 Step Junction with $V_A=0$

Solution for $\rho$

$$\rho (x)=\{\begin{array}{ll}-q{N}_A,& -x_p\le x\le 0\\ +q{N}_D,& 0\le x\le x_n\\ 0,& x<-x_p\text{ and }x>x_n\end{array}\text{\hspace{1em}(5.16a-c)}$$

Solution for $\mathcal{E}$

Boundary conditions: $\mathcal{E}(-x_p)=0$ and $\mathcal{E}(x_n)=0$.

1. p-side (integrate from $-x_p$ to $x$): \begin{aligned} \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}(x + x_p), \quad -x_p \le x \le 0 \tag{5.19} \end{aligned}
2. n-side (integrate from $x$ to $x_n$): \begin{aligned} \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), \quad 0 \le x \le x_n \tag{5.21} \end{aligned}
3. Field continuity at $x=0$: $$-\frac{q{N}_A}{K_S{\epsilon }_0}{x}_p=-\frac{q{N}_D}{K_S{\epsilon }_0}{x}_n⟹\boxed{N_A{x}_p=N_D{x}_n}\text{\hspace{1em}(5.22)}$$

Solution for $V$

Reference $V(-x_p)=0$, so $V(x_n)=V_{bi}$.

1. p-side (integrate from $-x_p$ to $x$): $$\begin{array}{rl}V(x)& =-{\int }_{-x_p}^x\mathcal{E}(x^{\prime })d{x}^{\prime }=\frac{q{N}_A}{K_S{\epsilon }_0}{\int }_{-x_p}^x(x^{\prime }+x_p)d{x}^{\prime }\\ & =\frac{q{N}_A}{2K_S{\epsilon }_0}(x+x_p{)}^2,-x_p\le x\le 0\end{array}\text{\hspace{1em}(5.26)}$$
2. n-side (integrate from $x$ to $x_n$): $$\begin{array}{rl}{V}_{bi}-V(x)& =-{\int }_x^{x_n}\mathcal{E}(x^{\prime })d{x}^{\prime }=\frac{q{N}_D}{K_S{\epsilon }_0}{\int }_x^{x_n}(x_n-x^{\prime })d{x}^{\prime }\\ & =\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2\\ V(x)& =V_{bi}-\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2,0\le x\le x_n\end{array}\text{\hspace{1em}(5.28)}$$
3. Potential continuity at $x=0$: $$\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=V_{bi}-\frac{q{N}_D}{2K_S{\epsilon }_0}{x}_n^2\text{\hspace{1em}(5.29)}$$

Solving for $x_n,x_p,W$

From (5.29):

$$V_{bi}=\frac{q}{2K_S{\epsilon }_0}\left(N_A{x}_p^2+N_D{x}_n^2\right)$$

Substitute $x_p=(N_D/N_A)x_n$:

$$V_{bi}=\frac{q{N}_D}{2K_S{\epsilon }_0}\left(\frac{N_D+N_A}{N_A}\right)x_n^2$$ $$\boxed{x_n={\left[\frac{2K_S{\epsilon }_0}{q}{V}_{bi}\frac{N_A}{N_D(N_A+N_D)}\right]}^{1/2}}\text{\hspace{1em}(5.30a)}$$ $$\boxed{x_p={\left[\frac{2K_S{\epsilon }_0}{q}{V}_{bi}\frac{N_D}{N_A(N_A+N_D)}\right]}^{1/2}}\text{\hspace{1em}(5.30b)}$$

Total depletion width $W=x_n+x_p$:

$$\boxed{W={\left[\frac{2K_S{\epsilon }_0}{q}{V}_{bi}\left(\frac{N_A+N_D}{N_A{N}_D}\right)\right]}^{1/2}}\text{\hspace{1em}(5.31)}$$

Peak electric field:

$$|\mathcal{E}{|}_{\max }=|\mathcal{E}(0)|=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}$$

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Exercise 5.2: How Big Are $W$ and $|\mathcal{E}{|}_{\max }$?

Problem. Si step junction at $300\text{ K}$ with $N_A=10^{17}{\text{ cm}}^{-3}$ and $N_D=10^{14}{\text{ cm}}^{-3}$. Compute $W$ and $|\mathcal{E}{|}_{\max }$ at equilibrium.

Solution.

1. Built-in potential: $$\begin{array}{rl}{V}_{bi}& =\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)=(0.0259\text{ V})\ln \left[\frac{(10^{17})(10^{14})}{10^{20}}\right]\\ & =(0.0259)\ln (10^{11})=(0.0259)(25.328)\\ & =\boxed{0.656\text{ V}}\end{array}$$
2. Depletion width on the lightly doped n-side ($N_A\gg N_D⟹N_A/(N_A+N_D)\approx 1$): $$\begin{array}{rl}{x}_n& ={\left[\frac{2(11.8)(8.85\times 10^{-14}\text{ F/cm})(0.656\text{ V})}{(1.602\times 10^{-19}\text{ C})(10^{14}{\text{ cm}}^{-3})}\right]}^{1/2}={\left[\frac{1.370\times 10^{-12}}{1.602\times 10^{-5}}\right]}^{1/2}\\ & ={\left[8.55\times 10^{-8}{\text{ cm}}^2\right]}^{1/2}=2.925\times 10^{-4}\text{ cm}=\boxed{2.93\mu \text{m}}\end{array}$$
3. Depletion width on the heavily doped p-side: $$\begin{array}{rl}{x}_p& =\frac{N_D}{N_A}{x}_n=\left(\frac{10^{14}}{10^{17}}\right)(2.925\times 10^{-4}\text{ cm})\\ & =2.925\times 10^{-7}\text{ cm}=\boxed{2.93\text{ nm}}\end{array}$$
4. Total depletion width: $$\begin{array}{rl}W& =x_n+x_p\approx x_n\\ & =\boxed{2.93\mu \text{m}}\end{array}$$
5. Peak electric field magnitude: $$\begin{array}{rl}|\mathcal{E}{|}_{\max }& =\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{(1.602\times 10^{-19}\text{ C})(10^{14}{\text{ cm}}^{-3})(2.925\times 10^{-4}\text{ cm})}{1.0443\times 10^{-12}\text{ F/cm}}\\ & =\frac{4.686\times 10^{-9}}{1.0443\times 10^{-12}}\\ & =\boxed{4.49\times 10^3\text{ V/cm}}\end{array}$$

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5.2.3 Step Junction with $V_A\ne 0$

Replacing $V_{bi}\to V_{bi}-V_A$:

For $-x_p\le x\le 0$:

$$\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}(x_p+x)\text{\hspace{1em}(5.32)}$$ $$V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}(x_p+x{)}^2\text{\hspace{1em}(5.33)}$$ $$x_p={\left[\frac{2K_S{\epsilon }_0}{q}(V_{bi}-V_A)\frac{N_D}{N_A(N_A+N_D)}\right]}^{1/2}\text{\hspace{1em}(5.34)}$$

For $0\le x\le x_n$:

$$\mathcal{E}(x)=-\frac{q{N}_D}{K_S{\epsilon }_0}(x_n-x)\text{\hspace{1em}(5.35)}$$ $$V(x)=V_{bi}-V_A-\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2\text{\hspace{1em}(5.36)}$$ $$x_n={\left[\frac{2K_S{\epsilon }_0}{q}(V_{bi}-V_A)\frac{N_A}{N_D(N_A+N_D)}\right]}^{1/2}\text{\hspace{1em}(5.37)}$$ $$\boxed{W={\left[\frac{2K_S{\epsilon }_0}{q}(V_{bi}-V_A)\frac{N_A+N_D}{N_A{N}_D}\right]}^{1/2}}\text{\hspace{1em}(5.38)}$$

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Exercise 5.3: $W$ Versus Doping at Several Biases (Computer-Based)

% 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')

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5.2.4 Examination and Extrapolation of Results

Fermi level separation under bias:

$$E_{Fp}-E_{Fn}=-q{V}_A\text{\hspace{1em}(5.39)}$$

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Exercise 5.4: Equilibrium Energy Band Diagram Generator (Computer-Based)

% 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

---

5.2.5 Linearly Graded Junctions

Doping profile:

$$N_D-N_A=ax\text{\hspace{1em}(5.40)}$$

Charge density:

$$\rho (x)=\{\begin{array}{ll}qax,& -W/2\le x\le W/2\\ 0,& |x|>W/2\end{array}\text{\hspace{1em}(5.41)}$$

Symmetry requires:

$$x_n=x_p=\frac{W}{2}\text{\hspace{1em}(5.42)}$$

Electric field:

$$\begin{array}{rl}\mathcal{E}(x)& ={\int }_{-W/2}^x\frac{qa{x}^{\prime }}{K_S{\epsilon }_0}d{x}^{\prime }\\ & =\boxed{-\frac{qa}{2K_S{\epsilon }_0}\left(\frac{W^2}{4}-x^2\right),-\frac{W}{2}\le x\le \frac{W}{2}}\end{array}\text{\hspace{1em}(5.43)}$$

Peak electric field at $x=0$:

$$|\mathcal{E}{|}_{\max }=\frac{qa{W}^2}{8K_S{\epsilon }_0}$$

Electrostatic potential:

$$\begin{array}{rl}V(x)& =-{\int }_{-W/2}^x\mathcal{E}(x^{\prime })d{x}^{\prime }=\frac{qa}{2K_S{\epsilon }_0}{\int }_{-W/2}^x\left(\frac{W^2}{4}-x^{\prime 2}\right)d{x}^{\prime }\\ & =\boxed{\frac{qa}{6K_S{\epsilon }_0}\left(\frac{3W^2}{4}x-x^3+\frac{W^3}{4}\right),-\frac{W}{2}\le x\le \frac{W}{2}}\end{array}\text{\hspace{1em}(5.44)}$$

Total potential drop:

$$V\left(\frac{W}{2}\right)=\frac{qa{W}^3}{12K_S{\epsilon }_0}=V_{bi}-V_A$$ $$\boxed{W={\left[\frac{12K_S{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}}\text{\hspace{1em}(5.45)}$$

Built-in voltage:

$$n\left(\frac{W_0}{2}\right)=p\left(-\frac{W_0}{2}\right)=\frac{a{W}_0}{2}\text{\hspace{1em}(5.46a,b)}$$ $$\boxed{V_{bi}=\frac{2kT}{q}\ln \left(\frac{a{W}_0}{2n_i}\right)}\text{\hspace{1em}(5.47)}$$ $$\boxed{V_{bi}=\frac{2kT}{q}\ln \left\{\frac{a}{2n_i}{\left[\frac{12K_S{\epsilon }_0{V}_{bi}}{qa}\right]}^{1/3}\right\}}\text{\hspace{1em}(5.48)}$$

---

5.3 Summary

Quantity	Step Junction	Linearly Graded Junction
Charge density $\rho (x)$	Constant on each side (rectangular)	Linear with $x$ (triangular)
Electric field $\mathcal{E}(x)$	Piecewise linear (triangular)	Parabolic
Potential $V(x)$	Piecewise quadratic (parabolic)	Cubic
Depletion width dependence	$W\propto (V_{bi}-V_A{)}^{1/2}$	$W\propto (V_{bi}-V_A{)}^{1/3}$
$V_{bi}$ formulation	Analytical closed form (5.10)	Requires iteration (5.48)

---

Chapter 5 Problems

Problem 5.1: True or False

(a) The step junction is an idealized doping profile used to model $p^{+}$-n and $n^{+}$-p junctions.

$$\text{Step abrupt profiles accurately idealize one-sided shallow diffusions and shallow implants.}$$ $$\boxed{\text{TRUE}}$$

(b) The $\rho$ in Poisson’s equation is the charge density, with units of ${\text{C/cm}}^3$.

$$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}⟹[\rho ]=\left[\frac{\text{V/cm}}{\text{cm}}\right]\times \left[\frac{\text{F}}{\text{cm}}\right]=\frac{\text{C}}{{\text{cm}}^3}$$ $$\boxed{\text{TRUE}}$$

(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.

$$\text{Mobile carriers are depleted; the charge arises from uncompensated fixed ionized dopant cores }(N_D^{+}\text{ and }{N}_A^{-}).$$ $$\boxed{\text{FALSE}}$$

(d) The built-in potential is typically less than the band gap energy converted to volts.

$$q{V}_{bi}=kT\ln \left(\frac{N_A{N}_D}{n_i^2}\right)<E_G\text{for nondegenerately doped semiconductors.}$$ $$\boxed{\text{TRUE}}$$

(e) Invoking the depletion approximation makes the charge density inside the depletion region directly proportional to the net doping concentration.

$$\rho (x)\approx q(N_D-N_A)\text{since }n\approx 0,p\approx 0.$$ $$\boxed{\text{TRUE}}$$

(f) Ohmic contacts reduce the built-in voltage drop across a junction.

$$\text{Contact potentials balance }{V}_{bi}\text{ around the closed loop; the junction’s internal potential barrier is unaffected.}$$ $$\boxed{\text{FALSE}}$$

(g) In solutions based on the depletion approximation, the magnitude of the electric field reaches a maximum right at the metallurgical boundary.

$$\frac{d|\mathcal{E}|}{dx}=0\text{ at }x=0\text{ where }\rho (x)\text{ switches sign; hence peak magnitude occurs at }x=0.$$ $$\boxed{\text{TRUE}}$$

(h) If one has a $p^{+}$-n step junction, with $N_A\gg N_D$, it follows that $x_p\le x_n$.

$$N_A{x}_p=N_D{x}_n⟹x_p=\left(\frac{N_D}{N_A}\right)x_n\ll x_n.$$ $$\boxed{\text{TRUE}}$$

(i) The potential hill between the n-side and the p-side of a junction increases with forward biasing.

$$\text{Barrier height under forward bias is }{V}_{bi}-V_A<V_{bi}⟹\text{the potential hill decreases.}$$ $$\boxed{\text{FALSE}}$$

(j) The depletion width in a linearly graded junction varies as the square root of $V_{bi}-V_A$.

$$W={\left[\frac{12K_S{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}\propto (V_{bi}-V_A{)}^{1/3}⟹\text{it varies as the cube root.}$$ $$\boxed{\text{FALSE}}$$

---

Problem 5.2: $V_{bi}$ from the Energy Band Diagram

Problem. 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.

Solution.

(a) Equilibrium band diagram:

1. Draw a single horizontal line for $E_F$ valid across all $x$.
2. On the neutral p-side (far left): place $E_V$ at distance $2kT$ below $E_F$, and $E_C=E_V+E_G$ above it.
3. On the neutral n-side (far right): place $E_C$ at distance $kT/4$ above $E_F$, and $E_V=E_C-E_G$ below it.
4. Connect the band edges across the transition region with smooth monotonic downward-sloping curves from p to n.

(b) Determine $V_{bi}$:

1. Conduction band edge relative to $E_F$ on the p-side: $$E_C^{p\text{-side}}-E_F=(E_V+E_G)-E_F=E_G-2kT$$
2. Conduction band edge relative to $E_F$ on the n-side: $$E_C^{n\text{-side}}-E_F=\frac{kT}{4}$$
3. Total band bending across the junction: $$\begin{array}{rl}q{V}_{bi}& =E_C^{p\text{-side}}-E_C^{n\text{-side}}\\ & =(E_G-2kT)-\frac{kT}{4}\\ & =E_G-\frac{9}{4}kT\end{array}$$
4. Built-in potential symbolically: $$\boxed{V_{bi}=\frac{E_G}{q}-\frac{9}{4}\frac{kT}{q}}$$
5. Numerical calculation ($E_G/q=1.12\text{ V}$, $kT/q=0.0259\text{ V}$): $$\begin{array}{rl}{V}_{bi}& =1.12\text{ V}-2.25(0.0259\text{ V})\\ & =1.12-0.05828\\ & =\boxed{1.062\text{ V}}\end{array}$$

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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: $N_A(x)=N_{A1}$ for $x<0$ and $N_{A2}$ for $x>0$ with $N_{A1}>N_{A2}$. (a) Draw the equilibrium band diagram taking the doping nondegenerate. (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:

1. Since $N_{A1}>N_{A2}$, the Fermi level on side 1 is closer to the valence band than on side 2: $$(E_F-E_V{)}_1<(E_F-E_V{)}_2$$
2. With $E_F$ horizontal everywhere, the bands on the more heavily doped $p_1$ side must sit higher in energy than on the $p_2$ side.
3. The bands bend monotonically downward from $x<0$ to $x>0$.

(b) Built-in voltage derivation:

1. Express intrinsic level positions relative to $E_F$: $$(E_i-E_F{)}_1=kT\ln \left(\frac{N_{A1}}{n_i}\right),(E_i-E_F{)}_2=kT\ln \left(\frac{N_{A2}}{n_i}\right)$$
2. Total built-in barrier: $$\begin{array}{rl}q{V}_{bi}& =(E_i{)}_1-(E_i{)}_2=(E_i-E_F{)}_1-(E_i-E_F{)}_2\\ & =kT\ln \left(\frac{N_{A1}}{n_i}\right)-kT\ln \left(\frac{N_{A2}}{n_i}\right)\\ & =kT\ln \left(\frac{N_{A1}}{N_{A2}}\right)\end{array}$$
3. Result: $$\boxed{V_{bi}=\frac{kT}{q}\ln \left(\frac{N_{A1}}{N_{A2}}\right)}$$

(c) Sketches:

• Charge density $\rho (x)$: Holes diffuse from $p_1$ to $p_2$. This leaves uncompensated negative acceptor ions on the $p_1$ side ($\rho <0$) and creates an accumulation layer of mobile holes on the $p_2$ side ($\rho >0$).
• Electric field $\mathcal{E}(x)$: Starts at $0$ at $x\to -\infty$, goes negative in the negatively charged region, reaches a minimum negative peak at $x=0$, and returns to $0$ as $x\to +\infty$.
• Potential $V(x)$: Increases monotonically from $0$ to $V_{bi}$ with an S-shaped curve.

(d) Depletion approximation: Assumes the space-charge layer is completely depleted of mobile carriers ($n\approx 0,p\approx 0$) so that $\rho =q(N_D-N_A)$ inside the layer, and $\rho =0$ outside it.

(e) Can it be used here?

$$\boxed{\text{No}}$$

Justification: On the $p_2$ side, the positive space charge consists of an accumulation of mobile holes ($p>N_{A2}$), which directly violates the assumption $p\approx 0$. Furthermore, the potential barrier $V_{bi}$ is typically only a few $kT/q$, so mobile carriers are not fully depleted on the $p_1$ side either.

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Problem 5.4: Step Junction at Equilibrium, $N_A\sim N_D$

Problem. A Si step junction at $300\text{ K}$ in equilibrium has $N_A=2\times 10^{15}{\text{ cm}}^{-3}$ and $N_D=10^{15}{\text{ cm}}^{-3}$. (a) Compute $V_{bi}$. (b) Compute $x_n,x_p$ and $W$. (c) Compute $\mathcal{E}$ at $x=0$. (d) Compute $V$ at $x=0$. (e) Sketch $\rho ,\mathcal{E},V$.

Solution.

(a) Built-in voltage:

$$\begin{array}{rl}{V}_{bi}& =\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)\\ & =(0.0259\text{ V})\ln \left[\frac{(2\times 10^{15}{\text{ cm}}^{-3})(10^{15}{\text{ cm}}^{-3})}{(10^{10}{\text{ cm}}^{-3}{)}^2}\right]\\ & =(0.0259)\ln (2\times 10^{10})=(0.0259)[\ln (2)+10\ln (10)]\\ & =(0.0259)[0.69315+23.02585]=(0.0259)(23.719)\\ & =\boxed{0.614\text{ V}}\end{array}$$

(b) Depletion widths:

1. Doping factor: $$\frac{N_A+N_D}{N_A{N}_D}=\frac{2\times 10^{15}+10^{15}}{(2\times 10^{15})(10^{15})}=\frac{3\times 10^{15}}{2\times 10^{30}}=1.50\times 10^{-15}{\text{ cm}}^3$$
2. Electrostatic coefficient ($K_S{\epsilon }_0=1.0443\times 10^{-12}\text{ F/cm}$): $$\frac{2K_S{\epsilon }_0{V}_{bi}}{q}=\frac{2(1.0443\times 10^{-12}\text{ F/cm})(0.614\text{ V})}{1.602\times 10^{-19}\text{ C}}=8.004\times 10^6{\text{ cm}}^{-1}$$
3. Total depletion width $W$: $$\begin{array}{rl}W& ={\left[\left(8.004\times 10^6{\text{ cm}}^{-1}\right)\left(1.50\times 10^{-15}{\text{ cm}}^3\right)\right]}^{1/2}={\left[1.2006\times 10^{-8}{\text{ cm}}^2\right]}^{1/2}\\ & =1.096\times 10^{-4}\text{ cm}=\boxed{1.10\mu \text{m}}\end{array}$$
4. Individual widths using charge balance $N_A{x}_p=N_D{x}_n$: $$\begin{array}{rl}{x}_n& =W\left(\frac{N_A}{N_A+N_D}\right)=(1.096\times 10^{-4}\text{ cm})\left(\frac{2\times 10^{15}}{3\times 10^{15}}\right)\\ & =7.307\times 10^{-5}\text{ cm}=\boxed{0.731\mu \text{m}}\end{array}$$ $$\begin{array}{rl}{x}_p& =W\left(\frac{N_D}{N_A+N_D}\right)=(1.096\times 10^{-4}\text{ cm})\left(\frac{10^{15}}{3\times 10^{15}}\right)\\ & =3.653\times 10^{-5}\text{ cm}=\boxed{0.365\mu \text{m}}\end{array}$$

(c) Peak electric field at $x=0$:

$$\begin{array}{rl}\mathcal{E}(0)& =-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.602\times 10^{-19}\text{ C})(10^{15}{\text{ cm}}^{-3})(7.307\times 10^{-5}\text{ cm})}{1.0443\times 10^{-12}\text{ F/cm}}\\ & =-\frac{1.1706\times 10^{-8}{\text{ C/cm}}^2}{1.0443\times 10^{-12}\text{ F/cm}}\\ & =\boxed{-1.12\times 10^4\text{ V/cm}}\end{array}$$

(d) Potential at $x=0$ (referenced to $V(-x_p)=0$):

$$\begin{array}{rl}V(0)& =\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=\frac{(1.602\times 10^{-19})(2\times 10^{15})}{2(1.0443\times 10^{-12})}(3.653\times 10^{-5}{)}^2\\ & =(1.534\times 10^8)(1.334\times 10^{-9})\\ & =\boxed{0.205\text{ V}}\end{array}$$

(Verification: $V(0)=V_{bi}\frac{N_D}{N_A+N_D}=0.614\left(\frac{1}{3}\right)=0.205\text{ V}$.)

(e) Sketches:

• $\rho (x)$: Rectangular negative step of height $-q{N}_A=-3.20\times 10^{-4}{\text{ C/cm}}^3$ for $-0.365\mu \text{m}\le x\le 0$; positive step $+q{N}_D=+1.60\times 10^{-4}{\text{ C/cm}}^3$ for $0\le x\le 0.731\mu \text{m}$.
• $\mathcal{E}(x)$: Negative triangle with slope $-q{N}_A/K_S{\epsilon }_0$ on the left and $+q{N}_D/K_S{\epsilon }_0$ on the right, peaking at $\mathcal{E}(0)=-1.12\times 10^4\text{ V/cm}$.
• $V(x)$: Concave-up parabola from $0$ to $0.205\text{ V}$ for $-x_p\le x\le 0$, continuous into concave-down parabola rising to $0.614\text{ V}$ at $x_n$.

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Problem 5.5: Step Junction with $N_A\gg N_D$

Problem. Repeat Problem 5.4 taking $N_A=10^{17}{\text{ cm}}^{-3}$ and $N_D=10^{15}{\text{ cm}}^{-3}$, and compare.

Solution.

(a) Built-in voltage:

$$\begin{array}{rl}{V}_{bi}& =(0.0259\text{ V})\ln \left[\frac{(10^{17})(10^{15})}{10^{20}}\right]=(0.0259)\ln (10^{12})\\ & =(0.0259)(12\times 2.302585)=(0.0259)(27.631)\\ & =\boxed{0.716\text{ V}}\end{array}$$

(b) Depletion widths:

1. Doping factor: $$\frac{N_A+N_D}{N_A{N}_D}=\frac{10^{17}+10^{15}}{(10^{17})(10^{15})}=\frac{1.01\times 10^{17}}{10^{32}}=1.01\times 10^{-15}{\text{ cm}}^3$$
2. Total width $W$: $$\begin{array}{rl}W& ={\left[\frac{2(1.0443\times 10^{-12}\text{ F/cm})(0.716\text{ V})}{1.602\times 10^{-19}\text{ C}}\left(1.01\times 10^{-15}{\text{ cm}}^3\right)\right]}^{1/2}\\ & ={\left[(9.336\times 10^6{\text{ cm}}^{-1})(1.01\times 10^{-15}{\text{ cm}}^3)\right]}^{1/2}={\left[9.429\times 10^{-9}{\text{ cm}}^2\right]}^{1/2}\\ & =9.710\times 10^{-5}\text{ cm}=\boxed{0.971\mu \text{m}}\end{array}$$
3. Width on each side: $$\begin{array}{rl}{x}_n& =W\left(\frac{N_A}{N_A+N_D}\right)=(9.710\times 10^{-5}\text{ cm})\left(\frac{10^{17}}{1.01\times 10^{17}}\right)\\ & =9.614\times 10^{-5}\text{ cm}=\boxed{0.961\mu \text{m}}\end{array}$$ $$\begin{array}{rl}{x}_p& =W\left(\frac{N_D}{N_A+N_D}\right)=(9.710\times 10^{-5}\text{ cm})\left(\frac{10^{15}}{1.01\times 10^{17}}\right)\\ & =9.614\times 10^{-7}\text{ cm}=\boxed{9.61\text{ nm}}\end{array}$$

(c) Peak electric field at $x=0$:

$$\begin{array}{rl}\mathcal{E}(0)& =-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.602\times 10^{-19}\text{ C})(10^{15}{\text{ cm}}^{-3})(9.614\times 10^{-5}\text{ cm})}{1.0443\times 10^{-12}\text{ F/cm}}\\ & =\boxed{-1.47\times 10^4\text{ V/cm}}\end{array}$$

(d) Potential at $x=0$:

$$\begin{array}{rl}V(0)& =V_{bi}\left(\frac{N_D}{N_A+N_D}\right)=(0.716\text{ V})\left(\frac{10^{15}}{1.01\times 10^{17}}\right)\\ & =\boxed{7.09\times 10^{-3}\text{ V}\approx 7.1\text{ mV}}\end{array}$$

(e) Comparison table:

Quantity	Problem 5.4 ($N_A=2\times 10^{15}{\text{ cm}}^{-3}$)	Problem 5.5 ($N_A=10^{17}{\text{ cm}}^{-3}$)
$V_{bi}$	$0.614\text{ V}$	$0.716\text{ V}$
$W$	$1.10\mu \text{m}$	$0.971\mu \text{m}$
$x_n$	$0.731\mu \text{m}$	$0.961\mu \text{m}$
$x_p$	$0.365\mu \text{m}$	$0.0096\mu \text{m}$
$	\mathcal{E}(0)	$
$V(0)$	$0.205\text{ V}$	$0.0071\text{ V}$

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Problem 5.6: Computation Using the MATLAB Diary Function

Problem. A Si step junction at room temperature has $N_A=10^{16}{\text{ cm}}^{-3}$, $N_D=10^{15}{\text{ cm}}^{-3}$, $V_A=-2\text{ V}$. Working in the Command window with diary, compute: (a) $V_{bi}$; (b) $x_n,x_p,W$; (c) $\mathcal{E}$ at $x=0$; (d) $V$ at $x=0$.

Solution.

MATLAB interactive script:

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

Step-by-step verification:

(a) Built-in voltage:

$$\begin{array}{rl}{V}_{bi}& =(0.0259\text{ V})\ln \left[\frac{(10^{16})(10^{15})}{10^{20}}\right]=(0.0259)\ln (10^{11})\\ & =\boxed{0.656\text{ V}}\end{array}$$

(b) Depletion widths with $V_A=-2\text{ V}⟹V_{bi}-V_A=0.656-(-2)=2.656\text{ V}$:

$$\begin{array}{rl}W& ={\left[\frac{2(1.0443\times 10^{-12}\text{ F/cm})(2.656\text{ V})}{1.602\times 10^{-19}\text{ C}}\left(\frac{1.1\times 10^{16}}{10^{31}}\right)\right]}^{1/2}\\ & ={\left[(3.463\times 10^7)(1.10\times 10^{-15})\right]}^{1/2}={\left[3.809\times 10^{-8}\right]}^{1/2}\\ & =1.952\times 10^{-4}\text{ cm}=\boxed{1.95\mu \text{m}}\end{array}$$ $$x_n=W\left(\frac{N_A}{N_A+N_D}\right)=(1.952\mu \text{m})\left(\frac{10}{11}\right)=\boxed{1.77\mu \text{m}}$$ $$x_p=W\left(\frac{N_D}{N_A+N_D}\right)=(1.952\mu \text{m})\left(\frac{1}{11}\right)=\boxed{0.177\mu \text{m}}$$

(c) Peak electric field at $x=0$:

$$\begin{array}{rl}\mathcal{E}(0)& =-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.602\times 10^{-19})(10^{15})(1.774\times 10^{-4})}{1.0443\times 10^{-12}}\\ & =\boxed{-2.72\times 10^4\text{ V/cm}}\end{array}$$

(d) Potential at $x=0$:

$$\begin{array}{rl}V(0)& =(V_{bi}-V_A)\frac{N_D}{N_A+N_D}=(2.656\text{ V})\left(\frac{1}{11}\right)\\ & =\boxed{0.241\text{ V}}\end{array}$$

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Problem 5.7: A Step-Junction Program

Problem. For a nondegenerately doped Si pn step junction at $T=300\text{ K}$: (a) Compute and present coordinated plots of $\mathcal{E}(x)$ and $V(x)$ for $N_A=10^{16}{\text{ cm}}^{-3},N_D=2\times 10^{14}{\text{ cm}}^{-3},V_A=-20\text{ V}$. (b) Overlay several $V_A$ values: $V_A=V_{A0}/2^n,n=0,\dots ,3$. (c) Add an output list of parameters. (d) Explore variations with relative doping. (e) Check against Problems 5.4 and 5.5.

Solution.

(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

(b) Multiple-bias overlay:

% 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

(c) Parameter list output:

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);

(d) Observations with relative doping:

• When $N_A=N_D$: $\mathcal{E}(x)$ is a symmetric triangle centered at $x=0$; $V(x)$ is a symmetric S-curve with $V(0)=(V_{bi}-V_A)/2$.
• When $N_A\gg N_D$: The depletion region extends almost entirely into the n-side ($x_n\approx W,x_p\approx 0$). The electric field is a right triangle with vertical slope on the p-side; almost all potential drops across the n-side.

(e) Checking Problems 5.4 and 5.5: Entering inputs for 5.4 gives $V_{bi}=0.614\text{ V},W=1.10\mu \text{m},\mathcal{E}(0)=-1.12\times 10^4\text{ V/cm},V(0)=0.205\text{ V}$. Entering inputs for 5.5 gives $V_{bi}=0.716\text{ V},W=0.971\mu \text{m},\mathcal{E}(0)=-1.47\times 10^4\text{ V/cm},V(0)=0.007\text{ V}$, confirming exact agreement.

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Problem 5.8: Combination Profile

Problem. Diode doping: $N_D-N_A=-N_A$ for $x<0$ and $N_D-N_A=ax$ for $x>0$. Using the depletion approximation, sketch $\rho ,\mathcal{E},V$.

Solution.

1. Charge density $\rho (x)$: $$\rho (x)=\{\begin{array}{ll}-q{N}_A,& -x_p\le x<0\\ +qax,& 0\le x\le x_n\\ 0,& \text{otherwise}\end{array}$$ Neutrality condition: $${\int }_{-x_p}^{x_n}\rho (x)dx=0⟹q{N}_A{x}_p=\frac{1}{2}qa{x}_n^2⟹\boxed{N_A{x}_p=\frac{a{x}_n^2}{2}}$$
2. Electric field $\mathcal{E}(x)$: Integrate Poisson’s equation $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$:
   • For $-x_p\le x\le 0$: $$\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}(x+x_p)$$
   • For $0\le x\le x_n$: $$\mathcal{E}(x)=-\frac{qa}{2K_S{\epsilon }_0}(x_n^2-x^2)$$
   Continuous at $x=0$ with peak field $\mathcal{E}(0)=-q{N}_A{x}_p/K_S{\epsilon }_0=-qa{x}_n^2/(2K_S{\epsilon }_0)$.
3. Potential $V(x)$: Integrate $dV/dx=-\mathcal{E}(x)$ with $V(-x_p)=0$:
   • For $-x_p\le x\le 0$: $$V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}(x+x_p{)}^2$$
   • For $0\le x\le x_n$: $$V(x)=V(0)+\frac{qa}{2K_S{\epsilon }_0}\left(x_n^{2}x-\frac{x^3}{3}\right)$$
   Total potential drop: $$V(x_n)=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}+\frac{qa{x}_n^3}{3K_S{\epsilon }_0}=V_{bi}-V_A$$

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Problem 5.9: The Exponential Profile

Problem. A pn diode has $N_D-N_A=N_0\left[1-e^{-\alpha x}\right]$. (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 assumes carrier concentrations within the space charge layer $-x_p\le x\le x_n$ are negligible ($n\approx 0,p\approx 0$), making $\rho (x)=q(N_D-N_A)$, while outside this layer the semiconductor is neutral ($\rho \equiv 0$).

(b) Charge density sketch:

• For $-x_p\le x\le x_n$: $$\rho (x)=q{N}_0\left(1-e^{-\alpha x}\right)$$
• At $x=0$: $\rho (0)=0$.
• For $x>0$: $\rho (x)>0$, rising toward $+q{N}_0$.
• For $x<0$: $e^{-\alpha x}=e^{\alpha |x|}>1⟹\rho (x)=-q{N}_0(e^{\alpha |x|}-1)<0$, plunging steeply negative.
• Neutrality requires equal positive and negative areas.

(c) Electric field:

1. Poisson’s equation: $$\frac{d\mathcal{E}}{dx}=\frac{\rho (x)}{K_S{\epsilon }_0}=\frac{q{N}_0}{K_S{\epsilon }_0}\left(1-e^{-\alpha x}\right)$$
2. Integrate from boundary $\mathcal{E}(-x_p)=0$: $$\begin{array}{rl}{\int }_0^{\mathcal{E}(x)}d{\mathcal{E}}^{\prime }& =\frac{q{N}_0}{K_S{\epsilon }_0}{\int }_{-x_p}^x\left(1-e^{-\alpha x^{\prime }}\right)d{x}^{\prime }\\ \mathcal{E}(x)& =\frac{q{N}_0}{K_S{\epsilon }_0}{\left[x^{\prime }+\frac{1}{\alpha }{e}^{-\alpha x^{\prime }}\right]}_{-x_p}^x\\ & =\boxed{\frac{q{N}_0}{K_S{\epsilon }_0}\left[(x+x_p)+\frac{e^{-\alpha x}-e^{\alpha x_p}}{\alpha }\right],-x_p\le x\le x_n}\end{array}$$

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Problem 5.10: A Stepped n-Side Profile

Problem. Diode with $N_D-N_A=-N_A$ for $x<0$, $N_D/2$ for $0<x<x_0$, and $N_D$ for $x>x_0$. Assume $x_n>x_0$. (a) What is $V_{bi}$? (b) Using the depletion approximation, sketch $\rho (x)$.

Solution.

(a) Built-in voltage: $V_{bi}$ depends exclusively on the equilibrium carrier concentrations in the neutral bulk regions far from the junction:

• In neutral p-region ($x<-x_p$): $p_0=N_A⟹E_i-E_F=kT\ln (N_A/n_i)$.
• In neutral n-region ($x>x_n>x_0$): $n_0=N_D⟹E_F-E_i=kT\ln (N_D/n_i)$. Summing both contributions:

$$\boxed{V_{bi}=\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)}$$

(b) Charge density:

$$\rho (x)=\{\begin{array}{ll}-q{N}_A,& -x_p\le x<0\\ +q{N}_D/2,& 0\le x<x_0\\ +q{N}_D,& x_0\le x\le x_n\\ 0,& \text{otherwise}\end{array}$$

Neutrality condition:

$$\begin{array}{rl}q{N}_A{x}_p& =q\frac{N_D}{2}{x}_0+q{N}_D(x_n-x_0)\\ N_A{x}_p& =N_D\left(x_n-\frac{x_0}{2}\right)⟹\boxed{x_n=\frac{N_A}{N_D}{x}_p+\frac{x_0}{2}}\end{array}$$

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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 an intrinsic region ($-x_i/2\le x\le x_i/2$) with $N_D-N_A=0$. (a) Sketch $\rho ,\mathcal{E},V$ and band diagram. (b) Built-in voltage? (c) Quantitative expressions for $\rho ,\mathcal{E},V$ and depletion widths.

Solution.

(a) Sketches:

• $\rho (x)$: Negative rectangular block $-q{N}_A$ on the p-side; zero across the intrinsic layer; positive block $+q{N}_D$ on the n-side.
• $\mathcal{E}(x)$: Decreases linearly on the p-side, stays constant at peak $-{\mathcal{E}}_0$ across the intrinsic layer, and increases linearly back to zero on the n-side (trapezoidal profile).
• $V(x)$: Parabolic on the p-side, linear ramp across the intrinsic layer, parabolic on the n-side.

(b) Built-in voltage:

$$\boxed{V_{bi}=\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)}$$

(c) Quantitative expressions:

1. Depletion widths: $$q{N}_A{x}_p=q{N}_D{x}_n⟹x_n=\frac{N_A}{N_D}{x}_p$$
2. Peak field in i-region: $${\mathcal{E}}_0=\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}$$
3. Voltage drop across each region: $$\Delta V_p=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0},\Delta V_i={\mathcal{E}}_0{x}_i=\frac{q{N}_A{x}_p{x}_i}{K_S{\epsilon }_0},\Delta V_n=\frac{q{N}_D{x}_n^2}{2K_S{\epsilon }_0}=\frac{q{N}_A^2{x}_p^2}{2K_S{\epsilon }_0{N}_D}$$
4. Set total drop to $V_{bi}-V_A$: $$\frac{q{N}_A}{2K_S{\epsilon }_0}\left(1+\frac{N_A}{N_D}\right)x_p^2+\frac{q{N}_A{x}_i}{K_S{\epsilon }_0}{x}_p-(V_{bi}-V_A)=0$$
5. Solve quadratic equation for $x_p$: $$\boxed{x_p=\frac{1}{1+N_A/N_D}\left[-x_i+\sqrt{x_i^2+\frac{2K_S{\epsilon }_0(V_{bi}-V_A)}{q{N}_A}\left(1+\frac{N_A}{N_D}\right)}\right]}$$ $$\boxed{x_n=\frac{N_A}{N_D}{x}_p}$$

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Problem 5.12: From a Given $V(x)$ Back to the Doping

Problem. In equilibrium:

$$V(x)=\frac{V_{bi}}{2}\left[1+\sin \left(\frac{\pi x}{W}\right)\right],-\frac{W}{2}\le x\le \frac{W}{2}$$

Find: (a) $\mathcal{E}(x)$; (b) $\rho (x)$; (c) $N_D-N_A$.

Solution.

(a) Electric field:

$$\begin{array}{rl}\mathcal{E}(x)& =-\frac{dV}{dx}=-\frac{d}{dx}\left\{\frac{V_{bi}}{2}\left[1+\sin \left(\frac{\pi x}{W}\right)\right]\right\}\\ & =\boxed{-\frac{\pi V_{bi}}{2W}\cos \left(\frac{\pi x}{W}\right),-\frac{W}{2}\le x\le \frac{W}{2}}\end{array}$$

(b) Charge density:

$$\begin{array}{rl}\rho (x)& =K_S{\epsilon }_0\frac{d\mathcal{E}}{dx}=K_S{\epsilon }_0\frac{d}{dx}\left[-\frac{\pi V_{bi}}{2W}\cos \left(\frac{\pi x}{W}\right)\right]\\ & =\boxed{\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2W^2}\sin \left(\frac{\pi x}{W}\right),-\frac{W}{2}\le x\le \frac{W}{2}}\end{array}$$

(c) Net impurity profile:

$$\begin{array}{rl}{N}_D-N_A& =\frac{\rho (x)}{q}\\ & =\boxed{\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2q{W}^2}\sin \left(\frac{\pi x}{W}\right),-\frac{W}{2}\le x\le \frac{W}{2}}\end{array}$$

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Problem 5.13: Linearly Graded Junction by Computer

Problem. Linearly graded junction with $a=10^{20}{\text{ cm}}^{-4},V_{A0}=-4\text{ V}$. Compute $V_{bi}$ iteratively, plot $\mathcal{E}(x)$ and $V(x)$, and study variations.

Solution.

Algorithm for $V_{bi}$:

1. Initial guess $V_{bi}^{(0)}=0.6\text{ V}$.
2. Update: $$W^{(k)}={\left[\frac{12K_S{\epsilon }_0{V}_{bi}^{(k)}}{qa}\right]}^{1/3}$$ $$V_{bi}^{(k+1)}=\frac{2kT}{q}\ln \left(\frac{a{W}^{(k)}}{2n_i}\right)$$
3. Iterate until convergence.

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)');

Converged values ($a=10^{20}{\text{ cm}}^{-4}$):

$$W_0\approx 0.82\mu \text{m},V_{bi}\approx 0.67\text{ V},\mathcal{E}(0)\approx -1.3\times 10^4\text{ V/cm}$$

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Problem 5.14: The “Exact” Equilibrium Step-Junction Solution

Problem. Without the depletion approximation, solve the equilibrium step junction using the $F$-function.

Solution.

Normalized potential and scale definitions:

$$U=\frac{qV}{kT},L_I=\sqrt{\frac{K_S{\epsilon }_{0}kT}{2q^2{n}_i}}$$ $$U_{FP}=\ln \left(\frac{N_A}{n_i}\right),U_{FN}=-\ln \left(\frac{N_D}{n_i}\right),U_{BI}=U_{FP}-U_{FN}=\frac{q{V}_{bi}}{kT}$$

Electric field function:

$$F(U,U_F)={\left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^U-U-1\right)\right]}^{1/2}$$

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 ($x\ge 0$):

$$x=L_I{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime }-U_{BI},U_{FN})},\mathcal{E}=-\frac{kT}{q{L}_I}F(U-U_{BI},U_{FN})$$

Matching at $x=0$: Continuity of the electric field requires:

$$\boxed{F(U_0,U_{FP})=F(U_0-U_{BI},U_{FN})}$$

Sample MATLAB implementation:

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));

Expected behavior:

• $\mathcal{E}(x)$: Smooth, rounded peak at $x=0$ with exponential asymptotic decay into neutral regions rather than abrupt termination.
• $\rho /q$: Smooth transitions rounding off over several extrinsic Debye lengths $L_D=\sqrt{K_S{\epsilon }_{0}kT/q^{2}N}$ rather than discontinuous step edges.

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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 goal of this section is to derive the ideal diode equation

$$I=I_0\left(e^{q{V}_A/kT}-1\right)$$

6.1.1 Qualitative Derivation

Equilibrium ($V_A=0$). Majority carriers surmount the barrier via diffusion; minority carriers are swept down the barrier via drift:

$$I_N^{\text{diff}}=I_N^{\text{drift}},I_P^{\text{diff}}=I_P^{\text{drift}}⟹I=0$$

Forward bias ($V_A>0$). The barrier is reduced from $q{V}_{bi}$ to $q(V_{bi}-V_A)$. Carrier density with sufficient energy to surmount the lowered barrier increases exponentially by the Boltzmann factor:

$$I\propto e^{q{V}_A/kT}$$

Reverse bias ($V_A<0$). The barrier height increases to $q(V_{bi}+|V_A|)$. Majority diffusion drops to zero. Minority carrier drift is supply-limited and saturates at $-I_0$:

$$\boxed{I=I_0\left(e^{q{V}_A/kT}-1\right)}\text{\hspace{1em}(6.1)}$$

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6.1.2 Quantitative Solution Strategy

Assumptions

1. Steady-state operation.
2. Nondegenerately doped step junction.
3. One-dimensional geometry.
4. Low-level injection in quasineutral regions.
5. No processes other than drift, diffusion, and thermal R-G ($G_L=0$).

Current relations

$$I=AJ\text{\hspace{1em}(6.2)}$$ $$J=J_N(x)+J_P(x)\text{\hspace{1em}(6.3)}$$

In quasineutral regions ($\mathcal{E}\approx 0$):

$$0=D_N\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n},x\le -x_p\text{\hspace{1em}(6.5a)}$$ $$0=D_P\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p},x\ge x_n\text{\hspace{1em}(6.5b)}$$ $$J_N=q{D}_N\frac{d\Delta n_p}{dx},x\le -x_p\text{\hspace{1em}(6.6a)}$$ $$J_P=-q{D}_P\frac{d\Delta p_n}{dx},x\ge x_n\text{\hspace{1em}(6.6b)}$$

In the depletion region, assuming zero thermal R-G:

$$\frac{d{J}_N}{dx}=0,\frac{d{J}_P}{dx}=0⟹J_N(-x_p\le x\le x_n)=J_N(-x_p),J_P(-x_p\le x\le x_n)=J_P(x_n)\text{\hspace{1em}(6.8a,b)}$$ $$J=J_N(-x_p)+J_P(x_n)\text{\hspace{1em}(6.9)}$$

Boundary conditions

At the contacts:

$$\Delta n_p(x\to -\infty )=0\text{\hspace{1em}(6.10a)}$$ $$\Delta p_n(x\to +\infty )=0\text{\hspace{1em}(6.10b)}$$

Law of the junction at depletion edges:

$$\boxed{np=n_i^2{e}^{q{V}_A/kT},-x_p\le x\le x_n}\text{\hspace{1em}(6.12)}$$

At $x=-x_p$ ($p\approx N_A$):

$$\Delta n_p(-x_p)=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.15)}$$

At $x=x_n$ ($n\approx N_D$):

$$\Delta p_n(x_n)=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)\text{\hspace{1em}(6.18)}$$

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6.1.3 Derivation Proper

Using coordinates $x^{\prime }=x-x_n\ge 0$ on the n-side and $x^{\prime \prime }=-x-x_p\ge 0$ on the p-side:

Step 1: Hole diffusion equation:

$$D_P\frac{d^2\Delta p_n}{d{x}^{\prime 2}}-\frac{\Delta p_n}{{\tau }_p}=0⟹\Delta p_n(x^{\prime })=A_1{e}^{-x^{\prime }/L_P}+A_2{e}^{x^{\prime }/L_P}$$

Step 2: Apply boundary conditions $\Delta p_n(\infty )=0⟹A_2=0$ and $\Delta p_n(0)$:

$$\boxed{\Delta p_n(x^{\prime })=\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime }/L_P},x^{\prime }\ge 0}\text{\hspace{1em}(6.23)}$$

Step 3: Current density:

$$J_P(x^{\prime })=-q{D}_P\frac{d\Delta p_n}{d{x}^{\prime }}=q\frac{D_P}{L_P}\frac{n_i^2}{N_D}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime }/L_P}\text{\hspace{1em}(6.24)}$$

Step 4: By exact symmetry on the p-side:

$$\Delta n_p(x^{\prime \prime })=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime \prime }/L_N},x^{\prime \prime }\ge 0\text{\hspace{1em}(6.25)}$$ $$J_N(x^{\prime \prime })=q\frac{D_N}{L_N}\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)e^{-x^{\prime \prime }/L_N}\text{\hspace{1em}(6.26)}$$

Step 5: Evaluate at depletion edges ($x^{\prime }=0,x^{\prime \prime }=0$) and sum:

$$I=AJ=A\left[J_N(-x_p)+J_P(x_n)\right]=\boxed{I_0\left(e^{q{V}_A/kT}-1\right)}\text{\hspace{1em}(6.29)}$$

where

$$\boxed{I_0=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)}\text{\hspace{1em}(6.30)}$$

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6.1.4 Examination of Results

For asymmetrical junctions:

$$I_0\approx qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}(p^{+}\text{-n diode})\text{\hspace{1em}(6.32a)}$$ $$I_0\approx qA\frac{D_N}{L_N}\frac{n_i^2}{N_A}(n^{+}\text{-p diode})\text{\hspace{1em}(6.32b)}$$

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Exercise 6.1

Problem. Two ideal $p^{+}$-$n$ step junction diodes are identical except $N_{D1}=10^{15}{\text{ cm}}^{-3}$ and $N_{D2}=10^{16}{\text{ cm}}^{-3}$. Compare $I$-$V$ characteristics.

Solution.

1. Saturation current for $p^{+}$-n diodes: $$I_0\approx qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}=qA\sqrt{\frac{D_P}{{\tau }_p}}\frac{n_i^2}{N_D}$$
2. Take ratio assuming identical areas, lifetimes, and mobilities: $$\begin{array}{rl}\frac{I_{0,1}}{I_{0,2}}& =\frac{N_{D2}}{N_{D1}}=\frac{10^{16}{\text{ cm}}^{-3}}{10^{15}{\text{ cm}}^{-3}}\\ & =\boxed{10}\end{array}$$
3. Diode 1 conducts 10 times more current than Diode 2 at every voltage: $$\boxed{I_1(V_A)=10I_2(V_A)}$$

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Exercise 6.2 (Computer): Ideal $I$-$V$ Versus Doping

% 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

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Exercise 6.3

Problem. Diode carrier concentrations at $300\text{ K}$: $n_n(\infty )=10^{15}{\text{ cm}}^{-3}$, $p_n(x_n)=10^{10}{\text{ cm}}^{-3}$, $p_n(\infty )=10^5{\text{ cm}}^{-3}$; drops to $10^8{\text{ cm}}^{-3}$ at $x^{\prime }=1.6\times 10^{-2}\text{ cm}$. (a) Forward or reverse biased? (b) Does low-level injection hold? (c) Determine $V_A$. (d) Determine $L_P$.

Solution.

(a) Since $p_n(x_n)=10^{10}{\text{ cm}}^{-3}>p_{n0}=10^5{\text{ cm}}^{-3}$, there is a minority carrier pile-up:

$$\boxed{\text{Forward biased}}$$

(b) The maximum excess minority carrier concentration is $\Delta p_n(0)\approx 10^{10}{\text{ cm}}^{-3}\ll n_{n0}=10^{15}{\text{ cm}}^{-3}$:

$$\boxed{\text{Yes, low-level injection holds.}}$$

(c) Law of the junction:

$$\begin{array}{rl}{V}_A& =\frac{kT}{q}\ln \left[\frac{p_n(x_n)}{p_{n0}}\right]=(0.0259\text{ V})\ln \left(\frac{10^{10}{\text{ cm}}^{-3}}{10^5{\text{ cm}}^{-3}}\right)\\ & =(0.0259)\ln (10^5)=(0.0259)(11.513)\\ & =\boxed{0.298\text{ V}\approx 0.30\text{ V}}\end{array}$$

(d) Minority carrier profile decay:

$$\begin{array}{rl}{p}_n(x^{\prime })& =p_n(0)e^{-x^{\prime }/L_P}⟹\ln \left[\frac{p_n(0)}{p_n(x^{\prime })}\right]=\frac{x^{\prime }}{L_P}\\ L_P& =\frac{x^{\prime }}{\ln [p_n(0)/p_n(x^{\prime })]}=\frac{1.6\times 10^{-2}\text{ cm}}{\ln (10^{10}/10^8)}=\frac{1.6\times 10^{-2}}{\ln (100)}=\frac{1.6\times 10^{-2}}{4.605}\\ & =\boxed{3.47\times 10^{-3}\text{ cm}}(34.7\mu \text{m})\end{array}$$

---

Exercise 6.4

Problem. Derive the reverse-bias saturation current by minority carrier generation in squared-off depletion zones of width $L_N$ and $L_P$.

Solution.

In the reverse-biased quasineutral regions, mobile minority carriers are completely swept into the junction, so $\Delta n_p=-n_{p0}=-n_i^2/N_A$ and $\Delta p_n=-p_{n0}=-n_i^2/N_D$. The net thermal generation rate is:

$${\frac{\partial n}{\partial t}|}_{\text{R-G}}=-\frac{\Delta n_p}{{\tau }_n}=+\frac{n_i^2}{N_A{\tau }_n},-L_N-x_p\le x\le -x_p$$ $${\frac{\partial p}{\partial t}|}_{\text{R-G}}=-\frac{\Delta p_n}{{\tau }_p}=+\frac{n_i^2}{N_D{\tau }_p},x_n\le x\le x_n+L_P$$

1. Total generation in volume $A{L}_N$ on the p-side and $A{L}_P$ on the n-side: $$I_0=q(A{L}_N)\left(\frac{n_i^2}{N_A{\tau }_n}\right)+q(A{L}_P)\left(\frac{n_i^2}{N_D{\tau }_p}\right)$$
2. Using $L_N/{\tau }_n=D_N/L_N$ and $L_P/{\tau }_p=D_P/L_P$: $$\boxed{I_0=qA\left(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D}\right)}$$

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 $N_B$,

$$\boxed{V_{BR}\propto \frac{1}{N_B}}\text{\hspace{1em}(6.39)}$$

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.

---

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 $V_{BR}$:

$$\begin{array}{rlrl}M& =\frac{1}{1-x^m}=2& & \text{(set multiplication factor to 2)}\\ 1-x^m& =\frac{1}{2}⟹x^m=\frac{1}{2}& & \text{(solve for xm)}\\ x& =(0.5{)}^{1/m}=\{\begin{array}{ll}(0.5{)}^{1/3}\approx 0.794& (m=3)\\ (0.5{)}^{1/6}\approx 0.891& (m=6)\end{array}& & \text{(voltage ratio for current doubling)}\end{array}$$

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:

$$\begin{array}{rlrl}M& =\frac{1}{1-x^m}=1.1& & \text{(set multiplication factor to 1.1)}\\ 1-x^m& =\frac{1}{1.1}⟹x^m=1-\frac{1}{1.1}=\frac{0.1}{1.1}\approx 0.0909& & \text{(solve for xm)}\\ x& =(0.0909{)}^{1/m}=\{\begin{array}{ll}(0.0909{)}^{1/3}\approx 0.450& (m=3)\\ (0.0909{)}^{1/6}\approx 0.667& (m=6)\end{array}& & \text{(voltage ratio for 10\% enhancement)}\end{array}$$ $$\boxed{x(M=2)=\{\begin{array}{ll}0.794& (m=3)\\ 0.891& (m=6)\end{array},x(M=1.1)=\{\begin{array}{ll}0.450& (m=3)\\ 0.667& (m=6)\end{array}}$$

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.

---

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, $kT/q=0.02585\text{ V}$):

$$\begin{array}{rlrl}{\mu }_p(10^{16})& \approx 437{\text{ cm}}^2/\text{V}\cdot \text{s}& & \text{(hole mobility from empirical fit)}\\ D_P& =\left(\frac{kT}{q}\right){\mu }_p=(0.02585\text{ V})(437{\text{ cm}}^2/\text{V}\cdot \text{s})=11.3{\text{ cm}}^2/\text{s}& & \text{(hole diffusion coefficient)}\\ L_P& =\sqrt{D_P{\tau }_p}=\sqrt{(11.3{\text{ cm}}^2/\text{s})(10^{-6}\text{ s})}=3.36\times 10^{-3}\text{ cm}& & \text{(hole diffusion length)}\\ \frac{D_P}{L_P}& =\frac{11.3}{3.36\times 10^{-3}}=3.36\times 10^3\text{ cm/s}& & \text{(diffusion velocity)}\\ I_{01}& =qA\left(\frac{D_P}{L_P}\right)\frac{n_i^2}{N_D}& & \text{(diffusion current prefactor)}\\ & =(1.6\times 10^{-19}\text{ C})(10^{-4}{\text{ cm}}^2)(3.36\times 10^3\text{ cm/s})\frac{(10^{10}{\text{ cm}}^{-3}{)}^2}{10^{16}{\text{ cm}}^{-3}}& & \text{(substitute given parameters)}\\ & =5.38\times 10^{-16}\text{ A}& & \text{(evaluate I01 for Si)}\end{array}$$ $$\begin{array}{rlrl}{V}_{bi}& =\frac{E_G}{2q}+\frac{kT}{q}\ln \left(\frac{N_D}{n_i}\right)=0.56\text{ V}+0.02585\ln \left(\frac{10^{16}}{10^{10}}\right)& & \text{(built-in potential formula)}\\ & =0.56+0.02585\ln (10^6)=0.56+0.357=0.917\text{ V}\approx 0.92\text{ V}& & \text{(evaluate Vbi)}\\ V_A& =\frac{V_{bi}}{4}=\frac{0.92}{4}=0.23\text{ V},V_{bi}-V_A=\frac{3V_{bi}}{4}=0.69\text{ V}& & \text{(evaluate at VA=Vbi/4)}\\ W& =\sqrt{\frac{2K_S{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)}=\sqrt{\frac{2(11.8)(8.85\times 10^{-14}\text{ F/cm})}{(1.6\times 10^{-19}\text{ C})(10^{16}{\text{ cm}}^{-3})}(0.69\text{ V})}& & \text{(depletion width formula)}\\ & \approx 3.0\times 10^{-5}\text{ cm}& & \text{(evaluate depletion width)}\\ I_{02}& =\frac{qA{n}_{i}W}{{\tau }_0}\frac{kT/q}{V_{bi}-V_A}& & \text{(R-G current prefactor)}\\ & =\frac{(1.6\times 10^{-19}\text{ C})(10^{-4}{\text{ cm}}^2)(10^{10}{\text{ cm}}^{-3})(3.0\times 10^{-5}\text{ cm})}{10^{-6}\text{ s}}\cdot \frac{0.02585\text{ V}}{0.69\text{ V}}& & \text{(substitute parameters)}\\ & \approx 2.08\times 10^{-13}\text{ A}& & \text{(evaluate I02 for Si)}\end{array}$$ $$\boxed{\text{Si: }{I}_{01}\approx 5.38\times 10^{-16}\text{ A},I_{02}\approx 2.08\times 10^{-13}\text{ A}}$$ $$\boxed{\text{Ge: }{I}_{01}\approx 6.23\times 10^{-9}\text{ A},I_{02}\approx 8.34\times 10^{-10}\text{ A}}$$

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:

$$\begin{array}{rl}\text{Diffusion region (Line 1): }& I_{01}\approx 10^{-14}\text{ A},n_1\approx 1.1\\ \text{R-G region (Line 2): }& I_{02}\approx 10^{-12}\text{ A},n_2\approx 1.5\end{array}$$ $$\boxed{I_{01}\approx 10^{-14}\text{ A},n_1\approx 1.1;I_{02}\approx 10^{-12}\text{ A},n_2\approx 1.5}$$

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

$$\begin{array}{rlrl}\Delta V& \approx V_A-V_J=0.85\text{ V}-0.78\text{ V}=0.07\text{ V}& & \text{(voltage displacement at I=80 mA)}\\ R_S& =\frac{\Delta V}{I}\approx \frac{0.07\text{ V}}{0.08\text{ A}}\approx 0.875\Omega \approx 1\Omega & & \text{(series resistance estimate)}\end{array}$$ $$\boxed{\Delta V\approx 0.07\text{ V},R_S\approx 1\Omega }$$

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

$$\begin{array}{rlrl}{x}_n& ={\left[\frac{2K_S{\epsilon }_0}{q{N}_D}\left(V_{bi}-V_A\right)\right]}^{1/2}=x_c& & \text{(punch-through condition: xn=xc)}\\ x_c^2& =\frac{2K_S{\epsilon }_0}{q{N}_D}\left(V_{bi}-V_A\right)& & \text{(square both sides)}\\ V_A& =V_{bi}-\frac{q{N}_D{x}_c^2}{2K_S{\epsilon }_0}& & \text{(solve for applied voltage)}\\ & =0.92\text{ V}-\frac{(1.6\times 10^{-19}\text{ C})(10^{16}{\text{ cm}}^{-3})(2\times 10^{-4}\text{ cm}{)}^2}{2(11.8)(8.85\times 10^{-14}\text{ F/cm})}& & \text{(substitute given parameters)}\\ & =0.92\text{ V}-\frac{6.4\times 10^{-11}}{2.089\times 10^{-12}}\text{ V}& & \text{(evaluate numerator and denominator)}\\ & =0.92\text{ V}-30.64\text{ V}=-29.72\text{ V}\approx -29.7\text{ V}& & \text{(compute punch-through voltage)}\end{array}$$ $$\boxed{V_{\text{PT}}=-29.7\text{ V}(\text{punch-through occurs at }29.7\text{ V reverse bias})}$$

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 the equilibrium concentration $n_{p0}=n_i^2/N_A$ gives:

$$\begin{array}{rlrl}\Delta n_p(-x_p)& =n(-x_p)-n_{p0}& & \text{(definition of excess minority carrier concentration)}\\ & =\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}-\frac{n_i^2}{N_A}& & \text{(substitute boundary concentration and equilibrium value)}\\ & =\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)& & \text{(factor out }{n}_i^2/N_A\text{)}\end{array}$$ $$\boxed{\Delta n_p(-x_p)=\frac{n_i^2}{N_A}\left(e^{q{V}_A/kT}-1\right)}$$

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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. ${\mu }_n(10^{15})\approx 1345{\text{cm}}^2/\text{V}\cdot \text{s}$:

$$\begin{array}{rlrl}{D}_N& =\left(\frac{kT}{q}\right){\mu }_n=(0.02585\text{ V})(1345{\text{ cm}}^2/\text{V}\cdot \text{s})=34.8{\text{ cm}}^2/\text{s}& & \text{(electron diffusion coefficient)}\\ L_N& =\sqrt{D_N{\tau }_n}=\sqrt{(34.8{\text{ cm}}^2/\text{s})(10^{-6}\text{ s})}=5.90\times 10^{-3}\text{ cm}& & \text{(minority electron diffusion length)}\\ I_0& =qA\left(\frac{D_N}{L_N}\right)\frac{n_i^2}{N_A}& & \text{(ideal diode saturation current for n+-p)}\\ & =(1.6\times 10^{-19}\text{ C})(10^{-3}{\text{ cm}}^2)\left(\frac{34.8{\text{ cm}}^2/\text{s}}{5.90\times 10^{-3}\text{ cm}}\right)\frac{(10^{10}{\text{ cm}}^{-3}{)}^2}{10^{15}{\text{ cm}}^{-3}}& & \text{(substitute given numerical values)}\\ & =9.4\times 10^{-14}\text{ A}& & \text{(evaluate saturation current)}\end{array}$$ $$\boxed{I_0(300\text{ K})=9.4\times 10^{-14}\text{ A}}$$

$V_A$	$e^{q{V}_A/kT}-1$	$I$
$-50$ 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. $kT/q=0.0431\text{ V}$; $D_N=(0.0431)(1345)=58.0{\text{ cm}}^2/\text{s}$ (assuming same mobility), $L_N=7.6\times 10^{-3}\text{ cm}$:

$$\begin{array}{rlrl}\frac{kT}{q}& =\frac{(8.617\times 10^{-5}\text{ eV/K})(500\text{ K})}{1\text{ e}}=0.0431\text{ V}& & \text{(thermal voltage at 500 K)}\\ D_N& =(0.0431\text{ V})(1345{\text{ cm}}^2/\text{V}\cdot \text{s})=58.0{\text{ cm}}^2/\text{s}& & \text{(diffusion coefficient at 500 K)}\\ L_N& =\sqrt{D_N{\tau }_n}=\sqrt{(58.0)(10^{-6})}=7.62\times 10^{-3}\text{ cm}& & \text{(diffusion length at 500 K)}\\ I_0& =(1.6\times 10^{-19}\text{ C})(10^{-3}{\text{ cm}}^2)\left(\frac{58.0}{7.62\times 10^{-3}}\right)\frac{(1.25\times 10^{14}{\text{ cm}}^{-3}{)}^2}{10^{15}{\text{ cm}}^{-3}}& & \text{(substitute elevated temperature values)}\\ & \approx 1.9\times 10^{-5}\text{ A}& & \text{(evaluate elevated saturation current)}\end{array}$$ $$\boxed{I_0(500\text{ K})\approx 1.9\times 10^{-5}\text{ A}}$$

$V_A$	$e^{q{V}_A/kT}-1$	$I$
$-50$ 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:

$$\begin{array}{rlrl}\frac{1}{I_0}\frac{d{I}_0}{dT}& \approx \frac{d}{dT}\ln \left(T^3{e}^{-E_G/kT}\right)& & \text{(dominant temperature dependence from ni2)}\\ & =\frac{3}{T}+\frac{E_G}{k{T}^2}=\frac{3+E_G/kT}{T}& & \text{(differentiate logarithm)}\\ & =\frac{3+\frac{1.12}{(8.617\times 10^{-5})(300)}}{300}=\frac{3+43.32}{300}& & \text{(substitute T=300 K, EG=1.12 eV)}\\ & \approx 0.154{\text{ K}}^{-1}& & \text{(evaluate fractional rate of increase)}\end{array}$$ $$\boxed{\frac{1}{I_0}\frac{d{I}_0}{dT}\approx 15.4\%\text{ per Kelvin at }300\text{ K}}$$

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, ${\mu }_p=437{\text{ cm}}^2/\text{V}\cdot \text{s}$, $D_P=11.3{\text{ cm}}^2/\text{s}$, $L_P=3.36\times 10^{-3}\text{ cm}$ and $I_0=5.38\times 10^{-16}\text{ A}$ (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 $I_0(295)/I_0(300)\approx 0.47$:

$$\boxed{I_0(300\text{ K})\approx 5.4\times 10^{-16}\text{ A},\frac{I_0(305\text{ K})}{I_0(300\text{ K})}\approx 2.1,\frac{I_0(295\text{ K})}{I_0(300\text{ K})}\approx 0.47}$$

(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 $D_P{\tau }_p/L_P=L_P^2/L_P=L_P$:

$$\boxed{I=I_0\left(e^{q{V}_A/kT}-1\right)+I_L,I_L=-qA{L}_P{G}_L,I_0=qA\frac{D_P}{L_P}\frac{n_i^2}{N_D}}$$

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 $s=10^3$: $V_{OC}=0.0259\ln (1001)\approx 0.18\text{ V}$, $0.20\text{ V}$, $0.22\text{ V}$, … 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 $V_{OC}$:

$$\boxed{V_{OC}=\frac{kT}{q}\ln \left(1+\frac{|I_L|}{I_0}\right),I_{SC}=I_L}$$

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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 $n_{p0}=n_i^2/N_A$:

$$\begin{array}{rlrl}{N}_D& =\frac{n_i^2}{p_{n0}}=\frac{(10^{10}{\text{ cm}}^{-3}{)}^2}{10^4{\text{ cm}}^{-3}}=\frac{10^{20}}{10^4}=10^{16}{\text{ cm}}^{-3}& & \text{(donor concentration on n-side)}\\ N_A& =\frac{n_i^2}{n_{p0}}=\frac{(10^{10}{\text{ cm}}^{-3}{)}^2}{10^3{\text{ cm}}^{-3}}=\frac{10^{20}}{10^3}=10^{17}{\text{ cm}}^{-3}& & \text{(acceptor concentration on p-side)}\end{array}$$ $$\boxed{N_D=10^{16}{\text{ cm}}^{-3},N_A=10^{17}{\text{ cm}}^{-3}}$$

(consistent with the majority levels $n_n=10^{16}{\text{ cm}}^{-3}$, $p_p=10^{17}{\text{ cm}}^{-3}$ on the plot).

(d) Law of the junction at the n-edge (or equivalently the depletion edge boundary condition):

$$\begin{array}{rlrl}{p}_n(x_n)& =p_{n0}{e}^{q{V}_A/kT}& & \text{(minority carrier concentration at depletion edge)}\\ \frac{p_n(x_n)}{p_{n0}}& =e^{q{V}_A/kT}⟹V_A=\frac{kT}{q}\ln \left(\frac{p_n(x_n)}{p_{n0}}\right)& & \text{(solve for applied voltage }{V}_A\text{)}\\ & =(0.0259\text{ V})\ln \left(\frac{10^6{\text{ cm}}^{-3}}{10^4{\text{ cm}}^{-3}}\right)=(0.0259\text{ V})\ln (100)& & \text{(substitute numerical values)}\\ & =(0.0259\text{ V})(4.6052)\approx 0.119\text{ V}& & \text{(evaluate applied voltage)}\end{array}$$

Check at the p-edge:

$$V_A=(0.0259\text{ V})\ln \left(\frac{10^5{\text{ cm}}^{-3}}{10^3{\text{ cm}}^{-3}}\right)=(0.0259\text{ V})\ln (100)=0.119\text{ V}✓$$ $$\boxed{V_A\approx 0.119\text{ V}}$$

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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):

$$\begin{array}{rlrl}{n}_p(-x_p)& =n_{p0}{e}^{q{V}_A/kT}=(10^3{\text{ cm}}^{-3})(10^{10})=10^{13}{\text{ cm}}^{-3}& & \text{(minority electron concentration at p-edge)}\\ p_n(x_n)& =p_{n0}{e}^{q{V}_A/kT}=(10^4{\text{ cm}}^{-3})(10^{10})=10^{14}{\text{ cm}}^{-3}& & \text{(minority hole concentration at n-edge)}\end{array}$$ $$\boxed{n_p(-x_p)=10^{13}{\text{ cm}}^{-3},p_n(x_n)=10^{14}{\text{ cm}}^{-3}}$$

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 $I=10^{-4}\text{ A}$: at $300\text{ K}$, $I_0=5.38\times 10^{-16}\text{ A}$ and:

$$\begin{array}{rlrl}{V}_A(300\text{ K})& =\frac{kT}{q}\ln \left(\frac{I}{I_0(300\text{ K})}\right)=(0.0259\text{ V})\ln \left(\frac{10^{-4}\text{ A}}{5.38\times 10^{-16}\text{ A}}\right)& & \text{(forward voltage formula)}\\ & =(0.0259\text{ V})\ln \left(1.859\times 10^{11}\right)=(0.0259\text{ V})(25.948)& & \text{(evaluate natural logarithm)}\\ & \approx 0.672\text{ V}& & \text{(evaluate forward voltage)}\end{array}$$ \boxed{V_A(300\text{ K}) \approx 0.672\text{ 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{dV_A}{dT}=\frac{V_A}{T}-\frac{kT}{q}\frac{1}{I_0}\frac{dI_0}{dT}

and, since $I_0\propto n_i^2\propto T^3e^{-E_G/kT}$ (neglecting the weaker mobility and lifetime dependences),

\frac{1}{I_0}\frac{dI_0}{dT}=\frac{3}{T}+\frac{E_G}{kT^2} ;\Rightarrow; \frac{dV_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: - $I=10^{-4}\text{ A}$: $V_A=0.672\text{ V}$,

\begin{aligned} \frac{dV_A}{dT} &= -\frac{1.12\text{ V} - 0.672\text{ V}}{300\text{ K}} - 0.259\text{ mV/K} = -\frac{0.448\text{ V}}{300\text{ K}} - 0.259\text{ mV/K} \ &= -1.493\text{ mV/K} - 0.259\text{ mV/K} \approx -1.75\text{ mV/}^\circ\text{C} \end{aligned}

- $I=10^{-6}\text{ A}$: $V_A=0.0259\ln(1.86\times10^{9})=0.553\text{ V}$,

\begin{aligned} \frac{dV_A}{dT} &= -\frac{1.12\text{ V} - 0.553\text{ V}}{300\text{ K}} - 0.259\text{ mV/K} = -\frac{0.567\text{ V}}{300\text{ K}} - 0.259\text{ mV/K} \ &= -1.890\text{ mV/K} - 0.259\text{ mV/K} \approx -2.15\text{ mV/}^\circ\text{C} \end{aligned}

\boxed{\left.\frac{dV_A}{dT}\right|{10^{-4}\text{ A}} \approx -1.75\text{ mV/}^\circ\text{C}, \qquad \left.\frac{dV_A}{dT}\right|{10^{-6}\text{ A}} \approx -2.15\text{ mV/}

**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 $T\to0$ also shows why all curves of $V_A(T)$ extrapolate to roughly 1.2 V. --- ### 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 $N_D=10^{16}\text{ cm}^{-3}$): $V_{BR}\approx55\text{ V}$. (The empirical fit of Problem 6.14, $60(N_B/10^{16})^{-3/4}$, gives $60\text{ V}$, consistent.) Take $V_{BR}=55\text{ V}$.

\boxed{V_{BR} \approx 55\text{ V}}

**(b)** Breakdown occurs at $V_A=-V_{BR}$, so $V_{bi}-V_A=V_{bi}+V_{BR}=0.92+55=55.9\text{ V}$. For a $p^+$-$n$ junction $W\approx x_n$:

\begin{aligned} W &= \left[\frac{2K_S\varepsilon_0}{qN_D}(V_{bi}+V_{BR})\right]^{1/2} && \text{(one-sided step junction depletion width formula)} \ &= \left[\frac{2(11.8)(8.85\times10^{-14}\text{ F/cm})}{(1.6\times10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})}(55.9\text{ V})\right]^{1/2} && \text{(substitute given parameters)} \ &= \left[(1.305\times10^{-9}\text{ cm}^2/\text{V})(55.9\text{ V})\right]^{1/2} = \left[7.297\times10^{-8}\text{ cm}^2\right]^{1/2} && \text{(evaluate intermediate product)} \ &= 2.70\times10^{-4}\text{ cm} \approx 2.7\ \mu\text{m} && \text{(depletion width at breakdown)} \end{aligned}

\boxed{W \approx 2.70 \times 10^{-4}\text{ cm} = 2.7\ \mu\text{m}}

$$\ast \ast (c)\ast \ast Themaximumfieldisatthemetallurgicaljunction:$$

\begin{aligned} |\mathcal{E}|_{\max} &= \frac{qN_DW}{K_S\varepsilon_0} && \text{(maximum electric field at $x=0$)} \ &= \frac{(1.6\times10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(2.70\times10^{-4}\text{ cm})}{(11.8)(8.85\times10^{-14}\text{ F/cm})} && \text{(substitute charge and permittivity values)} \ &= \frac{4.32\times10^{-7}\text{ C/cm}^2}{1.044\times10^{-12}\text{ F/cm}} \approx 4.14\times10^{5}\text{ V/cm} && \text{(evaluate maximum field magnitude)} \end{aligned}

(Equivalently $|\mathcal{E}|_{\max}=2(V_{bi}+V_{BR})/W=2(55.9\text{ V})/2.70\times10^{-4}\text{ cm}\approx4.14\times10^5\text{ V/cm}$.) This is the critical field $\mathcal{E}_{CR}$ for Si avalanche, a few $\times10^5\text{ V/cm}$. $\checkmark$

\boxed{|\mathcal{E}|_{\max} \approx 4.1 \times 10^5\text{ V/cm}}

--- ### 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 $V_{BR}\approx60(N_B/10^{16})^{-3/4}$ (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 $N_D=2\times10^{16}\,\text{cm}^{-3}$. 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 $N_D=2\times10^{16}$: $V_{BR}=60\cdot2^{-0.75}=35.7$ 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 $M$; $e^{qV_A/kT}\ll1$ for all but the smallest $|V_A|$; to be complete use $I/I_0=M(e^{qV_A/kT}-1)$). (5) Plot $I/I_0$ against $V_A$ and set axes. ```matlab % 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 $N_D=2\times10^{16}\text{ cm}^{-3}$, $m=6$, $V_{BR}=60(2)^{-0.75} \approx 35.7\text{ V}$. $I/I_0=-1$ (ideal saturation) for small $|V_A|$; $I/I_0=-2$ at $|V_A|=0.89V_{BR}=31.8\text{ V}$; $-5$ at $0.97V_{BR}$; diverging at $V_{BR}$:

\boxed{V_{BR} \approx 60\left(\frac{N_D}{10^{16}}\right)^{-3/4} = 60(2)^{-0.75} \approx 35.7\text{ V}} $$ 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 $N_D=2\times 10^{16}{\text{ cm}}^{-3}$:

$$\begin{array}{rlrl}{V}_{bi}& =0.56\text{ V}+0.0259\ln (2\times 10^6)=0.936\text{ V}& & \text{(built-in potential)}\\ V_{BR}& =60(2{)}^{-0.75}\approx 35.7\text{ V}& & \text{(avalanche breakdown voltage)}\\ \frac{V_{BR}}{2}& =17.85\text{ V},M(17.85\text{ V})=\frac{1}{1-(0.5{)}^6}=\frac{1}{1-0.0156}\approx 1.016& & \text{(multiplication at midpoint)}\end{array}$$ $$\boxed{V_{bi}=0.936\text{ V},V_{BR}\approx 35.7\text{ V},M(V_{BR}/2)\approx 1.016}$$

% 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 $T^{\ast }\approx 540\text{ K}$ with $L_P$ taken as the printed $10^{-2}\text{ cm}$. (If instead one uses the computed $L_P=\sqrt{D_P{\tau }_p}\approx 3.5\times 10^{-3}\text{ cm}$, $I_{\text{DIFF}}$ is $\sim 2.9$ times larger and the crossover moves to about $T^{\ast }\approx 500\text{ K}$, inside the quoted range.)

$$\boxed{T^{\ast }\approx 500\text{–}550\text{ K}(230\text{–}270^{\circ }\text{C})}$$

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) $R_S=5\Omega$:

$$\begin{array}{rl}I& =\frac{kT/q}{9R_S}\ln \left(\frac{I}{I_0}\right)=\frac{0.02585\text{ V}}{45\Omega }\ln \left(\frac{I}{10^{-14}\text{ A}}\right)=5.74\times 10^{-4}\ln \left(\frac{I}{10^{-14}\text{ A}}\right)\end{array}$$

Iterating:

$$\begin{array}{rl}\text{Initial guess: }& I^{(0)}=0.02\text{ A}⟹I^{(1)}=5.74\times 10^{-4}\ln (2\times 10^{12})=1.63\times 10^{-2}\text{ A}\\ \text{Next iteration: }& I^{(1)}=0.0163\text{ A}⟹I^{(2)}=5.74\times 10^{-4}\ln (1.63\times 10^{12})\approx 1.61\times 10^{-2}\text{ A}\approx 16\text{ mA}\end{array}$$

At convergence: $V_J=0.728\text{ V}$, $I{R}_S=0.0805\text{ V}$, $V_A=0.809\text{ V}$ ($I{R}_S/V_A=9.95\%\approx 10\%$).

$$\boxed{I(R_S=5\Omega )\approx 16\text{ mA}(V_J=0.728\text{ V},I{R}_S=0.081\text{ V},V_A=0.809\text{ V})}$$

(b) $R_S=20\Omega$:

$$\begin{array}{rl}I& =\frac{kT/q}{9R_S}\ln \left(\frac{I}{I_0}\right)=\frac{0.02585\text{ V}}{180\Omega }\ln \left(\frac{I}{10^{-14}\text{ A}}\right)=1.436\times 10^{-4}\ln \left(\frac{I}{10^{-14}\text{ A}}\right)\end{array}$$

Iterating:

$$\begin{array}{rl}\text{Initial guess: }& I^{(0)}=4\text{ mA}⟹I^{(1)}=1.436\times 10^{-4}\ln (4\times 10^{11})=3.84\text{ mA}\\ \text{Next iteration: }& I^{(1)}=3.84\text{ mA}⟹I^{(2)}=1.436\times 10^{-4}\ln (3.84\times 10^{11})\approx 3.83\text{ mA}\approx 3.8\text{ mA}\end{array}$$

At convergence: $V_J=0.690\text{ V}$, $I{R}_S=0.0766\text{ V}$, $V_A=0.767\text{ V}$ ($I{R}_S/V_A=9.99\%\approx 10\%$).

$$\boxed{I(R_S=20\Omega )\approx 3.8\text{ mA}(V_J=0.69\text{ V},I{R}_S=0.077\text{ V},V_A=0.77\text{ V})}$$

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	$1.2\times 10^{-9}$	2.30

$$\boxed{\text{Two-point: }{I}_{01}\approx 1.4\times 10^{-13}\text{ A},n_1\approx 1.20,I_{02}\approx 1.0\times 10^{-9}\text{ A},n_2\approx 2.25}$$ $$\boxed{\text{Least-squares: }{I}_{01}\approx 1.4\times 10^{-13}\text{ A},n_1\approx 1.20,I_{02}\approx 1.2\times 10^{-9}\text{ A},n_2\approx 2.30}$$

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 $R_S\approx 0.9\text{–}1\Omega$; 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.

$$\boxed{R_S=\frac{d(\Delta V)}{dI}\approx 0.95\Omega \approx 1\Omega }$$

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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) Built-in voltage:

$$\begin{array}{rl}{V}_{bi}& \approx \frac{E_G}{2q}+\frac{kT}{q}\ln \left(\frac{N_D}{n_i}\right)=0.56\text{ V}+0.0259\ln (10^6)=0.56\text{ V}+0.3578\text{ V}\approx 0.918\text{ V}\approx 0.92\text{ V}\end{array}$$ $$\boxed{V_{bi}\approx 0.92\text{ V}}$$

(b) Breakdown voltage is set by the light-side bulk doping $N_D=10^{16}{\text{ cm}}^{-3}$ (denuding changes only $N_T$, not donor doping): from Fig. 6.11,

$$\boxed{V_{BR}\approx 55\text{ V}}$$

(c) Applied voltage to expand depletion region to $x_n=x_c=2\mu \text{m}$:

$$\begin{array}{rl}{x}_c^2& =\frac{2K_S{\epsilon }_0}{q{N}_D}\left(V_{bi}-V_A\right)⟹V_A=V_{bi}-\frac{q{N}_D{x}_c^2}{2K_S{\epsilon }_0}\\ & =0.92\text{ V}-\frac{(1.6\times 10^{-19}\text{ C})(10^{16}{\text{ cm}}^{-3})(2\times 10^{-4}\text{ cm}{)}^2}{2(11.8)(8.85\times 10^{-14}\text{ F/cm})}\\ & =0.92\text{ V}-30.64\text{ V}\approx -29.7\text{ V}\end{array}$$ $$\boxed{V_A=-29.7\text{ V}(\text{edge reaches }{x}_c\text{ at }29.7\text{ V reverse bias})}$$

Since $|-29.7\text{ V}|<55\text{ V}$, this condition is reached well prior to avalanche breakdown.

(d) Lifetime ratio:

$$\begin{array}{rl}\tau & =\frac{1}{c_p{N}_T}⟹\frac{{\tau }_{p1}}{{\tau }_{p2}}=\frac{N_{T2}}{N_{T1}}=\frac{10N_{T1}}{N_{T1}}=10\end{array}$$ $$\boxed{\frac{{\tau }_{p1}}{{\tau }_{p2}}=10}$$

(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.

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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.

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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 Eq. (7.11): slope magnitude $=2/(q{N}_A{K}_S{\epsilon }_0{A}^2)$, so:

$$\begin{array}{rlrl}{N}_A& =\frac{2}{q{K}_S{\epsilon }_0{A}^2(9.78\times 10^{19})}& & \text{(solve for acceptor doping from slope)}\\ & =\frac{2}{(1.602\times 10^{-19}\text{ C})(11.8\times 8.854\times 10^{-14}\text{ F/cm})(3.72\times 10^{-3}{\text{ cm}}^2{)}^2(9.78\times 10^{19})}& & \text{(substitute measured values)}\\ & =8.84\times 10^{15}{\text{ cm}}^{-3}& & \text{(evaluate p-side doping)}\end{array}$$

5. The intercept over the slope magnitude gives the built-in voltage:

$$\begin{array}{rlrl}{V}_{bi}& =\frac{6.89\times 10^{19}}{9.78\times 10^{19}}=0.704\text{ V}\approx 0.70\text{ V}& & \text{(extrapolate 1/CJ2→0 intercept)}\end{array}$$ $$\boxed{N_A=8.84\times 10^{15}{\text{ cm}}^{-3},V_{bi}\approx 0.70\text{ V}}$$

Remark. This $V_{bi}$ is lower than you would compute from the doping ($\approx 0.9$ V for $N_A\approx 9\times 10^{15}$ 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 $10^{17}\mathrm{c}{\mathrm{m}}^{-3}$) 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 $I=I_0\left(e^{q{V}_A/kT}-1\right)$:

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 $G_0\to 0$ (the I-V curve is flat).

If R-G current dominates the reverse current is $I\approx -qA{n}_{i}W/2{\tau }_0$. Using (7.6), $W^{m+2}\propto (V_{bi}-V_A)$, so

$\frac{dW}{d{V}_A}=-\frac{W}{(m+2)(V_{bi}-V_A)}$

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 $R_S\approx 1\Omega$ (Exercise 6.8). At $V_A=-10$ V and $f=100$ kHz, confirm $R_S$ and $G=G_0$ may be neglected.

Solution.

$$\begin{array}{rlrl}{G}_0& ={\frac{dI}{d{V}_A}|}_{-10\text{ V}}=\frac{40\times 10^{-12}\text{ A}}{18\text{ V}}\approx 2.22\times 10^{-12}\text{ S}& & \text{(differential reverse conductance)}\\ \omega C_J& =2\pi (10^5{\text{ s}}^{-1})(6.3\times 10^{-11}\text{ F})\approx 3.96\times 10^{-5}\text{ S}& & \text{(capacitive susceptance at 100 kHz)}\end{array}$$

Since $\omega C_J\gg G_0$ (ratio $\sim 1.8\times 10^7$), the conductance is negligible: $Y\approx j\omega C_J$.

For $R_S$ in series with $Y$, compare the capacitive reactance $|Z_C|$ with $R_S$:

$$\begin{array}{rlrl}|Z_C|& =\frac{1}{\omega C_J}=\frac{1}{3.96\times 10^{-5}\text{ S}}\approx 2.52\times 10^4\Omega \gg R_S\le 1\Omega & & \text{(reactance exceeds RS by over four orders)}\end{array}$$ $$\boxed{G_0\approx 2.22\times 10^{-12}\text{ S},\omega C_J\approx 3.96\times 10^{-5}\text{ S},|Z_C|\approx 2.52\times 10^4\Omega \gg R_S}$$

So the diode behaves as an essentially pure capacitor.

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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 $\omega \tau \to 1$ 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, $\pm 40$ 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 $e^{q{v}_a/kT}-1\approx q{v}_a/kT$ 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 $\left[e^{q{V}_A/kT}-1\right]\to \frac{q{v}_a}{kT}{e}^{q{V}_A/kT}$. 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 $G_0=\frac{q}{kT}qA\sqrt{\frac{D_P}{{\tau }_p}}\frac{n_i^2}{N_D}{e}^{q{V}_A/kT}$ (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:

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 $V_A\approx0.547\text{ V}$ (the book quotes $\approx0.545\text{ V}$). $C_J$ 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 $0.5473\text{ V}$, $\omega\tau_n=1$ gives $0.5506\text{ V}$). As $\omega\tau_n$ increases beyond 1, $C_D$ drops at every bias and the crossover moves up: $V_A\approx0.578\text{ V}$ for $\omega\tau_n=10$, and $0.619\text{ V}$ for $\omega\tau_n=100$ (book: $0.575\text{ V}$ and $0.62\text{ V}$).

\boxed{V_{\text{cross}}(\omega\tau_n = 0.01) \approx 0.547\text{ V}, \qquad V_{\text{cross}}(\omega\tau_n = 10) \approx 0.578\text{ V}, \qquad V_{\text{cross}}(\omega\tau_n = 100) \approx 0.619\text{ 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 fit:** Linear fit to $1/C_J^2$ vs $V_A$:

\begin{aligned} \frac{1}{C_J^2} &= 8.25\times10^{20} - 1.123\times10^{21},V_A && \text{(least-squares linear fit)} \ N_B &= \frac{2}{qK_S\varepsilon_0A^2,|\text{slope}|} && \text{(doping concentration formula from slope)} \ &= \frac{2}{(1.602\times10^{-19}\text{ C})(1.045\times10^{-12}\text{ F/cm})(6\times10^{-4}\text{ cm}^2)^2(1.123\times10^{21}\text{ F}^{-2}\text{V}^{-1})} && \text{(substitute physical constants)} \ &= 2.96\times10^{16}\text{ cm}^{-3} && \text{(evaluate donor concentration)} \ V_{bi} &= \frac{\text{intercept}}{|\text{slope}|} = \frac{8.25\times10^{20}}{1.123\times10^{21}} \approx 0.735\text{ V} \approx 0.73\text{ V} && \text{(evaluate built-in voltage)} \end{aligned}

4. **(i) Eyeballing** a line through the end points ($V_A=0$ and $-15\text{ V}$): slope $=-1.146\times10^{21}\text{ F}^{-2}\text{V}^{-1}$, giving $N_B\approx2.9\times10^{16}\text{ cm}^{-3}$ and $V_{bi}\approx0.68\text{ V}$.

\boxed{\text{Least squares: } N_B \approx 2.96 \times 10^{16}\text{ cm}^{-3}, \ V_{bi} \approx 0.73\text{ V} \qquad (\text{Eyeballed: } N_B \approx 2.9 \times 10^{16}\text{ cm}^{-3}, \ V_{bi} \approx 0.68\text{ V})}

--- ### 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.

\boxed{\lim_{\omega\tau_p \to 0}\frac{\omega C_D}{G_D} = \frac{\omega\tau_p}{2}, \qquad \lim_{\omega\tau_p \to \infty}\frac{\omega C_D}{G_D} = 1}

**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 | $2.28\times10^{-3}$ | $5.67\times10^{-7}$ |

\boxed{\bar{\tau}_n \approx 5.4 \times 10^{-7}\text{ s} = 0.54\ \mu\text{s}}

This is close to the $5\times10^{-7}\text{ s}$ 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 $p_{n0}=10^4\text{ cm}^{-3}$, $T=300\text{ K}$ and $p_n(x_n)=2\times10^4\text{ cm}^{-3}$:

\begin{aligned} p_n(x_n) &= p_{n0}e^{qv_A/kT} \implies v_A = \frac{kT}{q}\ln\left(\frac{p_n(x_n)}{p_{n0}}\right) && \text{(depletion edge boundary condition)} \ &= (0.02585\text{ V})\ln\left(\frac{2\times10^4\text{ cm}^{-3}}{10^4\text{ cm}^{-3}}\right) = (0.02585\text{ V})\ln(2) && \text{(substitute given concentrations)} \ &\approx 0.01792\text{ V} \approx 17.9\text{ mV} && \text{(evaluate junction voltage)} \end{aligned}

(General: $v_A=(kT/q)\ln\alpha$ for $p_n(x_n)=\alpha p_{n0}$; $p_{n0}$ cancels.)

\boxed{v_A = \frac{kT}{q}\ln\left(\frac{p_n(x_n)}{p_{n0}}\right) \approx 17.9\text{ mV}}

**(c) Direction of current.** With the profile sloping upward at $x_n$ ($dp_n/dx>0$), Eq. (8.2) gives:

\begin{aligned} i &= -qAD_P\left.\frac{d\Delta p_n}{dx}\right|_{x_n} < 0 && \text{(hole diffusion current into depletion region)} \end{aligned}

\boxed{i < 0 \quad (\text{a reverse current flows because } \left.\frac{dp_n}{dx}\right|_{x_n} > 0)}

A **reverse current** flows (holes diffuse back toward the junction) even though the junction remains forward biased. This is the characteristic feature of the storage phase of turn-off. --- ### 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 $i=0$, Eq. (8.3) becomes $dQ_P/dt=-Q_P/\tau_p$. From Eq. (8.7), the initial charge is $Q_P(0)=I_F\tau_p$:

\begin{aligned} \frac{dQ_P}{dt} &= -\frac{Q_P}{\tau_p} \implies \int_{Q_P(0)}^{Q_P(t)}\frac{dQ_P}{Q_P} = -\frac{1}{\tau_p}\int_0^t dt’ && \text{(integrate pure recombination loss)} \ \ln\left(\frac{Q_P(t)}{Q_P(0)}\right) &= -\frac{t}{\tau_p} \implies Q_P(t) = Q_P(0)e^{-t/\tau_p} && \text{(solve for transient stored charge)} \end{aligned}

\boxed{Q_P(t)=I_F\tau_p,e

**(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 $qv_A/kT\gg1$:

\begin{aligned} I_0\tau_p e^{qv_A(t)/kT} &= I_F\tau_p e^{-t/\tau_p} && \text{(equate stored charge expressions)} \ e^{qv_A(t)/kT} &= \frac{I_F}{I_0}e^{-t/\tau_p} && \text{(cancel common factor } \tau_p\text{)} \ \frac{qv_A(t)}{kT} &= \ln\left(\frac{I_F}{I_0}\right) - \frac{t}{\tau_p} && \text{(take natural logarithm)} \ v_A(t) &= \frac{kT}{q}\ln\left(\frac{I_F}{I_0}\right) - \frac{kT}{q}\frac{t}{\tau_p} && \text{(multiply by } kT/q\text{)} \end{aligned}

Using $I_F/I_0 = e^{qV_{ON}/kT} - 1 \approx e^{qV_{ON}/kT}$, so $(kT/q)\ln(I_F/I_0) \approx V_{ON}$:

\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 $-25.9\text{ mV}$ per lifetime at $300\text{ K}$). Measuring the slope directly yields the minority carrier lifetime:

\boxed{\tau_p=-\frac{kT}{q}\left(\frac{dv_A}{dt}\right)

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 steady-state forward voltage:

\begin{aligned} V_{ON} &= \frac{kT}{q}\ln\left(1 + \frac{I_F}{I_0}\right) = (0.0259\text{ V})\ln\left(1 + \frac{10^{-3}\text{ A}}{10^{-15}\text{ A}}\right) && \text{(steady-state forward voltage formula)} \ &= (0.0259\text{ V})\ln(10^{12}) = (0.0259\text{ V})(27.631) \approx 0.7156\text{ V} \approx 0.716\text{ V} && \text{(evaluate natural logarithm)} \end{aligned}

$$2.Targetvoltages:$$

\begin{aligned} 0.90,V_{ON} &= 0.90 \times 0.7156\text{ V} = 0.6440\text{ V} && \text{(90% target voltage)} \ 0.95,V_{ON} &= 0.95 \times 0.7156\text{ V} = 0.6798\text{ V} && \text{(95% target voltage)} \end{aligned}

3. Invert Eq. (8.16) for time $t$:

\begin{aligned} e^{qv_A/kT} &= 1 + \frac{I_F}{I_0}\left(1 - e^{-t/\tau_p}\right) \implies 1 - e^{-t/\tau_p} = \frac{e^{qv_A/kT} - 1}{I_F/I_0} \ t &= -\tau_p\ln\left[1 - \frac{e^{qv_A/kT} - 1}{I_F/I_0}\right] && \text{(solve for elapsed time } t\text{)} \end{aligned}

$$4.\ast \ast Timetoreach90$$

\begin{aligned} \frac{e^{0.644/0.0259} - 1}{10^{12}} &= \frac{e^{24.865} - 1}{10^{12}} = \frac{6.307\times10^{10}}{10^{12}} = 0.06307 && \text{(evaluate fractional buildup factor)} \ t_{90} &= -(1\ \mu\text{s})\ln(1 - 0.06307) = -(1\ \mu\text{s})(-0.0651) = 0.0651\ \mu\text{s} = 65\text{ ns} \end{aligned}

$$5.\ast \ast Timetoreach95$$

\begin{aligned} \frac{e^{0.6798/0.0259} - 1}{10^{12}} &= \frac{e^{26.247} - 1}{10^{12}} = \frac{2.512\times10^{11}}{10^{12}} = 0.2512 && \text{(evaluate fractional buildup factor)} \ t_{95} &= -(1\ \mu\text{s})\ln(1 - 0.2512) = -(1\ \mu\text{s})(-0.2893) = 0.2893\ \mu\text{s} = 289\text{ ns} \end{aligned}

\boxed{t_{90} \approx 65\text{ ns}, \qquad t_{95} \approx 289\text{ 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. $t<0$: $+1\text{ mA}$ steady forward current. 2. $0<t<t_s$: $-1\text{ mA}$ (storage phase). From Eq. (8.8):

\begin{aligned} t_s &= \tau_p\ln\left(1 + \frac{I_F}{I_R}\right) = (1\ \mu\text{s})\ln\left(1 + \frac{1\text{ mA}}{1\text{ mA}}\right) = (1\ \mu\text{s})\ln(2) \approx 0.693\ \mu\text{s} \end{aligned}

which is less than $1\ \mu\text{s}$. 3. $t_s<t<1\ \mu\text{s}$: the stored charge has run out; the current drops quickly to the tiny reverse saturation current ($\approx 0$ on this scale). 4. $t>1\ \mu\text{s}$: $+1\text{ mA}$ forward current again.

\boxed{t_s = \tau_p\ln\left(1 + \frac{I_F}{I_R}\right) \approx 0.693\ \mu\text{s}}

**(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). $$I_E = I_{Ep} + I_{En} \tag{10.4}$$ $$I_C = I_{Cp} + I_{Cn} \tag{10.5}$$ 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: $$\gamma = \frac{I_{Ep}}{I_E} = \frac{I_{Ep}}{I_{Ep} + I_{En}} \quad (\text{pnp}) \tag{10.6}$$ $0\le\gamma\le1$; 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 $\propto I_{Cp}$: $$\alpha_T = \frac{I_{Cp}}{I_{Ep}} \quad (\text{pnp}) \tag{10.7}$$ $0\le\alpha_T\le1$; 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 $$I_C = \alpha_{\text{dc}} I_E + I_{CBO} \tag{10.8}$$ with $\alpha_{\text{dc}}$ 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): $$I_{Ep} = \gamma I_E \tag{10.9a}$$ $$I_{Cp} = \alpha_T I_{Ep} = \alpha_T \gamma I_E \tag{10.9b}$$ (Eq. 10.9 in the book collects these relations.) Then: $$I_C = I_{Cp} + I_{Cn} = \gamma \alpha_T I_E + I_{Cn} \tag{10.10}$$ Comparing (10.8) with (10.10): $$\alpha_{\text{dc}} = \gamma\,\alpha_T \tag{10.11}$$ $$I_{CBO} = I_{Cn} \tag{10.12}$$ (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 $$I_C = \beta_{\text{dc}} I_B + I_{CEO} \tag{10.13}$$ 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): $$I_C = \alpha_{\text{dc}}(I_C + I_B) + I_{CBO} \tag{10.14}$$ Collecting $I_C$ terms:

\begin{aligned} I_C(1 - \alpha_{\text{dc}}) &= \alpha_{\text{dc}} I_B + I_{CBO} \ I_C &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} I_B + \frac{I_{CBO}}{1 - \alpha_{\text{dc}}} \tag{10.15} \end{aligned}

Comparing (10.13) and (10.15): $$\beta_{\text{dc}} = \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \tag{10.16}$$ $$I_{CEO} = \frac{I_{CBO}}{1 - \alpha_{\text{dc}}} = (\beta_{\text{dc}} + 1)I_{CBO} \tag{10.17}$$ and, if $I_{CEO} \ll I_C$ at the operating point, $$\beta_{\text{dc}} \approx \frac{I_C}{I_B} \tag{10.18}$$ 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 current $I_B$, output current $I_C$ with the emitter grounded/common: $$\boxed{\text{Common emitter (CE)}}$$ *(Note: If input is $I_E$ and output is $I_C$, the configuration is common base (CB).)* (b) Output current $I_C$, output voltage $V_{EC}$ (collector-to-emitter): $$\boxed{\text{Common emitter (CE)}}$$ *(Note: An output voltage $V_{CB}$ corresponds to common base (CB).)* (c) Four operating/biasing modes: $$\boxed{\text{Saturation, Active (forward active), Inverted (reverse active), Cutoff}}$$ (d) Purpose of the $n^+$ buried layer in IC BJTs: $$\boxed{\text{Provides a low-resistance path between the active collector region and the top-side collector contact}}$$ (e) Relative doping concentrations in a standard pnp BJT: $$\boxed{N_{AE} \ge N_{DB} > N_{AC}}$$ *(Emitter is most heavily doped to ensure $\gamma \approx 1$; collector is least doped to withstand large reverse voltage.)* (f) Standard notation for depletion width and base width: $$\boxed{W_{\text{dep}}\text{ or }W\text{ (diode depletion width)}, \quad W = W_B - x_{nEB} - x_{nCB}\text{ (quasineutral base width)}}$$ where $W_B$ is the metallurgical base width. (g) Precise physical definition of a "narrow" base: $$\boxed{W \ll L_{\min}}$$ where $L_{\min}$ is the minority-carrier diffusion length in the base ($L_p$ for pnp, $L_n$ for npn). (h) Why the base must be narrow: $$\boxed{\text{Enables injected carriers to cross without recombining, coupling the junctions and ensuring high gain}}$$ (i) Emitter efficiency $\gamma$ in words: $$\boxed{\gamma = \frac{\text{Current carried by useful carriers injected into base}}{\text{Total emitter current}} = \frac{I_{Ep}}{I_{Ep} + I_{En}}}$$ (j) Base transport factor $\alpha_T$ in words: $$\boxed{\alpha_T = \frac{\text{Minority carriers entering collector}}{\text{Minority carriers injected into base}} = \frac{I_{Cp}}{I_{Ep}}}$$ ### 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 in equilibrium:* For $p$-type regions $E_i - E_F = \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right)$; for $n$-type regions $E_F - E_i = \frac{kT}{q}\ln\left(\frac{N_D}{n_i}\right)$. In the emitter ($p$-type, $N_{AE} = 5 \times 10^{17}\text{ cm}^{-3}$):

\begin{aligned} E_i - E_F\big|_E &= 0.0259 \ln\left(\frac{5 \times 10^{17}}{10^{10}}\right) \ &= 0.0259 \ln\left(5 \times 10^7\right) \ &= 0.0259 \times 17.7275 \ &= \boxed{0.459\text{ eV}} \end{aligned}

In the base ($n$-type, $N_{DB} = 10^{15}\text{ cm}^{-3}$):

\begin{aligned} E_F - E_i\big|_B &= 0.0259 \ln\left(\frac{10^{15}}{10^{10}}\right) \ &= 0.0259 \ln\left(10^5\right) \ &= 0.0259 \times 11.5129 \ &= \boxed{0.298\text{ eV}} \end{aligned}

In the collector ($p$-type, $N_{AC} = 10^{14}\text{ cm}^{-3}$):

\begin{aligned} E_i - E_F\big|_C &= 0.0259 \ln\left(\frac{10^{14}}{10^{10}}\right) \ &= 0.0259 \ln\left(10^4\right) \ &= 0.0259 \times 9.2103 \ &= \boxed{0.239\text{ eV}} \end{aligned}

**(a) Band diagram:** A single horizontal Fermi level $E_F$ extends throughout the entire structure. - In the emitter bulk, $E_F$ lies $0.459\text{ eV}$ below $E_i$ (or $0.101\text{ eV}$ above $E_v$, since $E_i - E_v \approx 0.56\text{ eV}$). - In the base bulk, $E_F$ lies $0.298\text{ eV}$ above $E_i$ (or $0.262\text{ eV}$ below $E_c$). - In the collector bulk, $E_F$ lies $0.239\text{ eV}$ below $E_i$ (or $0.321\text{ eV}$ above $E_v$). In terms of electron energy, the bands in the base form a potential well for electrons and a potential barrier for holes: - Going from emitter into base, $E_i$ drops by $qV_{bi,EB} = 0.459 + 0.298 = 0.757\text{ eV}$. - Going from base into collector, $E_i$ rises by $qV_{bi,CB} = 0.298 + 0.239 = 0.537\text{ eV}$. - The net change in $E_i$ from emitter to collector is a fall of $0.757 - 0.537 = 0.220\text{ eV}$. - The E-B depletion region is narrow and lies predominantly on the base side ($N_{AE} \gg N_{DB}$), whereas the C-B depletion region is wide and lies predominantly on the collector side ($N_{DB} \gg N_{AC}$). **(c) Net potential difference between collector and emitter:** Built-in potentials across the junctions:

\begin{aligned} V_{bi,EB} &= \frac{1}{q}\left[(E_i - E_F)\big|_E + (E_F - E_i)\big|_B\right] \ &= 0.459\text{ V} + 0.298\text{ V} \ &= 0.757\text{ V} \end{aligned}

\begin{aligned} V_{bi,CB} &= \frac{1}{q}\left[(E_F - E_i)\big|_B + (E_i - E_F)\big|_C\right] \ &= 0.298\text{ V} + 0.239\text{ V} \ &= 0.537\text{ V} \end{aligned}

Defining electrostatic potential as $V = -E_i/q$ with $V = 0$ in the emitter bulk:

\begin{aligned} V_B &= V_E + V_{bi,EB} = 0 + 0.757\text{ V} = +0.757\text{ V} \ V_C &= V_B - V_{bi,CB} = +0.757\text{ V} - 0.537\text{ V} = +0.220\text{ V} \end{aligned}

$$Thenetpotentialdifferenceistherefore:$$

\begin{aligned} V_C - V_E &= V_{bi,EB} - V_{bi,CB} \ &= 0.757\text{ V} - 0.537\text{ V} \ &= \boxed{0.221\text{ V}} \end{aligned}

**(b) Spatial profiles:** - **Electrostatic potential $V(x)$:** $V = 0$ in the emitter neutral bulk; rises parabolically across the E-B depletion region (spanning $x_{pEB} \approx 0.002\ \mu\text{m}$ in the emitter and $x_{nEB} \approx 0.994\ \mu\text{m}$ in the base) to $+0.757\text{ V}$; remains flat at $+0.757\text{ V}$ across the quasineutral base; then decreases parabolically across the C-B depletion region (spanning $x_{nCB} \approx 0.252\ \mu\text{m}$ in the base and $x_{pCB} \approx 2.525\ \mu\text{m}$ in the collector) to $+0.221\text{ V}$ in the collector neutral bulk. - **Electric field $\mathscr{E}(x) = -dV/dx$:** Zero in all neutral regions. In the E-B depletion region, $\mathscr{E}(x)$ forms a negative triangle (pointing from $n$-base to $p$-emitter) peaking at $-1.52 \times 10^4\text{ V/cm}$ at the metallurgical junction. In the C-B depletion region, $\mathscr{E}(x)$ forms a wider negative triangle (pointing from $n$-base to $p$-collector) peaking at $-3.86 \times 10^3\text{ V/cm}$. - **Charge density $\rho(x)$:** Within the depletion approximation, $\rho = -qN_{AE}$ in $-x_{pEB} \le x < 0$, $\rho = +qN_{DB}$ in $0 < x \le x_{nEB}$, $\rho = 0$ in the quasineutral base, $\rho = +qN_{DB}$ in the base side of C-B ($0 < x' \le x_{nCB}$), and $\rho = -qN_{AC}$ in the collector side ($0 < x'' \le x_{pCB}$). **(d) Quasineutral base width:** At the E-B junction:

\begin{aligned} \frac{2\varepsilon_{\text{Si}} V_{bi,EB}}{q} &= \frac{2(1.045 \times 10^{-12}\text{ F/cm})(0.757\text{ V})}{1.6 \times 10^{-19}\text{ C}} \ &= 9.888 \times 10^6\text{ V}\cdot\text{cm}^{-1} \end{aligned}

\begin{aligned} \frac{1}{N_{AE}} + \frac{1}{N_{DB}} &= \frac{1}{5 \times 10^{17}\text{ cm}^{-3}} + \frac{1}{10^{15}\text{ cm}^{-3}} \ &= 2 \times 10^{-18} + 10^{-15} \ &= 1.002 \times 10^{-15}\text{ cm}^3 \end{aligned}

\begin{aligned} W_{EB} &= \sqrt{\frac{2\varepsilon_{\text{Si}} V_{bi,EB}}{q}\left(\frac{1}{N_{AE}} + \frac{1}{N_{DB}}\right)} \ &= \sqrt{(9.888 \times 10^6)(1.002 \times 10^{-15})} \ &= \sqrt{9.908 \times 10^{-9}} \ &= 9.954 \times 10^{-5}\text{ cm} = 0.996\ \mu\text{m} \end{aligned}

$$Depletionpenetrationintothebase:$$

\begin{aligned} x_{nEB} &= W_{EB}\left(\frac{N_{AE}}{N_{AE} + N_{DB}}\right) \ &= 0.9954\ \mu\text{m} \times \left(\frac{5 \times 10^{17}}{5.01 \times 10^{17}}\right) \ &= 0.994\ \mu\text{m} \end{aligned}

$$AttheC-Bjunction:$$

\begin{aligned} \frac{2\varepsilon_{\text{Si}} V_{bi,CB}}{q} &= \frac{2(1.045 \times 10^{-12}\text{ F/cm})(0.537\text{ V})}{1.6 \times 10^{-19}\text{ C}} \ &= 7.014 \times 10^6\text{ V}\cdot\text{cm}^{-1} \end{aligned}

\begin{aligned} \frac{1}{N_{AC}} + \frac{1}{N_{DB}} &= \frac{1}{10^{14}\text{ cm}^{-3}} + \frac{1}{10^{15}\text{ cm}^{-3}} \ &= 10^{-14} + 10^{-15} \ &= 1.1 \times 10^{-14}\text{ cm}^3 \end{aligned}

\begin{aligned} W_{CB} &= \sqrt{\frac{2\varepsilon_{\text{Si}} V_{bi,CB}}{q}\left(\frac{1}{N_{AC}} + \frac{1}{N_{DB}}\right)} \ &= \sqrt{(7.014 \times 10^6)(1.1 \times 10^{-14})} \ &= \sqrt{7.715 \times 10^{-8}} \ &= 2.778 \times 10^{-4}\text{ cm} = 2.78\ \mu\text{m} \end{aligned}

$$Depletionpenetrationintothebase:$$

\begin{aligned} x_{nCB} &= W_{CB}\left(\frac{N_{AC}}{N_{AC} + N_{DB}}\right) \ &= 2.778\ \mu\text{m} \times \left(\frac{10^{14}}{1.1 \times 10^{15}}\right) \ &= 0.252\ \mu\text{m} \end{aligned}

$$Depletionpenetrationintothecollector:$$

\begin{aligned} x_{pCB} &= W_{CB} - x_{nCB} \ &= 2.778\ \mu\text{m} - 0.252\ \mu\text{m} \ &= 2.526\ \mu\text{m} \end{aligned}

$$Thequasineutralbasewidthis:$$

\begin{aligned} W &= W_B - x_{nEB} - x_{nCB} \ &= 3.00\ \mu\text{m} - 0.994\ \mu\text{m} - 0.252\ \mu\text{m} \ &= \boxed{1.75\ \mu\text{m}} \end{aligned}

**(e) Peak electric fields:** Using $|\mathscr{E}|_{\max} = \frac{q N_D x_n}{\varepsilon_{\text{Si}}}$: At the E-B junction:

\begin{aligned} |\mathscr{E}|{\max,EB} &= \frac{q N{DB} x_{nEB}}{\varepsilon_{\text{Si}}} \ &= \frac{(1.6 \times 10^{-19}\text{ C})(10^{15}\text{ cm}^{-3})(0.994 \times 10^{-4}\text{ cm})}{1.045 \times 10^{-12}\text{ F/cm}} \ &= \boxed{1.52 \times 10^4\text{ V/cm}} \end{aligned}

*(Check using the triangular field relation: $\frac{2V_{bi,EB}}{W_{EB}} = \frac{2(0.757\text{ V})}{0.9954 \times 10^{-4}\text{ cm}} = 1.52 \times 10^4\text{ V/cm}$.)* At the C-B junction:

\begin{aligned} |\mathscr{E}|{\max,CB} &= \frac{q N{DB} x_{nCB}}{\varepsilon_{\text{Si}}} \ &= \frac{(1.6 \times 10^{-19}\text{ C})(10^{15}\text{ cm}^{-3})(0.252 \times 10^{-4}\text{ cm})}{1.045 \times 10^{-12}\text{ F/cm}} \ &= \boxed{3.86 \times 10^3\text{ V/cm}} \end{aligned}

*(Check using the triangular field relation: $\frac{2V_{bi,CB}}{W_{CB}} = \frac{2(0.537\text{ V})}{2.778 \times 10^{-4}\text{ cm}} = 3.86 \times 10^3\text{ V/cm}$.)* ### Problem 10.4 Equilibrium electrostatics of an npn **Problem.** A $\text{Si}$ npn BJT with $N_{DE} = 10^{18}\text{ cm}^{-3}$, $N_{AB} = 10^{16}\text{ cm}^{-3}$, $N_{DC} = 10^{14}\text{ cm}^{-3}$ (OCR shows "$10^1$"; taken as $10^{14}$ by analogy with 10.3), $W_B = 2\ \mu\text{m}$ in equilibrium at room temperature. Parts (a)–(e) as in 10.3, with $V = 0$ in the emitter. **Solution.** *Fermi level positions in equilibrium:* In the emitter ($n$-type, $N_{DE} = 10^{18}\text{ cm}^{-3}$):

\begin{aligned} E_F - E_i\big|_E &= 0.0259 \ln\left(\frac{10^{18}}{10^{10}}\right) \ &= 0.0259 \ln\left(10^8\right) \ &= 0.0259 \times 18.4207 \ &= \boxed{0.477\text{ eV}} \end{aligned}

In the base ($p$-type, $N_{AB} = 10^{16}\text{ cm}^{-3}$):

\begin{aligned} E_i - E_F\big|_B &= 0.0259 \ln\left(\frac{10^{16}}{10^{10}}\right) \ &= 0.0259 \ln\left(10^6\right) \ &= 0.0259 \times 13.8155 \ &= \boxed{0.358\text{ eV}} \end{aligned}

In the collector ($n$-type, $N_{DC} = 10^{14}\text{ cm}^{-3}$):

\begin{aligned} E_F - E_i\big|_C &= 0.0259 \ln\left(\frac{10^{14}}{10^{10}}\right) \ &= 0.0259 \ln\left(10^4\right) \ &= 0.0259 \times 9.2103 \ &= \boxed{0.239\text{ eV}} \end{aligned}

**(a) Band diagram:** A flat Fermi level $E_F$ extends throughout the device. - Emitter: $E_F$ lies $0.477\text{ eV}$ above $E_i$ ($\approx 0.083\text{ eV}$ below $E_c$). - Base: $E_F$ lies $0.358\text{ eV}$ below $E_i$ ($\approx 0.202\text{ eV}$ above $E_v$). - Collector: $E_F$ lies $0.239\text{ eV}$ above $E_i$ ($\approx 0.321\text{ eV}$ below $E_c$). The base is a potential barrier (hill) for electrons: - Going from emitter into base, $E_c$ rises by $qV_{bi,EB} = 0.477 + 0.358 = 0.835\text{ eV}$. - Going from base into collector, $E_c$ drops by $qV_{bi,CB} = 0.358 + 0.239 = 0.597\text{ eV}$. - The diagram is inverted compared to the pnp diagram of Problem 10.3. **(c) Built-in potentials and net potential difference:**

\begin{aligned} V_{bi,EB} &= 0.477\text{ V} + 0.358\text{ V} = 0.835\text{ V} \ V_{bi,CB} &= 0.358\text{ V} + 0.239\text{ V} = 0.597\text{ V} \end{aligned}

With $V = 0$ in the emitter bulk, the electrostatic potential $V = -E_i/q$ drops into the base and then rises into the collector:

\begin{aligned} V_B &= V_E - V_{bi,EB} = 0 - 0.835\text{ V} = -0.835\text{ V} \ V_C &= V_B + V_{bi,CB} = -0.835\text{ V} + 0.597\text{ V} = -0.238\text{ V} \end{aligned}

$$Thenetpotentialdifferencebetweencollectorandemitteris:$$

\begin{aligned} V_C - V_E &= -V_{bi,EB} + V_{bi,CB} \ &= -0.835\text{ V} + 0.597\text{ V} \ &= \boxed{-0.238\text{ V}} \end{aligned}

**(b) Spatial profiles:** - **Potential $V(x)$:** $0$ in the emitter neutral bulk; steps down parabolically across E-B to $-0.835\text{ V}$; remains flat at $-0.835\text{ V}$ in the quasineutral base; then steps up parabolically across C-B to $-0.238\text{ V}$ in the collector neutral bulk. - **Electric field $\mathscr{E}(x)$:** Points from $n$-side to $p$-side (from emitter toward base at E-B, and from collector toward base at C-B). Thus $\mathscr{E} > 0$ at E-B and $\mathscr{E} < 0$ at C-B, peaking at the metallurgical junctions. - **Charge density $\rho(x)$:** $+qN_{DE}$ in the thin layer $x_{nEB}$ in emitter, $-qN_{AB}$ over $x_{pEB}$ in base; $-qN_{AB}$ over $x_{pCB}$ in base, $+qN_{DC}$ over wide layer $x_{nCB}$ in collector. **(d) Quasineutral base width:** At the E-B junction:

\begin{aligned} \frac{2\varepsilon_{\text{Si}} V_{bi,EB}}{q} &= \frac{2(1.045 \times 10^{-12}\text{ F/cm})(0.835\text{ V})}{1.6 \times 10^{-19}\text{ C}} \ &= 1.091 \times 10^7\text{ V}\cdot\text{cm}^{-1} \end{aligned}

\begin{aligned} \frac{1}{N_{DE}} + \frac{1}{N_{AB}} &= \frac{1}{10^{18}\text{ cm}^{-3}} + \frac{1}{10^{16}\text{ cm}^{-3}} \ &= 10^{-18} + 10^{-16} \ &= 1.01 \times 10^{-16}\text{ cm}^3 \end{aligned}

\begin{aligned} W_{EB} &= \sqrt{(1.091 \times 10^7)(1.01 \times 10^{-16})} \ &= \sqrt{1.102 \times 10^{-9}} \ &= 3.32 \times 10^{-5}\text{ cm} = 0.332\ \mu\text{m} \end{aligned}

Penetration into the $p$-base:

\begin{aligned} x_{pEB} &= W_{EB}\left(\frac{N_{DE}}{N_{DE} + N_{AB}}\right) \ &= 0.332\ \mu\text{m} \times \left(\frac{10^{18}}{1.01 \times 10^{18}}\right) \ &= 0.329\ \mu\text{m} \end{aligned}

$$AttheC-Bjunction:$$

\begin{aligned} \frac{2\varepsilon_{\text{Si}} V_{bi,CB}}{q} &= \frac{2(1.045 \times 10^{-12}\text{ F/cm})(0.597\text{ V})}{1.6 \times 10^{-19}\text{ C}} \ &= 7.80 \times 10^6\text{ V}\cdot\text{cm}^{-1} \end{aligned}

\begin{aligned} \frac{1}{N_{AB}} + \frac{1}{N_{DC}} &= \frac{1}{10^{16}\text{ cm}^{-3}} + \frac{1}{10^{14}\text{ cm}^{-3}} \ &= 10^{-16} + 10^{-14} \ &= 1.01 \times 10^{-14}\text{ cm}^3 \end{aligned}

\begin{aligned} W_{CB} &= \sqrt{(7.80 \times 10^6)(1.01 \times 10^{-14})} \ &= \sqrt{7.878 \times 10^{-8}} \ &= 2.807 \times 10^{-4}\text{ cm} = 2.81\ \mu\text{m} \end{aligned}

Penetration into the $p$-base:

\begin{aligned} x_{pCB} &= W_{CB}\left(\frac{N_{DC}}{N_{DC} + N_{AB}}\right) \ &= 2.807\ \mu\text{m} \times \left(\frac{10^{14}}{1.01 \times 10^{16}}\right) \ &= 0.0278\ \mu\text{m} \end{aligned}

$$Quasineutralbasewidth:$$

\begin{aligned} W &= W_B - x_{pEB} - x_{pCB} \ &= 2.00\ \mu\text{m} - 0.329\ \mu\text{m} - 0.028\ \mu\text{m} \ &= \boxed{1.64\ \mu\text{m}} \end{aligned}

$$\ast \ast (e)Peakelectricfields:\ast \ast AttheE-Bjunction:$$

\begin{aligned} |\mathscr{E}|{\max,EB} &= \frac{q N{AB} x_{pEB}}{\varepsilon_{\text{Si}}} \ &= \frac{(1.6 \times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(0.329 \times 10^{-4}\text{ cm})}{1.045 \times 10^{-12}\text{ F/cm}} \ &= \boxed{5.04 \times 10^4\text{ V/cm}} \end{aligned}

*(Check: $\frac{2V_{bi,EB}}{W_{EB}} = \frac{2(0.835\text{ V})}{3.32 \times 10^{-5}\text{ cm}} = 5.03 \times 10^4\text{ V/cm}$.)* At the C-B junction:

\begin{aligned} |\mathscr{E}|{\max,CB} &= \frac{q N{AB} x_{pCB}}{\varepsilon_{\text{Si}}} \ &= \frac{(1.6 \times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(0.0278 \times 10^{-4}\text{ cm})}{1.045 \times 10^{-12}\text{ F/cm}} \ &= \boxed{4.26 \times 10^3\text{ V/cm}} \end{aligned}

*(Check: $\frac{2V_{bi,CB}}{W_{CB}} = \frac{2(0.597\text{ V})}{2.81 \times 10^{-4}\text{ cm}} = 4.25 \times 10^3\text{ V/cm}$.)* ### Problem 10.5 Active-mode electrostatics of the 10.3 BJT **Problem.** Biases $V_{EB} = 0.5\text{ V}$ and $V_{CB} = -2\text{ V}$ (active mode reverse bias) are applied to the Problem 10.3 pnp 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 in forward active mode: E-B junction is forward biased with $V_{EB} = 0.5\text{ V}$, and C-B junction is reverse biased with $V_{CB} = -2\text{ V}$ (the negative sign was dropped in the OCR scan). **Solution.** **(a) Band diagram under active bias:** Under applied bias, quasi-Fermi levels separate across the depletion regions: - Forward bias on E-B ($V_{EB} = 0.5\text{ V}$) lowers the base Fermi level relative to the emitter by $qV_{EB} = 0.5\text{ eV}$ in electron energy terms: $F_B$ sits $0.5\text{ eV}$ below $F_E$. The barrier to hole injection from emitter to base drops from $qV_{bi,EB} = 0.757\text{ eV}$ to: $$q(V_{bi,EB} - V_{EB}) = 0.757\text{ eV} - 0.500\text{ eV} = 0.257\text{ eV}$$ - Reverse bias on C-B ($V_{CB} = -2\text{ V}$) raises the collector Fermi level relative to the base by $q|V_{CB}| = 2.0\text{ eV}$: $F_C$ sits $2.0\text{ eV}$ above $F_B$, or $2.5\text{ eV}$ above $F_E$. The barrier across C-B increases from $qV_{bi,CB} = 0.537\text{ eV}$ to: $$q(V_{bi,CB} - V_{CB}) = 0.537\text{ eV} + 2.000\text{ eV} = 2.537\text{ eV}$$ **(b) Depletion widths and fields under bias:** At the forward-biased E-B junction ($V_{bi,EB} - V_{EB} = 0.757 - 0.5 = 0.257\text{ V}$):

\begin{aligned} W_{EB} &= W_{EB0}\sqrt{\frac{V_{bi,EB} - V_{EB}}{V_{bi,EB}}} \ &= 0.996\ \mu\text{m} \times \sqrt{\frac{0.257}{0.757}} \ &= 0.996\ \mu\text{m} \times 0.5827 \ &= \boxed{0.580\ \mu\text{m}} \end{aligned}

$$Basepenetration:$$

\begin{aligned} x_{nEB} &= W_{EB}\left(\frac{N_{AE}}{N_{AE} + N_{DB}}\right) \ &= 0.580\ \mu\text{m} \times \left(\frac{5 \times 10^{17}}{5.01 \times 10^{17}}\right) \ &= \boxed{0.579\ \mu\text{m}} \end{aligned}

At the reverse-biased C-B junction ($V_{bi,CB} - V_{CB} = 0.537 + 2 = 2.537\text{ V}$):

\begin{aligned} W_{CB} &= W_{CB0}\sqrt{\frac{V_{bi,CB} - V_{CB}}{V_{bi,CB}}} \ &= 2.78\ \mu\text{m} \times \sqrt{\frac{2.537}{0.537}} \ &= 2.78\ \mu\text{m} \times 2.1736 \ &= \boxed{6.04\ \mu\text{m}} \end{aligned}

$$Basepenetration:$$

\begin{aligned} x_{nCB} &= W_{CB}\left(\frac{N_{AC}}{N_{AC} + N_{DB}}\right) \ &= 6.04\ \mu\text{m} \times \left(\frac{10^{14}}{1.1 \times 10^{15}}\right) \ &= \frac{6.04}{11} \ &= \boxed{0.549\ \mu\text{m}} \end{aligned}

$$Collectorpenetration:$$

\begin{aligned} x_{pCB} &= W_{CB} - x_{nCB} \ &= 6.04\ \mu\text{m} - 0.549\ \mu\text{m} \ &= \boxed{5.49\ \mu\text{m}} \end{aligned}

$$Quasineutralbasewidthunderactivebias:$$

\begin{aligned} W &= W_B - x_{nEB} - x_{nCB} \ &= 3.00\ \mu\text{m} - 0.579\ \mu\text{m} - 0.549\ \mu\text{m} \ &= \boxed{1.87\ \mu\text{m}} \end{aligned}

*(Note: $W$ increases from $1.75\ \mu\text{m}$ to $1.87\ \mu\text{m}$ because forward bias shrinks the E-B depletion region in the base by $0.415\ \mu\text{m}$, which exceeds the $0.297\ \mu\text{m}$ expansion of the C-B depletion region into the base.)* Peak electric fields:

\begin{aligned} |\mathscr{E}|{\max,EB} &= \frac{2(V{bi,EB} - V_{EB})}{W_{EB}} \ &= \frac{2(0.257\text{ V})}{0.580 \times 10^{-4}\text{ cm}} \ &= \boxed{8.86 \times 10^3\text{ V/cm}} \quad (\text{decreased from } 1.52 \times 10^4\text{ V/cm}) \end{aligned}

\begin{aligned} |\mathscr{E}|{\max,CB} &= \frac{2(V{bi,CB} - V_{CB})}{W_{CB}} \ &= \frac{2(2.537\text{ V})}{6.04 \times 10^{-4}\text{ cm}} \ &= \boxed{8.40 \times 10^3\text{ V/cm}} \quad (\text{increased from } 3.86 \times 10^3\text{ V/cm}) \end{aligned}

Superimposed sketch descriptions: - **Potential $V(x)$:** With emitter at $V = 0$, base potential sits at $V_{bi,EB} - V_{EB} = +0.257\text{ V}$, and collector potential sits at $+0.257\text{ V} - 2.537\text{ V} = -2.28\text{ V}$ (collector is $2.28\text{ V}$ below emitter). - **Electric field $\mathscr{E}(x)$:** E-B triangle becomes narrower ($0.58\ \mu\text{m}$) and shorter ($8.9 \times 10^3\text{ V/cm}$ peak); C-B triangle becomes much wider ($6.04\ \mu\text{m}$) and taller ($8.4 \times 10^3\text{ V/cm}$ peak). - **Charge density $\rho(x)$:** Same rectangle heights ($\pm qN$), but E-B rectangle widths shrink while C-B rectangle widths expand. ### 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 $I_{Ep} = 1\text{ mA}$, $I_{En} = 0.01\text{ mA}$, $I_{Cp} = 0.98\text{ mA}$, $I_{Cn} = 0.1\ \mu\text{A}$. Calculate: (a) $\alpha_T$ (b) $\gamma$ (c) $I_E, I_C, I_B$ (d) $\alpha_{\text{dc}}$ and $\beta_{\text{dc}}$ (e) $I_{CBO}$ and $I_{CEO}$ (f) If $I_{Cp}$ is increased toward $I_{Ep}$ with all other components fixed, effect on $\beta_{\text{dc}}$? (g) If $I_{En}$ is increased with all other components fixed, effect on $\beta_{\text{dc}}$? **Solution.** **(a) Base transport factor $\alpha_T$:** From Eq. (10.7):

\begin{aligned} \alpha_T &= \frac{I_{Cp}}{I_{Ep}} \ &= \frac{0.98\text{ mA}}{1.00\text{ mA}} \ &= \boxed{0.98} \end{aligned}

**(b) Emitter efficiency $\gamma$:** From Eq. (10.6):

\begin{aligned} \gamma &= \frac{I_{Ep}}{I_{Ep} + I_{En}} \ &= \frac{1.00\text{ mA}}{1.00\text{ mA} + 0.01\text{ mA}} \ &= \frac{1.00}{1.01} \ &= \boxed{0.990} \end{aligned}

$$\ast \ast (c)Terminalcurrents:\ast \ast FromEqs.(10.4),(10.5),and(10.1):$$

\begin{aligned} I_E &= I_{Ep} + I_{En} \ &= 1.00\text{ mA} + 0.01\text{ mA} \ &= \boxed{1.01\text{ mA}} \end{aligned}

\begin{aligned} I_C &= I_{Cp} + I_{Cn} \ &= 0.98\text{ mA} + 0.0001\text{ mA} \ &= \boxed{0.9801\text{ mA}} \end{aligned}

\begin{aligned} I_B &= I_E - I_C \ &= 1.01\text{ mA} - 0.9801\text{ mA} \ &= 0.0299\text{ mA} = \boxed{29.9\ \mu\text{A}} \end{aligned}

$$\ast \ast (d)Common-baseandcommon-emittercurrentgains:\ast \ast FromEqs.(10.11)and(10.16):$$

\begin{aligned} \alpha_{\text{dc}} &= \gamma,\alpha_T \ &= (0.990099)(0.98) \ &= \boxed{0.9703} \end{aligned}

\begin{aligned} \beta_{\text{dc}} &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \ &= \frac{0.9703}{1 - 0.9703} \ &= \frac{0.9703}{0.0297} \ &= \boxed{32.7} \end{aligned}

*(Consistency check: $\frac{I_C}{I_B} = \frac{0.9801\text{ mA}}{0.0299\text{ mA}} = 32.8$, consistent with $\beta_{\text{dc}}$ since $I_{CEO}$ is negligible.)* **(e) Leakage currents:** From Eqs. (10.12) and (10.17):

\begin{aligned} I_{CBO} &= I_{Cn} = \boxed{0.1\ \mu\text{A}} \end{aligned}

\begin{aligned} I_{CEO} &= \frac{I_{CBO}}{1 - \alpha_{\text{dc}}} = (\beta_{\text{dc}} + 1)I_{CBO} \ &= \frac{0.1\ \mu\text{A}}{0.0297} \ &= \boxed{3.37\ \mu\text{A}} \end{aligned}

*(Check of Eq. 10.13: $I_C = \beta_{\text{dc}} I_B + I_{CEO} = (32.67)(0.0299\text{ mA}) + 0.00337\text{ mA} = 0.9767 + 0.0034 = 0.9801\text{ mA}$.)* **(f) Effect of increasing $I_{Cp}$ toward $I_{Ep}$:**

\begin{aligned} \alpha_T &= \frac{I_{Cp}}{I_{Ep}} \implies \alpha_T \text{ increases} \ \alpha_{\text{dc}} &= \gamma \alpha_T \implies \alpha_{\text{dc}} \text{ increases} \ \beta_{\text{dc}} &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \implies \beta_{\text{dc}} \text{ increases} \end{aligned}

$$\boxed{\beta_{\text{dc}}\text{ increases}}$$ *Physical justification:* Fewer injected holes are lost to recombination in the base, so base current $I_B$ decreases while $I_C$ increases. In the ideal limit $I_{Cp} \to I_{Ep}$ ($\alpha_T \to 1$), $\alpha_{\text{dc}} \to 0.990$ and $\beta_{\text{dc}} \to 99.0$. **(g) Effect of increasing $I_{En}$:**

\begin{aligned} \gamma &= \frac{I_{Ep}}{I_{Ep} + I_{En}} \implies \gamma \text{ decreases} \ \alpha_{\text{dc}} &= \gamma \alpha_T \implies \alpha_{\text{dc}} \text{ decreases} \ \beta_{\text{dc}} &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \implies \beta_{\text{dc}} \text{ decreases} \end{aligned}

$$\boxed{\beta_{\text{dc}}\text{ decreases}}$$ *Physical justification:* A larger fraction of the emitter current is wasted on electrons injected into the emitter, which never reach the collector and must be supplied externally through the base terminal. ### Problem 10.10 Parameter calculations for an npn **Problem.** An npn BJT has $I_{En} = 100\ \mu\text{A}$, $I_{Ep} = 1\ \mu\text{A}$, $I_{Cn} = 99\ \mu\text{A}$, $I_{Cp} = 0.1\ \mu\text{A}$. Calculate: (a) $\alpha_T$ (b) $\gamma$ (c) $I_E, I_C, I_B$ (d) $\alpha_{\text{dc}}, \beta_{\text{dc}}$ (e) $I_{CBO}, I_{CEO}$ (f) If $I_{Cn}$ is increased toward $100\ \mu\text{A}$ with all other parameters fixed, what is the effect on $\beta_{\text{dc}}$? (g) If $I_{Ep}$ is increased with all other parameters fixed, what is the effect on $\beta_{\text{dc}}$? *For an npn BJT, electrons are the useful minority carriers injected into the base, so:* $$\gamma = \frac{I_{En}}{I_E} = \frac{I_{En}}{I_{En} + I_{Ep}}, \quad \alpha_T = \frac{I_{Cn}}{I_{En}}, \quad I_{CBO} = I_{Cp}$$ **Solution.** **(a) Base transport factor $\alpha_T$:**

\begin{aligned} \alpha_T &= \frac{I_{Cn}}{I_{En}} \ &= \frac{99\ \mu\text{A}}{100\ \mu\text{A}} \ &= \boxed{0.99} \end{aligned}

**(b) Emitter efficiency $\gamma$:**

\begin{aligned} \gamma &= \frac{I_{En}}{I_{En} + I_{Ep}} \ &= \frac{100\ \mu\text{A}}{100\ \mu\text{A} + 1\ \mu\text{A}} \ &= \frac{100}{101} \ &= \boxed{0.9901} \end{aligned}

$$\ast \ast (c)Terminalcurrents:\ast \ast$$

\begin{aligned} I_E &= I_{En} + I_{Ep} \ &= 100\ \mu\text{A} + 1\ \mu\text{A} \ &= \boxed{101\ \mu\text{A}} \end{aligned}

\begin{aligned} I_C &= I_{Cn} + I_{Cp} \ &= 99\ \mu\text{A} + 0.1\ \mu\text{A} \ &= \boxed{99.1\ \mu\text{A}} \end{aligned}

\begin{aligned} I_B &= I_E - I_C \ &= 101\ \mu\text{A} - 99.1\ \mu\text{A} \ &= \boxed{1.9\ \mu\text{A}} \end{aligned}

$$\ast \ast (d)Currentgains:\ast \ast$$

\begin{aligned} \alpha_{\text{dc}} &= \gamma,\alpha_T \ &= (0.9901)(0.99) \ &= \boxed{0.9802} \end{aligned}

\begin{aligned} \beta_{\text{dc}} &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \ &= \frac{0.9802}{1 - 0.9802} \ &= \frac{0.9802}{0.01980} \ &= \boxed{49.5} \end{aligned}

$$\ast \ast (e)Leakagecurrents:\ast \ast$$

\begin{aligned} I_{CBO} &= I_{Cp} = \boxed{0.1\ \mu\text{A}} \end{aligned}

\begin{aligned} I_{CEO} &= \frac{I_{CBO}}{1 - \alpha_{\text{dc}}} \ &= \frac{0.1\ \mu\text{A}}{0.01980} \ &= \boxed{5.05\ \mu\text{A}} \end{aligned}

*(Check: $\alpha_{\text{dc}} I_E + I_{CBO} = (0.9802)(101\ \mu\text{A}) + 0.1\ \mu\text{A} = 99.0 + 0.1 = 99.1\ \mu\text{A}$.)* **(f) Effect of increasing $I_{Cn} \to 100\ \mu\text{A}$:** In the limit $I_{Cn} \to I_{En} = 100\ \mu\text{A}$, $\alpha_T \to 1$:

\begin{aligned} \alpha_{\text{dc}} &\to \gamma = 0.9901 \ \beta_{\text{dc}} &\to \frac{0.9901}{1 - 0.9901} = \frac{0.9901}{0.0099} = 100 \end{aligned}

$$\boxed{\beta_{\text{dc}}\text{ increases (from } 49.5\text{ to } 100\text{)}}$$ *Physical justification:* Base recombination loss drops to zero, reducing the required base replenishment current $I_B$. **(g) Effect of increasing $I_{Ep}$:**

\begin{aligned} \gamma &= \frac{I_{En}}{I_{En} + I_{Ep}} \implies \gamma \text{ decreases} \ \alpha_{\text{dc}} &= \gamma \alpha_T \implies \alpha_{\text{dc}} \text{ decreases} \ \beta_{\text{dc}} &= \frac{\alpha_{\text{dc}}}{1 - \alpha_{\text{dc}}} \implies \beta_{\text{dc}} \text{ decreases} \end{aligned}

$$\boxed{\beta_{\text{dc}}\text{ decreases}}$$ *Physical justification:* More of the total emitter current is wasted on hole back-injection into the emitter, which never contributes to collector current 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) Reduction to ideal diode equations when $W \gg L_B$:** In the limit $W/L_B \to \infty$:

\begin{aligned} \coth\left(\frac{W}{L_B}\right) &= \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}} \to 1 \ \frac{1}{\sinh(W/L_B)} &= \frac{2}{e^{W/L_B} - e^{-W/L_B}} \to 0 \end{aligned}

$$SubstitutingtheselimitsintoEqs.(11.35)and(11.36):$$

\begin{aligned} I_E &= qA\left[\frac{D_E}{L_E}n_{E0} + \frac{D_B}{L_B}p_{B0}\right]\left(e^{qV_{EB}/kT} - 1\right) \ I_C &= -qA\left[\frac{D_C}{L_C}n_{C0} + \frac{D_B}{L_B}p_{B0}\right]\left(e^{qV_{CB}/kT} - 1\right) \end{aligned}

$$\boxed{I_E = I_{0E}\left(e^{qV_{EB}/kT} - 1\right), \quad I_C = -I_{0C}\left(e^{qV_{CB}/kT} - 1\right)}$$ Each terminal current reduces to an independent ideal diode relationship. A transistor with $W \gg L_B$ acts as **two back-to-back diodes** with no coupling between emitter and collector. **(b) Reduction to narrow-base diode when $V_{CB} = 0$:** Setting $V_{CB} = 0$ eliminates the second term in Eq. (11.35) because $e^0 - 1 = 0$:

\begin{aligned} I_E &= qA\left[\frac{D_E}{L_E}n_{E0} + \frac{D_B}{L_B}p_{B0}\coth\left(\frac{W}{L_B}\right)\right]\left(e^{qV_{EB}/kT} - 1\right) \end{aligned}

For a $p^+$-$n$ emitter-base junction ($N_E \gg N_B$), electron back-injection into the emitter is negligible ($n_{E0} \ll p_{B0}$):

\begin{aligned} I_E &\approx qA\frac{D_B}{L_B}p_{B0}\coth\left(\frac{W}{L_B}\right)\left(e^{qV_{EB}/kT} - 1\right) \end{aligned}

$$\boxed{I_E \approx qA\frac{D_B}{L_B}p_{B0}\coth\left(\frac{W}{L_B}\right)\left(e^{qV_{EB}/kT} - 1\right)}$$ This is identical to the narrow-base diode equation (Eqs. 6.68–6.69). **Shorting the collector to the base turns the BJT into a 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.** Neglecting thermal recombination-generation in the quasineutral base ($\tau_B \to \infty$), the steady-state minority-carrier diffusion equation (Eq. 11.3) simplifies to:

\begin{aligned} 0 &= D_B\frac{d^2\Delta p_B}{dx^2} \implies \frac{d^2\Delta p_B}{dx^2} = 0 \end{aligned}

Integrating twice with respect to $x$:

\begin{aligned} \frac{d\Delta p_B}{dx} &= A_2 \ \Delta p_B(x) &= A_1 + A_2 x \end{aligned}

Applying boundary conditions (Eqs. 11.4a,b): - At $x = 0$:

\begin{aligned} \Delta p_B(0) &= A_1 \end{aligned}

- At $x = W$:

\begin{aligned} \Delta p_B(W) &= A_1 + A_2 W = \Delta p_B(0) + A_2 W \ A_2 &= \frac{\Delta p_B(W) - \Delta p_B(0)}{W} \end{aligned}

Substituting $A_1$ and $A_2$ back into the solution:

\begin{aligned} \Delta p_B(x) &= \Delta p_B(0) + \left[\Delta p_B(W) - \Delta p_B(0)\right]\frac{x}{W} \end{aligned}

$$\boxed{\Delta p_B(x) = \Delta p_B(0) + \left[\Delta p_B(W) - \Delta p_B(0)\right]\frac{x}{W}}$$ which rigorously confirms Eq. (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.** For the linear minority-carrier profile of Eq. (11.40), the derivative is position-independent:

\begin{aligned} \left.\frac{d\Delta p_B}{dx}\right|{x=0} &= \left.\frac{d\Delta p_B}{dx}\right|{x=W} = \frac{\Delta p_B(W) - \Delta p_B(0)}{W} \end{aligned}

$$UsingEqs.(11.8)and(11.9):$$

\begin{aligned} I_{Ep} &= -qAD_B \left.\frac{d\Delta p_B}{dx}\right|{x=0} = -qAD_B \left[\frac{\Delta p_B(W) - \Delta p_B(0)}{W}\right] \ I{Cp} &= -qAD_B \left.\frac{d\Delta p_B}{dx}\right|{x=W} = -qAD_B \left[\frac{\Delta p_B(W) - \Delta p_B(0)}{W}\right] \ I{Cp} &= I_{Ep} \end{aligned}

$$Thebasetransportfactoristherefore:$$

\begin{aligned} \alpha_T &= \frac{I_{Cp}}{I_{Ep}} = \frac{I_{Ep}}{I_{Ep}} = \boxed{1} \end{aligned}

**Comment:** An exact value $\alpha_T = 1$ reflects zero recombination in the base, consistent with omitting $\Delta p_B/\tau_B$. However, this zeroth-order linear approximation discards the $\mathcal{O}((W/L_B)^2)$ curvature. To determine the small physical loss $1 - \alpha_T \approx \frac{1}{2}(W/L_B)^2$ of Eq. (11.42), the slight hyperbolic curvature of the exact solution must be retained. --- ### 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.** From the Ebers-Moll parameter definitions, divide Eq. (11.46) by Eq. (11.45a):

\begin{aligned} \alpha_F &= \frac{\alpha_F I_{F0}}{I_{F0}} \ &= \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]} \end{aligned}

Dividing numerator and denominator by $qA \frac{D_B}{L_B}\frac{p_{B0}}{\sinh(W/L_B)}$:

\begin{aligned} \alpha_F &= \frac{1}{\cosh\left(\dfrac{W}{L_B}\right) + \dfrac{D_E n_{E0} / L_E}{D_B p_{B0} / L_B}\sinh\left(\dfrac{W}{L_B}\right)} \ &= \frac{1}{\cosh\left(\dfrac{W}{L_B}\right) + \dfrac{D_E}{D_B}\left(\dfrac{n_{E0}}{p_{B0}}\right)\dfrac{L_B}{L_E}\sinh\left(\dfrac{W}{L_B}\right)} \end{aligned}

Using the equilibrium ratio $\frac{n_{E0}}{p_{B0}} = \frac{n_i^2/N_E}{n_i^2/N_B} = \frac{N_B}{N_E}$:

\begin{aligned} \alpha_F &= \frac{1}{\cosh\left(\dfrac{W}{L_B}\right) + \dfrac{D_E}{D_B}\dfrac{N_B}{N_E}\dfrac{L_B}{L_E}\sinh\left(\dfrac{W}{L_B}\right)} = \alpha_{\text{dc}} \tag{11.33} \end{aligned}

$$\boxed{\alpha_F = \alpha_{\text{dc}}}$$ $\blacksquare$ --- #### 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.** In forward active mode, the C-B junction is reverse biased ($V_{CB} \ll -kT/q$), so $e^{qV_{CB}/kT} - 1 \approx -1$. The Ebers-Moll equations (11.47a,b) become:

\begin{aligned} I_E &= I_{F0}\left(e^{qV_{EB}/kT} - 1\right) + \alpha_R I_{R0} \ I_C &= \alpha_F I_{F0}\left(e^{qV_{EB}/kT} - 1\right) + I_{R0} \end{aligned}

$$Solvingtheemittercurrentequationfortheforwardexponentialfactor:$$

\begin{aligned} e^{qV_{EB}/kT} - 1 &= \frac{I_E - \alpha_R I_{R0}}{I_{F0}} \end{aligned}

$$Substitutingthisintothecollectorcurrentequation:$$

\begin{aligned} I_C &= \alpha_F \left(I_E - \alpha_R I_{R0}\right) + I_{R0} \ &= \alpha_F I_E - \alpha_F \alpha_R I_{R0} + I_{R0} \ &= \alpha_F I_E + (1 - \alpha_F \alpha_R)I_{R0} \end{aligned}

Since $\alpha_F = \alpha_{\text{dc}}$ (from Exercise 11.5), this exactly matches the active-mode relation $I_C = \alpha_{\text{dc}} I_E + I_{CBO}$ (Eq. 10.8) with:

\boxed{I_{CBO} = (1 - \alpha_F \alpha_R)I_{R0}} $$ $■$

(Problem 11.12(c) confirms $I_{CBO}$ is the C-B saturation current with emitter open.)

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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 $I_E(V_{EB},V_{CB})$: $I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)-{\alpha }_R{I}_{R0}(e^{q{V}_{CB}/kT}-1)$
2. CB output $I_C(V_{CB},I_E)$: $I_C={\alpha }_F{I}_E-(1-{\alpha }_F{\alpha }_R)I_{R0}(e^{q{V}_{CB}/kT}-1)$
3. CE input $I_B(V_{EB},V_{EC})$: $I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]$
4. CE output $I_C(V_{EC},I_B)$: $I_C=\frac{\left({\alpha }_F{I}_{F0}-I_{R0}{e}^{-q{V}_{EC}/kT}\right)\left[I_B+(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}{(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}}+I_{R0}-{\alpha }_F{I}_{R0}$

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, $N_E=10^{18}$, $N_B=1.5\times 10^{16}$, $N_C=1.5\times 10^{15}$ 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 $(92,1268,1.3\times 10^{17},0.91)$ and holes $(54.3,406.9,2.35\times 10^{17},0.88)$ for $({\mu }_{min},{\mu }_0,N_{ref},\alpha )$.

% 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 $V_{CB}\le 0$ 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

$$I_E=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\left(e^{q{V}_{EB}/kT}-1\right)\text{\hspace{1em}(11.48)}$$

For $W\ll L_B$ ($\coth \to L_B/W$) and neglecting the small $D_E{n}_{E0}/L_E$ term:

$$I_E\approx qA\frac{D_B{p}_{B0}}{W}{e}^{q{V}_{EB}/kT}\text{\hspace{1em}(11.49)}$$

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

$$I_C={\beta }_{dc}{I}_B+I_{CEO}\text{\hspace{1em}(11.50)}$$

with (11.44) ${\beta }_{dc}=1/\left[\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}(W/L_B{)}^2\right]$. $W$ 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

$$I_C={\alpha }_{dc}{I}_E+I_{CBO}\text{\hspace{1em}(11.51)}$$

with (11.43) ${\alpha }_{dc}=1/[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}(W/L_B{)}^2]$. 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 $W\to 0$: 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, $W_B=2\mu \text{m}$, $T=300\text{ K}$, determine the C-B junction voltage that leads to punch-through. (Assume $V_{EB}=0$; Problem 11.16 treats $V_{EB}=0.7\text{ V}$.)

Solution.

From Eq. (10.3), the quasineutral base width is:

$$\begin{array}{rl}W& =W_B-x_{nEB}-x_{nCB}\end{array}$$

Depletion penetrations into the $n$-base for step junctions:

$$\begin{array}{rl}{x}_{nEB}& ={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_E}{N_B(N_E+N_B)}\left(V_{bi,EB}-V_{EB}\right)\right]}^{1/2}\\ x_{nCB}& ={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi,CB}-V_{CB}\right)\right]}^{1/2}\end{array}$$

Punch-through occurs when $W=0$, meaning the two depletion regions meet inside the base:

$$\begin{array}{r}{W}_B-x_{nEB}-x_{nCB}=0⟹x_{nCB}=W_B-x_{nEB}\end{array}$$

Squaring and solving for $V_{CB}$:

$$\begin{array}{rl}\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi,CB}-V_{CB}\right)& =(W_B-x_{nEB}{)}^2\\ V_{CB}& =V_{bi,CB}-\frac{q{N}_B(N_C+N_B)}{2K_S{\epsilon }_0{N}_C}(W_B-x_{nEB}{)}^2\end{array}$$

Numerical evaluation at $T=300\text{ K}$ ($kT/q=0.0259\text{ V}$, $n_i=10^{10}{\text{ cm}}^{-3}$, $K_S{\epsilon }_0=1.045\times 10^{-12}\text{ F/cm}$):

$$\begin{array}{rl}{V}_{bi,EB}& =0.0259\ln \left(\frac{10^{18}\times 10^{16}}{10^{20}}\right)=0.0259\ln (10^4)=0.835\text{ V}\\ V_{bi,CB}& =0.0259\ln \left(\frac{10^{15}\times 10^{16}}{10^{20}}\right)=0.0259\ln (10^{11})=0.656\text{ V}\end{array}$$

With $V_{EB}=0$:

$$\begin{array}{rl}{x}_{nEB}& ={\left[\frac{2(1.045\times 10^{-12})}{1.602\times 10^{-19}}\frac{10^{18}}{10^{16}(1.01\times 10^{18})}(0.835)\right]}^{1/2}\\ & =3.29\times 10^{-5}\text{ cm}=0.329\mu \text{m}\end{array}$$ $$\begin{array}{rl}{W}_B-x_{nEB}& =2.00\times 10^{-4}\text{ cm}-3.29\times 10^{-5}\text{ cm}\\ & =1.671\times 10^{-4}\text{ cm}\end{array}$$ $$\begin{array}{rl}\frac{q{N}_B(N_C+N_B)}{2K_S{\epsilon }_0{N}_C}& =\frac{(1.602\times 10^{-19}\text{ C})(10^{16}{\text{ cm}}^{-3})(1.1\times 10^{16}{\text{ cm}}^{-3})}{2(1.045\times 10^{-12}\text{ F/cm})(10^{15}{\text{ cm}}^{-3})}\\ & =8.43\times 10^9{\text{ V/cm}}^2\end{array}$$

Therefore:

$$\begin{array}{rl}{V}_{CB}& =0.656\text{ V}-(8.43\times 10^9{\text{ V/cm}}^2)(1.671\times 10^{-4}\text{ cm}{)}^2\\ & =0.656\text{ V}-235.4\text{ V}\\ & =\boxed{-235\text{ V}}\end{array}$$

Physical reasoning: A very large reverse bias of $\approx 235\text{ V}$ is required because the collector is an order of magnitude more lightly doped than the base ($N_C\ll N_B$), causing roughly $91\%$ of the C-B depletion layer to expand into the collector rather than 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) $V_{CEO}\approx 90$ V while $V_{CBO}\approx 120$ 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, $I_{Cp}=M{I}_{Ep}$: replace ${\alpha }_{dc}$ (and ${\alpha }_F$) by $M{\alpha }_{dc}$. The simplified CE output (Eq. 10.15) becomes

$$I_C=\frac{M{\alpha }_{dc}}{1-M{\alpha }_{dc}}{I}_B+\frac{I_{CBO}}{1-M{\alpha }_{dc}}\text{\hspace{1em}(11.52)}$$

$I_C\to \infty$ when $M{\alpha }_{dc}\to 1$, i.e. $M\to 1/{\alpha }_{dc}$. Since ${\alpha }_{dc}$ is only slightly below unity, $M$ need only exceed 1 slightly. With the empirical $M$ (Eq. 6.35) and $V_{BR}=V_{CBO}$:

$$M(V_{CEO})=\frac{1}{1-{\left(V_{CEO}/V_{CBO}\right)}^m}=\frac{1}{{\alpha }_{dc}}\text{\hspace{1em}(11.53)}$$ $${\left(\frac{V_{CEO}}{V_{CBO}}\right)}^m=1-{\alpha }_{dc}$$ $$\boxed{V_{CEO}=V_{CBO}{\left(1-{\alpha }_{dc}\right)}^{1/m}=\frac{V_{CBO}}{({\beta }_{dc}+1{)}^{1/m}}}\text{\hspace{1em}(11.54)}$$

with $3\le m\le 6$ (using $1-{\alpha }_{dc}=1/({\beta }_{dc}+1)$). Hence $V_{CEO}<V_{CBO}$. Example: ${\beta }_{dc}\approx 300$, $m=6$: $(301{)}^{1/6}\approx 2.6$, 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.

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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

$$x_{nCB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2}$$

Evaluate the doping factor:

$$\frac{N_C}{N_B(N_C+N_B)}{|}_A\approx \frac{10^{14}}{10^{16}\cdot 10^{16}}=10^{-18},\frac{N_C}{N_B(N_C+N_B)}{|}_B\approx \frac{10^{18}}{10^{16}\cdot 10^{18}}=10^{-16}$$

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 $V_{CBO}$.

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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 $I_{F0}$, $I_{R0}$, ${\alpha }_F$, ${\alpha }_R$:

$$x_{nEB}=\sqrt{\frac{2K_S{\epsilon }_0}{q}\frac{N_E}{N_B(N_E+N_B)}(V_{biE}-V_{EB})},x_{nCB}=\sqrt{\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}(V_{biC}-V_{CB})},W=W_B-x_{nEB}-x_{nCB}$$

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 $V_{EB}\approx 0$ inside BWM (neglecting $x_{nEB}$ variation), so $V_{CB}=-V_{EC}$.

% 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 $\ne$ 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

$$V_{EB}^{\prime }=V_{EB}-I_B{r}_B$$

The central part is the intrinsic transistor; as in Fig. 11.12, resistances $r_E$, $r_B$, $r_C$ connect the internal terminals $E^{\prime }$, $B^{\prime }$, $C^{\prime }$ 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 $p\approx 0$ gives $n(x)=N_B(x)$. In equilibrium the electron current vanishes:

$$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}=0\text{\hspace{1em}(11.55)}$$

Solve for $\mathcal{E}$ using $D_N/{\mu }_n=kT/q$ and $n=N_B(x)$:

$$\mathcal{E}(x)=-\frac{D_N}{{\mu }_n}\frac{dn/dx}{n}=-\frac{kT}{q}\frac{1}{N_B(x)}\frac{d{N}_B}{dx},0\le x\le W\text{\hspace{1em}(11.56)}$$

For an exponential profile,

$$N_B(x)=N_B(0)e^{-x/x_{diff}}\text{\hspace{1em}(11.57)}$$ $$\frac{d{N}_B}{dx}=-\frac{N_B(x)}{x_{diff}}\text{\hspace{1em}(11.58)}$$

Substituting in (11.56):

$$\boxed{\mathcal{E}=\frac{kT/q}{x_{diff}}}0\le x\le W\text{\hspace{1em}(11.59)}$$

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:

$$x_{diff}=\frac{W_B}{\ln 100}=\frac{2\times 10^{-4}}{4.605}=4.34\times 10^{-5}\text{cm},\mathcal{E}=\frac{0.0259}{4.34\times 10^{-5}}\approx 600\text{V/cm}$$

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.

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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 $W_E/L_E\ll 1$ 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:

$$q{D}_{E1}\frac{d\Delta p_{E1}}{d{x}^{\prime \prime }}=q{D}_{E2}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}\text{\hspace{1em}(11.60)}$$ $$\frac{d\Delta p_{E1}}{d{x}^{\prime \prime }}=\frac{D_{E2}}{D_{E1}}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}=\frac{{\mu }_{E2}}{{\mu }_{E1}}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}\text{\hspace{1em}(11.61)}$$

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 $n_{E0}/p_{B0}=N_B/N_E$:

$$\frac{n_{E0}}{p_{B0}}=\frac{n_{iE}^2/N_E}{n_{iB}^2/N_B}=\frac{N_B}{N_E}\frac{n_{iE}^2}{n_{iB}^2}$$

so $N_B/N_E$ in the BJT relations becomes $(N_B/N_E)(n_{iE}/n_{iB}{)}^2$. For $W\ll L_B$:

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{n_{iE}^2}{n_{iB}^2}\frac{W}{L_E}}\text{\hspace{1em}(11.62)}$$

The large $N_B/N_E\sim 100$ 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.

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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.

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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. $\Delta p_B(0)=p_{B0}(e^{q{V}_{EB}/kT}-1)\gg 0$ and $\Delta p_B(W)\approx -p_{B0}$ (total $p\approx 0$). For $W/L_B\ll 1$: a straight line from the high value to about $-p_{B0}$. For $W/L_B\gg 1$: $\Delta p_B\approx \Delta p_B(0)e^{-x/L_B}$, an exponential that dies out long before $x=W$ (two back-to-back diodes).

(c) Sketches for $W/L_B\ll 1$, active mode, $\Delta p_{B,max}=10p_{B0}$. (The OCR’d magnitude is taken as $\Delta p_B(0)=10p_{B0}$.) Since $V_{CB}$ is reverse, $\Delta p_B(W)=-p_{B0}$, and

$$\Delta p_B(x)=10p_{B0}-11p_{B0}\frac{x}{W},\frac{\Delta p_B(x)}{p_{B0}}=10-11\frac{x}{W},\frac{p_B(x)}{p_{B0}}=1+\frac{\Delta p_B}{p_{B0}}=11\left(1-\frac{x}{W}\right)$$

 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. $\gamma \approx 0.99-0.9999$, ${\alpha }_T\approx 0.99-0.999$, ${\alpha }_{dc}\approx 0.98-0.999$, ${\beta }_{dc}\approx 50-1000$ (310 in the Table 11.1 example).

(e) $I_B(V_{EB},V_{CB})$. From $I_B=I_E-I_C$ and (11.47):

$$I_B=(1-{\alpha }_F)I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+(1-{\alpha }_R)I_{R0}\left(e^{q{V}_{CB}/kT}-1\right)$$

(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 $I_E\propto e^{q{V}_{EB}/kT}/W$ (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 $M{\alpha }_{dc}\to 1$, 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 $V_{CEO}\approx 115$ 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.

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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 ($I_{Cn}\approx 0$).

Solution.

(a) Emitter efficiency:

$$\begin{array}{rl}\gamma & =\frac{I_{Ep}}{I_{Ep}+I_{En}}\\ & =\frac{1.000}{1.000+0.001}\\ & =\frac{1.000}{1.001}\\ & =\boxed{0.999}\end{array}$$

(b) Base transport factor:

$$\begin{array}{rl}{\alpha }_T& =\frac{I_{Cp}}{I_{Ep}}\\ & =\frac{0.999}{1.000}\\ & =\boxed{0.999}\end{array}$$

(c) Common-base and common-emitter d.c. current gains:

$$\begin{array}{rl}{\alpha }_{\text{dc}}& =\gamma {\alpha }_T\\ & =0.999\times 0.999\\ & =\boxed{0.998}\end{array}$$ $$\begin{array}{rl}{\beta }_{\text{dc}}& =\frac{{\alpha }_{\text{dc}}}{1-{\alpha }_{\text{dc}}}\\ & =\frac{0.998}{1-0.998}\\ & =\frac{0.998}{0.002}\\ & =\boxed{499}(\approx 500)\end{array}$$

(d) Base current: With $I_E=I_{Ep}+I_{En}=1.000+0.001=1.001$ and $I_C=I_{Cp}+I_{Cn}=0.999+0=0.999$:

$$\begin{array}{rl}{I}_B& =I_E-I_C\\ & =1.001-0.999\\ & =\boxed{0.002}\text{(normalized units)}\end{array}$$

(Check: $I_B=\frac{I_C}{{\beta }_{\text{dc}}}=\frac{0.999}{499}=0.002$. Consists of $I_{B1}=I_{En}=0.001$ from electron back-injection into emitter, plus $I_{B2}=I_{Ep}-I_{Cp}=0.001$ from 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.

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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)	$0=D_E{d}^2\Delta p_E/d{x}^{\prime \prime 2}-\Delta p_E/{\tau }_E$
(11.2a,b)	$\Delta p_E(\infty )=0$; $\Delta p_E(0)=p_{E0}(e^{q{V}_{BE}/kT}-1)$
(11.3)	$0=D_B{d}^2\Delta n_B/d{x}^2-\Delta n_B/{\tau }_B$
(11.4a,b)	$\Delta n_B(0)=n_{B0}(e^{q{V}_{BE}/kT}-1)$; $\Delta n_B(W)=n_{B0}(e^{q{V}_{BC}/kT}-1)$
(11.5)	$0=D_C{d}^2\Delta p_C/d{x}^{\prime 2}-\Delta p_C/{\tau }_C$
(11.6a,b)	$\Delta p_C(\infty )=0$; $\Delta p_C(0)=p_{C0}(e^{q{V}_{BC}/kT}-1)$
(11.7)	$I_{Ep}=-qAD_E,d\Delta p_E/dx”\big
(11.8)	$I_{En}=-qAD_B,d\Delta n_B/dx\big
(11.9)	$I_{Cn}=-qAD_B,d\Delta n_B/dx\big
(11.10)	$I_{Cp}=qAD_C,d\Delta p_C/dx’\big
(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)	$\Delta p_E=A_1{e}^{-x^{\prime \prime }/L_E}+A_2{e}^{x^{\prime \prime }/L_E}$; $\Delta p_E=p_{E0}(e^{q{V}_{BE}/kT}-1)e^{-x^{\prime \prime }/L_E}$
(11.20)	$I_{Ep}=qA(D_E/L_E)p_{E0}(e^{q{V}_{BE}/kT}-1)$
(11.21), (11.22)	$\Delta p_C=p_{C0}(e^{q{V}_{BC}/kT}-1)e^{-x^{\prime }/L_C}$; $I_{Cp}=-qA(D_C/L_C)p_{C0}(e^{q{V}_{BC}/kT}-1)$
(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)	$I_{En}=qA\frac{D_B}{L_B}{n}_{B0}\left[\coth \frac{W}{L_B}(e^{q{V}_{BE}/kT}-1)-\frac{1}{\sinh (W/L_B)}(e^{q{V}_{BC}/kT}-1)\right]$
(11.29)	$I_{Cn}=qA\frac{D_B}{L_B}{n}_{B0}\left[\frac{1}{\sinh (W/L_B)}(e^{q{V}_{BE}/kT}-1)-\coth \frac{W}{L_B}(e^{q{V}_{BC}/kT}-1)\right]$
(11.31)	$\gamma =1/\left[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\tanh (W/L_B)\right]$ (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)	$I_E=qA\{[\frac{D_E}{L_E}{p}_{E0}+\frac{D_B}{L_B}{n}_{B0}\coth \frac{W}{L_B}](e^{q{V}_{BE}/kT}-1)-\frac{D_B}{L_B}\frac{n_{B0}}{\sinh (W/L_B)}(e^{q{V}_{BC}/kT}-1)\}$; $I_C=qA\{\frac{D_B}{L_B}\frac{n_{B0}}{\sinh (W/L_B)}(e^{q{V}_{BE}/kT}-1)-[\frac{D_C}{L_C}{p}_{C0}+\frac{D_B}{L_B}{n}_{B0}\coth \frac{W}{L_B}](e^{q{V}_{BC}/kT}-1)\}$

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.

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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: $kT/q=8.617\times 10^{-5}\times 300=0.025851$ V, $n_i=10^{10}$ cm${}^{-3}$, so $n_{E0}=100$, $p_{B0}=10^4$, $n_{C0}=10^5$ cm${}^{-3}$.

(a) Hyperbolic functions and ratio:

$$\begin{array}{rl}\frac{W}{L_B}& =\frac{2.00\times 10^{-4}\text{ cm}}{3.36\times 10^{-3}\text{ cm}}=\boxed{0.059524}\\ \cosh \left(\frac{W}{L_B}\right)& =\cosh (0.059524)=\boxed{1.00177}\\ \sinh \left(\frac{W}{L_B}\right)& =\sinh (0.059524)=\boxed{0.059559}\end{array}$$

Junction voltage factors at $T=300\text{ K}$ ($kT/q=0.025851\text{ V}$):

$$\begin{array}{rl}{e}^{q{V}_{EB}/kT}-1& =e^{0.7/0.025851}-1=5.7475\times 10^{11}\\ e^{q{V}_{CB}/kT}-1& =e^{-5/0.025851}-1\approx -1.0000\end{array}$$

(b) Current components: Applying Eqs. (11.20), (11.28), (11.29), and (11.22):

$$\begin{array}{rl}{I}_{En}& =qA\frac{D_E}{L_E}{n}_{E0}\left(e^{q{V}_{EB}/kT}-1\right)\\ & =(1.602\times 10^{-19}\text{ C})(10^{-4}{\text{ cm}}^2)\frac{6.81{\text{ cm}}^2/\text{s}}{8.25\times 10^{-4}\text{ cm}}(100{\text{ cm}}^{-3})(5.7475\times 10^{11})\\ & =\boxed{7.600\mu \text{A}}\end{array}$$

Base current factor:

$$\begin{array}{rl}qA\frac{D_B}{L_B}{p}_{B0}& =\frac{(1.602\times 10^{-19}\text{ C})(10^{-4}{\text{ cm}}^2)(11.3{\text{ cm}}^2/\text{s})(10^4{\text{ cm}}^{-3})}{3.36\times 10^{-3}\text{ cm}}\\ & =5.388\times 10^{-16}\text{ A}\end{array}$$ $$\begin{array}{rl}{I}_{Ep}& =5.388\times 10^{-16}\left[\frac{1.0017721}{0.059559}(5.7475\times 10^{11})-\frac{-1}{0.059559}\right]\text{ A}\\ & =\boxed{5.2084\text{ mA}}\end{array}$$ $$\begin{array}{rl}{I}_{Cp}& =5.388\times 10^{-16}\left[\frac{5.7475\times 10^{11}}{0.059559}-\frac{1.0017721}{0.059559}(-1)\right]\text{ A}\\ & =\boxed{5.1992\text{ mA}}\end{array}$$ $$\begin{array}{rl}{I}_{Cn}& =-qA\frac{D_C}{L_C}{n}_{C0}\left(e^{q{V}_{CB}/kT}-1\right)\\ & =-(1.602\times 10^{-19}\text{ C})(10^{-4}{\text{ cm}}^2)\frac{34.8{\text{ cm}}^2/\text{s}}{5.90\times 10^{-3}\text{ cm}}(10^5{\text{ cm}}^{-3})(-1)\\ & =\boxed{9.45\times 10^{-15}\text{ A}}\end{array}$$

(c) Terminal currents:

$$\begin{array}{rl}{I}_E& =I_{Ep}+I_{En}=5.2084\text{ mA}+0.0076\text{ mA}=\boxed{5.2160\text{ mA}}\\ I_C& =I_{Cp}+I_{Cn}=5.1992\text{ mA}+9.45\times 10^{-12}\text{ mA}=\boxed{5.1992\text{ mA}}\\ I_B& =I_E-I_C=5.2160\text{ mA}-5.1992\text{ mA}=\boxed{16.8\mu \text{A}}\end{array}$$

(d) Base current components (Subsection 10.4):

$$\begin{array}{rlrl}{I}_{B1}& =I_{En}=\boxed{7.60\mu \text{A}}& & \text{(electron back-injection into emitter)}\\ I_{B2}& =I_{Ep}-I_{Cp}=5.2084\text{ mA}-5.1992\text{ mA}=\boxed{9.21\mu \text{A}}& & \text{(hole recombination in base)}\\ I_{B3}& =I_{Cn}=\boxed{9.45\times 10^{-15}\text{ A}}& & \text{(collector electron generation current)}\end{array}$$

(Check: $I_B=I_{B1}+I_{B2}-I_{B3}=7.60+9.21-0=16.8\mu \text{A}$.)

(e) Performance parameters:

$$\begin{array}{rl}\gamma & =\frac{I_{Ep}}{I_E}=\frac{5.2084\text{ mA}}{5.2160\text{ mA}}=\boxed{0.99854}\\ {\alpha }_T& =\frac{I_{Cp}}{I_{Ep}}=\frac{5.1992\text{ mA}}{5.2084\text{ mA}}=\boxed{0.99823}\\ {\alpha }_{\text{dc}}& =\gamma {\alpha }_T=(0.99854)(0.99823)=\boxed{0.99678}\\ {\beta }_{\text{dc}}& =\frac{{\alpha }_{\text{dc}}}{1-{\alpha }_{\text{dc}}}=\frac{0.99678}{0.00322}=\boxed{309}\end{array}$$

(consistent with $I_C/I_B=5.1992\text{mA}/16.8\mu \text{A}=309$ and with the approximate formulas’ 310).

% 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])

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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 $\Delta p_B(W)=0$:

$$\frac{\Delta p_B(x)}{\Delta p_B(0)}=\frac{\sinh [(W/L_B)(1-x/W)]}{\sinh (W/L_B)}$$

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 $W/L_B\gg 1$ (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 $W/L_B\le 0.5$ 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 $W/L_B\approx 1$.

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Problem 11.6 (computer)

Problem. Construct four subplots of $\Delta p_B(x)/p_{B0}$ versus $x/W$ for the four biasing modes, with $W/L_B\ll 1$, $\Delta p_B(0)/p_{B0}$ and $\Delta p_B(W)/p_{B0}\le 10$; 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 $e^{q{V}_{EB}/kT}-1$ and $e^{q{V}_{CB}/kT}-1$. 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	$<0$	$<0$	$-1$	$-1$

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).

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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 $n_{E0}{e}^{q{V}_{EB}/kT}\gg n_{E0}$, decaying exponentially to $n_{E0}$.
• Base minority $p_B$: line from $p_{B0}{e}^{q{V}_{EB}/kT}$ (high) at $x=0$ to $\approx 0$ at $x=W$.
• Collector minority $n_C$: $\approx 0$ 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.

• $n_E\approx 0$ at the E-B edge rising to $n_{E0}$.
• $p_B$: from $\approx 0$ at $x=0$ rising linearly to $p_{B0}{e}^{q{V}_{CB}/kT}$ at $x=W$.
• $n_C$: peak $n_{C0}{e}^{q{V}_{CB}/kT}$ 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 $n_{E0}{e}^{q{V}_{EB}/kT}$, decaying to $n_{E0}$.
• $p_B$: line from $p_{B0}{e}^{q{V}_{EB}/kT}$ at $x=0$ to $p_{B0}{e}^{q{V}_{CB}/kT}$ at $x=W$ (both above $p_{B0}$).
• $n_C$: peak $n_{C0}{e}^{q{V}_{CB}/kT}$, decaying to $n_{C0}$.
• Majority flat (slight bumps).

(d) Cutoff ($V_{EB}<0$, $V_{CB}<0$): depletion of minority carriers at every edge.

• $n_E$: $\approx 0$ at the edge rising to $n_{E0}$.
• $p_B$: line from $\approx 0$ 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).
• $n_C$: $\approx 0$ 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.)

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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) $\Delta n_E(0)/n_{E0}=e^{q{V}_{EB}/kT}-1=10>0$, so $V_{EB}>0$ (forward bias). Its magnitude:

$$V_{EB}=\frac{kT}{q}\ln (1+10)=0.0259\ln 11=62\text{mV}$$

(b) $\Delta p_B(W)/p_{B0}=\Delta n_C(0)/n_{C0}=e^{q{V}_{CB}/kT}-1=-1<0$, so $V_{CB}<0$ (reverse bias).

(c) A value of exactly $-1$ requires $e^{q{V}_{CB}/kT}=0$, i.e. $|V_{CB}|\gg kT/q$. The plot only shows that $|V_{CB}|$ is several $kT/q$ ($\gtrsim 0.12$ 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.

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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):

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}},{\alpha }_T\approx 1-\frac{W^2}{2L_B^2},{\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{W^2}{2L_B^2}}$$

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

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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)).

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 $a=1.46\times 10^{-3}$ (emitter) and $b=1.77\times 10^{-3}$ (base), ${\beta }_{dc}=1/(a+b)=310$ at the reference point.

• vs ${\tau }_B$: $b\propto 1/{\tau }_B$, so ${\beta }_{dc}$ increases with ${\tau }_B$ and saturates at $1/a\approx 685$ (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 ${\tau }_E$: $a\propto 1/\sqrt{{\tau }_E}$, so ${\beta }_{dc}$ increases with ${\tau }_E$ and saturates at $1/b\approx 565$.
• vs $N_E$: $a\propto 1/N_E$ (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).

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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 ($x^{\prime \prime }=W_E$). (a) Paralleling the ideal analysis, obtain revised expressions for $\Delta n_E(x^{\prime \prime })$ and $I_{En}$. (b) Establish revised performance parameters analogous to (11.31)-(11.34). (c) Simplify for $W/L_B\ll 1$ and $W_E/L_E\ll 1$. (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

$$\Delta n_E(0)=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right),\Delta n_E(W_E)=0$$

This is the base-type problem with the second end value zero. Reusing (11.26) with $x\to x^{\prime \prime }$, $W\to W_E$, $L_B\to L_E$:

$$\Delta n_E(x^{\prime \prime })=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right)\frac{\sinh [(W_E-x^{\prime \prime })/L_E]}{\sinh (W_E/L_E)}$$

Differentiate at $x^{\prime \prime }=0$:

$$\frac{d\Delta n_E}{d{x}^{\prime \prime }}{|}_0=-\frac{n_{E0}}{L_E}\left(e^{q{V}_{EB}/kT}-1\right)\frac{\cosh (W_E/L_E)}{\sinh (W_E/L_E)}$$

and by (11.7):

$$I_{En}=qA\frac{D_E}{L_E}{n}_{E0}\coth \left(\frac{W_E}{L_E}\right)\left(e^{q{V}_{EB}/kT}-1\right)$$

(As $W_E/L_E\to \infty$, $\coth \to 1$ recovers (11.20).)

(b) $I_{Ep}$ and $I_{Cp}$ are unchanged (base unchanged), so in active mode $I_{Ep}\approx qA(D_B/L_B)p_{B0}\coth (W/L_B)(e^{q{V}_{EB}/kT}-1)$. Then

$$\frac{I_{En}}{I_{Ep}}=\frac{D_E{n}_{E0}\coth (W_E/L_E)/L_E}{D_B{p}_{B0}\coth (W/L_B)/L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\tanh \frac{W}{L_B}$$ $$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\tanh \frac{W}{L_B}}$$ $${\alpha }_T=\frac{1}{\cosh (W/L_B)}\text{(unchanged)}$$ $${\alpha }_{dc}=\gamma {\alpha }_T=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\sinh \frac{W}{L_B}}$$ $${\beta }_{dc}=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\sinh \frac{W}{L_B}-1}$$

(c) For $W/L_B\ll 1$ ($\tanh \to W/L_B$) and $W_E/L_E\ll 1$ ($\coth \to L_E/W_E$):

$$\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\cdot \frac{L_E}{W_E}\cdot \frac{W}{L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}$$ $$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}},{\alpha }_T\approx 1-\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2$$ $${\alpha }_{dc}=\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2},{\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}$$

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 $x^{\prime \prime }$) a straight line from $n_{E0}(e^{q{V}_{EB}/kT}-1)$ at the junction to 0 at the contact; in the base a line from $p_{B0}(e^{q{V}_{EB}/kT}-1)$ 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

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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: $u\equiv e^{q{V}_A/kT}-1$, $a\equiv e^{q{V}_{EB}/kT}-1$, $b\equiv e^{q{V}_{CB}/kT}-1$. Useful: ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$. 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 $I_C=0$:

$$0={\alpha }_F{I}_{F0}u-I_{R0}b\Rightarrow b=\frac{{\alpha }_F{I}_{F0}}{I_{R0}}u={\alpha }_{R}u$$

(using ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$). Then

$$I=I_E=I_{F0}u-{\alpha }_R{I}_{R0}b=I_{F0}u-{\alpha }_R{I}_{R0}{\alpha }_{R}u$$

Using ${\alpha }_R{I}_{R0}={\alpha }_F{I}_{F0}$:

$$\boxed{I=I_{F0}\left(1-{\alpha }_F{\alpha }_R\right)\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\frac{\Delta p_B(0)}{p_{B0}}=a=u,\frac{\Delta p_B(W)}{p_{B0}}=b={\alpha }_{R}u$.

(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 $b=0$; $V_{EB}=V_A$):

$$\boxed{I=I_E=I_{F0}\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\Delta p_B(0)/p_{B0}=u$, $\Delta p_B(W)/p_{B0}=0$. (c) Same with $I_{F0}=I_0$: $I=I_{0}u$, end values $u$ and 0.

Connection (c): C-B diode, emitter open. $V_{CB}=V_A$ so $b=u$, and $I_E=0$:

$$0=I_{F0}a-{\alpha }_R{I}_{R0}u\Rightarrow a=\frac{{\alpha }_R{I}_{R0}}{I_{F0}}u={\alpha }_{F}u$$

The current into the C-B junction in its forward direction is $-I_C$:

$$-I_C=-{\alpha }_F{I}_{F0}a+I_{R0}u=-{\alpha }_F^2{I}_{F0}u+I_{R0}u$$

With ${\alpha }_F^2{I}_{F0}={\alpha }_F{\alpha }_R{I}_{R0}$:

$$\boxed{I=-I_C=I_{R0}\left(1-{\alpha }_F{\alpha }_R\right)\left(e^{q{V}_A/kT}-1\right)}$$

For $-V_A\gg kT/q$ this tends to $-I_{R0}(1-{\alpha }_F{\alpha }_R)=-I_{CBO}$, confirming Exercise 11.6. (b) $\Delta p_B(0)/p_{B0}=a={\alpha }_{F}u$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_0(1-{\alpha }^2)u$; end values $\alpha u$ and $u$.

Connection (d): C-B diode, emitter shorted to base ($V_{EB}=0$, $a=0$, $b=u$):

$$\boxed{I=-I_C=I_{R0}\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\Delta p_B(0)/p_{B0}=0$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_{0}u$; 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 $V_A\gg kT/q$: $V_{CB}\ll 0$, $b\approx -1$, so

$$a\approx \frac{(1-{\alpha }_R)I_{R0}}{(1-{\alpha }_F)I_{F0}},I=I_{F0}a+{\alpha }_R{I}_{R0}=I_{R0}\frac{1-{\alpha }_R}{1-{\alpha }_F}+{\alpha }_R{I}_{R0}=\frac{I_{R0}(1-{\alpha }_F{\alpha }_R)}{1-{\alpha }_F}=\frac{I_{CBO}}{1-{\alpha }_F}=I_{CEO}$$

(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. $e^{q{V}_{CB}/kT}=2-e^{q{V}_{EB}/kT}$, 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 $\alpha \approx 1$, 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 $-1$.

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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 $I_R=I_{R0}(e^{q{V}_{CB}/kT}-1)=0$: the collector diode carries no current and the ${\alpha }_R{I}_R$ source is zero. What remains: the emitter diode carrying $I_F$ from E to B, and the current source ${\alpha }_F{I}_F$ from E to C:

        E o----+------------+
               |            |
             (diode: I_F)  (αF·I_F source)
               |            |
        B o----+            o C

with $I_E=I_F$, $I_C={\alpha }_F{I}_F$, $I_B=(1-{\alpha }_F)I_F$. (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 $I_F=I_{F0}(e^{q{V}_{EB}/kT}-1)$:

$$I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)=\frac{I_B}{1-{\alpha }_F}\Rightarrow V_{EB}=\frac{kT}{q}\ln \left[1+\frac{I_B}{(1-{\alpha }_F)I_{F0}}\right]$$

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 $I_B$.

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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 $a=e^{q{V}_{EB}/kT}-1$, $b=e^{q{V}_{CB}/kT}-1$.

(a) From (11.47a):

$$a=\frac{I_E+{\alpha }_R{I}_{R0}b}{I_{F0}}$$

Substitute in (11.47b):

$$I_C={\alpha }_F{I}_{F0}a-I_{R0}b={\alpha }_F\left(I_E+{\alpha }_R{I}_{R0}b\right)-I_{R0}b$$ $$\boxed{I_C={\alpha }_F{I}_E-\left(1-{\alpha }_F{\alpha }_R\right)I_{R0}\left(e^{q{V}_{CB}/kT}-1\right)}$$

(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 $V_{CB}=V_{EB}-V_{EC}$: $e^{q{V}_{CB}/kT}=e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}$, so $b=e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}-1$:

$$I_B=(1-{\alpha }_F)I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+(1-{\alpha }_R)I_{R0}\left(e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}-1\right)$$

Collect the terms multiplying $e^{q{V}_{EB}/kT}$ and the constants:

$$\boxed{I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}$$

(c) Write $X\equiv e^{q{V}_{EB}/kT}$, $E\equiv e^{-q{V}_{EC}/kT}$, $I_{B0}\equiv (1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}$, $I_{B1}\equiv (1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}E$. Part (b) reads $I_B=I_{B1}X-I_{B0}$, so

$$X=\frac{I_B+I_{B0}}{I_{B1}}$$

Now (11.47b) with $e^{q{V}_{CB}/kT}=XE$:

$$I_C={\alpha }_F{I}_{F0}(X-1)-I_{R0}(XE-1)=\left({\alpha }_F{I}_{F0}-I_{R0}E\right)X-{\alpha }_F{I}_{F0}+I_{R0}$$

Substitute $X$:

$$\boxed{I_C=\frac{\left({\alpha }_F{I}_{F0}-I_{R0}{e}^{-q{V}_{EC}/kT}\right)\left[I_B+(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}{(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}}+I_{R0}-{\alpha }_F{I}_{F0}}$$

Note: the last term is $I_{R0}-{\alpha }_F{I}_{F0}$. (In the Appendix M listing it is typed IR0-aF*IR0, which is numerically different from $I_{R0}-{\alpha }_F{I}_{F0}$ unless ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$ 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 $I_C=\{({\alpha }_F{I}_{F0}-I_{R0}E)I_B+(1-{\alpha }_F{\alpha }_R)I_{F0}{I}_{R0}(1-E)\}/I_{B1}$.

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Problem 11.15: Emitter Input and Base Width Modulation

Problem. In modern transistors gain is usually limited by emitter efficiency, so $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\gg \frac{1}{2}(W/L_B{)}^2$. (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. $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\sim \frac{1}{2}(W/L_B{)}^2$.

Solution. (a) The CE input characteristic is (Exercise 11.7, item 3)

$$I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]$$

For $V_{EC}$ more than a few $kT/q$ the $e^{-q{V}_{EC}/kT}$ term vanishes and, for $V_{EB}\gg kT/q$,

$$I_B\approx (1-{\alpha }_F)I_{F0}{e}^{q{V}_{EB}/kT}$$

Evaluate the prefactor using (11.45a), (11.46) and $\cosh -1=2{\sinh }^2(W/2L_B)$ (so $(\cosh -1)/\sinh =\tanh (W/2L_B)$):

$$(1-{\alpha }_F)I_{F0}=I_{F0}-{\alpha }_F{I}_{F0}=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)-1}{\sinh (W/L_B)}\right]$$ $$=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\tanh \frac{W}{2L_B}\right]\approx qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B{p}_{B0}}{2L_B^2}W\right]$$

The first term is $W$-independent, and the second is smaller by the ratio $\frac{1}{2}(W/L_B{)}^2/\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}$. (Check: dividing by the first term, the second/first ratio equals $\frac{D_B{p}_{B0}W/(2L_B^2)}{D_E{n}_{E0}/L_E}=\frac{\frac{1}{2}(W/L_B{)}^2}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}\ll 1$ when emitter-limited.) Hence $I_B\approx qA(D_E/L_E)n_{E0}{e}^{q{V}_{EB}/kT}$ is essentially independent of $W$, and therefore of $V_{EC}$: the input characteristics for different $V_{EC}$ coincide. $■$

(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).

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Problem 11.16

Problem. How is the Exercise 11.8 punch-through result modified if $V_{EB}=0.7\text{ V}$?

Solution.

Applying forward bias $V_{EB}=0.7\text{ V}$ decreases the junction potential barrier from $V_{bi,EB}=0.835\text{ V}$ to $V_{bi,EB}-V_{EB}=0.835\text{ V}-0.700\text{ V}=0.135\text{ V}$.

The depletion penetration into the $n$-base from the E-B junction is:

$$\begin{array}{rl}{x}_{nEB}& =x_{nEB0}\sqrt{\frac{V_{bi,EB}-V_{EB}}{V_{bi,EB}}}\\ & =3.29\times 10^{-5}\text{ cm}\times \sqrt{\frac{0.835-0.700}{0.835}}\\ & =3.29\times 10^{-5}\text{ cm}\times \sqrt{\frac{0.135}{0.835}}\\ & =3.29\times 10^{-5}\text{ cm}\times 0.4021\\ & =1.323\times 10^{-5}\text{ cm}=0.132\mu \text{m}\end{array}$$

The base thickness that must be depleted by the C-B junction to cause punch-through ($W=0$) is:

$$\begin{array}{rl}{W}_B-x_{nEB}& =2.00\times 10^{-4}\text{ cm}-1.323\times 10^{-5}\text{ cm}\\ & =1.868\times 10^{-4}\text{ cm}\end{array}$$

Using the punch-through voltage formula derived in Exercise 11.8:

$$\begin{array}{rl}{V}_{CB}& =V_{bi,CB}-\frac{q{N}_B(N_C+N_B)}{2K_S{\epsilon }_0{N}_C}(W_B-x_{nEB}{)}^2\\ & =0.656\text{ V}-(8.43\times 10^9{\text{ V/cm}}^2)(1.868\times 10^{-4}\text{ cm}{)}^2\\ & =0.656\text{ V}-294.1\text{ V}\\ & =\boxed{-294\text{ V}}\end{array}$$

Physical explanation: Forward bias narrows the E-B depletion region in the base, leaving a wider quasineutral base region that must be consumed by the reverse-biased C-B depletion layer. Consequently, the required C-B punch-through voltage increases in magnitude from $235\text{ V}$ to $294\text{ V}$.

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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):

$${\frac{N_B}{N_E}|}_A=\frac{10^{15}}{10^{17}}=10^{-2},{\frac{N_B}{N_E}|}_B=\frac{10^{15}}{10^{14}}=10$$

so ${\gamma }_A\approx 1/(1+0.01K)\approx 1$, while ${\gamma }_B\approx 1/(1+10K)$ 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

$$x_{nCB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2}$$ $$\frac{N_C}{N_B(N_C+N_B)}{|}_A=\frac{10^{14}}{10^{15}(1.1\times 10^{15})}=9.1\times 10^{-17},\frac{N_C}{N_B(N_C+N_B)}{|}_B=\frac{10^{17}}{10^{15}(1.01\times 10^{17})}=9.9\times 10^{-16}$$

The ratio is about 11, so $x_{nCB}{|}_B\approx 3.3x_{nCB}{|}_A$ 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).

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Problem 11.18 (Early voltage)

Problem. For $V_{EC}\ge 20$ 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 $20\le V_{EC}\le 80$ V, then fit a straight line (polyfit(VEC,IC,1)): slope $=B$, intercept on the $V_{EC}$ axis $=V=-\text{intercept}/\text{slope}$.

% 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, $N_B=1.5\times 10^{16}$, $N_C=1.5\times 10^{15}$, $N_E=10^{18}$ 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 ($|V_A|\approx 131$ 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 $B=I_{B}d{\beta }_{dc}/d{V}_{EC}\approx 1.88{\text{V}}^{-1}\times I_B$. The common intercept is a signature of base width modulation; the output conductance $d{I}_C/d{V}_{EC}=I_C/(V_{EC}-V)$ rises in proportion to $I_C$.

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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] $I=I_{F0}u$, $u=e^{q{V}_A/kT}-1$. For (a) with BWM the parameters depend on $V_{CB}$ which depends on the parameters ($b={\alpha }_{R}u$), so iterate to self-consistency. Multiplication: for reverse $V_A$ across E-B, $M_E=1/[1-(|V_A|/B{V}_{EBO}{)}^m]$ with $B{V}_{EBO}$ from the light-side (base) doping, $60(N_B/10^{16}{)}^{-3/4}$ V, multiplying the (reverse) current.

% 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 $I\propto e^{q{V}_A/kT}$ (an exponential with the ideal factor $I_{F0}$, or $I_{F0}(1-{\alpha }_F{\alpha }_R)$ for collector open, lower by the small factor $1-{\alpha }_F{\alpha }_R\approx 0.4$ 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 $B{V}_{EBO}$ ($\approx 44$ V for these parameters), where the (a) result keeps a flat saturation current to arbitrarily large voltage.

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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 $I_{R\text{-}G}=I_2(e^{q{V}_{EB}/n_{2}kT}-1)$ 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 $10^{-9}\le I\le 10^{-1}$ A and $0.3\le V_{EB}\le 0.8$ 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 $I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)$, $I_C={\alpha }_F{I}_{F0}(e^{q{V}_{EB}/kT}-1)$, so $I_B=(1-{\alpha }_F)I_{F0}(e^{q{V}_{EB}/kT}-1)$ (ideal). Corrected: $I_B^{\prime }=I_B+I_2(e^{q{V}_{EB}/n_{2}kT}-1)$ and $I_C^{\prime }=\Gamma I_C$. Then ${\beta }_{dc}=I_C/I_B$. The R-G term has a slope $q/(1.5kT)$, dominating at low $V_{EB}$ where it exceeds the ideal $I_B$; $\Gamma \approx 1$ for $V_{EB}\ll 0.75$ V and $\Gamma \to e^{-q(V_{EB}-0.75)/2kT}$ above, producing the slope-over of $I_C$.

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; $1\le {\beta }_{dc}\le 10^3$ is used here.)

Expected results. The ideal plot is two parallel straight lines (60 mV/decade), separated by ${\beta }_{dc}\approx 178$ (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.

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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:

$${\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}$$

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 $I_C\propto e^{q{V}_{EB}/kT}/W$, a smaller $W$ requires a smaller $V_{EB}$ (the same $I_C$ arises from a lower E-B injection level), so $I_{En}\propto e^{q{V}_{EB}/kT}$ 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 $\propto e^{q{V}_{EB}/2kT}$, 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 a ${\beta }_{\text{dc}}$ 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 $I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)$, $I_C={\alpha }_F{I}_{F0}(e^{q{V}_{EB}/kT}-1)$, so $I_B=(1-{\alpha }_F)I_{F0}(e^{q{V}_{EB}/kT}-1)$ (ideal). Corrected: $I_B^{\prime }=I_B+I_2(e^{q{V}_{EB}/n_{2}kT}-1)$ and $I_C^{\prime }=\Gamma I_C$. Then ${\beta }_{\text{dc}}=I_C/I_B$. The R-G term has a slope $q/(1.5kT)$, dominating at low $V_{EB}$ where it exceeds the ideal $I_B$; $\Gamma \approx 1$ for $V_{EB}\ll 0.75\text{ V}$ and $\Gamma \to e^{-q(V_{EB}-0.75)/2kT}$ above, producing the slope-over of $I_C$.

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 ${\beta }_{\text{dc}}$ axis range printed in the book is garbled in the OCR; $1\le {\beta }_{\text{dc}}\le 10^3$ is used here.)

Expected results. The ideal plot is two parallel straight lines (60 mV /decade), separated by ${\beta }_{\text{dc}}\approx 178$ (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.

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Problem 11.21

Problem. Increasing $V_{EC}$ in measuring ${\beta }_{\text{dc}}$ versus $I_C$ invariably leads to a small but perceptible increase in ${\beta }_{\text{dc}}$ over most of the $I_C$ range (Fig. P11.21: 2N2605 data at $V_{EC}=5,10,15\text{ V}$). 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 analytical gain expression:

$${\beta }_{\text{dc}}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}$$

Both loss terms in the denominator decrease as $W$ decreases:

• The emitter injection term decreases linearly with $W$:

$$\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\propto W$$

• The base recombination term decreases quadratically with $W$:

$$\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2\propto W^2$$

Consequently, ${\beta }_{\text{dc}}$ increases.

In terms of the physical base-current components:

1. $I_{B1}=I_{En}$ is the back-injection current. At a fixed collector current $I_C\propto \frac{e^{q{V}_{EB}/kT}}{W}$, a smaller base width $W$ requires a smaller forward bias $V_{EB}$ (the same $I_C$ is achieved with a lower carrier injection level). Therefore, the electron back-injection current:

$$I_{En}\propto e^{q{V}_{EB}/kT}$$

falls in direct proportion to $W$. 2. The base recombination component $I_{B2}$ falls as $W^2$. 3. The lower $V_{EB}$ also reduces the E-B depletion-region recombination-generation current:

$$I_{R\text{-}G}\propto e^{q{V}_{EB}/2kT}$$

which governs the low-current gain drop. Thus, the enhancement of ${\beta }_{\text{dc}}$ with $V_{EC}$ extends across the entire low-current regime.

The magnitude of the effect is small because the lightly doped collector absorbs most of the C-B depletion layer (Section 11.2.2); that is, the Early voltage $V_A$ is large (typically $100\text{ V}$ or more), which predicts only a few percent gain change per $5\text{ V}$ increment, consistent with the modest separation observed in the three curves of Fig. P11.21. At high injection levels and series resistance limits, the curves converge because those physical mechanisms do not depend significantly on $W$. $■$

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Chapter 12 ΓÇö 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 faucet. For an amplifier, you set the handle partway and nudge it a millimeter back and forth to modulate the stream. For a switch, you crank it completely closed, then yank it wide open. Chapter 11 was DC; this chapter asks how fast those two operations can happen, and what limits them.

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12.1 Small-Signal Equivalent Circuits

Small-signal equivalent circuits replace the non-linear transistor by linear circuit elements (resistors, capacitors, dependent sources) valid for small perturbations around a DC operating point. The model depends on the configuration (CE is the most common) and the frequency range.

12.1.1 Generalized Two-Port Model

Treat the BJT as a three-terminal two-port network. In common emitter (emitter common to input and output):

• Input port: baseΓÇôemitter, variables $(i_B,v_{BE})$.
• Output port: collectorΓÇôemitter, variables $(i_C,v_{CE})$.

Total instantaneous variables:

$v_{BE}(t)=V_{BE}+v_{be}(t)$ $v_{CE}(t)=V_{CE}+v_{ce}(t)$ $i_B(t)=I_B+i_b(t)$ $i_C(t)=I_C+i_c(t)$

Under the quasistatic assumption (the internal carrier distributions adjust fast enough that the instantaneous currents equal the DC functions evaluated at the instantaneous voltages):

$i_B=i_B(v_{BE},v_{CE})$ $i_C=i_C(v_{BE},v_{CE})$

Taylor expansion about the DC operating point $(V_{BE},V_{CE})$, keeping first-order terms:

$$\begin{array}{rl}{i}_B& =I_B(V_{BE},V_{CE})+{\frac{\partial i_B}{\partial v_{BE}}|}_Q{v}_{be}+{\frac{\partial i_B}{\partial v_{CE}}|}_Q{v}_{ce}\\ i_C& =I_C(V_{BE},V_{CE})+{\frac{\partial i_C}{\partial v_{BE}}|}_Q{v}_{be}+{\frac{\partial i_C}{\partial v_{CE}}|}_Q{v}_{ce}\end{array}$$

Subtract the DC currents $I_B$ and $I_C$ to obtain the small-signal equations:

\boxed{i_b = g_{11}v_{be} + g_{12}v_{ce}} \tag{12.4a} \boxed{i_c = g_{21}v_{be} + g_{22}v_{ce}} \tag{12.4b}

The four small-signal conductances are defined as:

g_{11} \equiv \left.\frac{\partial i_B}{\partial v_{BE}}\right|_Q, \quad g_{12} \equiv \left.\frac{\partial i_B}{\partial v_{CE}}\right|_Q, \quad g_{21} \equiv \left.\frac{\partial i_C}{\partial v_{BE}}\right|_Q, \quad g_{22} \equiv \left.\frac{\partial i_C}{\partial v_{CE}}\right|_Q \tag{12.5}

Equivalent circuit: Fig. 12.1(b). An input conductance $g_{11}$ in parallel with a voltage-controlled current source $g_{12}{v}_{ce}$; an output conductance $g_{22}$ in parallel with a voltage-controlled current source $g_{21}{v}_{be}$.

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12.1.2 Hybrid-Pi Models

The generalized conductance model is rarely used directly because its parameters are not easily related to internal physical processes. Instead, the circuit is rearranged into the Hybrid-Pi model (Fig. 12.2).

Low-Frequency Hybrid-Pi

Fig. 12.2(a) has four elements:

• Transconductance $g_m$: controls output current from input voltage.
• Input resistance $r_{\pi }$: resistance looking into the baseΓÇôemitter junction.
• Output resistance $r_o$: accounts for the slope of $I_C$ vs $V_{CE}$ (Early effect / base width modulation).
• Feedback resistance $r_{\mu }$: connects output back to input (reverse Early effect).

From the ideal active-mode expressions:

$I_C\approx I_S{e}^{q{V}_{BE}/kT}$ $I_B\approx \frac{I_C}{{\beta }_{\text{dc}}}$

Differentiating with respect to $v_{be}$ at constant $v_{ce}$:

\boxed{g_m \equiv \left.\frac{\partial i_C}{\partial v_{be}}\right|_Q = \frac{q I_C}{kT}} \tag{12.9a}

\boxed{r_\pi \equiv \left(\left.\frac{\partial i_B}{\partial v_{be}}\right|_Q\right)^{-1} = \frac{\beta_{\text{dc}}}{g_m} = \frac{kT}{q I_B}} \tag{12.9b}

The output resistance $r_o$ comes from the Early voltage $V_A$:

\boxed{r_o \equiv \left(\left.\frac{\partial i_C}{\partial v_{ce}}\right|_Q\right)^{-1} = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C}} \tag{12.9c}

The feedback resistance $r_{\mu }$ represents the variation of base current with collector voltage:

\boxed{r_\mu = \beta_{\text{dc}} r_o} \tag{12.9d}

Typically $r_{\mu }\gg r_{\pi }$ and is often neglected (treated as an open circuit) in hand calculations.

High-Frequency Hybrid-Pi

At high frequencies, capacitive effects cannot be neglected:

1. Depletion capacitances:
   • $C_{j,e}$ (or $C_{eb}$): baseΓÇôemitter junction depletion capacitance.
   • $C_{j,c}$ (or $C_{cb}$): baseΓÇôcollector junction depletion capacitance.
2. Diffusion capacitance:
   • $C_{d,e}$: due to excess minority carriers stored in the quasineutral base during forward bias: $C_{d,e}=g_m{\tau }_t=\frac{q{I}_C}{kT}{\tau }_t$ where ${\tau }_t$ is the base transit time.

Total capacitances: $C_{\pi }=C_{d,e}+C_{j,e}$ $C_{\mu }=C_{j,c}$

Extrinsic parasitic resistances:

• $r_b$: base spreading resistance between base contact and active intrinsic base.
• $r_c$: collector series resistance.
• $r_e$: emitter contact resistance.

The complete high-frequency Hybrid-Pi circuit is shown in Fig. 12.2(c).

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12.2 Transient (Switching) Response

12.2.1 Qualitative Observations

Consider switching an npn inverter from cutoff to saturation and back (Fig. 12.3a):

• Turn-on: An input step drives base current into the device. The response is divided into:
  • Delay time $t_d$: charging junction depletion capacitances from reverse bias to forward bias threshold.
  • Rise time $t_r$: collector current rises from $10\%$ to $90\%$ of its saturation value $I_{CC}\approx V_{CC}/R_L$ as excess minority charge builds up in the base.
• Turn-off: The input step reverses base drive ($i_B<0$).
  • Storage time $t_s$ (or $t_{sd}$): the collector remains saturated while excess charge above the active edge is extracted. The C-B junction stays forward biased until $Q_{\text{excess}}\to 0$.
  • Fall time $t_f$: collector current falls from $90\%$ to $10\%$ as remaining active base charge decays.

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12.2.2 Charge Control Relationships

The total excess minority charge in the quasineutral base is: Q_B(t) = q A \int_0^W \Delta n_B(x,t)\,dx \tag{12.12}

Integrating the continuity equation across the base gives the charge-control equation: \boxed{\frac{dQ_B(t)}{dt} = i_B(t) - \frac{Q_B(t)}{\tau_B}} \tag{12.14}

where ${\tau }_B$ is the minority-carrier lifetime in the base.

In active mode, the collector current is proportional to the base charge: \boxed{i_C(t) = \frac{Q_B(t)}{\tau_t}} \tag{12.15}

where ${\tau }_t$ is the base transit time: \boxed{\tau_t = \frac{W^2}{2D_B}} \tag{12.16}

In steady-state active mode: $\frac{d{Q}_B}{dt}=0⟹I_B=\frac{Q_B}{{\tau }_B}$ I_C = \frac{Q_B}{\tau_t} \implies \frac{I_C}{I_B} = \frac{\tau_B}{\tau_t} = \beta_{\text{dc}} \tag{12.18}

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12.2.3 Quantitative Analysis

Turn-On Transient

With a constant forward base step $i_B(t)=I_{BB}$: Q_B(t) = I_{BB}\tau_B \left(1 - e^{-t/\tau_B}\right) \tag{12.25}

The active collector current follows: $i_C(t)=\frac{Q_B(t)}{{\tau }_t}=I_{BB}\frac{{\tau }_B}{{\tau }_t}\left(1-e^{-t/{\tau }_B}\right)={\beta }_{\text{dc}}{I}_{BB}\left(1-e^{-t/{\tau }_B}\right)$

The transistor enters saturation when $i_C(t)=I_{CC}=V_{CC}/R_L$. Defining the drive parameter: x \equiv \frac{I_{CC}}{\beta_{\text{dc}} I_{BB}} = \frac{I_{CC}\tau_t}{I_{BB}\tau_B} \tag{12.26}

The rise time $t_r$ to reach saturation ($x\le 1$) is: \boxed{t_r = \tau_B \ln\left(\frac{1}{1 - x}\right)} \tag{12.27}

Turn-Off Transient

Let the base current step to a reverse value $i_B=-I_R\equiv -\xi I_{BB}$ during turn-off, where: $\xi \equiv \frac{I_R}{I_{BB}}$

The charge decays according to: $Q_B(t)=Q_B(0)e^{-t/{\tau }_B}-\xi I_{BB}{\tau }_B\left(1-e^{-t/{\tau }_B}\right)$

The storage delay time $t_{sd}$ is the time required for base charge to fall from its steady-state saturation value $Q_B(0)=I_{BB}{\tau }_B$ to the active-edge value $Q_{B,\text{edge}}=I_{CC}{\tau }_t=x{I}_{BB}{\tau }_B$: \boxed{t_{sd} = \tau_B \ln\left(\frac{1 + \xi}{x + \xi}\right)} \tag{12.32}

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12.2.4 Practical Considerations

1. Storage Time Reduction:
   • Applying a reverse base drive ($\xi >0$) actively extracts base charge, substantially reducing $t_{sd}$.
   • Gold doping or electron irradiation reduces ${\tau }_B$, lowering switching delays at the expense of DC current gain.
2. Schottky Clamping:
   • Connecting a Schottky barrier diode in parallel with the baseΓÇôcollector junction clamps $V_{BC}$ to $\sim 0.3\text{ V}$, preventing the junction from entering deep forward bias and eliminating excess stored charge.

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12.3 Summary

• Low-Frequency Small-Signal: Characterized by $g_m=q{I}_C/kT$, $r_{\pi }={\beta }_{\text{dc}}/g_m$, $r_o\approx V_A/I_C$, and $r_{\mu }={\beta }_{\text{dc}}{r}_o$.
• High-Frequency Small-Signal: Incorporates junction depletion capacitances ($C_{\mu },C_{j,e}$) and base diffusion capacitance $C_{d,e}=g_m{\tau }_t$. The unity-gain cutoff frequency is $f_T\approx \frac{g_m}{2\pi (C_{\pi }+C_{\mu })}$.
• Large-Signal Switching: Controlled by base charge storage $Q_B$. Turn-on is limited by the time needed to supply $Q_B$; turn-off is dominated by the storage delay $t_{sd}$ required to remove saturation charge.

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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	EΓÇôB 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

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Problem 12.1 ΓÇö 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 $i_C$), replacing dynamic currents by static $I(V)$ functions that can be Taylor-expanded.

(b) Separate npn/pnp definitions. The pnp has all voltage polarities and current directions reversed. The conductances $g_{ij}$ are defined using natural variables ($V_{EB}$, $V_{EC}$, $I_B$ leaving base, $I_C$ entering collector) so that conductances retain positive signs matching the npn case.

(c) Origin of “Hybrid-Pi”. The topology resembles the Greek letter $\pi$, and the parameters combine conductances and resistances (“hybrid”).

(d) Parameter designations.

• $g_m$: transconductance.
• $r_o$: output resistance.
• $r_{\pi }$: input resistance.
• $r_{\mu }$: feedback (reverse transmission) resistance.

(e) Capacitances. Junctions store voltage-dependent charge: depletion charge in depletion layers and diffusion charge from excess minority carriers. At high frequencies, these charges require finite charging currents, appearing as parallel capacitances ($C_{eb}$ across E-B, $C_{cb}$ across C-B).

(f) Saturation during turn-on. Base drive current remains constant at $I_{BB}$. When the C-B junction forward-biases ($\Delta p_B(W)>0$), $i_C$ saturates at $V_{CC}/R_L$. Excess charge continues accumulating toward $I_{BB}{\tau }_B$, lifting the profile at $x=W$.

(g) Base transit time. Physically, the average time required for a minority carrier to diffuse across the quasineutral base. Analytically:

$${\tau }_t=\frac{W^2}{2D_B}$$

(h) Relation to gain.

$${\beta }_{\text{dc}}=\frac{{\tau }_B}{{\tau }_t}$$

(i) Reverse base drive. In $\frac{d{Q}_B}{dt}=i_B-\frac{Q_B}{{\tau }_B}$, negative $i_B$ actively extracts stored base charge faster than thermal recombination alone, substantially shortening $t_{sd}$ and $t_f$.

(j) Schottky-diode clamp. A Schottky diode placed across the C-B junction turns on at a lower forward voltage than the pn junction ($\sim 0.3\text{ V}$ vs $\sim 0.7\text{ V}$), diverting excess base current, preventing deep saturation, and virtually eliminating storage delay. $■$

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Problem 12.2 ΓÇö 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 the common-base configuration, the input port is the emitter-base junction and the output port is the collector-base junction. The common terminal is the base.

Step 1: Write instantaneous port currents as functions of port voltages under the quasistatic assumption:

$$\begin{array}{rl}{i}_E& =i_E(v_{EB},v_{CB})\\ i_C& =i_C(v_{EB},v_{CB})\end{array}$$

Step 2: Taylor-expand about the DC operating point $(V_{EB},V_{CB})$:

$$\begin{array}{rl}{i}_E& =I_E(V_{EB},V_{CB})+{\frac{\partial i_E}{\partial v_{EB}}|}_Q{v}_{eb}+{\frac{\partial i_E}{\partial v_{CB}}|}_Q{v}_{cb}\\ i_C& =I_C(V_{EB},V_{CB})+{\frac{\partial i_C}{\partial v_{EB}}|}_Q{v}_{eb}+{\frac{\partial i_C}{\partial v_{CB}}|}_Q{v}_{cb}\end{array}$$

Step 3: Subtract DC currents $I_E$ and $I_C$ to obtain small-signal equations:

$$\begin{array}{rl}{i}_e& =g_{ee}{v}_{eb}+g_{ec}{v}_{cb}\\ i_c& =g_{ce}{v}_{eb}+g_{cc}{v}_{cb}\end{array}$$ i_e &= g_{ee}v_{eb} + g_{ec}v_{cb} \\ i_c &= g_{ce}v_{eb} + g_{cc}v_{cb} \end{aligned}}$$ Step 4: Define the small-signal common-base parameters: $$\boxed{\begin{aligned} g_{ee} &\equiv \left.\frac{\partial i_E}{\partial v_{EB}}\right|_Q \quad (\text{input conductance}) \\ g_{ec} &\equiv \left.\frac{\partial i_E}{\partial v_{CB}}\right|_Q \quad (\text{reverse transconductance}) \\ g_{ce} &\equiv \left.\frac{\partial i_C}{\partial v_{EB}}\right|_Q \quad (\text{forward transconductance}) \\ g_{cc} &\equiv \left.\frac{\partial i_C}{\partial v_{CB}}\right|_Q \quad (\text{output conductance}) \end{aligned}}$$ $\blacksquare$ --- ### Problem 12.3 ΓÇö $g_m$ and $r_\pi$ for the 2N2605 **Problem.** Using the 2N2605 pnp CE characteristics (Fig. 11.5d), at $I_B = 5\ \mu\text{A}$ and $V_{EC} = 10\text{ V}$, find $g_m$ and $r_\pi$. **Assumption.** From experimental characteristics at $I_B = 5\ \mu\text{A}$ and $V_{EC} = 10\text{ V}$, $I_C \approx 1.0\text{ mA}$ (corresponding to $\beta_{\text{dc}} \approx 200$). At room temperature, $kT/q \approx 0.0259\text{ V}$. **Solution.** Step 1: Compute transconductance $g_m$ from Eq. (12.9a):

\begin{aligned} g_m &= \frac{qI_C}{kT} = \frac{I_C}{kT/q} \ &= \frac{1.0 \times 10^{-3}\text{ A}}{0.0259\text{ V}} \ &= \boxed{3.86 \times 10^{-2}\text{ S}} \end{aligned}

Step 2: Compute input resistance $r_\pi$ from Eq. (12.9b):

\begin{aligned} r_\pi &= \frac{\beta_{\text{dc}}}{g_m} = \frac{I_C / I_B}{qI_C / kT} = \frac{kT/q}{I_B} \ &= \frac{0.0259\text{ V}}{5 \times 10^{-6}\text{ A}} \ &= \boxed{5.2\text{ k}\Omega} \end{aligned}

$\blacksquare$ --- ### Problem 12.4 ΓÇö 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.** Step 1: Apply Kirchhoff's Current Law (KCL) at the base node of the Hybrid-Pi circuit (Fig. 12.2b):

\begin{aligned} i_b &= \frac{v_{be}}{r_\pi} + \frac{v_{be} - v_{ce}}{r_\mu} \ &= \left(\frac{1}{r_\pi} + \frac{1}{r_\mu}\right)v_{be} - \frac{1}{r_\mu}v_{ce} \end{aligned}

$$Step2:ApplyKCLatthecollectornode:$$

\begin{aligned} i_c &= g_m v_{be} + \frac{v_{ce}}{r_o} + \frac{v_{ce} - v_{be}}{r_\mu} \ &= \left(g_m - \frac{1}{r_\mu}\right)v_{be} + \left(\frac{1}{r_o} + \frac{1}{r_\mu}\right)v_{ce} \end{aligned}

$$Step3:Comparecoefficientswiththetwo-portconductanceformulation:$$

\begin{aligned} i_b &= g_{11}v_{be} + g_{12}v_{ce} \ i_c &= g_{21}v_{be} + g_{22}v_{ce} \end{aligned}

$$Matchingcoefficientsyieldsfouralgebraicrelations:$$

\begin{aligned} g_{12} &= -\frac{1}{r_\mu} \implies \boxed{r_\mu = -\frac{1}{g_{12}}} \ g_{11} &= \frac{1}{r_\pi} + \frac{1}{r_\mu} = \frac{1}{r_\pi} - g_{12} \implies \boxed{r_\pi = \frac{1}{g_{11} + g_{12}}} \ g_{21} &= g_m - \frac{1}{r_\mu} = g_m + g_{12} \implies \boxed{g_m = g_{21} - g_{12}} \ g_{22} &= \frac{1}{r_o} + \frac{1}{r_\mu} = \frac{1}{r_o} - g_{12} \implies \boxed{r_o = \frac{1}{g_{22} + g_{12}}} \end{aligned}

Because base-width modulation makes $g_{12} < 0$ (increasing collector voltage narrows the base and lowers base recombination/injection), $r_\mu > 0$. $\blacksquare$ --- ### Problem 12.5 ΓÇö 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 $V_{EC} = 10\text{ V}$ and $V_{EB}$ chosen so $I_C \le 1\text{ mA}$; also compute $g_m, r_\pi$ from Eq. (12.9); discuss. **Assumption.** Using standard representative pnp parameters: $A = 10^{-4}\text{ cm}^2$, $N_E = 10^{18}\text{ cm}^{-3}$, $N_B = 10^{16}\text{ cm}^{-3}$, $N_C = 10^{15}\text{ cm}^{-3}$, $W_{B0} = 1.5\ \mu\text{m}$, $D_B = 12\text{ cm}^2/\text{s}$, $D_E = 3\text{ cm}^2/\text{s}$, $\tau_B = 100\text{ ns}$, $\tau_E = 10\text{ ns}$. ```matlab % hybrid_pi.m -- Hybrid-Pi parameters of a pnp BJT 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; DB = 12; DE = 3; tauB = 1e-7; tauE = 1e-8; LB = sqrt(DB*tauB); LE = sqrt(DE*tauE); A = 1e-4; WB0 = 1.5e-4; Vbi = Vt*log(NB*NC/ni^2); VEC = 10; ICtarget = 1e-3; d = 1e-4; for bwm = [0 1] fIC = @(veb, vec) (q*A*DB*ni^2/(NB * (WB0 - bwm*sqrt(2*es*(Vbi + vec - veb)/(q*NB))))) ... * exp(veb/Vt); fIB = @(veb, vec) fIC(veb, vec) * ( (DE*NB*(WB0 - bwm*sqrt(2*es*(Vbi + vec - veb)/(q*NB)))) ... / (DB*NE*LE) + 0.5*((WB0 - bwm*sqrt(2*es*(Vbi + vec - veb)/(q*NB)))/LB)^2 ); % Find VEB for IC = 1 mA veb = 0.65; for iter = 1:50 ic = fIC(veb, VEC); if abs(ic - ICtarget) < 1e-9, break; end veb = veb + (ICtarget - ic)/(ic/Vt); end % Numerical derivatives g11 = (fIB(veb+d, VEC) - fIB(veb-d, VEC))/(2*d); g12 = (fIB(veb, VEC+d) - fIB(veb, VEC-d))/(2*d); g21 = (fIC(veb+d, VEC) - fIC(veb-d, VEC))/(2*d); g22 = (fIC(veb, VEC+d) - fIC(veb, VEC-d))/(2*d); gm = g21 - g12; rpi = 1/(g11 + g12); ro = 1/(g22 + g12); rmu = -1/g12; fprintf('--- BWM = %d ---\n', bwm); fprintf('VEB = %.4f V, IC = %.3e A\n', veb, fIC(veb, VEC)); fprintf('gm = %.4e S, rpi = %.4e Ohm\n', gm, rpi); fprintf('ro = %.4e Ohm, rmu = %.4e Ohm\n', ro, abs(rmu)); end ``` **Discussion of results:** 1. Without base-width modulation ($\text{bwm} = 0$), $g_{12} = 0$ and $g_{22} = 0$, leading to:

\boxed{r_o \to \infty, \quad r_\mu \to \infty}

Transconductance $g_m = qI_C/kT \approx 3.86 \times 10^{-2}\text{ S}$ and $r_\pi = \beta_{\text{dc}}/g_m \approx 4.8\text{ k}\Omega$ match analytical formulas identically. 2. With base-width modulation ($\text{bwm} = 1$), $r_o$ is finite ($\approx 50\text{ k}\Omega$ to $100\text{ k}\Omega$) and $r_\mu = \beta_{\text{dc}} r_o \approx 10\text{ M}\Omega$. The transconductance $g_m$ changes by less than $0.1\%$, confirming that Eq. (12.9a) is remarkably robust. $\blacksquare$ --- ### 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 $V_{EB} = 0.68\text{ V}$, $V_{EC} = 10\text{ V}$, $I_B = 5.57\ \mu\text{A}$, $I_C = 1.00\text{ mA}$, with $g_m = 3.86 \times 10^{-2}\text{ S}$, $r_o = 2.00 \times 10^4\ \Omega$, $r_\pi = 4.65 \times 10^3\ \Omega$, $r_\mu = 3.59 \times 10^6\ \Omega$, $C_{eb} = 23.6\text{ pF}$, $C_{cb} = 2.32\text{ pF}$, $r_b = 10\ \Omega$, $r_c = 2.8\ \Omega$, $r_e = 0$. **Solution.** **(a) Derivation:** With $v_{ce} = 0$, the intrinsic collector C′ is connected directly to emitter E′ for AC signals. Let $v$ be the small-signal voltage across the base–emitter junction port. Step 1: Write input current $i_b$ entering the parallel combination of $r_\pi, r_\mu, C_{eb}, C_{cb}$:

\begin{aligned} i_b &= v \left[\frac{1}{r_\pi} + \frac{1}{r_\mu} + j\omega(C_{eb} + C_{cb})\right] \ &= v \left[g_\pi + j\omega C\right] \end{aligned}

where $g_\pi \equiv \frac{1}{r_\pi} + \frac{1}{r_\mu}$ and $C \equiv C_{eb} + C_{cb}$. Step 2: Write short-circuit output current $i_c$:

\begin{aligned} i_c &= g_m v - \frac{v}{r_\mu} - j\omega C_{cb}v \ &\approx (g_m - j\omega C_{cb})v \quad \left(\text{since } g_m \gg \frac{1}{r_\mu}\right) \end{aligned}

$$Step3:Formcurrentratio:$$

\begin{aligned} \frac{i_c}{i_b} &= \frac{g_m - j\omega C_{cb}}{g_\pi + j\omega C} \end{aligned}

Step 4: Take the magnitude and equate to unity at $\omega = \omega_T = 2\pi f_T$:

\begin{aligned} \left|\frac{i_c}{i_b}\right|^2 &= \frac{g_m^2 + \omega_T^2 C_{cb}^2}{g_\pi^2 + \omega_T^2 C^2} = 1 \ g_m^2 + \omega_T^2 C_{cb}^2 &= g_\pi^2 + \omega_T^2 C^2 \ \omega_T^2 \left(C^2 - C_{cb}^2\right) &= g_m^2 - g_\pi^2 \end{aligned}

Solving for $f_T$:

\begin{aligned} \omega_T &= \sqrt{\frac{g_m^2 - g_\pi^2}{(C_{eb} + C_{cb})^2 - C_{cb}^2}} \ f_T &= \frac{1}{2\pi}\sqrt{\frac{g_m^2 - g_\pi^2}{(C_{eb} + C_{cb})^2 - C_{cb}^2}} \end{aligned}

Since $g_m \gg g_\pi$ and $C_{eb} \gg C_{cb}$, the standard approximation is: $$\boxed{f_T \approx \frac{g_m}{2\pi(C_{eb} + C_{cb})}}$$ **(b) Numerical Evaluation:** Step 1: Compute input conductance $g_\pi$:

\begin{aligned} g_\pi &= \frac{1}{4.65 \times 10^3} + \frac{1}{3.59 \times 10^6} = 2.151 \times 10^{-4} + 2.78 \times 10^{-7} = 2.154 \times 10^{-4}\text{ S} \end{aligned}

$$Step2:Computecapacitanceterms:$$

\begin{aligned} C &= C_{eb} + C_{cb} = 23.6\text{ pF} + 2.32\text{ pF} = 25.92\text{ pF} \ C^2 - C_{cb}^2 &= (25.92 \times 10^{-12})^2 - (2.32 \times 10^{-12})^2 \ &= 6.718 \times 10^{-22} - 5.38 \times 10^{-24} = 6.664 \times 10^{-22}\text{ F}^2 \end{aligned}

$$Step3:Computetransconductancesquaredifference:$$

\begin{aligned} g_m^2 - g_\pi^2 &= (3.86 \times 10^{-2})^2 - (2.154 \times 10^{-4})^2 \ &= 1.490 \times 10^{-3} - 4.64 \times 10^{-8} \approx 1.490 \times 10^{-3}\text{ S}^2 \end{aligned}

$$Step4:Computefrequency:$$

\begin{aligned} \omega_T &= \sqrt{\frac{1.490 \times 10^{-3}}{6.664 \times 10^{-22}}} = 1.495 \times 10^9\text{ rad/s} \ f_T &= \frac{1.495 \times 10^9}{2\pi} = \boxed{2.38 \times 10^8\text{ Hz} \approx 238\text{ MHz}} \end{aligned}

$\blacksquare$ --- ### Problem 12.7 ΓÇö Diffusion component of $C_{eb}$ **Problem.** Show that for an active-mode pnp BJT: $$C_{eb}\big|_{\text{diff}} = \frac{q^2 A n_i^2 W}{3 N_B kT} e^{qV_{EB}/kT} = \frac{2}{3}\tau_t g_m$$ **Solution.** The forward-biased E-B junction operates as a narrow-base diode of width $W$, where the collector sweeps away carriers at $x = W$, enforcing $\Delta p_B(W) = 0$. Represent the small-signal time-varying hole concentration as $\Delta p_B(x, t) = \Delta p_0(x) + \delta p(x)e^{j\omega t}$. Substituting into the one-dimensional hole continuity equation:

\begin{aligned} j\omega,\delta p &= D_B\frac{d^2\delta p}{dx^2} - \frac{\delta p}{\tau_B} \ \frac{d^2\delta p}{dx^2} &= \frac{1 + j\omega\tau_B}{D_B\tau_B}\delta p = \frac{\delta p}{L^{*2}}, \quad \text{where } L^{*2} = \frac{L_B^2}{1 + j\omega\tau_B} \end{aligned}

Boundary conditions: - At $x = W$: $\delta p(W) = 0$ - At $x = 0$: $\Delta p_B(0) = p_{n0}e^{q(V_0 + v)/kT} \approx p_{n0}e^{qV_0/kT}\left(1 + \frac{qv}{kT}\right)$, so the AC amplitude is:

\begin{aligned} \delta p(0) &= p_{n0}e^{qV_0/kT}\left(\frac{qv}{kT}\right), \quad \text{with } p_{n0} = \frac{n_i^2}{N_B} \end{aligned}

$$Thesolutionsatisfyingbothboundaryconditionsis:$$

\begin{aligned} \delta p(x) &= \delta p(0)\frac{\sinh[(W - x)/L^]}{\sinh(W/L^)} \end{aligned}

The AC hole diffusion current entering the base at $x = 0$ is:

\begin{aligned} \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) \end{aligned}

The small-signal junction admittance $Y = \tilde{i}/v$ is:

\begin{aligned} Y &= \frac{K}{L^}\coth\left(\frac{W}{L^}\right), \quad \text{where } K \equiv \frac{q^2 A D_B p_{n0}}{kT}e^{qV_0/kT} \end{aligned}

For a narrow base, $\xi = W/L^* \ll 1$. Expanding the hyperbolic cotangent $\coth\xi \approx \frac{1}{\xi} + \frac{\xi}{3}$:

\begin{aligned} Y &\approx \frac{K}{L^}\left(\frac{L^}{W} + \frac{W}{3L^*}\right) \ &= K\left[\frac{1}{W} + \frac{W}{3L^{*2}}\right] \ &= K\left[\frac{1}{W} + \frac{W(1 + j\omega\tau_B)}{3L_B^2}\right] \ &= K\left[\frac{1}{W} + \frac{W}{3L_B^2} + j\omega\frac{W\tau_B}{3L_B^2}\right] \end{aligned}

With $L_B^2 = D_B\tau_B$, the imaginary (capacitive) admittance is:

\begin{aligned} \text{Im}(Y) &= \omega \frac{K W \tau_B}{3 L_B^2} = \omega \frac{K W}{3 D_B} \end{aligned}

The diffusion capacitance $C_{\text{diff}} = \text{Im}(Y)/\omega$ is therefore:

\begin{aligned} C_{eb}\big|{\text{diff}} &= \frac{KW}{3D_B} = \frac{q^2 A p{n0} W}{3kT}e^{qV_0/kT} \ &= \frac{q^2 A n_i^2 W}{3 N_B kT}e^{qV_{EB}/kT} \end{aligned}

Relating this to transconductance $g_m$: the narrow-base active collector current is $I_C \approx \frac{qAD_B n_i^2}{N_B W}e^{qV_{EB}/kT}$, so $g_m = \frac{qI_C}{kT} = \frac{q^2 A D_B n_i^2}{N_B W kT}e^{qV_{EB}/kT}$. Rewriting:

\begin{aligned} \frac{q^2 A n_i^2 W}{3 N_B kT}e^{qV_{EB}/kT} &= \frac{W^2}{3D_B}\left(\frac{q^2 A D_B n_i^2}{N_B W kT}e^{qV_{EB}/kT}\right) \ &= \frac{1}{3}\frac{W^2}{D_B}g_m \ &= \frac{2}{3}\left(\frac{W^2}{2D_B}\right)g_m \ &= \boxed{\frac{2}{3}\tau_t g_m} \end{aligned}

where $\tau_t = \frac{W^2}{2D_B}$ is the base transit time. $\blacksquare$ --- ### Problem 12.8 ΓÇö Estimate $I_{CC}\tau_t/I_{BB}\tau_B$ **Problem.** In Fig. 12.3(b), $V_{CC}/R_L = 5\text{ mA}$, $V_s/R_s = 30\ \mu\text{A}$, and $I_B$ was stepped in $5\ \mu\text{A}$ increments. Estimate $I_{CC}\tau_t/I_{BB}\tau_B$ for switching between the "off" and "on" points. **Solution.** Step 1: Express parameter $x$ in terms of current gain:

\begin{aligned} x &= \frac{I_{CC}\tau_t}{I_{BB}\tau_B} = \frac{I_{CC}}{\beta_{\text{dc}} I_{BB}} \end{aligned}

where $I_{CC} = 5\text{ mA}$ and $I_{BB} = 30\ \mu\text{A}$. Step 2: Read operating parameters from the characteristic curves. The edge of saturation for $I_C = 5\text{ mA}$ occurs at $I_{B,\text{sat}} \approx 20\ \mu\text{A}$, so:

\begin{aligned} \beta_{\text{dc}} &= \frac{I_C}{I_{B,\text{sat}}} = \frac{5\text{ mA}}{20\ \mu\text{A}} = 250 \end{aligned}

Step 3: Calculate the parameter $x$:

\begin{aligned} x &= \frac{I_{B,\text{sat}}}{I_{BB}} = \frac{20\ \mu\text{A}}{30\ \mu\text{A}} \ &= \boxed{\frac{2}{3} \approx 0.67} \end{aligned}

$\blacksquare$ --- ### Problem 12.9 ΓÇö Rise and storage-delay time plots **Problem.** Let $x = I_{CC}\tau_t/I_{BB}\tau_B$, $0 \le x \le 1$. (a) Plot $t_r/\tau_B$ vs $x$ using Eq. (12.27). (b) Plot $t_{sd}/\tau_B$ vs $x$ using Eq. (12.32) for $\xi = 0$ and $\xi = 1$. **Solution.** From Eqs. (12.27) and (12.32):

\begin{aligned} \frac{t_r}{\tau_B} &= -\ln(1 - x) \ \frac{t_{sd}}{\tau_B} &= \ln\left(\frac{1 + \xi}{\xi + x}\right) \end{aligned}

For $\xi = 0$:

\begin{aligned} \frac{t_{sd}}{\tau_B} &= -\ln x \end{aligned}

For $\xi = 1$:

\begin{aligned} \frac{t_{sd}}{\tau_B} &= \ln\left(\frac{2}{1 + x}\right) \end{aligned}

Table of values: | $x$ | $t_r/\tau_B$ | $t_{sd}/\tau_B$ ($\xi=0$) | $t_{sd}/\tau_B$ ($\xi=1$) | |---|---|---|---| | 0.05 | 0.051 | 3.000 | 0.646 | | 0.10 | 0.105 | 2.303 | 0.598 | | 0.20 | 0.223 | 1.609 | 0.511 | | 0.30 | 0.357 | 1.204 | 0.431 | | 0.40 | 0.511 | 0.916 | 0.357 | | 0.50 | 0.693 | 0.693 | 0.288 | | 0.60 | 0.916 | 0.511 | 0.223 | | 0.70 | 1.204 | 0.357 | 0.163 | | 0.80 | 1.609 | 0.223 | 0.105 | | 0.90 | 2.303 | 0.105 | 0.051 | | 0.99 | 4.605 | 0.010 | 0.005 | | 1.00 | $\infty$ | 0 | 0 | ```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 ΓÇö Fall time **Problem.** (a) Derive an analytical expression for the fall time $t_f$ using the measurement-based definition (Fig. 12.6: $i_C$ falling from $0.9I_{CC}$ to $0.1I_{CC}$) and Eq. (12.31b). (b) Plot $t_f/\tau_B$ versus $x$ for $\xi = 0$ and $\xi = 1$, and discuss the results. **Solution.** **(a) Derivation of $t_f$:** In the active-mode turn-off interval ($t \ge t_{sd}$), from Eq. (12.31b) with $\frac{I_{BB}\tau_B}{\tau_t} = \frac{I_{CC}}{x}$:

\begin{aligned} i_C(t) &= \frac{I_{CC}}{x}\left[(1 + \xi)e^{-t/\tau_B} - \xi\right] \end{aligned}

Setting $i_C(t_1) = 0.9I_{CC}$:

\begin{aligned} \frac{I_{CC}}{x}\left[(1 + \xi)e^{-t_1/\tau_B} - \xi\right] &= 0.9I_{CC} \ (1 + \xi)e^{-t_1/\tau_B} - \xi &= 0.9x \ e^{-t_1/\tau_B} &= \frac{\xi + 0.9x}{1 + \xi} \ t_1 &= \tau_B \ln\left(\frac{1 + \xi}{\xi + 0.9x}\right) \end{aligned}

Setting $i_C(t_2) = 0.1I_{CC}$:

\begin{aligned} \frac{I_{CC}}{x}\left[(1 + \xi)e^{-t_2/\tau_B} - \xi\right] &= 0.1I_{CC} \ (1 + \xi)e^{-t_2/\tau_B} - \xi &= 0.1x \ e^{-t_2/\tau_B} &= \frac{\xi + 0.1x}{1 + \xi} \ t_2 &= \tau_B \ln\left(\frac{1 + \xi}{\xi + 0.1x}\right) \end{aligned}

The fall time $t_f$ is the difference:

\begin{aligned} t_f &= t_2 - t_1 \ &= \tau_B \left[\ln\left(\frac{1 + \xi}{\xi + 0.1x}\right) - \ln\left(\frac{1 + \xi}{\xi + 0.9x}\right)\right] \ &= \tau_B \ln\left(\frac{\xi + 0.9x}{\xi + 0.1x}\right) \end{aligned}

$$\boxed{t_f = \tau_B \ln\left(\frac{\xi + 0.9x}{\xi + 0.1x}\right)}$$ **(b) Numerical evaluation and discussion:** - For $\xi = 0$ (open-base turn-off, $i_B = 0$):

\begin{aligned} t_f &= \tau_B \ln\left(\frac{0.9x}{0.1x}\right) = \tau_B \ln(9) \approx \boxed{2.197,\tau_B} \end{aligned}

which is strictly independent of $x$ (stored charge decays purely by thermal recombination with time constant $\tau_B$). - For $\xi = 1$ (reverse base drive, $i_B = -I_{BB}$):

\begin{aligned} \frac{t_f}{\tau_B} &= \ln\left(\frac{1 + 0.9x}{1 + 0.1x}\right) \end{aligned}

Table of values: | $x$ | $t_f/\tau_B$ ($\xi = 0$) | $t_f/\tau_B$ ($\xi = 1$) | |---|---|---| | 0.05 | 2.197 | 0.044 | | 0.10 | 2.197 | 0.076 | | 0.20 | 2.197 | 0.145 | | 0.40 | 2.197 | 0.268 | | 0.60 | 2.197 | 0.373 | | 0.80 | 2.197 | 0.463 | | 1.00 | 2.197 | 0.547 | ```matlab x = linspace(0.01, 1.0, 500); tf_0 = log(9) * ones(size(x)); tf_1 = log((1 + 0.9*x) ./ (1 + 0.1*x)); figure; plot(x, tf_0, 'r--', x, tf_1, 'b-', 'LineWidth', 1.5); grid on; xlabel('x = I_{CC}\tau_t / (I_{BB}\tau_B)'); ylabel('t_f / \tau_B'); legend('\xi = 0 (i_B = 0)', '\xi = 1 (i_B = -I_{BB})'); title('Normalized Fall Time t_f / \tau_B'); ``` $\blacksquare$ --- # Chapter 13 ΓÇö 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: $J_{12}$ (P1ΓÇôN2), $J_{23}$ (N2ΓÇôP3), $J_{34}$ (P3ΓÇôN4). - Outer P region (P1) and its contact: **anode (A)**. - Outer N region (N4) and its contact: **cathode (K)**. - Inner regions N2 and P3 are the **bases**. - 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 anode, $I_G$ current into gate. **Fabrication and doping profile (Fig. 13.1b).** Starts with a thinned, lightly doped n-type Si wafer ($N_D \sim 5\times10^{13}\text{ cm}^{-3}$), 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 heavily doped N4 cathode. The base regions N2 and P3 have *moderate electrical widths*, e.g. $W_2 \sim 250\ \mu\text{m}$ and $W_3 \sim 25\ \mu\text{m}$, comparable to minority diffusion lengths. The junctions are close enough to interact via carrier transport, but not so close as to punch through at low voltage. **Characteristics (Fig. 13.2).** - $V_{AK} < 0$: independent of $I_G$, like a reverse-biased pn diode; low leakage current until avalanche breakdown at $-V_{BR}$ (reverse blocking voltage). - $V_{AK} > 0$ with $I_G = 0$: low leakage current until $V_{AK}$ reaches the **forward blocking voltage** $V_{BF}$. Beyond $V_{BF}$, the device switches through negative resistance into a low-impedance conducting mode with $V_{AK} \approx 1\text{ V}$. - $I_G > 0$ lowers $V_{BF}$, triggering the conducting mode at lower anode voltages. --- ## 13.2 SCR Operational Theory ### The diode model Model the SCR as three pn diodes in series (Fig. 13.3a): - $V_{AK} < 0$: $J_{23}$ is forward-biased, while $J_{12}$ and $J_{34}$ are reverse-biased. $J_{12}$ (lightly doped on N2 side) sustains almost all the reverse blocking voltage. - $V_{AK} > 0$: $J_{12}$ and $J_{34}$ are forward-biased, while central junction $J_{23}$ is reverse-biased, supporting the applied voltage until breakdown or switching occurs. ### The two-transistor model Bisect the PNPN structure into two interconnected transistors: - A pnp transistor $T_1$ (P1ΓÇôN2ΓÇôP3). - An npn transistor $T_2$ (N2ΓÇôP3ΓÇôN4). The collector of $T_1$ feeds the base of $T_2$, and the collector of $T_2$ feeds the base of $T_1$. Total current at the central junction $J_{23}$:

I_{AK} = \alpha_1 I_{AK} + \alpha_2 (I_{AK} + I_G) + I_{RO}

where $\alpha_1, \alpha_2$ are the common-base current gains, and $I_{RO}$ is the reverse leakage current of $J_{23}$. Solving for $I_{AK}$: $$\boxed{I_{AK} = \frac{\alpha_2 I_G + I_{RO}}{1 - (\alpha_1 + \alpha_2)}}$$ **Regenerative Switching Criterion:** As $V_{AK}$ or $I_G$ increases, carrier injection increases, causing gains $\alpha_1$ and $\alpha_2$ to rise. When: $$\boxed{\alpha_1 + \alpha_2 \to 1}$$ the denominator approaches zero, initiating regenerative positive feedback that latches the SCR into the high-current on-state. --- ## 13.3 Practical Turn-On/Turn-Off Considerations ### 13.3.1 Circuit Operation Once turned on, the gate loses control. The device can only be turned off by reducing $I_{AK}$ below the **holding current** $I_H$ (natural commutation or forced commutation). ### 13.3.2 Additional Triggering Mechanisms 1. **Thermal triggering**: High temperature increases leakage $I_{RO}$, raising $\alpha_1 + \alpha_2$ to unity. 2. **Optical triggering**: Photons generate electron-hole pairs in the depletion region of $J_{23}$ (Light-Activated SCR or LASCR). 3. **Overvoltage breakdown**: Large $V_{AK}$ triggers avalanche multiplication $M$ at $J_{23}$. ### 13.3.3 Shorted-Cathode Configuration A resistive short between cathode metallization and P3 base shunts low-level leakage current away from $J_{34}$, preventing premature triggering and stabilizing $V_{BF}$. ### 13.3.4 di/dt and dv/dt Effects - **$di/dt$ limitation**: Carrier conduction begins near the gate and spreads laterally with finite velocity ($\sim 0.1\text{ mm}/\mu\text{s}$). Excessively rapid current rise causes localized hot spots and device destruction. - **$dv/dt$ triggering**: Rapid rise of $V_{AK}$ drives displacement current $C_{j2}\frac{dv}{dt}$ through $J_{23}$, which acts like gate current and may cause false triggering. ### 13.3.5 Triggering Time Turn-on delay time is governed by base transit times: $$t_{\text{on}} \approx \frac{W_2 W_3}{2\sqrt{D_p D_n}}$$ ### 13.3.6 Switching Advantages/Disadvantages (SCR vs BJT) - **Advantages**: Handles much larger voltages and currents; latching requires only a brief trigger pulse. - **Disadvantages**: Cannot be turned off from the gate (except GTO); slower switching speed. --- ## 13.4 Other PNPN Devices - **Gate Turn-Off Thyristor (GTO)**: Highly interdigitated gate structure allowing turn-off via a large negative gate current pulse. - **TRIAC**: Bidirectional thyristor capable of conducting current in both directions, triggered by either positive or negative gate pulses. - **DIAC**: Two-terminal bidirectional trigger diode without gate. - **Silicon Controlled Switch (SCS)**: Four-terminal device with both anode gate and cathode gate. --- ## 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; $t_{\text{on}}\approx W_2W_3/2\sqrt{D_pD_n}$. - 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: ction 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 $3.66\text{ eV}$ ($\text{Mg}$) to $5.15\text{ eV}$ ($\text{Ni}$). - **Semiconductor work function** $\Phi_s$ has two pieces:

\Phi_s = \chi + (E_c - E_F)_{\text{FB}} \tag{14.1}

- **Electron affinity** $\chi = (E_0 - E_c)_{\text{surface}}$: invariant property of the semiconductor. $\chi = 4.00,\ 4.03,\ 4.07\text{ eV}$ for $\text{Ge}$, $\text{Si}$, $\text{GaAs}$. - $(E_c - E_F)_{\text{FB}}$: 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})} \tag{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 $E_c - E_F = (E_c - E_F)_{\text{FB}}$. - (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 systematic construction recipe: 1. Draw the surface-included diagrams of the isolated parts. 2. Align both to the common vacuum level $E_0$ 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 smoothly to their bulk positions. 5. Erase the construction lines. *Result (described):* - For $\Phi_M < \Phi_s$: In the $p$-type semiconductor, $E_F$ lies low (near $E_v$), while $E_{FM}$ sits higher. Electrons transfer from the metal into the semiconductor, recombining with holes near the surface to leave uncompensated ionized acceptors ($N_A^-$). This forms a depletion layer where the energy bands bend *downward* toward the interface, creating an electrostatic barrier for majority-carrier holes. - For $\Phi_M > \Phi_s$: Since $E_{FM}$ lies below $E_{Fs}$, electrons transfer from the semiconductor to the metal, causing holes to accumulate in the surface layer. The energy bands bend *upward* toward the interface, bringing $E_v$ closer to or above $E_F$ (accumulation layer). **(c)** Verification of rectifying versus ohmic behavior: Empty electron states in the metal below $E_{FM}$ act effectively as hole states. - For $\Phi_M < \Phi_s$: At thermal equilibrium, an energy barrier opposes hole transport in both directions. Under forward bias ($V_A < 0$, metal biased negatively relative to semiconductor), $E_{FM}$ is raised relative to $E_{Fs}$, decreasing the potential barrier for holes crossing from semiconductor to metal. The resulting hole current increases exponentially with the band separation. Under reverse bias ($V_A > 0$), the barrier height increases, blocking semiconductor hole injection and leaving only a minute, saturating reverse hole current from metal to semiconductor. Hence, the contact is **rectifying**. - For $\Phi_M > \Phi_s$: There exists no potential barrier for holes moving from semiconductor to metal. The minute barrier opposing hole transport in the reverse direction is readily eliminated by a small applied bias. Hence, the contact exhibits **ohmic-like** conduction. **(d)** Step-by-step derivation of the barrier height $\Phi_B$: Step 1: Write the barrier height $\Phi_B$ as the energy separation between $E_{FM}$ and the valence band edge $E_v$ evaluated at the metallurgical interface:

\Phi_B = E_{FM} - E_v\big|_{\text{interface}}

Step 2: Express $E_v\big|_{\text{interface}}$ in terms of the conduction band edge $E_c\big|_{\text{interface}}$ and the bandgap $E_G$:

E_v\big|{\text{interface}} = E_c\big|{\text{interface}} - E_G

Step 3: Substitute this relation into the expression for $\Phi_B$:

\begin{aligned} \Phi_B &= E_{FM} - \left(E_c\big|{\text{interface}} - E_G\right) \ &= E_G - \left(E_c\big|{\text{interface}} - E_{FM}\right) \end{aligned}

Step 4: Align energies relative to the common interface vacuum level $E_0$:

E_0 - E_c\big|_{\text{interface}} = \chi

E_0 - E_{FM} = \Phi_M

$$Step5:Subtractthetwovacuum-referencedenergyexpressions:$$

\left(E_0 - E_{FM}\right) - \left(E_0 - E_c\big|_{\text{interface}}\right) = \Phi_M - \chi

E_c\big|{\text{interface}} - E{FM} = \Phi_M - \chi

Step 6: Substitute $E_c\big|_{\text{interface}} - E_{FM}$ into the equation for $\Phi_B$:

\begin{aligned} \Phi_B &= E_G - (\Phi_M - \chi) \ &= E_G + \chi - \Phi_M \end{aligned}

\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, an electrostatic potential difference develops across the MS diode at equilibrium. Referring to Fig. 14.4(a), where the interface barrier is $\Phi_B$ and the neutral bulk conduction band sits $(E_c - E_F)_{\text{FB}}$ above $E_F$, the total band bending is:

V_{bi} = \frac{1}{q}\Big[\Phi_B - (E_c - E_F)_{\text{FB}}\Big] \tag{14.3}

with $\Phi_B = \Phi_M - \chi$ for an ideal $n$-type contact. #### Charge density Electrons are depleted adjacent to the interface, leaving an uncompensated donor space-charge region of width $W$ inside the semiconductor. The compensating negative charge resides in an ultra-thin sheet of electrons inside the metal at the interface, approximated as a Dirac delta distribution $\delta(x)$. Using the depletion approximation:

\rho(x) = \begin{cases} qN_D, & 0 \le x \le W \tag{14.4a} \ 0, & x > W \tag{14.4b} \end{cases}

Here $x_n \to W$ since the depletion region lies entirely in the semiconductor. Inside the metal, $\mathcal{E} = 0$ and $V = \text{const}$. #### Electric Field Poisson's equation in the depletion region:

\frac{d\mathcal{E}}{dx} = \frac{\rho(x)}{K_s\varepsilon_0} = \frac{qN_D}{K_s\varepsilon_0}, \quad 0 \le x \le W \tag{14.5}

Separating variables and integrating from $x$ to $W$, subject to the boundary condition $\mathcal{E}(W) = 0$:

\int_{\mathcal{E}(x)}^{0} d\mathcal{E}’ = \frac{qN_D}{K_s\varepsilon_0}\int_x^W dx’ \tag{14.6}

0 - \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} \tag{14.7}

The electric field points in the $-x$ direction. The maximum field magnitude occurs at $x = 0$:

|\mathcal{E}|_{\max} = \frac{qN_D W}{K_s\varepsilon_0}

#### Electrostatic Potential Using $\mathcal{E} = -dV/dx$:

\frac{dV}{dx} = -\mathcal{E}(x) = \frac{qN_D}{K_s\varepsilon_0}(W - x) \tag{14.8}

Integrating from $x$ to $W$ with reference $V(W) \equiv 0$:

\int_{V(x)}^{0} dV’ = \frac{qN_D}{K_s\varepsilon_0}\int_x^W (W - x’),dx’ \tag{14.9}

-V(x) = \frac{qN_D}{K_s\varepsilon_0}\left[-\frac{(W - x’)^2}{2}\right]_x^W = \frac{qN_D}{2K_s\varepsilon_0}(W - x)

\boxed{V(x) = -\frac{qN_D}{2K_s\varepsilon_0}(W - x)^2, \quad 0 \le x \le W} \tag{14.10}

At equilibrium, $V(0) = -V_{bi}$. Under applied bias $V_A$, $V(0) = -(V_{bi} - V_A)$. #### Depletion Width Evaluating Eq. (14.10) at $x = 0$:

-(V_{bi} - V_A) = -\frac{qN_D W^2}{2K_s\varepsilon_0} \tag{14.11}

Solving for $W$:

\boxed{W = \left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi} - V_A)\right]^{1/2}} \tag{14.12}

--- ### Exercise 14.2 **Problem.** Copper is deposited on a carefully prepared $n$-type silicon substrate to make an ideal Schottky diode. $\Phi_M \approx 4.65\text{ eV}$, $\chi = 4.03\text{ eV}$, $N_D = 10^{16}\text{ cm}^{-3}$, $T = 300\text{ K}$. Find: - (a) $\Phi_B$ - (b) $V_{bi}$ - (c) $W$ at $V_A = 0$ - (d) $|\mathcal{E}|_{\max}$ at $V_A = 0$ **Solution.** **(a) Calculation of $\Phi_B$:** Step 1: Apply the ideal barrier height formula for an $n$-type contact:

\Phi_B = \Phi_M - \chi

$$Step2:Substitutethenumericalvalues:$$

\begin{aligned} \Phi_B &= 4.65\text{ eV} - 4.03\text{ eV} \ &= 0.62\text{ eV} \end{aligned}

\boxed{\Phi_B = 0.62\text{ eV}}

**(b) Calculation of $V_{bi}$:** Step 1: Compute $(E_c - E_F)_{\text{FB}}$ in the bulk semiconductor assuming $E_i \approx \text{midgap}$ with $E_G = 1.12\text{ eV}$ and $n_i = 10^{10}\text{ cm}^{-3}$:

(E_c - E_F)_{\text{FB}} = \frac{E_G}{2} - kT\ln\left(\frac{N_D}{n_i}\right)

Step 2: Evaluate using $kT = 0.0259\text{ eV}$:

\begin{aligned} (E_c - E_F)_{\text{FB}} &= 0.56\text{ eV} - (0.0259\text{ eV})\ln\left(\frac{10^{16}\text{ cm}^{-3}}{10^{10}\text{ cm}^{-3}}\right) \ &= 0.56\text{ eV} - (0.0259\text{ eV})\ln(10^6) \ &= 0.56\text{ eV} - (0.0259)(13.8155)\text{ eV} \ &= 0.56\text{ eV} - 0.3578\text{ eV} \ &\approx 0.202\text{ eV} \approx 0.20\text{ eV} \end{aligned}

Step 3: Calculate the built-in potential $V_{bi}$:

\begin{aligned} V_{bi} &= \frac{1}{q}\left[\Phi_B - (E_c - E_F)_{\text{FB}}\right] \ &= \frac{1}{q}[0.62\text{ eV} - 0.20\text{ eV}] \ &= 0.42\text{ V} \end{aligned}

\boxed{V_{bi} = 0.42\text{ V}}

**(c) Calculation of $W$ at $V_A = 0$:** Step 1: Use the depletion width relation Eq. (14.12) with $V_A = 0$:

W = \left[\frac{2K_s\varepsilon_0 V_{bi}}{qN_D}\right]

Step 2: Substitute $K_s = 11.8$, $\varepsilon_0 = 8.85\times 10^{-14}\text{ F/cm}$, $q = 1.6\times 10^{-19}\text{ C}$, $N_D = 10^{16}\text{ cm}^{-3}$, and $V_{bi} = 0.42\text{ V}$:

\begin{aligned} W &= \left[\frac{2(11.8)(8.85\times 10^{-14}\text{ F/cm})(0.42\text{ V})}{(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})}\right]^{1/2} \ &= \left[\frac{8.7702\times 10^{-13}}{1.6\times 10^{-3}}\right]^{1/2}\text{ cm} \ &= \left[5.4814\times 10^{-10}\right]^{1/2}\text{ cm} \ &= 2.341\times 10^{-5}\text{ cm} = 0.234\ \mu\text{m} \end{aligned}

\boxed{W = 2.34\times 10^{-5}\text{ cm} = 0.234\ \mu\text{m}}

**(d) Calculation of $|\mathcal{E}|_{\max}$ at $V_A = 0$:** Step 1: Compute the maximum field magnitude at the interface $x = 0$:

|\mathcal{E}|_{\max} = \frac{qN_D W}{K_s\varepsilon_0}

$$Step2:Substitutenumericalvalues:$$

\begin{aligned} |\mathcal{E}|_{\max} &= \frac{(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(2.341\times 10^{-5}\text{ cm})}{(11.8)(8.85\times 10^{-14}\text{ F/cm})} \ &= \frac{3.7456\times 10^{-8}\text{ C/cm}^2}{1.0443\times 10^{-12}\text{ F/cm}} \ &= 3.587\times 10^{4}\text{ V/cm} \approx 3.59\times 10^{4}\text{ V/cm} \end{aligned}

\boxed{|\mathcal{E}|_{\max} = 3.59\times 10^{4}\text{ V/cm}}

--- ### 14.2.2 $I$-$V$ Characteristics In an MS ($n$-type) diode, **electron injection from the semiconductor into the metal routinely dominates** over recombination or minority-carrier injection. The MS diode is a **majority carrier device**. The majority-carrier current is governed by **thermionic emission**. #### Derivation of the Thermionic Emission Current **Step 1: Velocity criterion for barrier crossing.** An electron entering the depletion region must possess sufficient kinetic energy directed along $-x$ to surmount the barrier:

\frac{1}{2} m_n^* v_x^2 \ge q(V_{bi} - V_A) \tag{14.13}

|v_x| \ge v_{\min} = \left[\frac{2q(V_{bi} - V_A)}{m_n^*}\right]^{1/2} \tag{14.14}

where $v_x$ is directed toward the interface ($v_x < 0$). **Step 2: Differential current contribution.**

I_{S\to M, x} = -qA,v_x,n(v_x) \tag{14.15}

$$\ast \ast Step3:Totalcurrentfromintegratedvelocitydistribution.\ast \ast$$

I_{S\to M} = -qA\int_{-\infty}^{-v_{\min}} v_x,n(v_x),dv_x \tag{14.16}

$$\ast \ast Step4:Maxwell-Boltzmannvelocitydistribution.\ast \ast$$

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} \tag{14.17}

**Step 5: Integration.** Let $a = m_n^*/2kT$. Evaluating the integral:

\int_{-\infty}^{-v_{\min}} v_x e^{-a v_x^2},dv_x = \left[-\frac{1}{2a}e^{-a v_x^2}\right]{-\infty}^{-v{\min}} = -\frac{1}{2a}e^{-a v_{\min}^2} = -\frac{kT}{m_n^}e^{-m_n^ v_{\min}

Since $\frac{1}{2}m_n^* v_{\min}^2 = q(V_{bi} - V_A)$:

\begin{aligned} 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^2 e^{-[(E_c - E_F){\text{FB}} + qV{bi} - qV_A]/kT} \end{aligned}

Using $(E_c - E_F)_{\text{FB}} + qV_{bi} = \Phi_B$:

\boxed{I_{S\to M} = A A^* T^2 e^{-\Phi_B/kT}e^{qV_A/kT}} \tag{14.18}

where the **Richardson constant** $A^*$ is:

A^* = \frac{4\pi q,m_n^* k^2}{h^3} = \left(\frac{m_n^*}{m_0}\right)\frac{4\pi q,m_0 k^2}{h^3} \tag{14.19}

\frac{4\pi q,m_0 k^2}{h^3} = 120\text{ A/(cm}^2\text{-K}^2\text{)} \tag{14.20}

**Step 6: Metal-to-semiconductor current.** Electrons in the metal face a bias-independent barrier $\Phi_B$:

I_{M\to S}(V_A) = I_{M\to S}(V_A = 0) \tag{14.21}

At zero bias, thermodynamic equilibrium requires $I(0) = 0$:

I_{M\to S}(0) = -I_{S\to M}(0) = -A A^* T^2 e^{-\Phi_B/kT} \tag{14.22}

$$\ast \ast Step7:Totalnetcurrent.\ast \ast$$

I = I_{S\to M} + I_{M\to S} \tag{14.23}

\boxed{I = I_S\left(e^{qV_A/kT} - 1\right)} \tag{14.24}

\boxed{I_S = A A^* T^2 e^{-\Phi_B/kT}} \tag{14.25}

#### Non-idealities in Reverse Bias 1. **Breakdown:** Avalanche breakdown limits the reverse blocking voltage. 2. **Schottky Barrier Lowering (Image Force Lowering):**

\Phi_B = \Phi_{B0} - \Delta\Phi_B \tag{14.26}

\Delta\Phi_B = \left[\frac{q,\mathcal{E}_s}{4\pi K_s\varepsilon_0}\right]^{1/2} \tag{14.27}

where $\mathcal{E}_s$ is the surface electric field magnitude. --- ### Exercise 14.3 **Problem.** Model the $0.10\text{ V} \le V_A \le 0.35\text{ V}$ portion of Fig. 14.6(a) as $I = I_0 e^{qV_A / n_1 kT}$. Using the data below, perform a least-squares fit for $I_0$ and $n_1$. | $V_A\text{ (V)}$ | $I\text{ (A)}$ | |---|---| | $0.10$ | $4.047\times 10^{-7}$ | | $0.15$ | $2.792\times 10^{-6}$ | | $0.20$ | $1.890\times 10^{-5}$ | | $0.25$ | $1.263\times 10^{-4}$ | | $0.30$ | $8.084\times 10^{-4}$ | | $0.35$ | $4.487\times 10^{-3}$ | **Solution.** Step 1: Linearize the relation by taking the natural logarithm:

\ln I = \ln I_0 + \left(\frac{q}{n_1 kT}\right)V_A

This represents a straight line $y = c_1 x + c_2$, with:

x = V_A, \quad y = \ln I, \quad c_1 = \frac{q}{n_1 kT}, \quad c_2 = \ln I_0

$$Step2:Extractthediodeparametersfromthefitparameters:$$

n_1 = \frac{q}{kT,c_1} = \frac{1}{0.0259,c_1}, \qquad I_0 = e

$$Step3:Executetheleast-squaresregressionviaMATLAB:‘‘‘matlabI=[4.047e-72.792e-61.890e-51.263e-48.084e-44.487e-3];VA=[\mathrm{0.100.150.200.250.300.35}];y=log(I);c=polyfit(VA,y,1);slope=c(1);formatcompactn1=1/(0.0259\ast slope)Is=exp(c(2))‘‘‘Step4:Analyticalcheckofresults:$$

\text{slope } c_1 = 37.47\text{ V}^{-1}, \quad c_2 = -18.398

n_1 = \frac{1}{0.0259 \times 37.47} = 1.03

I_0 = e^{-18.398} = 1.02\times 10^{-8}\text{ A}

\boxed{n_1 = 1.03, \quad I_0 = I_S = 1.02\times 10^{-8}\text{ A}}

$$Sanitycheckusingboundaryendpoints:$$

\begin{aligned} \text{slope} &\approx \frac{\ln(4.487\times 10^{-3}) - \ln(4.047\times 10^{-7})}{0.35 - 0.10} \ &= \frac{-5.4066 - (-14.7200)}{0.25} = \frac{9.3134}{0.25} = 37.25\text{ V}^{-1} \end{aligned}

n_1 \approx \frac{1}{0.0259 \times 37.25} \approx 1.04

--- ### 14.2.3 a.c. Response Small-signal voltage variations modulate the depletion width $W$, yielding a junction capacitance equivalent to a parallel-plate capacitor:

C = \frac{K_s\varepsilon_0 A}{W} \tag{14.28}

Substituting Eq. (14.12) for $W$:

\boxed{C = \frac{K_s\varepsilon_0 A}{\left[\dfrac{2K_s\varepsilon_0}{qN_D}(V_{bi} - V_A)\right]^{1/2}}} \tag{14.29}

$$Invertingandsquaring:$$

\frac{1}{C^2} = \frac{2(V_{bi} - V_A)}{qN_D K_s\varepsilon_0 A^2} \tag{14.30}

Because minority carrier storage is negligible, **the MS diode exhibits no diffusion capacitance or diffusion conductance**. --- ### 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, $A = 1.5\times 10^{-3}\text{ cm}^2$, room temperature. **Solution.** Step 1: Determine the intercept on the $V_A$ voltage axis where $1/C^2 = 0$:

V_{bi} \approx 0.6\text{ V}

$$Step2:Determinethemagnitudeoftheslopefromthelinearplot:$$

\begin{aligned} |\text{slope}| &= \frac{\Delta(1/C^2)}{\Delta V_A} \ &\approx \frac{6\times 10^{-3}\text{ pF}^{-2}}{10.3\text{ V} - (-0.6\text{ V})} = \frac{6\times 10^{21}\text{ F}^{-2}}{10.9\text{ V}} \ &\approx 5.50\times 10^{20}\text{ F}^{-2}\text{V}^{-1} \end{aligned}

Step 3: Relate the slope to donor doping $N_D$ via Eq. (14.30):

|\text{slope}| = \frac{2}{qK_s\varepsilon_0 A^2 N_D}

\begin{aligned} N_D &= \frac{2}{qK_s\varepsilon_0 A^2 |\text{slope}|} \ &= \frac{2}{(1.6\times 10^{-19}\text{ C})(11.8)(8.85\times 10^{-14}\text{ F/cm})(1.5\times 10^{-3}\text{ cm}^2)^2(5.50\times 10^{20}\text{ F}^{-2}\text{V}^{-1})} \ &= \frac{2}{(1.6\times 10^{-19})(1.0443\times 10^{-12})(2.25\times 10^{-6})(5.50\times 10^{20})} \ &= \frac{2}{2.0677\times 10^{-16}}\text{ cm}^{-3} \ &= 9.67\times 10^{15}\text{ cm}^{-3} \approx 9.7\times 10^{15}\text{ cm}^{-3} \end{aligned}

Step 4: Compute the Fermi level position $(E_c - E_F)_{\text{FB}}$:

\begin{aligned} (E_c - E_F)_{\text{FB}} &= \frac{E_G}{2} - kT\ln\left(\frac{N_D}{n_i}\right) \ &= 0.56\text{ eV} - (0.0259\text{ eV})\ln\left(\frac{9.7\times 10^{15}}{10^{10}}\right) \ &\approx 0.20\text{ eV} \end{aligned}

Step 5: Compute the barrier height $\Phi_B$ using Eq. (14.3):

\begin{aligned} \Phi_B &= qV_{bi} + (E_c - E_F)_{\text{FB}} \ &\approx 0.6\text{ eV} + 0.2\text{ eV} \ &= 0.8\text{ eV} \end{aligned}

\boxed{\Phi_B \approx 0.8\text{ eV}}

--- ### 14.2.4 Transient Response MS diodes switch rapidly because they are majority-carrier devices with negligible minority carrier storage delay. Switching speed is governed primarily by the device $RC$ time constant ($C_J$ and series resistance $R_S$). --- ## 14.3 Practical Contact Considerations ### 14.3.1 Rectifying Contacts In real contacts, surface states cause **Fermi-level pinning**, making $\Phi_B$ largely independent of $\Phi_M$. For GaAs: - $\Phi_B \approx \frac{2}{3}E_G \approx 0.95\text{ eV}$ ($n$-type) - $\Phi_B \approx \frac{1}{3}E_G \approx 0.47\text{ eV}$ ($p$-type) ### 14.3.2 Ohmic Contacts Practical ohmic contacts are fabricated by **heavy doping** ($N_D > 10^{19}\text{ cm}^{-3}$) of the surface region beneath the contact, reducing the depletion width:

W \propto N_D

This allows carriers to traverse the barrier via quantum mechanical tunneling (**field emission**). To prevent aluminum **spiking** through shallow junctions, refractory metal **silicides** (e.g., $\text{TiSi}_2$) are used in modern integrated circuits. --- ## 14.4 Summary An ideal metal-semiconductor contact produces a rectifying barrier when $\Phi_M > \Phi_s$ ($n$-type) or $\Phi_M < \Phi_s$ ($p$-type), and ohmic behavior otherwise. Current transport is dominated by majority-carrier thermionic emission. The absence of minority carrier storage eliminates diffusion capacitance, enabling picosecond switching times. In practical contacts, Fermi-level pinning fixes $\Phi_B$, and ohmic contacts are achieved by heavy surface doping to promote field emission. --- ## Chapter 14 Problems **Problem information table** | 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 synonymous terms designating a rectifying metal-semiconductor contact. - A "hot carrier diode" is a Schottky diode with an ultra-small junction area ($C \lesssim 1\text{ pF}$) designed for sub-nanosecond switching, named after the energetic majority carriers injected into the metal. **(b) Ideal contact with $\Phi_M = \Phi_s$ boundary:** - For an $n$-type semiconductor: ohmic if $\Phi_M < \Phi_s$, rectifying if $\Phi_M > \Phi_s$. - For a $p$-type semiconductor: ohmic if $\Phi_M > \Phi_s$, rectifying if $\Phi_M < \Phi_s$. **(c) Electrostatics differences:** 1. Compensating charge in the MS diode is a delta sheet inside the metal at the interface rather than an extended depletion layer ($x_p \to 0$, $x_n \to W$). 2. The reference $V = 0$ is defined at the semiconductor bulk depletion edge $x = W$. 3. $V_{bi} = [\Phi_B - (E_c - E_F)_{\text{FB}}]/q$ is determined directly by the interface barrier $\Phi_B$ rather than two bulk dopings. **(d) Dominant current component:** Thermionic emission of majority carriers over the interface barrier. **(e) Determination of $I_{M\to S}$ without metal injection modeling:** Metal electrons face an invariant barrier $\Phi_B$ irrespective of bias $V_A$. Under zero bias, thermodynamic equilibrium requires $I = I_{S\to M}(0) + I_{M\to S}(0) = 0$, yielding:

I_{M\to S}(V_A) = I_{M\to S}(0) = -I_{S\to M}(0)

**(f) Absence of diffusion capacitance:** Diffusion capacitance arises from stored excess minority carriers in neutral bulk regions. In an MS diode, minority injection is negligible, eliminating excess carrier storage. **(g) Fast switching mechanism:** Zero minority carrier storage eliminates the reverse recovery storage delay ($t_s = 0$), leaving device speed limited solely by the junction capacitance $RC$ charging time. **(h) Schottky diode clamp:** An MS diode connected across the base-collector junction of an $npn$ BJT. Turning on at a lower forward voltage than the Si $pn$ junction, it diverts excess base current, preventing deep transistor saturation. **(i) Practical ohmic contact procedure:** Heavy donor or acceptor doping of the surface layer ($N \gtrsim 10^{19}\text{ cm}^{-3}$) to narrow the depletion barrier, followed by metal deposition and thermal annealing. **(j) Spiking:** Localized penetration of aluminum through a shallow junction caused by non-uniform Al-Si interdiffusion during thermal annealing, resulting in an electrical short. --- ### Problem 14.2 **Problem.** Determine the electrical nature (rectifying or ohmic) and parameters for combinations AΓÇôF at $300\text{ K}$ with $N_D = 10^{16}\text{ cm}^{-3}$ or $N_A = 10^{16}\text{ cm}^{-3}$. Use $kT = 0.0259\text{ eV}$. Flat-band reference relations:

(E_c - E_F)_{\text{FB}} = \frac{E_G}{2} - kT\ln\left(\frac{N_D}{n_i}\right) \quad (n\text{-type})

(E_F - E_v)_{\text{FB}} = \frac{E_G}{2} - kT\ln\left(\frac{N_A}{n_i}\right) \quad (p\text{-type})

\Phi_s = \chi + (E_c - E_F){\text{FB}} \quad (n\text{-type}), \qquad \Phi_s = \chi + E_G - (E_F - E_v){\text{FB}} \quad (p\text{-type})

#### Combination A: $\text{Ge}$, $n$-type, $N_D = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.75\text{ eV}$ $\chi = 4.00\text{ eV}$, $E_G = 0.66\text{ eV}$, $n_i = 2.4\times 10^{13}\text{ cm}^{-3}$. Step 1: Compute flat-band energy:

\begin{aligned} (E_c - E_F)_{\text{FB}} &= 0.33 - 0.0259\ln\left(\frac{10^{16}}{2.4\times 10^{13}}\right) \ &= 0.33 - 0.0259(6.032) = 0.33 - 0.156 = 0.174\text{ eV} \end{aligned}

Step 2: Calculate $\Phi_s$:

\Phi_s = 4.00 + 0.174 = 4.174\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.75\text{ eV} > \Phi_s = 4.174\text{ eV}$, the contact is **rectifying**.

\Phi_B = \Phi_M - \chi = 4.75 - 4.00 = 0.75\text{ eV}

qV_{bi} = \Phi_B - (E_c - E_F)_{\text{FB}} = 0.75 - 0.174 = 0.576\text{ eV} \approx 0.58\text{ eV}

\boxed{\text{Combination A: Rectifying, } \Phi_B = 0.75\text{ eV}, \ V_{bi} = 0.58\text{ V}}

#### Combination B: $\text{Ge}$, $p$-type, $N_A = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.75\text{ eV}$ Step 1: Compute flat-band position:

(E_F - E_v)_{\text{FB}} = 0.174\text{ eV}

Step 2: Compute $\Phi_s$:

\Phi_s = 4.00 + 0.66 - 0.174 = 4.486\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.75\text{ eV} > \Phi_s = 4.486\text{ eV}$ for a $p$-type semiconductor, the contact is **ohmic**.

\boxed{\text{Combination B: Ohmic}}

#### Combination C: $\text{Si}$, $n$-type, $N_D = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.00\text{ eV}$ $\chi = 4.03\text{ eV}$, $E_G = 1.12\text{ eV}$, $n_i = 10^{10}\text{ cm}^{-3}$. Step 1: Compute flat-band energy:

(E_c - E_F)_{\text{FB}} = 0.56 - 0.0259\ln(10^6) = 0.56 - 0.358 = 0.202\text{ eV}

Step 2: Compute $\Phi_s$:

\Phi_s = 4.03 + 0.202 = 4.232\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.00\text{ eV} < \Phi_s = 4.232\text{ eV}$, the contact is **ohmic**.

\boxed{\text{Combination C: Ohmic}}

#### Combination D: $\text{Si}$, $p$-type, $N_A = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.25\text{ eV}$ Step 1: Compute flat-band energy:

(E_F - E_v)_{\text{FB}} = 0.202\text{ eV}

Step 2: Compute $\Phi_s$:

\Phi_s = 4.03 + 1.12 - 0.202 = 4.948\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.25\text{ eV} < \Phi_s = 4.948\text{ eV}$ for $p$-type, the contact is **rectifying**.

\Phi_B = E_G + \chi - \Phi_M = 1.12 + 4.03 - 4.25 = 0.90\text{ eV}

qV_{bi} = \Phi_B - (E_F - E_v)_{\text{FB}} = 0.90 - 0.202 = 0.698\text{ eV} \approx 0.70\text{ eV}

\boxed{\text{Combination D: Rectifying, } \Phi_B = 0.90\text{ eV}, \ V_{bi} = 0.70\text{ V}}

#### Combination E: $\text{GaAs}$, $n$-type, $N_D = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.75\text{ eV}$ $\chi = 4.07\text{ eV}$, $E_G = 1.42\text{ eV}$, $n_i = 1.8\times 10^6\text{ cm}^{-3}$. Step 1: Compute flat-band position:

\begin{aligned} (E_c - E_F)_{\text{FB}} &= 0.71 - 0.0259\ln\left(\frac{10^{16}}{1.8\times 10^6}\right) \ &= 0.71 - 0.0259(22.438) = 0.71 - 0.581 = 0.129\text{ eV} \approx 0.13\text{ eV} \end{aligned}

Step 2: Compute $\Phi_s$:

\Phi_s = 4.07 + 0.13 = 4.20\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.75\text{ eV} > \Phi_s = 4.20\text{ eV}$, the contact is **rectifying**.

\Phi_B = \Phi_M - \chi = 4.75 - 4.07 = 0.68\text{ eV}

qV_{bi} = \Phi_B - (E_c - E_F)_{\text{FB}} = 0.68 - 0.13 = 0.55\text{ eV}

\boxed{\text{Combination E: Rectifying, } \Phi_B = 0.68\text{ eV}, \ V_{bi} = 0.55\text{ V}}

#### Combination F: $\text{GaAs}$, $p$-type, $N_A = 10^{16}\text{ cm}^{-3}$, $\Phi_M = 4.75\text{ eV}$ Step 1: Compute flat-band energy:

(E_F - E_v)_{\text{FB}} = 0.13\text{ eV}

Step 2: Compute $\Phi_s$:

\Phi_s = 4.07 + 1.42 - 0.13 = 5.36\text{ eV}

Step 3: Compare $\Phi_M$ and $\Phi_s$: Since $\Phi_M = 4.75\text{ eV} < \Phi_s = 5.36\text{ eV}$, the contact is **rectifying**.

\Phi_B = E_G + \chi - \Phi_M = 1.42 + 4.07 - 4.75 = 0.74\text{ eV}

qV_{bi} = \Phi_B - (E_F - E_v)_{\text{FB}} = 0.74 - 0.13 = 0.61\text{ eV}

\boxed{\text{Combination F: Rectifying, } \Phi_B = 0.74\text{ eV}, \ V_{bi} = 0.61\text{ V}}

--- ### Problem 14.3 **Problem.** Gold on Si Schottky diode, $\Phi_M = 5.10\text{ eV}$, $\chi = 4.03\text{ eV}$, $N_D = 10^{16}\text{ cm}^{-3}$. Find (a) $\Phi_B$, (b) $V_{bi}$, (c) $W$, and (d) $|\mathcal{E}|_{\max}$ at zero bias. **Solution.** **(a) Barrier height $\Phi_B$:**

\begin{aligned} \Phi_B &= \Phi_M - \chi \ &= 5.10\text{ eV} - 4.03\text{ eV} = 1.07\text{ eV} \end{aligned}

\boxed{\Phi_B = 1.07\text{ eV}}

**(b) Built-in potential $V_{bi}$:**

(E_c - E_F)_{\text{FB}} = 0.56 - 0.0259\ln(10^6) = 0.202\text{ eV}

\begin{aligned} V_{bi} &= \frac{1}{q}\left[\Phi_B - (E_c - E_F)_{\text{FB}}\right] \ &= 1.07\text{ V} - 0.202\text{ V} = 0.868\text{ V} \end{aligned}

\boxed{V_{bi} = 0.868\text{ V}}

**(c) Depletion width $W$:**

\begin{aligned} W &= \left[\frac{2K_s\varepsilon_0 V_{bi}}{qN_D}\right]^{1/2} \ &= \left[\frac{2(1.0443\times 10^{-12}\text{ F/cm})(0.868\text{ V})}{(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})}\right]^{1/2} \ &= \left[\frac{1.8129\times 10^{-12}}{1.6\times 10^{-3}}\right]^{1/2}\text{ cm} \ &= \left[1.133\times 10^{-9}\right]^{1/2}\text{ cm} \ &= 3.366\times 10^{-5}\text{ cm} \approx 0.337\ \mu\text{m} \end{aligned}

\boxed{W = 3.37\times 10^{-5}\text{ cm} = 0.337\ \mu\text{m}}

**(d) Maximum field $|\mathcal{E}|_{\max}$:**

\begin{aligned} |\mathcal{E}|_{\max} &= \frac{qN_D W}{K_s\varepsilon_0} \ &= \frac{(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(3.366\times 10^{-5}\text{ cm})}{1.0443\times 10^{-12}\text{ F/cm}} \ &= \frac{5.386\times 10^{-8}}{1.0443\times 10^{-12}}\text{ V/cm} = 5.157\times 10^4\text{ V/cm} \approx 5.16\times 10^4\text{ V/cm} \end{aligned}

\boxed{|\mathcal{E}|_{\max} = 5.16\times 10^4\text{ V/cm}}

--- ### Problem 14.4 **Problem.** Plot equilibrium $W$ versus $N_D$ for Si MS diodes at $300\text{ K}$, $10^{14} \le N_D \le 10^{17}\text{ cm}^{-3}$, for $\Phi_B = 0.5, 0.6, 0.7\text{ eV}$. **Solution.** At thermal equilibrium:

W = \left[\frac{2K_s\varepsilon_0 V_{bi}}{qN_D}\right]^{1/2}, \quad V_{bi} = \Phi_B - \left[0.56 - kT\ln\left(\frac{N_D}{n_i}\right)\right]

With $\frac{2K_s\varepsilon_0}{q} = \frac{2(1.0443\times 10^{-12})}{1.6\times 10^{-19}} = 1.305\times 10^7\text{ cm}^{-1}\text{V}^{-1}$: | $N_D\text{ (cm}^{-3}\text{)}$ | $(E_c - E_F)_{\text{FB}}\text{ (eV)}$ | $W\ (\mu\text{m}),\ \Phi_B = 0.5\text{ eV}$ | $\Phi_B = 0.6\text{ eV}$ | $\Phi_B = 0.7\text{ eV}$ | |---|---|---|---|---| | $10^{14}$ | $0.322$ | $1.53$ | $1.91$ | $2.22$ | | $10^{15}$ | $0.262$ | $0.56$ | $0.66$ | $0.76$ | | $10^{16}$ | $0.202$ | $0.20$ | $0.23$ | $0.26$ | | $10^{17}$ | $0.143$ | $0.068$ | $0.077$ | $0.085$ | ```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.** Set $E_F = 0$. For $n$-type with barrier $\Phi_B$:

V_{bi} = \Phi_B - (E_c - E_F)_{\text{FB}}

$$Depletionwidth:$$

W = \left[\frac{2K_s\varepsilon_0 V_{bi}}{qN_D}\right]

Potential energy for $0 \le x \le W$:

E_c(x) = (E_c - E_F){\text{FB}} + V{bi}\left(1 - \frac{x}{W}\right)

E_v(x) = E_c(x) - E_G, \qquad E_i(x) = E_c(x) - \frac{E_G}{2}

For $p$-type:

E_v(x) = -(E_F - E_v){\text{FB}} - V{bi}\left(1 - \frac{x}{W}\right)

E_c(x) = E_v(x) + E_G, \qquad E_i(x) = E_v(x) + \frac{E_G}{2}

```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 ``` --- ### Problem 14.6 Derivation of Eq. (14.18) **Problem.** Fill in the intermediate steps leading to Eq. (14.18) starting from Eqs. (14.16) and (14.17). **Solution.** Step 1: Write down the starting integral expression:

I_{S\to M} = -qA\int_{-\infty}^{-v_{\min}} v_x,n(v_x),dv_x

Step 2: Substitute the velocity distribution $n(v_x)$:

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

I_{S\to M} = -qA\left(\frac{4\pi kT,m_n^{2}}{h^3}\right)e^{(E_F - E_c)/kT}\int_{-\infty}^{-v_{\min}} v_x e^{-(m_n^/2kT)v_x

Step 3: Define $a = \frac{m_n^*}{2kT}$ and integrate:

\int_{-\infty}^{-v_{\min}} v_x e^{-a v_x^2},dv_x = \left[-\frac{1}{2a}e^{-a v_x^2}\right]{-\infty}^{-v{\min}} = -\frac{1}{2a}e^{-a v_{\min}^2} = -\frac{kT}{m_n^}e^{-m_n^ v_{\min}^2 / 2kT}

$$Step4:Combineprefactors:$$

\begin{aligned} I_{S\to M} &= -qA\left(\frac{4\pi kT,m_n^{2}}{h^3}\right)e^{(E_F - E_c)/kT}\left(-\frac{kT}{m_n^}e^{-m_n^* v_{\min}^2 / 2kT}\right) \ &= qA\left(\frac{4\pi k^2 T^2 m_n^}{h^3}\right)e^{(E_F - E_c)/kT}e^{-m_n^ v_{\min}^2 / 2kT} \ &= A\left(\frac{4\pi q m_n^* k^2}{h^3}\right)T^2 e^{(E_F - E_c)/kT}e^{-m_n^* v_{\min}^2 / 2kT} \ &= A A^* T^2 e^{(E_F - E_c)/kT}e^{-m_n^* v_{\min}^2 / 2kT} \end{aligned}

$$Step5:Relatekineticcutoffenergytopotentialbarrier:$$

\frac{1}{2}m_n^* v_{\min}^2 = q(V_{bi} - V_A)

e^{-m_n^* v_{\min}^2 / 2kT} = e^{-q(V_{bi} - V_A)/kT} = e^{-qV_{bi}/kT} e

$$Step6:Combineexponentterms:$$

e^{(E_F - E_c)/kT} e^{-qV_{bi}/kT} = e^{-[(E_c - E_F){\text{FB}} + qV{bi}]/kT}

Using $(E_c - E_F)_{\text{FB}} + qV_{bi} = \Phi_B$:

e^{-[(E_c - E_F){\text{FB}} + qV{bi}]/kT} = e

\boxed{I_{S\to M} = A A^* T^2 e^{-\Phi_B/kT} e

--- ### Problem 14.7 MS Photodiode **(a) Short circuit ($V_A = 0$):** Electron-hole pairs generated in the depletion layer and within a diffusion length $L_p$ in the neutral semiconductor are separated by the built-in electric field. Electrons drift into the bulk semiconductor ($+x$), while holes drift toward the interface to recombine with electrons in the metal. Conventional current flows from semiconductor to metal. Under the standard convention ($I > 0$ metal to semiconductor):

\boxed{I_{\text{SC}} = -I_L < 0}

**(b) Open circuit ($I = 0$):** Under open-circuit conditions, accumulated holes in the metal charge it positive relative to the semiconductor, establishing an open-circuit photovoltage $V_{\text{OC}} > 0$. This lowers the band bending to $q(V_{bi} - V_{\text{OC}})$, increasing the forward thermionic emission current until it balances the photocurrent. **(c) Photocurrent relation:** Collecting all carriers generated in $W$ and within $L_p$:

I_L = q A G_L (W + L_p)

W = \left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi} - V_A)\right]

$$Totalterminalcurrent:$$

\boxed{I = I_S\left(e^{qV_A/kT} - 1\right) - q A G_L (W + L_p)}

**(d) Open-circuit voltage:** Setting $I = 0$:

0 = I_S\left(e^{qV_{\text{OC}}/kT} - 1\right) - I_L \implies e^{qV_{\text{OC}}/kT} = 1 + \frac{I_L}{I_S}

\boxed{V_{\text{OC}} = \frac{kT}{q}\ln\left(1 + \frac{I_L}{I_S}\right)}

--- ### Problem 14.8 Effect of series resistance **Problem.** Plot forward $I$-$V$ for $R_S = 0, 0.1, 1, 10\ \Omega$, $0 \le V_A \le 0.6\text{ V}$, $10^{-9} \le I \le 10^{-1}\text{ A}$ using $I_S = 10^{-10}\text{ A}$. **Solution.** Terminal voltage relates to junction voltage $V_j$ via:

V_A = V_j + I R_S = \frac{kT}{q}\ln\left(\frac{I}{I_S} + 1\right) + I R_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') ``` --- ### Problem 14.9 Minority carrier injection ratio **Problem.** Estimate $I_{\text{DIFF}} / I_{\text{TE}}$ for Si with $A^* = 140\text{ A/(cm}^2\text{-K}^2\text{)}$, $\Phi_B = 0.72\text{ eV}$, $N_D = 10^{16}\text{ cm}^{-3}$, $\tau_p = 10^{-6}\text{ s}$, $T = 300\text{ K}$, $\mu_p \approx 400\text{ cm}^2/\text{V}\cdot\text{s}$. **Solution.** Step 1: Compute thermionic emission saturation current density:

\begin{aligned} J_{\text{TE}, S} &= A^* T^2 e^{-\Phi_B/kT} \ &= (140)(300)^2 \exp\left(-\frac{0.72}{0.0259}\right) \ &= (1.26\times 10^7)e^{-27.799} \ &= (1.26\times 10^7)(8.454\times 10^{-13}) \ &= 1.065\times 10^{-5}\text{ A/cm}^2 \approx 1.07\times 10^{-5}\text{ A/cm}^2 \end{aligned}

$$Step2:Computeholediffusionconstantanddiffusionlength:$$

D_p = \left(\frac{kT}{q}\right)\mu_p = (0.0259\text{ V})(400\text{ cm}^2/\text{V}\cdot\text{s}) = 10.36\text{ cm}

L_p = \sqrt{D_p \tau_p} = \sqrt{(10.36\text{ cm}^2/\text{s})(10^{-6}\text{ s})} = 3.219\times 10^{-3}\text{ cm}

$$Step3:Computeminority-carrierdiffusionsaturationcurrentdensity:$$

\begin{aligned} J_{\text{DIFF}, S} &= q\frac{D_p}{L_p}\frac{n_i^2}{N_D} \ &= (1.6\times 10^{-19}\text{ C})\left(\frac{10.36\text{ cm}^2/\text{s}}{3.219\times 10^{-3}\text{ cm}}\right)\left(\frac{10^{20}\text{ cm}^{-6}}{10^{16}\text{ cm}^{-3}}\right) \ &= (1.6\times 10^{-19})(3.218\times 10^3)(10^4) \ &= 5.15\times 10^{-12}\text{ A/cm}^2 \approx 5.2\times 10^{-12}\text{ A/cm}^2 \end{aligned}

$$Step4:Computetheinjectionratio:$$

\frac{I_{\text{DIFF}}}{I_{\text{TE}}} = \frac{J_{\text{DIFF}, S}}{J_{\text{TE}, S}} = \frac{5.15\times 10^{-12}\text{ A/cm}^2}{1.065\times 10^{-5}\text{ A/cm}^2} \approx 4.84\times 10

\boxed{\frac{I_{\text{DIFF}}}{I_{\text{TE}}} \approx 5\times 10

--- ### Problem 14.10 Schottky barrier lowering **Problem.** Compute surface field $\mathcal{E}_s$, barrier lowering $\Delta\Phi_B$, and current ratio for $N_D = 10^{16}\text{ cm}^{-3}$, $V_{bi} = 0.6\text{ V}$. **Solution.** **(a) Mathematical relations and evaluations:**

\mathcal{E}s = \left[\frac{2qN_D(V{bi} - V_A)}{K_s\varepsilon_0}\right]

\Delta\Phi_B = \left[\frac{q\mathcal{E}_s}{4\pi K_s\varepsilon_0}\right]

Sample computation at $V_A = -50\text{ V}$:

\begin{aligned} \mathcal{E}_s &= \left[\frac{2(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(0.6 - (-50)\text{ V})}{1.0443\times 10^{-12}\text{ F/cm}}\right]^{1/2} \ &= \left[\frac{3.2\times 10^{-3}(50.6)}{1.0443\times 10^{-12}}\right]^{1/2} = 3.938\times 10^5\text{ V/cm} \end{aligned}

\begin{aligned} \Delta\Phi_B &= \left[\frac{(1.6\times 10^{-19}\text{ C})(3.938\times 10^5\text{ V/cm})}{4\pi(1.0443\times 10^{-12}\text{ F/cm})}\right]^{1/2} \ &= \left[\frac{6.301\times 10^{-14}}{1.3123\times 10^{-11}}\right]^{1/2} = [4.801\times 10^{-3}]^{1/2} = 0.0693\text{ V} \end{aligned}

| $V_A\text{ (V)}$ | $\mathcal{E}_s\text{ (V/cm)}$ | $\Delta\Phi_B\text{ (V)}$ | $I_S(V_A)/I_S(0)$ | |---|---|---|---| | $0$ | $4.29\times 10^4$ | $0.0229$ | $1$ | | $-1$ | $7.00\times 10^4$ | $0.0292$ | $1.27$ | | $-10$ | $1.80\times 10^5$ | $0.0469$ | $2.5$ | | $-50$ | $3.94\times 10^5$ | $0.0693$ | $6.0$ | ```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) Reverse saturation current ratio:**

\frac{I(V_A)}{I(V_A = 0)} = \exp\left[\frac{\Delta\Phi_B(V_A) - \Delta\Phi_B(0)}{kT}\right]

**(c) Justification of reverse non-saturation:** Because reverse current depends exponentially on barrier height, a modest lowering $\Delta\Phi_B \sim 70\text{ mV}$ produces a 6-fold increase in current across $50\text{ V}$, explaining the lack of true reverse saturation. --- ### Problem 14.11 $C$-$V$ data analysis **Problem.** Least-squares fit of Table P14.11 for $V_{bi}$, $N_D$, $\Phi_B$, with $A = 1.5\times 10^{-3}\text{ cm}^2$. | $-V_A\text{ (V)}$ | $1/C^2\ (10^{21}\text{ F}^{-2})$ | |---|---| | $1.09$ | $0.953$ | | $2.08$ | $1.494$ | | $3.07$ | $2.035$ | | $4.06$ | $2.579$ | | $5.05$ | $3.125$ | | $6.04$ | $3.673$ | | $7.03$ | $4.217$ | | $8.02$ | $4.763$ | | $9.01$ | $5.320$ | | $10.00$ | $5.890$ | **Solution.** Step 1: Write linear regression equation $y = a + bx$ where $x = -V_A$ and $y = 1/C^2$:

\frac{1}{C^2} = \frac{2V_{bi}}{qK_s\varepsilon_0 A^2 N_D} + \frac{2}{qK_s\varepsilon_0 A^2 N_D}(-V_A) = a + bx

Step 2: Compute statistical sums ($n = 10$):

\bar{x} = 5.545\text{ V}, \quad \bar{y} = 3.405\times 10^{21}\text{ F}

S_{xx} = \sum (x_i - \bar{x})^2 = 80.86\text{ V}

S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = 44.71\times 10^{21}\text{ F}

Step 3: Extract slope $b$ and intercept $a$:

b = \frac{S_{xy}}{S_{xx}} = \frac{44.71\times 10^{21}}{80.86} = 0.5529\times 10^{21}\text{ F}^{-2}\text{V}

a = \bar{y} - b\bar{x} = 3.405\times 10^{21} - (0.5529\times 10^{21})(5.545) = 0.339\times 10^{21}\text{ F}

Step 4: Compute $V_{bi}$:

V_{bi} = \frac{a}{b} = \frac{0.339}{0.5529} = 0.613\text{ V}

Step 5: Compute $N_D$:

\begin{aligned} N_D &= \frac{2}{qK_s\varepsilon_0 A^2 b} \ &= \frac{2}{(1.6\times 10^{-19}\text{ C})(1.0443\times 10^{-12}\text{ F/cm})(1.5\times 10^{-3}\text{ cm}^2)^2(0.5529\times 10^{21}\text{ F}^{-2}\text{V}^{-1})} \ &= \frac{2}{2.078\times 10^{-16}} = 9.62\times 10^{15}\text{ cm}^{-3} \end{aligned}

Step 6: Compute barrier height $\Phi_B$:

(E_c - E_F)_{\text{FB}} = 0.56 - 0.0259\ln\left(\frac{9.62\times 10^{15}}{10^{10}}\right) = 0.56 - 0.357 = 0.203\text{ eV}

\Phi_B = qV_{bi} + (E_c - E_F)_{\text{FB}} = 0.613 + 0.203 = 0.816\text{ eV} \approx 0.82\text{ eV}

\boxed{V_{bi} = 0.61\text{ V}, \quad N_D = 9.6\times 10^{15}\text{ cm}^{-3}, \quad \Phi_B = 0.82\text{ 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.** $N_D(x) = ax$. Derive (a) $\rho, \mathcal{E}, V, W$; (b) determination of $V_{bi}$; (c) junction capacitance $C$. **Solution.** **(a) Derivation of electrostatics:** Step 1: Depletion space-charge density:

\rho(x) = qax, \quad 0 \le x \le W

Step 2: Integrate Poisson's equation for $\mathcal{E}(x)$ 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^W x’,dx’

0 - \mathcal{E}(x) = \frac{qa}{2K_s\varepsilon_0}(W^2 - x

\boxed{\mathcal{E}(x) = -\frac{qa}{2K_s\varepsilon_0}(W^2 - x^2), \quad 0 \le x \le W}

Step 3: Integrate field for electrostatic potential $V(x)$ with $V(W) = 0$:

\frac{dV}{dx} = -\mathcal{E}(x) = \frac{qa}{2K_s\varepsilon_0}(W^2 - x

\begin{aligned} \int_{V(x)}^{0} dV’ &= \frac{qa}{2K_s\varepsilon_0}\int_x^W (W^2 - x’^2),dx’ \ -V(x) &= \frac{qa}{2K_s\varepsilon_0}\left[W^2(W - x) - \frac{W^3 - x^3}{3}\right] \ &= \frac{qa}{6K_s\varepsilon_0}\left(2W^3 - 3W^2 x + x^3\right) \end{aligned}

$$Factoring:$$

\boxed{V(x) = -\frac{qa}{6K_s\varepsilon_0}(W - x)^2(2W + x)}

Step 4: Solve for $W$ evaluating at $x = 0$ with $V(0) = -(V_{bi} - V_A)$:

-(V_{bi} - V_A) = -\frac{qa}{6K_s\varepsilon_0}(2W^3) = -\frac{qaW

\boxed{W = \left[\frac{3K_s\varepsilon_0}{qa}(V_{bi} - V_A)\right]

**(b) Determination of $V_{bi}$:** At the depletion edge $x = W$, the donor concentration is $N_D = aW$:

(E_c - E_F)\Big|_{x=W} = kT\ln\left(\frac{N_c}{aW}\right)

\boxed{V_{bi} = \frac{1}{q}\left[\Phi_B - kT\ln\left(\frac{N_c}{aW_{\text{eq}}}\right)\right], \quad W_{\text{eq}} = \left[\frac{3K_s\varepsilon_0 V_{bi}}{qa}\right]

This transcendental system in $(V_{bi}, W_{\text{eq}})$ is solved iteratively to self-consistency. **(c) Capacitance:**

\begin{aligned} C &= \frac{K_s\varepsilon_0 A}{W} \ &= K_s\varepsilon_0 A\left[\frac{qa}{3K_s\varepsilon_0(V_{bi} - V_A)}\right]^{1/3} \end{aligned}

\boxed{C = A\left[\frac{qa(K_s\varepsilon_0)^2}{3(V_{bi} - V_A)}\right]

A plot of $1/C^3$ versus $V_A$ yields a straight line with slope proportional to $qa$. --- # Part II Supplement and Review ## 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 ### Junction diode basics (1ΓÇô17) | # | 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, $I_0 \propto 1/W_n$ 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 $\rho \approx 0$, $\mathcal{E}$ small. | | 6 | breakdown | Sudden large reverse current at $V_A = -V_{\text{BR}}$. | | 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 (18ΓÇô37) | # | 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, $C_J = K_s\varepsilon_0 A/W$. | | 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 | $FF = P_{\max}/(I_{\text{SC}}V_{\text{OC}})$, 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 (38ΓÇô68) | # | 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 $V_{\text{BE}}$. | | 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 (69ΓÇô100) | # | 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 | $g_m = dI_C/dV_{\text{BE}} = qI_C/kT$. | | 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 | $\Phi_M = E_0 - E_{\text{FM}}$. | | 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. | --- ## Problem Set A ### Problem A1 A $pn$ junction has doping profile: $N_D - N_A = -N_B$ for $x < 0$, $0$ for $0 \le x \le x_i$, and $+N_B$ for $x > x_i$. Neglect mobile carriers ($n = 0, p = 0$) in the region $0 \le x \le x_i$. **(a) Built-in voltage $V_{bi}$:** Step 1: Write the general built-in voltage formula across neutral semiconductor regions:

V_{bi} = \frac{kT}{q}\ln\left[\frac{n(x_n)}{n(-x_p)}\right]

$$Step2:Determinemajorityandminorityelectronconcentrationsinthequasineutralbulk:$$

n(x_n) = N_B

n(-x_p) = \frac{n_i

$$Step3:Substituteconcentrationsintothelogarithmicratio:$$

\begin{aligned} V_{bi} &= \frac{kT}{q}\ln\left[\frac{N_B}{n_i^2/N_B}\right] \ &= \frac{kT}{q}\ln\left(\frac{N_B^2}{n_i^2}\right) \ &= \frac{2kT}{q}\ln\left(\frac{N_B}{n_i}\right) \end{aligned}

\boxed{V_{bi} = \frac{2kT}{q}\ln\left(\frac{N_B}{n_i}\right)}

*Justification:* The total built-in potential depends solely on the positions of the bulk Fermi levels in the neutral regions outside the depletion layer, independent of the intervening spatial profile. **(b) Charge density $\rho(x)$:** Under the depletion approximation:

\boxed{\rho(x) = \begin{cases} -qN_B, & -x_p \le x < 0 \ 0, & 0 \le x \le x_i \ +qN_B, & x_i < x \le x_n \ 0, & \text{otherwise} \end{cases}}

**(c) Electric field $\mathcal{E}(x)$:** Step 1: For $-x_p \le x \le 0$, integrate Poisson's equation with $\mathcal{E}(-x_p) = 0$:

\frac{d\mathcal{E}}{dx} = -\frac{qN_B}{K_s\varepsilon_0} \implies \boxed{\mathcal{E}(x) = -\frac{qN_B}{K_s\varepsilon_0}(x + x_p)}

At $x = 0$:

\mathcal{E}(0) = -\frac{qN_B x_p}{K_s\varepsilon_0}

Step 2: For $0 \le x \le x_i$, $\rho = 0 \implies d\mathcal{E}/dx = 0$:

\boxed{\mathcal{E}(x) = \mathcal{E}(0) = -\frac{qN_B x_p}{K_s\varepsilon_0} = \text{const}}

Step 3: For $x_i \le x \le x_n$, integrate with $\mathcal{E}(x_n) = 0$:

\int_{\mathcal{E}(x)}^{0} d\mathcal{E}’ = \frac{qN_B}{K_s\varepsilon_0}\int_x^{x_n} dx’ \implies -\mathcal{E}(x) = \frac{qN_B}{K_s\varepsilon_0}(x_n - x)

\boxed{\mathcal{E}(x) = -\frac{qN_B}{K_s\varepsilon_0}(x_n - x)}

**(d) Relation between $x_n$ and $x_p$:** Step 1: Apply global space-charge neutrality $\int \rho(x)\,dx = 0$:

(-qN_B)x_p + (qN_B)(x_n - x_i) = 0

$$Step2:Simplify:$$

x_p = x_n - x_i

\boxed{x_n = x_p + x_i}

**(e) Equilibrium band diagram:** The Fermi level $E_F$ is a constant horizontal line throughout. The energy bands bend parabolically over $-x_p \le x \le 0$, proceed linearly with constant slope (constant electric field) over the intrinsic spacer $0 \le x \le x_i$, and bend parabolically with opposite curvature over $x_i \le x \le x_n$, accumulating a total potential shift of $qV_{bi}$. --- ### Problem A2 Si $p^+$-$n$ diodes at $300\text{ K}$: Diode #1 has $N_D = 10^{14}\text{ cm}^{-3}$, Diode #2 has $N_D = 10^{16}\text{ cm}^{-3}$. **(a) Larger $V_{bi}$:**

qV_{bi} = (E_i - E_F)_{p^+} + kT\ln\left(\frac{N_D}{n_i}\right)

Because $N_{D2} > N_{D1}$, the logarithmic term is larger for Diode #2.

\boxed{\text{Diode #2}}

**(b) Larger $V_{\text{BR}}$:** Avalanche breakdown voltage scales inversely with doping ($V_{\text{BR}} \propto N_D^{-3/4}$).

\boxed{\text{Diode #1}}

**(c) Larger $C_J$ at given reverse bias $|V_A| \gg V_{bi}$:**

C_J = A\left[\frac{q K_s \varepsilon_0 N_D}{2(|V_A| + V_{bi})}\right]^{1/2} \propto N_D

Because $N_{D2} > N_{D1}$, Diode #2 has a thinner depletion region and larger capacitance.

\boxed{\text{Diode #2}}

**(d) Larger $|I|$ if ideal (reverse bias):**

I_{\text{diff}} = -qA\sqrt{\frac{D_p}{\tau_p}}\frac{n_i

Because $N_{D1} \ll N_{D2}$ and mobility $\mu_p$ is larger in lightly doped material:

\boxed{\text{Diode #1}}

**(e) Larger $|I|$ in real Si diodes (generation dominated):**

I_{\text{R-G}} = -qA\frac{n_i W}{2\tau_0} \propto W \propto N_D

Because $N_{D1} < N_{D2}$, $W_1 > W_2$, generating more current.

\boxed{\text{Diode #1}}

$$\ast \ast (f)Largerdiffusioncapacitanceatgivenforwardbias:\ast \ast$$

C_D = \frac{q I_{\text{diff}} \tau_p}{2kT} \propto I_0 \propto \frac{1}{N_D}

At a fixed $V_A$, Diode #1 conducts higher ideal current and stores more minority charge.

\boxed{\text{Diode #1}}

**(g) Larger storage delay time $t_s$ for identical $I_F/I_R$:**

t_s \approx \tau_p \ln\left(1 + \frac{I_F}{I_R}\right)

Assuming equal carrier lifetimes $\tau_p$:

\boxed{\text{Same for both}}

--- ### Problem A3 **(a) Inverted-mode carrier activity in a $pnp$ BJT:** In inverted mode, the collector-base junction is forward biased and the emitter-base junction is reverse biased. The lightly doped collector injects holes into the base. Holes diffuse across the base from collector to emitter and are collected by the reverse-biased emitter junction. Because $N_C < N_B$, substantial electron back-injection from base to collector occurs, resulting in poor injection efficiency. **(b)(i) Reason for lower gain in backward connection:** Normal emitter efficiency is near unity because $N_E \gg N_B$. In reverse operation, the collector acts as the emitter; since $N_C < N_B$, emitter efficiency $\gamma$ drops drastically, suppressing common-base and common-emitter gains ($\alpha_{dc}, \beta_{dc}$). **(b)(ii) Reason for greater sensitivity to base-width modulation:** The reverse-biased junction carries the dominant depletion layer. In reverse operation, this is the E-B junction where $N_E \gg N_B$. Consequently, the depletion layer expands almost entirely into the base, causing large fractional variations in the quasineutral base width $W_B$. --- ## Problem Set B ### Problem B1 $pn$ diode doping profile: $N_D - N_A = N_0\left(e^{\alpha x} - 1\right)$. **(a) Depletion approximation definition:** Space charge is assumed to equal uncompensated dopant density $\rho(x) = q[N_D(x) - N_A(x)]$ within $-x_p \le x \le x_n$, while mobile carriers are neglected ($n \approx 0, p \approx 0$). Outside this region, net charge density is identically zero ($\rho = 0$). **(b) Charge density expression:**

\boxed{\rho(x) = \begin{cases} qN_0\left(e^{\alpha x} - 1\right), & -x_p \le x \le x_n \ 0, & \text{otherwise} \end{cases}}

$$\ast \ast (c)Electricfieldderivation:\ast \ast Step1:SetupPoisso{n}^{\prime }sequation:$$

\frac{d\mathcal{E}}{dx} = \frac{qN_0}{K_s\varepsilon_0}\left(e^{\alpha x} - 1\right)

Step 2: Integrate from $-x_p$ with boundary condition $\mathcal{E}(-x_p) = 0$:

\begin{aligned} \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 \ &= \frac{qN_0}{K_s\varepsilon_0}\left[\left(\frac{e^{\alpha x}}{\alpha} - x\right) - \left(\frac{e^{-\alpha x_p}}{\alpha} + x_p\right)\right] \end{aligned}

\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) Steps to complete the electrostatics:** 1. Determine potential $V(x)$ by integrating $-\mathcal{E}(x)$ from $-x_p$ using reference $V(-x_p) = 0$:

V(x) = -\int_{-x_p}^x \mathcal{E}(x’),dx’

2. Apply boundary conditions at the edge $x = x_n$: $\mathcal{E}(x_n) = 0$ (overall charge neutrality) and $V(x_n) = V_{bi} - V_A$. 3. Formulate $V_{bi}$ self-consistently from the doping evaluated at the depletion edges:

V_{bi} = \frac{kT}{q}\ln\left[\frac{N(x_n)|N(-x_p)|}{n_i

4. Solve the coupled transcendental equations numerically for $x_n$ and $x_p$. --- ### Problem B2 Si step junction at $300\text{ K}$: $W = x_n + x_p = 2\times 10^{-4}\text{ cm}$, $A = 10^{-3}\text{ cm}^2$, $\tau_n = \tau_p = \tau_0 = 10^{-6}\text{ s}$, $\mu_n = 1352\text{ cm}^2/\text{V}\cdot\text{s}$, $\mu_p = 459\text{ cm}^2/\text{V}\cdot\text{s}$, $K_s = 11.8$, $\varepsilon_0 = 8.85\times 10^{-14}\text{ F/cm}$, $n_i = 10^{10}\text{ cm}^{-3}$, $kT/q = 0.0259\text{ V}$, $E_G = 1.12\text{ eV}$. In neutral regions: $E_i - E_{Fp} = E_G/4$, $E_{Fn} - E_i = E_G/4$. **(a) Reverse bias $V_A$:** The Fermi level split reflects the applied voltage:

\begin{aligned} V_A &= -\frac{1}{q}(E_{Fp} - E_{Fn}) = -\frac{1}{q}\left[\frac{E_G}{4} + \frac{E_G}{4}\right] = -\frac{E_G}{2q} \ &= -\frac{1.12\text{ eV}}{2q} = -0.56\text{ V} \end{aligned}

\boxed{V_A = -0.56\text{ V}}

**(b) Built-in voltage $V_{bi}$:**

\begin{aligned} V_{bi} &= \frac{1}{q}\left[(E_i - E_F)_p + (E_F - E_i)_n\right] \ &= \frac{1}{q}\left[\frac{E_G}{4} + \frac{E_G}{4}\right] = 0.56\text{ V} \end{aligned}

\boxed{V_{bi} = 0.56\text{ V}}

$$\ast \ast (c)R-Greversecurrent:\ast \ast$$

\begin{aligned} I_{\text{R-G}} &= -qA\frac{n_i W}{2\tau_0} \ &= -\frac{(1.6\times 10^{-19}\text{ C})(10^{-3}\text{ cm}^2)(10^{10}\text{ cm}^{-3})(2\times 10^{-4}\text{ cm})}{2(10^{-6}\text{ s})} \ &= -\frac{3.2\times 10^{-16}}{2\times 10^{-6}}\text{ A} = -1.6\times 10^{-10}\text{ A} \end{aligned}

\boxed{I_{\text{R-G}} = -1.6\times 10^{-10}\text{ A}}

$$\ast \ast (d)Idealdiffusioncurrent:\ast \ast Step1:Computedopingconcentrations:$$

\begin{aligned} N_A = N_D &= n_i e^{E_G / 4kT} = (10^{10}\text{ cm}^{-3})\exp\left(\frac{1.12}{4\times 0.0259}\right) \ &= (10^{10})e^{10.811} = (10^{10})(4.956\times 10^4) = 4.96\times 10^{14}\text{ cm}^{-3} \end{aligned}

$$Step2:Computediffusioncoefficientsandlengths:$$

D_N = \left(\frac{kT}{q}\right)\mu_n = (0.0259\text{ V})(1352\text{ cm}^2/\text{V}\cdot\text{s}) = 35.02\text{ cm}

L_N = \sqrt{D_N \tau_n} = \sqrt{(35.02)(10^{-6})} = 5.918\times 10^{-3}\text{ cm}

D_P = \left(\frac{kT}{q}\right)\mu_p = (0.0259\text{ V})(459\text{ cm}^2/\text{V}\cdot\text{s}) = 11.89\text{ cm}

L_P = \sqrt{D_P \tau_p} = \sqrt{(11.89)(10^{-6})} = 3.448\times 10^{-3}\text{ cm}

$$Step3:Evaluatereversesaturationdiffusioncurrent:$$

\begin{aligned} I_{\text{DIFF}} &= -qA n_i^2\left[\frac{D_N}{L_N N_A} + \frac{D_P}{L_P N_D}\right] \ &= -(1.6\times 10^{-19})(10^{-3})(10^{20})\left[\frac{35.02}{(5.918\times 10^{-3})(4.96\times 10^{14})} + \frac{11.89}{(3.448\times 10^{-3})(4.96\times 10^{14})}\right] \ &= -(1.6\times 10^{-2})\left[\frac{35.02}{2.935\times 10^{12}} + \frac{11.89}{1.710\times 10^{12}}\right] \ &= -(1.6\times 10^{-2})\left[1.193\times 10^{-11} + 6.953\times 10^{-12}\right] \ &= -(1.6\times 10^{-2})(1.888\times 10^{-11})\text{ A} = -3.02\times 10^{-13}\text{ A} \end{aligned}

\boxed{I_{\text{DIFF}} = -3.02\times 10^{-13}\text{ A}}

$$\ast \ast (e)Junctioncapacitance:\ast \ast$$

\begin{aligned} C_J &= \frac{K_s\varepsilon_0 A}{W} \ &= \frac{(11.8)(8.85\times 10^{-14}\text{ F/cm})(10^{-3}\text{ cm}^2)}{2\times 10^{-4}\text{ cm}} \ &= \frac{1.0443\times 10^{-15}}{2\times 10^{-4}}\text{ F} = 5.22\times 10^{-12}\text{ F} = 5.22\text{ pF} \end{aligned}

\boxed{C_J = 5.22\text{ pF}}

**(f) Minority carrier profile:** Under reverse bias ($V_A = -0.56\text{ V} \ll -kT/q$), minority carrier densities drop to near zero at the edges: $n_p(-x_p) \approx 0$ and $p_n(x_n) \approx 0$. They recover exponentially to equilibrium densities $n_{p0} = p_{n0} = n_i^2/N_D = 2.0\times 10^5\text{ cm}^{-3}$ over characteristic lengths $L_N$ on the $p$-side and $L_P$ on the $n$-side. **(g) Significant diffusion capacitance:**

\boxed{\text{No}}

$$Diffusioncapacitancerequiresstoredexcessminoritycarriers;underreversebias,acarrierdeficitexists.\ast \ast (h)Storagedelaytime:\ast \ast$$

\boxed{\text{No}}

The junction is already reverse biased; there is no excess stored charge to extract. --- ### Problem B3 Two Si $pnp$ BJTs: identical except $W_{B1} > W_{B2}$. $N_E \gg N_B > N_C$, $W_B \le L_B$. **(a) Larger emitter efficiency $\gamma$:**

\gamma = \frac{1}{1 + \dfrac{D_E N_B W_B}{D_B N_E L_E}}

$$Anarrowerbaseminimizesthedenominator,raisingefficiency.$$

\boxed{\text{BJT #2}}

**(b) Larger base transport factor $\alpha_T$:**

\alpha_T \approx \frac{1}{1 + \frac{1}{2}(W_B/L_B)

$$Fewercarriersrecombinecrossingthenarrowerbase.$$

\boxed{\text{BJT #2}}

**(c) Larger $\beta_{dc}$:** Gain $\beta_{dc} = \frac{\gamma\alpha_T}{1 - \gamma\alpha_T}$ increases monotonically with both $\gamma$ and $\alpha_T$.

\boxed{\text{BJT #2}}

**(d) Greater sensitivity to base-width modulation:** Fractional change in base width is $\frac{\Delta W_B}{W_B}$, which is larger for smaller $W_B$.

\boxed{\text{BJT #2}}

**(e) Larger punch-through voltage:** Depleting the base completely requires extending the depletion region across the entire width $W_B$; the wider base tolerates larger $V_{CB}$.

\boxed{\text{BJT #1}}

**(f) Larger $V_{\text{CBO}}$:** $V_{\text{CBO}}$ is determined by the avalanche breakdown voltage of the C-B junction, which depends exclusively on doping levels.

\boxed{\text{Same for both}}

**(g) Larger $V_{\text{CEO}}$:**

V_{\text{CEO}} = V_{\text{CBO}}\left(\frac{1}{\beta_{dc} + 1}\right)

Because $\beta_{dc1} < \beta_{dc2}$, BJT #1 exhibits higher breakdown.

\boxed{\text{BJT #1}}

--- ## Problem Set C ### Problem C1 **(a) Assumption not invoked in deriving the ideal diode equation:** (iii) Narrow-base diode. (The classical derivation assumes wide quasineutral regions: $W \gg L$).

\boxed{\text{(iii)}}

**(b) Dominant current in Si at $300\text{ K}$ under reverse and small forward bias:** (ii) Thermal R-G generation/recombination in the depletion layer.

\boxed{\text{(ii)}}

**(c) Incorrect statement regarding junction breakdown:** (iv) Tunneling (Zener) dominates at low breakdown voltages ($V_{\text{BR}} \le 4.5\text{ V}$), whereas avalanche multiplication dominates at higher voltages ($V_{\text{BR}} \ge 6\text{ V}$).

\boxed{\text{(iv)}}

**(d) Correct statement about junction capacitance $C_J$:** The magnitude of $C_J$ decreases with increasing reverse bias ($C_J \propto (V_{bi} - V_A)^{-1/2}$).

\boxed{\text{Magnitude of } C_J \text{ decreases with increasing reverse bias}}

--- ### Problem C2 **(a) Bias state of both diodes:** Minority carrier boundary concentrations exceed equilibrium values ($p_n(x_n) > p_{n0}$).

\boxed{\text{(i) forward biased}}

**(b) Applied forward voltage comparison:** Since $p_n(x_n) = p_{n0}e^{qV_A/kT}$ and Diode B has a lower boundary value:

\boxed{\text{(iii) smaller than } (V_{\text{AB}} < V_{\text{AA}})}

**(c) Terminal forward current comparison:** $I \propto \left.\frac{dp_n}{dx}\right|_{x_n} \propto [p_n(x_n) - p_{n0}]$, which is substantially lower in Diode B.

\boxed{\text{(iii) significantly smaller than}}

**(d) Reverse breakdown voltage comparison:** $V_{\text{BR}}$ depends primarily on the donor doping $N_D$ of the lightly doped side, which is identical in both diodes.

\boxed{\text{(ii) roughly the same as}}

**(e) Storage delay time $t_s$ comparison:** $t_s \approx \tau_p \ln(1 + I_F/I_R)$. With identical lifetimes $\tau_p$ and switching ratios:

\boxed{\text{(iii) essentially the same for both}}

--- ### Problem C3 **(a) State of bias:** The diode is **forward biased** because minority densities at the space-charge edges are elevated above their bulk equilibrium values ($p_n(x_n) > p_{n0}$ and $n_p(-x_p) > n_{p0}$). **(b) Verification of low-level injection:** Low-level injection holds because the elevated minority concentrations remain orders of magnitude lower than the majority dopant concentrations ($p_n \ll N_D$ and $n_p \ll N_A$). **(c) Link between minority pile-up and $C_D$:** Diffusion capacitance represents the voltage derivative of stored excess minority charge:

C_D = \frac{dQ_{\text{stored}}}{dV_A} \approx \frac{q I \tau_p}{kT}

A larger minority accumulation corresponds directly to a larger diffusion capacitance. **(d) Link between pile-up and storage time $t_s$:** During turn-off, excess minority charge stored in the neutral regions must be swept out before the junction can support reverse voltage, setting the storage delay:

t_s \approx \tau_p \ln\left(1 + \frac{I_F}{I_R}\right) = \tau_p \ln\left(1 + \frac{Q_{\text{stored}}}{I_R \tau_p}\right)

--- ### Problem C4 **(a)ΓÇô(c) Effect of doubling $N_D$:** - (a) $I$-$V_A$: Depletion width contracts as $W \propto N_D^{-1/2}$. Since $I_{\text{R-G}} \propto W$, the reverse current magnitude **decreases by a factor of $1/\sqrt{2} \approx 0.71$**. - (b) $C$-$V_A$: Capacitance $C_J = K_s\varepsilon_0 A / W \propto N_D^{1/2}$ **increases by a factor of $\sqrt{2} \approx 1.41$**. - (c) $i(t)$: **No first-order change** in storage delay $t_s$, which depends on lifetime $\tau_p$ and the circuit ratio $I_F/I_R$. **(d)ΓÇô(f) Effect of doubling lifetimes $\tau_p$ and $\tau_0$:** - (d) $I$-$V_A$: Reverse generation current $I_{\text{R-G}} = -qAn_iW/(2\tau_0)$ **decreases by a factor of $2$**. - (e) $C$-$V_A$: **No effect**; junction capacitance depends only on doping and voltage. - (f) $i(t)$: The storage plateau length $t_s \approx \tau_p \ln(1 + I_F/I_R)$ **doubles**. --- # Part III: Field Effect Devices (Opener) Parts I and II treated junction devices where current is controlled by minority carrier injection across potential barriers. Part III treats **field effect devices**, where current is modulated by an electric field applied perpendicular to the conducting channel, squeezing or widening the majority carrier transport path. --- # Chapter 15: Field Effect Introduction: The J-FET and MESFET ## 15.1 General Introduction The **field effect** is the modulation of the electrical conductivity of a semiconductor channel by an electric field applied normal to the semiconductor surface. ### A timeline of field-effect devices | Year | Event | |---|---| | 1920sΓÇô30s | 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 $\text{SiO}_2$ gate insulator, inversion 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 | ### Roadmap of Part III - **Chapter 15:** J-FET and MESFET. - **Chapters 16ΓÇô17:** MOS Capacitor and long-channel MOSFET. - **Chapters 18ΓÇô19:** Non-ideal MOS effects, short-channel phenomena, and advanced FET structures. --- ## 15.2 J-FET ### 15.2.1 Introduction The J-FET controls majority carrier current flowing between **source** and **drain** by modulating the depletion widths of reverse-biased gate $pn$ junctions. It is a **unipolar transistor** because transport involves only one carrier polarity. ### 15.2.2 Qualitative Theory of Operation Standard biasing for an $n$-channel symmetrical J-FET:

V_G \le 0, \qquad V_D \ge 0

1. **Linear Region (Small $V_D$):** The channel behaves as an ohmic resistor with $I_D$ increasing linearly with $V_D$. 2. **Channel Pinch-Off:** As $V_D$ increases, the junction reverse bias increases toward the drain, funnelling the channel. At $V_D = V_{Dsat} = V_G - V_P$, the depletion regions touch at the drain end (**pinch-off**). 3. **Current Saturation ($V_D > V_{Dsat}$):** Any voltage in excess of $V_{Dsat}$ drops across a narrow depleted section $\Delta L$ at the drain. The potential drop across the remaining channel length remains pinned at $V_{Dsat}$, holding $I_D$ constant at $I_{Dsat}$. 4. **Gate Cutoff ($V_G \le V_P$):** Depletion regions touch along the entire channel even at $V_D = 0$, reducing $I_D$ to zero for all drain voltages. --- ### 15.2.3 Quantitative $I_D$-$V_D$ Relationships #### Step 1: Current density in the channel

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 Using channel width $Z$ and symmetry about the channel centerline $x = a$:

I_D = -Z\int_{W(y)}^{2a-W(y)} J_{N_y},dx = 2qZ\mu_n N_D [a - W(y)]\frac{dV}{dy} \tag{15.3b}

#### Step 3: Integrate over channel length

\int_0^L I_D,dy = 2qZ\mu_n N_D a \int_0^{V_D} \left[1 - \frac{W(V)}{a}\right]dV \tag{15.4}

I_D = \frac{2qZ\mu_n N_D a}{L}\int_0^{V_D} \left[1 - \frac{W(V)}{a}\right]dV \tag{15.5}

#### Step 4: Gradual channel approximation Assuming $L \gg a$, depletion width $W(V)$ at channel potential $V(y)$ is:

W(V) = \left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi} + V - V_G)\right]^{1/2} \tag{15.6}

#### Step 5: Pinch-off voltage Pinch-off occurs when $W \to a$ at $V_D = 0$ and $V_G = V_P$:

a = \left[\frac{2K_s\varepsilon_0}{qN_D}(V_{bi} - V_P)\right]^{1/2} \tag{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: Integration of $I_D$ Defining $G_0 = \frac{2qZ\mu_n N_D a}{L}$:

\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}} \tag{15.9}

#### Step 7: Current saturation Pinch-off at the drain implies $W(L) = a$ at $V(L) = V_{Dsat}$:

\boxed{V_{Dsat} = V_G - V_P} \tag{15.12}

Substituting $V_{Dsat}$ into Eq. (15.9):

\boxed{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}

$$\ast \ast Square-lawapproximation:\ast \ast$$

I_{Dsat} \approx I_{D0}\left(1 - \frac{V_G}{V_P}\right)^2, \qquad I_{D0} = I_{Dsat}\big|_{V_G = 0} \tag{15.14}

--- ### Exercise 15.1: Maximum half-width $a$ **Problem.** An $n$-channel Si J-FET has $N_D = 10^{16}\text{ cm}^{-3}$. At room temperature, find the maximum half-width $a$. **Solution.** Step 1: Set condition for non-destructive operation: channel pinch-off must occur before avalanche breakdown ($a \le W_{\text{BR}}$):

a_{\max} = W_{\text{BR}} = \left[\frac{2K_s\varepsilon_0(V_{bi} + V_{\text{BR}})}{qN_D}\right]

Step 2: Obtain breakdown and built-in potentials for $N_D = 10^{16}\text{ cm}^{-3}$:

V_{\text{BR}} \approx 55\text{ V}, \quad V_{bi} = 0.92\text{ V}

$$Step3:Evaluatenumericalexpression:$$

\begin{aligned} a_{\max} &= \left[\frac{2(11.8)(8.85\times 10^{-14}\text{ F/cm})(0.92\text{ V} + 55\text{ V})}{(1.6\times 10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})}\right]^{1/2} \ &= \left[\frac{(2.0886\times 10^{-12})(55.92)}{1.6\times 10^{-3}}\right]^{1/2}\text{ cm} \ &= \left[\frac{1.1679\times 10^{-10}}{1.6\times 10^{-3}}\right]^{1/2}\text{ cm} \ &= \left[7.299\times 10^{-8}\right]^{1/2}\text{ cm} \ &= 2.702\times 10^{-4}\text{ cm} = 2.70\ \mu\text{m} \end{aligned}

\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 Eq. (15.13) and Eq. (15.14) for $V_{bi}/V_P = -0.4$. **Solution.** Step 1: Define normalized variables $x = V_G/V_P$ and $z = V_{bi}/V_P$. Step 2: Evaluate Eq. (15.13) in normalized form:

\frac{I_{Dsat}}{I_{D0}} = \frac{(x - 1) - \frac{2}{3}(z - 1)\left[1 - \left(\dfrac{z - x}{z - 1}\right)^{3/2}\right]}{-1 - \frac{2}{3}(z - 1)\left[1 - \left(\dfrac{z}{z - 1}\right)

$$Step3:Evaluatesquare-lawapproximation:$$

\frac{I_{Dsat}}{I_{D0}} = (1 - x)

```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') ``` --- ### Exercise 15.3: Series source and drain resistance **Problem.** Show that extrinsic series resistances $R_S$ and $R_D$ are accounted for by replacing $V_D \to V_D - I_D(R_S + R_D)$ and $V_G \to V_G - I_D R_S$. **Solution.** Step 1: Write internal channel potentials at source and drain ends:

V(0) = I_D R_S, \qquad V(L) = V_D - I_D R_D

$$Step2:IntegrateEq.(15.5)betweentheinternalboundaries:$$

I_D = G_0 \int_{I_D R_S}^{V_D - I_D R_D} \left[1 - \left(\frac{V_{bi} + V - V_G}{V_{bi} - V_P}\right)^{1/2}\right] dV

Step 3: Perform variable substitution $V' = V - I_D R_S$:

dV’ = dV

At $V = I_D R_S \implies V' = 0$. At $V = V_D - I_D R_D \implies V' = V_D - I_D(R_S + R_D)$. Step 4: Rewrite the integrand argument:

V_{bi} + V - V_G = V_{bi} + V’ - (V_G - I_D R_S)

$$Step5:Defineeffectiveterminalvoltages:$$

V_D’ = V_D - I_D(R_S + R_D), \qquad V_G’ = V_G - I_D R_S

$$TheequationmapsidenticallytotheidealJ-FETformulationunderprimedcoordinates:$$

I_D = G_0 \int_0^{V_D’} \left[1 - \left(\frac{V_{bi} + V’ - V_G’}{V_{bi} - V_P}\right)^{1/2}\right] dV’

--- ### 15.2.4 a.c. Response Small-signal parameters are defined from Taylor expansion:

i_d = g_d v_d + g_m v_g \tag{15.19}

g_d \equiv \left.\frac{\partial I_D}{\partial V_D}\right|{V_G}, \qquad g_m \equiv \left.\frac{\partial I_D}{\partial V_G}\right|{V_D} \tag{15.18a, b}

**Table 15.1: J-FET small-signal parameters** | Parameter | Below pinch-off ($V_D \le V_{Dsat}$) | Above pinch-off ($V_D \ge V_{Dsat}$) | |---|---|---| | $g_d$ | $G_0\left[1 - \left(\dfrac{V_D + V_{bi} - V_G}{V_{bi} - V_P}\right)^{1/2}\right]$ | $0$ | | $g_m$ | $G_0\,\dfrac{(V_D + V_{bi} - V_G)^{1/2} - (V_{bi} - V_G)^{1/2}}{(V_{bi} - V_P)^{1/2}}$ | $G_0\left[1 - \left(\dfrac{V_{bi} - V_G}{V_{bi} - V_P}\right)^{1/2}\right]$ | --- ### Exercise 15.4: Effective transconductance with $R_S, R_D$ **Problem.** Show that $R_S$ reduces transconductance in saturation to $g_m' = g_m / (1 + g_m R_S)$. **Solution.** Step 1: Write the small-signal current relation using internal gate-to-source voltage $v_{gs}'$:

i_d = g_m v_{gs}’

$$Step2:Relateinternalvoltagetoterminalvoltages:$$

v_{gs}’ = v_g - i_d R_S

$$Step3:Substituteintothecurrentequation:$$

i_d = g_m(v_g - i_d R_S)

Step 4: Solve for $i_d$:

i_d(1 + g_m R_S) = g_m v_g

i_d = \left(\frac{g_m}{1 + g_m R_S}\right)v_g \equiv g_m’ v_g

\boxed{g_m’ = \frac{g_m}{1 + g_m R_S}}

--- ## 15.3 MESFET ### 15.3.1 General Information A MESFET uses a Schottky barrier gate contact directly on an $n$-type GaAs channel. - **D-MESFET (Depletion-mode):** $V_T < 0$, normally on at $V_G = 0$. - **E-MESFET (Enhancement-mode):** $V_T > 0$, built-in field completely depletes the channel at $V_G = 0$ (normally off). ### 15.3.2 Short-Channel Considerations #### Variable Mobility Model Velocity-field dependence:

v = \frac{\mu_0 \mathcal{E}}{1 + \mu_0 \mathcal{E} / v_{sat}} \tag{15.20}

$$Integratingacrossthechannelyields:$$

\boxed{I_D = \frac{I_D(\text{long-channel})}{1 + \dfrac{\mu_0 V_D}{v_{sat} L}}} \tag{15.22}

#### Saturated Velocity Model When carriers reach saturation velocity $v_{sat}$ prior to physical pinch-off:

\boxed{I_{Dsat} \approx q v_{sat} N_D Z a\left[1 - \left(\frac{V_{bi} - V_G}{V_{bi} - V_P}\right)^{1/2}\right]} \tag{15.24}

#### Two-Region Model Channel is divided into an ohmic low-field region ($y \le y_1$) and a velocity-saturated domain ($y_1 \le y \le L$). Current saturates when the drain field reaches $\mathcal{E}_{sat} = v_{sat} / \mu_0$:

\mathcal{E}{sat} L = \frac{(V{bi} - V_P)\left[3(u^2 - w^2) - 2(u^3 - w^3)\right]}{3(u - 1)} \tag{15.26}

u = \left(\frac{V_{Dsat} + V_{bi} - V_G}{V_{bi} - V_P}\right)^{1/2}, \qquad w = \left(\frac{V_{bi} - V_G}{V_{bi} - V_P}\right)

--- ## 15.4 Summary The J-FET and MESFET modulate majority carrier conduction via depletion regions established by reverse-biased $pn$ junctions or Schottky gates. In long channels, current saturation is governed by geometric pinch-off. In short-channel MESFETs, velocity saturation dominates, compressing $V_{Dsat}$ and limiting transconductance. --- ## Problems ### Problem 15.1: Quick quiz **(a) Definition of "field effect":** The modulation of the electrical conductivity of a semiconductor channel by an electric field applied perpendicular to the semiconductor surface. **(b) Terminal names and unipolar designation:** - **Source:** Terminal where carriers enter the semiconductor channel. - **Drain:** Terminal where carriers exit the channel. - **Gate:** Terminal controlling channel cross-section. - **Unipolar:** Current is conducted exclusively by a single carrier polarity (majority carriers). **(c) Current flow in a $p$-channel J-FET:** Under normal biasing ($V_G \ge 0, V_D \le 0$), holes travel from source to drain. Conventional current flows **out of the drain terminal**. **(d) Current constancy above pinch-off:** Additional drain voltage ($V_D - V_{Dsat}$) drops across the depleted drain boundary $\Delta L$. The channel region between source and pinch-off maintains an invariant potential profile ($V_{Dsat}$), supporting a constant saturation current. **(e) Pinch-off definition:** Complete depletion of the conducting channel width at a localized point, occurring at $V_D = V_{Dsat}$ at the drain end, or across the full channel when $V_G = V_P$. **(f) Definitions of $g_d$ and $g_m$:**

\boxed{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) Small-signal equivalent circuit in saturation:** Open circuit at the input gate terminal. At the output, a voltage-controlled current source $g_m v_{gs}$ connects drain to source (with $g_d \approx 0$). **(h) Acronym definitions:** - **MESFET:** Metal-Semiconductor Field-Effect Transistor. - **D-MESFET:** Depletion-mode MESFET (normally on). - **E-MESFET:** Enhancement-mode MESFET (normally off). **(i) Channel field concern in short-channel MESFETs:** High lateral fields ($\bar{\mathcal{E}} \sim V_D/L \ge 10^4\text{ V/cm}$) cause carrier drift velocity to saturate at $v_{sat}$, invalidating constant-mobility Ohm's law assumptions. **(j) Long-channel vs. two-region theory:** Long-channel theory presumes current saturates strictly via geometric pinch-off under constant mobility. Two-region theory accounts for velocity saturation prior to pinch-off, causing current saturation at lower voltages. --- ### Problem 15.2: The composite transistor **(a) Common-base BJT operation ($d \ll L_p$, $V_{EB} > 0, V_{CB} < 0$):** With E-B forward biased and C-B reverse biased, the structure acts as a lateral $pnp$ bipolar transistor. Holes injected from E cross the base $d$ without recombining ($d \ll L_p$). The characteristics are flat, horizontal lines representing collected current:

I_C \approx \alpha I_E \approx I_E

**(b) Parallel diode operation (E-C tied, D-B tied):** With E and C tied together, both $p^+$-$n$ junctions share the same potential relative to the $n$-substrate, operating as a single junction of twice the area:

I = I_0\left(e^{qV/kT} - 1\right)

**(c) J-FET operation (E-C tied as gate, $V_{EB} < 0, V_{DB} > 0$):** The reverse-biased $p^+$ regions act as symmetrical gates modulating the channel $d$. The characteristics reproduce the standard J-FET family: linear increase, knee roll-over, and saturation at $V_{DBsat} = V_{EB} - V_P$. --- ### 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 $V_G = 0$, $V_D = 5\text{ V}$, $V_{bi} = 1\text{ V}$, $V_P = -8\text{ V}$, find $y/L$ for $V(y) = 1, 2, 3, 4\text{ V}$ and compare with Fig. 15.11(c). **Solution.** **(a) Derivation of $y/L$:** Step 1: Write the integrated current equation from $0$ to an arbitrary point $y$:

I_D y = 2qZ\mu_n N_D a \int_0^{V(y)} \left[1 - \frac{W(V)}{a}\right] dV

Step 2: Write the current integrated over the entire channel length $L$:

I_D L = 2qZ\mu_n N_D a \int_0^{V_D} \left[1 - \frac{W(V)}{a}\right] dV

$$Step3:Dividethetwoexpressions:$$

\frac{y}{L} = \frac{\displaystyle\int_0^{V(y)} \left[1 - \frac{W(V)}{a}\right] dV}{\displaystyle\int_0^{V_D} \left[1 - \frac{W(V)}{a}\right] dV}

Step 4: Define the integral function $F(V)$ from Eq. (15.9):

F(V) = V - \frac{2}{3}\frac{(V + V_{bi} - V_G)^{3/2} - (V_{bi} - V_G)^{3/2}}{(V_{bi} - V_P)

\boxed{\frac{y}{L} = \frac{F(V(y))}{F(V_D)} = \frac{V(y) - \dfrac{2}{3}\dfrac{[V(y) + V_{bi} - V_G]^{3/2} - (V_{bi} - V_G)^{3/2}}{(V_{bi} - V_P)^{1/2}}}{V_D - \dfrac{2}{3}\dfrac{[V_D + V_{bi} - V_G]^{3/2} - (V_{bi} - V_G)^{3/2}}{(V_{bi} - V_P)

$$\ast \ast (b)Numericalevaluation:\ast \ast Givenparameters:$$

V_G = 0, \quad V_D = 5\text{ V}, \quad V_{bi} = 1\text{ V}, \quad V_P = -8\text{ V}

V_{bi} - V_P = 1 - (-8) = 9\text{ V} \implies (V_{bi} - V_P)^{1/2} = 3\text{ V}

V_{bi} - V_G = 1\text{ V}

$$Thefunctionalformsimplifiesto:$$

F(V) = V - \frac{2}{9}\left[(V + 1)^{3/2} - 1\right]

Step 1: Compute denominator $F(V_D) = F(5)$:

\begin{aligned} F(5) &= 5 - \frac{2}{9}\left[(5 + 1)^{3/2} - 1\right] \ &= 5 - \frac{2}{9}\left[6^{3/2} - 1\right] \ &= 5 - \frac{2}{9}(14.6969 - 1) \ &= 5 - \frac{2}{9}(13.6969) \ &= 5 - 3.0438 = 1.9562 \end{aligned}

Step 2: Evaluate $F(V)$ and $y/L$ for $V(y) = 1, 2, 3, 4\text{ V}$: - For $V(y) = 1\text{ V}$:

\begin{aligned} F(1) &= 1 - \frac{2}{9}\left[2^{3/2} - 1\right] = 1 - \frac{2}{9}(2.8284 - 1) = 1 - 0.4063 = 0.5937 \ \frac{y}{L} &= \frac{0.5937}{1.9562} = 0.3035 \approx 0.304 \end{aligned}

- For $V(y) = 2\text{ V}$:

\begin{aligned} F(2) &= 2 - \frac{2}{9}\left[3^{3/2} - 1\right] = 2 - \frac{2}{9}(5.1962 - 1) = 2 - 0.9325 = 1.0675 \ \frac{y}{L} &= \frac{1.0675}{1.9562} = 0.5457 \approx 0.546 \end{aligned}

- For $V(y) = 3\text{ V}$:

\begin{aligned} F(3) &= 3 - \frac{2}{9}\left[4^{3/2} - 1\right] = 3 - \frac{2}{9}(8 - 1) = 3 - 1.5556 = 1.4444 \ \frac{y}{L} &= \frac{1.4444}{1.9562} = 0.7384 \approx 0.738 \end{aligned}

- For $V(y) = 4\text{ V}$:

\begin{aligned} F(4) &= 4 - \frac{2}{9}\left[5^{3/2} - 1\right] = 4 - \frac{2}{9}(11.1803 - 1) = 4 - 2.2623 = 1.7377 \ \frac{y}{L} &= \frac{1.7377}{1.9562} = 0.8883 \approx 0.888 \end{aligned}

| $V(y)\text{ (V)}$ | $F(V)$ | $y/L$ | | :---: | :---: | :---: | | $1$ | $0.594$ | $0.304$ | | $2$ | $1.068$ | $0.546$ | | $3$ | $1.444$ | $0.738$ | | $4$ | $1.738$ | $0.888$ | | $5$ | $1.956$ | $1.000$ | *Comparison with Fig. 15.11(c):* In a uniform linear resistor, the voltages $1, 2, 3, 4\text{ V}$ occur at uniform spatial increments $y/L = 0.2, 0.4, 0.6, 0.8$. In the J-FET, progressive depletion layer widening constricts the conducting channel toward the drain, increasing incremental channel resistance per unit length. Consequently, the local field $|dV/dy|$ steepens toward the drain end, compressing the equipotential contours toward $y = L$. ### 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 the first derivative of $I_D$ with respect to $V_D$ equal to zero:

\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

$$Equatingtheinnertermto1:$$

\frac{(V_D + V_{bi} - V_G)^{1/2}}{(V_{bi} - V_P)^{1/2}} = 1

Multiplying both sides by $(V_{bi} - V_P)^{1/2}$:

(V_D + V_{bi} - V_G)^{1/2} = (V_{bi} - V_P)

$$Squaringbothsides:$$

V_D + V_{bi} - V_G = V_{bi} - V_P

Subtracting $V_{bi}$ and isolating $V_D = V_{Dsat}$:

V_{Dsat} = V_G - V_P

\boxed{V_{Dsat} = V_G - V_P}

$$Toconfirmitisamaximum,evaluatethesecondderivative:$$

\frac{d^2I_D}{dV_D^2} = -\frac{G_0}{2}\frac{1}{(V_D + V_{bi} - V_G)^{1/2}(V_{bi} - V_P)^{1/2}} < 0

Since $\dfrac{d^2I_D}{dV_D^2} < 0$, the extremum is indeed a maximum. --- ### 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 $V_G \equiv V_{GT} \le 0$, $V_D \ge 0$. 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) $V_P = -8\text{ V}$ (gates tied), $V_{bi} = 1\text{ V}$. Find the top-gate pinch-off voltage $V_{PT}$ with $V_{GB} = 0$, $V_D = 0$.** Determine the half-channel thickness $a$ and bottom depletion width $W_B(0)$:

a = \sqrt{A(V_{bi} - V_P)} = \sqrt{A \cdot [1 - (-8)]} = \sqrt{A \cdot 9} = 3\sqrt{A}

W_B(0) = \sqrt{A \cdot V_{bi}} = \sqrt{A \cdot 1} = \sqrt{A}

Apply the pinch-off condition across the total thickness $2a$:

W_T = 2a - W_B(0) = 2(3\sqrt{A}) - \sqrt{A} = 6\sqrt{A} - \sqrt{A} = 5\sqrt{A}

Substitute the expression for $W_T$ in terms of $V_{PT}$:

\sqrt{A(V_{bi} - V_{PT})} = 5\sqrt{A}

Divide by $\sqrt{A}$ and square both sides:

V_{bi} - V_{PT} = 25

Solve for $V_{PT}$:

V_{PT} = V_{bi} - 25 = 1 - 25 = -24\text{ V}

\boxed{V_{PT} = -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 $-8\text{ V}$ where each layer spans only $a$ (depletion $\propto \sqrt{V}$, so $25/9$ more voltage-equivalent). Assumes no breakdown at $-24\text{ V}$. **(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 occurs at $y = L$ where $V = V_{Dsat}$ such that $W_T + W_B = 2a$:

\sqrt{A(V_{bi} + V_{Dsat} - V_{GT})} + \sqrt{A(V_{bi} + V_{Dsat} - 0)} = 2\sqrt{A(V_{bi} - V_P)}

Dividing through by $\sqrt{A}$:

\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}

$$Definethefunctionrepresentingtheleft-handsideofpart(d):$$

f(u) = \sqrt{u - V_{GT}} + \sqrt{u}

$f(u)$ is monotonically increasing with $u$. Evaluate $f(u)$ at $u'$:

\begin{aligned} f(u’) &= \sqrt{(V_{bi} - V_P + V_{GT}) - V_{GT}} + \sqrt{u’} \ &= \sqrt{V_{bi} - V_P} + \sqrt{u’} \end{aligned}

Because $V_{GT} < 0$, it follows that $u' < V_{bi} - V_P$, and therefore:

f(u’) < \sqrt{V_{bi} - V_P} + \sqrt{V_{bi} - V_P} = 2\sqrt{V_{bi} - V_P}

Hence, to satisfy $f(u) = 2\sqrt{V_{bi} - V_P}$, the true root must satisfy $u > u'$:

\boxed{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) $I_D(V_D, V_{GT})$.** The channel thickness is $2a - W_T(V) - W_B(V)$:

I_D L = q Z \mu_n N_D \int_0^{V_D} \left[2a - W_T(V) - W_B(V)\right] dV

where $W_T(V) = \sqrt{A(V_{bi} + V - V_{GT})}$ and $W_B(V) = \sqrt{A(V_{bi} + V)}$. Integrating with respect to $V$:

I_D = \frac{q Z \mu_n N_D}{L}\left{2a V_D - \frac{2}{3}\sqrt{A}\left[(V_D + V_{bi} - V_{GT})^{3/2} - (V_{bi} - V_{GT})^{3/2} + (V_D + V_{bi})^{3/2} - V_{bi}

Using $\sqrt{A} = \dfrac{a}{\sqrt{V_{bi} - V_P}}$ and $G_0 = \dfrac{2q Z \mu_n N_D a}{L}$:

\frac{q Z \mu_n N_D}{L}\cdot \frac{2}{3}\sqrt{A} = \frac{G_0}{2a}\cdot \frac{2}{3}\frac{a}{(V_{bi} - V_P)^{1/2}} = \frac{G_0}{3(V_{bi} - V_P)

$$Substitutinggives:$$

\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)

valid for $V_D \le V_{Dsat}$ of part (d). Check: if both gates were at $V_{GT}$, the bracket doubles and $\frac{1}{3} \to \frac{2}{3}$, 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 $a \le x \le 2a$, where $\kappa$ is the grade constant ($\text{cm}^{-4}$). **(a) Depletion width.** One side is $p^+$, so depletion lies entirely in the $n$ side. Poisson's equation on $0 \le x \le W$ (junction at $x = 0$):

\frac{d\mathcal{E}}{dx} = \frac{q\kappa x}{K_S\varepsilon_0}

Integrating with the boundary condition $\mathcal{E}(W) = 0$:

\mathcal{E}(x) = -\frac{q\kappa}{2K_S\varepsilon_0}(W^2 - x

$$Thevoltagedropacrossthejunctionis:$$

\begin{aligned} V_{bi} - V_A &= \int_0^W |\mathcal{E}|,dx \ &= \frac{q\kappa}{2K_S\varepsilon_0}\int_0^W (W^2 - x^2),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} \end{aligned}

Solving for $W$:

\boxed{W = \left[\frac{3K_S\varepsilon_0}{q\kappa}(V_{bi} - V_A)\right]

(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:

\frac{W}{a} = \left(\frac{V_{bi} + V - V_G}{V_{bi} - V_P}\right)

where $a^3 = \dfrac{3K_S\varepsilon_0}{q\kappa}(V_{bi} - V_P)$ defines $V_P$. 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

(neglecting the doping dependence of $\mu_n$). Then:

I_D = q Z \mu_n \frac{dV}{dy}\kappa(a^2 - W

Integrating along the channel of length $L$:

I_D L = q Z \mu_n \kappa a^2 \int_0^{V_D}\left[1 - \left(\frac{W}{a}\right)^2\right] dV

$$where:$$

\left(\frac{W}{a}\right)^2 = \left(\frac{V_{bi} - V_G + V}{V_{bi} - V_P}\right)

$$Evaluatingtheintegral:$$

\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)

$$Thus:$$

\boxed{I_D = G_0’\left{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}}\right}, \quad G_0’ = \frac{q Z \mu_n \kappa a

$G_0'$ 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 $G_0R_S = G_0R_D = 0, 0.1, 0.5$, $V_{bi} = 1\text{ V}$, $V_P = -2.5\text{ V}$, with $I_{D0} = I_{Dsat}\big|_{V_G=0}$ from (15.13) (the intrinsic device). **Method.** Let:

F(V_1, V_2; V_G) = \int_{V_1}^{V_2}\left[1 - \frac{W}{a}\right] dV = (V_2 - V_1) - \frac{2}{3}\frac{(V_2 + V_{bi} - V_G)^{3/2} - (V_1 + V_{bi} - V_G)^{3/2}}{\sqrt{V_{bi} - V_P}}

Then $I_D = G_0 F$ with:

V_1 = I_D R_S, \qquad V_2 = \min{V_D - I_D R_D, , V_G - V_P}

(the minimum 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 $I_D R_S = j f_0 (G_0 R_S)$, so only $G_0 R_S$ and $G_0 R_D$ appear. The equation to solve per bias point is:

h(j) = j f_0 - F\big(j f_0 g_{rs}, , \min{V_D - j f_0 g_{rd}, , V_G - V_P}; , V_G\big) = 0

with $g_{rs} = G_0 R_S$, $g_{rd} = G_0 R_D$. $h(j)$ is strictly increasing in $j$ (increasing $j$ raises $V_1$ and lowers $V_2$, both reducing $F$), with $h(0) \le 0$ and $h(1) \ge 0$. 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; $V_D = 5\text{ V}$; columns correspond to $V_G = 0, -0.5, -1, -1.5, -2\text{ V}$): | $G_0R$ | $I_D/I_{D0}$ at $V_D = 5\text{ V}$ | | :---: | :--- | | 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 $V_D = 0.5, 1, 2, 5\text{ V}$ 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: 1. The linear-region slope drops (resistances add in series with the channel). 2. $I_{Dsat}$ is lower, most notably for $V_G = 0$ (high current) and negligibly for near-pinch-off curves (low current, minimal $IR$ drop). 3. 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, forming 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 - \frac{2}{3}\frac{(V + V_{bi})^{3/2} - V_{bi}^{3/2}}{(V_{bi} - V_P)

$$Thed.c.conductanceis:$$

G = \frac{I}{V}

\boxed{G = \frac{I}{V} = G_0\left{1 - \frac{2}{3}\frac{(V + V_{bi})^{3/2} - V_{bi}^{3/2}}{V(V_{bi} - V_P)

The a.c. conductance (this is $g_d$ at $V_G = 0$) is:

g = \frac{dI}{dV} = G_0\left{1 - \frac{2}{3}\cdot \frac{3}{2}\frac{(V + V_{bi})^{1/2}}{(V_{bi} - V_P)

\boxed{g = \frac{dI}{dV} = G_0\left[1 - \left(\frac{V + V_{bi}}{V_{bi} - V_P}\right)

For $V > -V_P$: $I = I_{D0}$ is constant, so $G = I_{D0}/V$ and $g = 0$. **(b) Numbers at $V = V_{Dsat}/2$.** Given $V_P = -2\text{ V}$, so $V_{Dsat} = -V_P = 2\text{ V}$ and $V = 1\text{ V}$. Parameters: $Z/L = 1$, $a = 0.5\ \mu\text{m} = 0.5 \times 10^{-4}\text{ cm}$, $N_D = 10^{16}\text{ cm}^{-3}$, $V_{bi} = 1\text{ V}$. Assumption: electron mobility is not given; take $\mu_n \approx 1250\text{ cm}^2/(\text{V}\cdot\text{s})$ (Si at $10^{16}\text{ cm}^{-3}$, about $300\text{ K}$). Also, the given $a$ and $N_D$ imply $V_{bi} - V_P = 1.9\text{ V}$, not $3\text{ V}$; use the stated $V_P = -2\text{ V}$ for the voltage dependence, giving $V_{bi} - V_P = 3\text{ V}$. Calculate $G_0$:

\begin{aligned} G_0 &= 2q\left(\frac{Z}{L}\right)\mu_n N_D a \ &= 2(1.6 \times 10^{-19}\text{ C})(1)(1250\text{ cm}^2/\text{V}\cdot\text{s})(10^{16}\text{ cm}^{-3})(0.5 \times 10^{-4}\text{ cm}) \ &= 2.0 \times 10^{-4}\text{ S} \end{aligned}

With $V_{bi} - V_P = 3\text{ V}$ and $V = 1\text{ V}$:

\begin{aligned} \frac{I}{G_0} &= 1 - \frac{2}{3}\frac{2^{3/2} - 1}{\sqrt{3}} \ &= 1 - \frac{2}{3}(1.0556) \ &= 0.2963\text{ V} \end{aligned}

Current $I$:

\begin{aligned} I &= (2.0 \times 10^{-4}\text{ S})(0.2963\text{ V}) \ &= 5.93 \times 10^{-5}\text{ A} \end{aligned}

$$Staticconductanceandresistance:$$

G = \frac{I}{V} = 5.93 \times 10^{-5}\text{ S}

R = \frac{1}{G} \approx 16.9\text{ k}\Omega

$$Differentialconductanceanddynamicresistance:$$

\begin{aligned} g &= 2.0 \times 10^{-4}\left[1 - \sqrt{\frac{2}{3}}\right] \ &= 2.0 \times 10^{-4}(0.1835) \ &= 3.67 \times 10^{-5}\text{ S} \end{aligned}

r = \frac{1}{g} \approx 27.2\text{ k}\Omega

\boxed{R \approx 16.9\text{ k}\Omega, \quad r \approx 27.2\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_G v_g

with magnitude $\omega C_G |v_g|$. The output (in saturation, where $g_d = 0$) delivers:

i_d = g_m v_g

The current gain $|i_d / i_g|$ is:

\left|\frac{i_d}{i_g}\right| = \frac{g_m}{\omega C_G}

Setting this gain to unity at the maximum operating frequency $\omega_{max} = 2\pi f_{max}$:

\omega_{max} C_G = g_m \implies 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 $g_m$ occurs at the no-depletion limit $G_0 = 2q Z \mu_n N_D a / L$ (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\frac{K_S\varepsilon_0 Z L}{a}

$$Substituteintothecutofffrequencyexpression:$$

\begin{aligned} f_{max} &= \frac{G_0}{2\pi C_G} \ &= \frac{2q Z \mu_n N_D a / L}{2\pi \cdot 2K_S\varepsilon_0 Z L / a} \end{aligned}

$$Cancelingcommonfactors:$$

\boxed{f_{max} = \frac{q \mu_n N_D a^2}{2\pi K_S\varepsilon_0 L

Short channels ($f \propto 1/L^2$), high mobility, and heavy channel doping give higher speed. **(c) Numbers.** For Si with $N_D = 10^{16}\text{ cm}^{-3}$, $a = 0.5\ \mu\text{m} = 0.5 \times 10^{-4}\text{ cm}$, $L = 5\ \mu\text{m} = 5 \times 10^{-4}\text{ cm}$, assuming $\mu_n = 1250\text{ cm}^2/(\text{V}\cdot\text{s})$:

\begin{aligned} f_{max} &= \frac{(1.6 \times 10^{-19})(1250)(10^{16})(0.5 \times 10^{-4})^2}{2\pi(11.8)(8.85 \times 10^{-14})(5 \times 10^{-4})^2} \ &= \frac{5.0 \times 10^{-9}}{1.64 \times 10^{-18}} \ &\approx 3.0 \times 10^9\text{ Hz} \end{aligned}

\boxed{f_{max} \approx 3\text{ GHz}}

--- ### Problem 15.10: Temperature Dependence of $g_m$ **Setup.** In saturation with $V_G = 0$:

g_m = G_0\left[1 - \sqrt{\frac{V_{bi}}{V_{bi} - V_P}}\right]

Neglecting dimensional changes, $V_{bi} - V_P = qN_D a^2 / (2K_S\varepsilon_0)$ is temperature independent:

V_{bi} - V_P = \frac{(1.6 \times 10^{-19})(10^{16})(0.6 \times 10^{-4})^2}{2(11.8)(8.85 \times 10^{-14})} = 2.76\text{ V}

Since $G_0 \propto \mu_n(T)$, and $V_{bi}(T) = \dfrac{kT}{q}\ln\left(\dfrac{N_A N_D}{n_i^2(T)}\right)$ with $N_A = 5 \times 10^{18}\text{ cm}^{-3}$, $N_D = 10^{16}\text{ cm}^{-3}$:

\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).** - $n_i(T) = 10^{10}(T/300)^{3/2}\exp\left[-\left(\dfrac{E_G(T)}{2kT} - \dfrac{E_G(300)}{2k\cdot 300}\right)\right]$, with $E_G(T) = 1.17 - 4.73 \times 10^{-4} T^2 / (T + 636)$. - $\mu_n$: Matthiessen's rule with lattice mobility $1430(T/300)^{-2.42}$ and ionized-impurity mobility $9900(T/300)^{3/2}\text{ cm}^2/(\text{V}\cdot\text{s})$ ($\mu_n(300) \approx 1250$). **(a) Results.** | $T\text{ (K)}$ | $V_{bi}\text{ (V)}$ | $\mu_n(T)/\mu_n(300)$ | $g_m(T)/g_m(300)$ | | :---: | :---: | :---: | :---: | | 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.** Fitting $g_m \propto T^{-n}$ to the log-log data yields $n \approx 1.65$. 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 carrier concentration grows), so the zero-bias depletion layer shrinks and the channel opens slightly, partially compensating for 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 $V_{biT}$, applied 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 long-channel theory ($v = \mu_0\mathcal{E}$) up to $V_{Dsat}$. With $A = 2K_S\varepsilon_0 / (qN_D)$:

I_D L = q Z \mu_0 N_D \int_0^{V_D} \left[2a - W_T - W_B\right] dV

where $W_T = \sqrt{A(V_{biT} + V - V_{GT})}$ and $W_B = \sqrt{A(V_{biB} + V - V_{GB})}$. Integrating:

I_D = \frac{q Z \mu_0 N_D}{L}\left{2a V_D - \frac{2}{3}\sqrt{A}\left[\Sigma_T + \Sigma_B\right]\right}

$$with:$$

\Sigma_T = (V_D + V_{biT} - V_{GT})^{3/2} - (V_{biT} - V_{GT})

\Sigma_B = (V_D + V_{biB} - V_{GB})^{3/2} - (V_{biB} - V_{GB})

Eliminate $\sqrt{A}$ using 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}

\frac{\sqrt{A}}{a} = \frac{2}{(V_{biT} - V_{PT})^{1/2} + V_{biB}

With $G_0 = 2q Z \mu_0 N_D a / L$, we have $\dfrac{q Z \mu_0 N_D}{L} = \dfrac{G_0}{2a}$, so:

\frac{G_0}{2a}\cdot \frac{2}{3}\sqrt{A} = \frac{2G_0}{3}\frac{1}{(V_{biT} - V_{PT})^{1/2} + V_{biB}

$$Thus:$$

\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}, \quad 0 \le V_D \le V_{Dsat}}

Above $V_{Dsat}$, $I_D = I_{Dsat} = I_D\big|_{V_{Dsat}}$, where $V_{Dsat}$ is found from the field condition $|\mathcal{E}(L)| = \mathcal{E}_{sat}$ (analogue of 15.26). Check: for a symmetric device with $V_{biT} = V_{biB}$, $V_{GT} = V_{GB}$, $(V_{biT} - V_{PT})^{1/2} + V_{biB}^{1/2} = 2\sqrt{V_{bi} - V_P}$ and $\Sigma_T = \Sigma_B$, which reproduces (15.9). (If the single-gate variant is desired, 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 electric-field-dependent mobility is:

\mu(\mathcal{E}) = \frac{\mu_0}{1 + \mu_0\mathcal{E}/v_{sat}}

$$Currentthroughthechannelisgivenby:$$

I_D = 2q Z N_D \mu(\mathcal{E})\mathcal{E}[a - W] = \frac{2q Z N_D \mu_0[a - W], dV/dy}{1 + (\mu_0/v_{sat}), dV/dy}

$$Multiplyingoutthedenominator:$$

I_D + I_D\frac{\mu_0}{v_{sat}}\frac{dV}{dy} = 2q Z \mu_0 N_D [a - W]\frac{dV}{dy}

Multiplying through by $dy$:

I_D,dy = 2q Z \mu_0 N_D [a - W],dV - \frac{I_D\mu_0}{v_{sat}},dV

Integrating along the channel for $0 \le y \le L$ (with $I_D$ constant) and $0 \le V \le V_D$:

I_D L = 2q Z \mu_0 N_D \int_0^{V_D}[a - W],dV - \frac{I_D\mu_0 V_D}{v_{sat}}

$$Rearrangingterms:$$

I_D\left(L + \frac{\mu_0 V_D}{v_{sat}}\right) = 2q Z \mu_0 N_D \int_0^{V_D}[a - W],dV = I_D(\text{long-channel})\cdot L

Dividing by $L\left(1 + \dfrac{\mu_0 V_D}{v_{sat} L}\right)$:

\boxed{I_D = \frac{I_D(\text{long-channel})}{1 + \mu_0 V_D / (v_{sat} L)}}

(For a 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), $I_D y = G_0 L F(V(y))$ with $F(V) = \int_0^V [1 - W/a]\,dV$ and $I_D = G_0 F(V_D)$. Differentiating with respect to $y$:

I_D = G_0 L\left[1 - \frac{W(V)}{a}\right]\frac{dV}{dy}

At the drain end for $V_D = V_{Dsat}$, transition to velocity saturation occurs when the magnitude of the field equals $|\mathcal{E}_{sat}|$ (with $\mathcal{E} = -dV/dy$, so $\mathcal{E}_{sat} < 0$):

\left.\frac{dV}{dy}\right|{y=L} = -\mathcal{E}{sat} = \frac{I_D}{G_0 L [1 - W(V_{Dsat})/a]} = \frac{F(V_{Dsat})}{L [1 - u]}

where $u = \dfrac{W(V_{Dsat})}{a} = \left[\dfrac{V_{Dsat} + V_{bi} - V_G}{V_{bi} - V_P}\right]^{1/2}$. Let $c = V_{bi} - V_P$ and $w = \sqrt{(V_{bi} - V_G)/c}$, such that $V_{Dsat} = c(u^2 - w^2)$:

\begin{aligned} F(V_{Dsat}) &= c(u^2 - w^2) - \frac{2}{3}c(u^3 - w^3) \ &= \frac{c}{3}\left[3(u^2 - w^2) - 2(u^3 - w^3)\right] \end{aligned}

$$Then:$$

\mathcal{E}{sat} L = -\frac{F(V{Dsat})}{1 - u} = \frac{F(V_{Dsat})}{u - 1}

Substituting $F(V_{Dsat})$:

\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)}}

$$Theunsimplifiedformis:$$

\mathcal{E}{sat} L = \frac{V{Dsat} - \frac{2}{3}\left[(V_{Dsat} + V_{bi} - V_G)^{3/2} - (V_{bi} - V_G)^{3/2}\right]/\sqrt{c}}{\sqrt{(V_{Dsat} + V_{bi} - V_G)/c} - 1}

which is 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 $\Phi(V_D) = \dfrac{F(V_D)}{1 - u(V_D)} - |\mathcal{E}_{sat}|L = 0$ on $(0, V_G - V_P)$. $\Phi$ rises monotonically from $-|\mathcal{E}_{sat}|L < 0$ to $+\infty$ as $u \to 1$, so there is exactly one root (found using `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. $f_0 = F(V_G = 0, V_D = -V_P) = -V_P - \dfrac{2}{3}(V_{bi} - V_P)\left[1 - \left(\dfrac{V_{bi}}{V_{bi} - V_P}\right)^{3/2}\right]$. 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** ($V_{bi} = 1\text{ V}$, $V_P = -2.5\text{ V}$, $|\mathcal{E}_{sat}| = 5 \times 10^3\text{ V/cm}$, $L = 1\ \mu\text{m}$, so $\mathcal{E}_{sat} L = 0.5\text{ V}$). Output: - $V_G = 0\text{ V}$: $V_{Dsat} = 0.437\text{ V}$, $I_{Dsat}/I_{D0} = 0.343$ - $V_G = -0.5\text{ V}$: $V_{Dsat} = 0.429\text{ V}$, $I_{Dsat}/I_{D0} = 0.246$ - $V_G = -1.0\text{ V}$: $V_{Dsat} = 0.412\text{ V}$, $I_{Dsat}/I_{D0} = 0.162$ - $V_G = -1.5\text{ V}$: $V_{Dsat} = 0.377\text{ V}$, $I_{Dsat}/I_{D0} = 0.089$ - $V_G = -2.0\text{ V}$: $V_{Dsat} = 0.291\text{ V}$, $I_{Dsat}/I_{D0} = 0.029$ This reproduces the shape of Fig. 15.23: low $V_{Dsat}$ (under $0.5\text{ V}$), maximum $I_{Dsat}/I_{D0}$ around $0.34$ (the plot axis tops at $0.5$), and a slope discontinuity at $V_{Dsat}$. **(b) (Assumed reading.)** Use $\mathcal{E}_{sat} = -10^4\text{ V/cm}$ with a long channel (e.g., $L = 100\ \mu\text{m}$, giving $\mathcal{E}_{sat} L = 100\text{ V}$), and compare with Fig. 15.16. Setting `Esat=1e4; L=100e-4` and extending `VD` to $5\text{ V}$, the curves converge to the long-channel set of Fig. 15.16: $V_{Dsat} \to V_G - V_P$ ($2.5\text{ V}$ at $V_G = 0$) and $I_{Dsat}/I_{D0}$ approaches the (15.13) values ($1.00, 0.61, 0.33, 0.14, 0.035$ at $V_G = 0, -0.5, -1, -1.5, -2\text{ V}$). Decreasing $L$ progressively lowers $V_{Dsat}$ and $I_{Dsat}$ (e.g., $\mathcal{E}_{sat} L = 5\text{ V}$ gives $V_{Dsat} = 1.85\text{ V}$, ratio $0.94$; $20\text{ V}$ gives $2.32\text{ V}$, ratio $0.996$). **(c) Where long-channel theory fails.** Define failure as $I_{Dsat}/I_{D0} < 0.95$ at $V_G = 0$. Solving $F(V_{Dsat})/f_0 = 0.95$ simultaneously with the $V_{Dsat}$ equation numerically gives:

V_{Dsat} \approx 1.90\text{ V}

\boxed{\mathcal{E}_{sat} L \approx 5.6\text{ V}}

The long-channel theory therefore holds if $|\mathcal{E}_{sat}|L \gtrsim 5.6\text{ V}$. Examples: - $L = 5.6 / (5 \times 10^3) = 1.1 \times 10^{-3}\text{ cm} = 11\ \mu\text{m}$ for $|\mathcal{E}_{sat}| = 5 \times 10^3\text{ V/cm}$ - $L = 5.6 / 10^4 = 5.6 \times 10^{-4}\text{ cm} = 5.6\ \mu\text{m}$ for $|\mathcal{E}_{sat}| = 10^4\text{ V/cm}$ Channels shorter than approximately $11\ \mu\text{m}$ (for $5 \times 10^3\text{ V/cm}$) or $5.6\ \mu\text{m}$ (for $10^4\text{ V/cm}$) 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: - $0.5\text{ V}$: $0.343$ - $1\text{ V}$: $0.551$ - $2\text{ V}$: $0.767$ - $3\text{ V}$: $0.864$ - $5\text{ V}$: $0.940$ - $10\text{ V}$: $0.983$ - $20\text{ V}$: $0.996$ --- # Chapter 16: MOS Fundamentals > Constants used throughout: $T = 300\text{ K}$, $kT/q = 0.0259\text{ V}$, $n_i = 10^{10}\text{ cm}^{-3}$, $K_S = 11.8$, $K_O = 3.9$, $\varepsilon_0 = 8.85 \times 10^{-14}\text{ F/cm}$, $q = 1.6 \times 10^{-19}\text{ C}$. 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 ($0.01\ \mu\text{m}$ to $1.0\ \mu\text{m}$) 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 $10^{18}\ \Omega\cdot\text{cm}$, 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. --- ## 16.2 Electrostatics ΓÇö Mostly Qualitative Before we crunch numbers, we want *pictures*. The MOS capacitor (metalΓÇôoxideΓÇôsemiconductor, "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 metalΓÇôgapΓÇôsemiconductor 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 metalΓÇôoxide and oxideΓÇôsemiconductor 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, $V_G \neq 0$ 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}

$$andthen:$$

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 **depletionΓÇôinversion 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 line plots graphically restate the physical regimes: For an ideal **n-type** device ($V_T < 0$):

\begin{aligned} \text{Inversion (INV):} \quad & V_G < V_T \ \text{Depletion (DEPL):} \quad & V_T < V_G < 0 \ \text{Accumulation (ACC):} \quad & V_G > 0 \end{aligned}

For an ideal **p-type** device ($V_T > 0$):

\begin{aligned} \text{Accumulation (ACC):} \quad & V_G < 0 \ \text{Depletion (DEPL):} \quad & 0 < V_G < V_T \ \text{Inversion (INV):} \quad & V_G > V_T \end{aligned}

On a $V_G$ axis:

\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}

\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 ($1\text{ \AA} = 10^{-8}\text{ cm}$) 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}

$$andakeymaterialparameter,the\ast \ast Fermipotential\ast \ast :$$

\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\left(\frac{N_A}{n_i}\right) \quad \text{(p-type)} \tag{16.8a}

\phi_F = -\frac{kT}{q}\ln\left(\frac{N_D}{n_i}\right) \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 depletionΓÇôinversion 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 boundary conditions:

\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 is determined by the sign of $\phi_F$ ($\phi_F > 0 \implies \text{p-type}$, $\phi_F < 0 \implies \text{n-type}$). The operating state follows by comparing $\phi_s$ to $0$ and $2\phi_F$: | Set | $\phi_F$ | $\phi_s$ | $2\phi_F$ | Doping | Condition | | :---: | :---: | :---: | :---: | :---: | :--- | | (i) | 12 | 12 | 24 | p | $0 < 12 < 24$: **Depletion** (mid-depletion, $\phi_s = \phi_F$) | | (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 | $\phi_s > 2\phi_F$ ($36 > 30$): **Inversion** | | (v) | $-15$ | 0 | $-30$ | n | $\phi_s = 0$: **Flat band** | Diagram descriptions: - **(i) p-type, depletion:** $E_F$ flat; $E_C, E_i, E_V$ bend downward toward the interface by $12\,kT$ (so $E_i$ remains above $E_F$ at the surface, i.e., depleted of holes). Charge: $+Q$ on the metal; $-Q$ as a wide block of exposed acceptors. - **(ii) n-type, accumulation:** bands bend downward by $3\,kT$ toward the interface, moving $E_C$ closer to $E_F$ (higher electron concentration). 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 $36\,kT$, exceeding $2\phi_F = 30\,kT$. Charge: $+Q$ on metal; $-Q$ as a thin electron inversion sheet at the interface plus a depletion block of exposed acceptors of width $\approx W_T$. - **(v) n-type, flat band:** all bands perfectly flat, $E_F$ level across the structure, zero net charge everywhere. --- ### 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 $\phi_F = 12\,kT/q$, $T = 300\text{ K}$, $kT/q = 0.0259\text{ V}$) guide the approximation. - **Fig. 16.8(a), accumulation** ($\phi_s = -6\,kT/q$): the accumulation-layer charge is squeezed into an extremely narrow region at the interface. - **Fig. 16.8(b), middle of depletion** ($\phi_s = \phi_F = 12\,kT/q$): the depletion region reaches deep (a fraction of a micron). - **Fig. 16.8(c), onset of inversion** ($\phi_s = 2\phi_F = 24\,kT/q$): wider depletion region with charge density $-qN_A$. - **Fig. 16.8(d), deep inversion** ($\phi_s = 2\phi_F + 6\,kT/q = 30\,kT/q$): 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)ΓÇô(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 $2\phi_F + 6\,kT/q$, the depletion width grows significantly 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}

$$Poisso{n}^{\prime }sequationreducesto:$$

\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 the boundary condition $\mathcal{E}(W) = 0$:

\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 $\phi(W) = 0$:

\phi(x) = \frac{qN_A}{2K_S\varepsilon_0}(W - x)^2 \qquad (0 \le x \le W) \tag{16.13}

Setting $\phi(0) = \phi_s$ determines $W$:

\phi_s = \frac{qN_A W^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), and (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 ($\phi_s = 2\phi_F$), reaching 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 $300\text{ K}$ over the typical MOS range (logΓÇôlog): it falls monotonically with doping, from several micrometres near $N_A \sim 10^{14}\text{ cm}^{-3}$ to a few hundredths of a micrometre near $10^{18}\text{ cm}^{-3}$. > **Sanity check:** for $N_A = 10^{16}\text{ cm}^{-3}$, $\phi_F = 0.0259\ln(10^6) = 0.358\text{ V}$ and $W_T = \sqrt{2(11.8)(8.854 \times 10^{-14})(0.716) / [(1.6 \times 10^{-19})(10^{16})]} \approx 0.31\ \mu\text{m}$. ### 16.3.2 Gate Voltage Relationship The results above are cast in terms of the surface potential $\phi_s$, an internal quantity. To connect to external terminals, we relate $V_G$ to $\phi_s$. The gate voltage in an ideal structure drops 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:

\Delta\phi_{semi} = \phi(x = 0) = \phi_s \tag{16.18}

$$Intheidealinsulator,withnofreecharge:$$

\frac{d\mathcal{E}_{ox}}{dx} = 0 \tag{16.19}

\mathcal{E}{ox} = -\frac{d\phi{ox}}{dx} = \text{constant} \tag{16.20}

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 surface field using the interface displacement boundary condition:

(D_{semi} - D_{ox})\big|{O\text{-}S} = Q{O\text{-}S} \tag{16.22}

where $D = K\varepsilon_0\mathcal{E}$ and $Q_{O\text{-}S} = 0$ for an ideal structure:

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}_s x_o \tag{16.25}

where $\mathcal{E}_s$ is the electric field in the semiconductor at the interface ($x = 0$). Substituting (16.18) and (16.25) into (16.17):

V_G = \phi_s + \frac{K_S}{K_O}\mathcal{E}_s x_o \tag{16.26}

In the delta-depletion solution, evaluating (16.12) at $x = 0$ with (16.15) yields:

\mathcal{E}_s = \left[\frac{2qN_A\phi_s}{K_S\varepsilon_0}\right]^{1/2} \tag{16.27}

Substituting $\mathcal{E}_s$ into (16.26):

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 \le 2\phi_F) \tag{16.28}

> **Footnote from the book:** strictly, the $\delta$-function layers of carriers at the OΓÇôS 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 $V_G$ΓÇô$\phi_s$ 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 $\mathcal{E}_{ox} \approx 3\,\mathcal{E}_s$ in an MOS system with no charge at the SiΓÇôSiO$_2$ 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 $x_o = 0.1\ \mu\text{m}$, $N_A = 10^{15}\text{ cm}^{-3}$, $T = 300\text{ K}$; delta-depletion vs. exact): - $\phi_s$ varies 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 drop almost entirely across the oxide. - The depletion-bias region is only slightly more than $1\text{ V}$ wide. Since the semiconductor's character changes drastically across this region, significant variation in electrical characteristics occurs over a 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 $T = 300\text{ K}$ has $x_o = 0.1\ \mu\text{m}$ 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:** $kT/q = 0.0259\text{ V}$, $n_i = 10^{10}\text{ cm}^{-3}$, $K_S = 11.8$, $K_O = 3.9$, $\varepsilon_0 = 8.85 \times 10^{-14}\text{ F/cm}$, $q = 1.6 \times 10^{-19}\text{ C}$, $x_o = 0.1\ \mu\text{m} = 10^{-5}\text{ cm}$. **(a)** For p-type Si, use (16.8a):

\frac{\phi_F}{kT/q} = \ln\left(\frac{N_A}{n_i}\right) = \ln\left(\frac{10^{15}}{10^{10}}\right) = \ln(10^5) = 11.51

\phi_F = 11.51 \times 0.0259\text{ V} = 0.298\text{ V}

\boxed{\phi_F = 11.51,kT/q = 0.298\text{ V}}

**(b)** Depletion width from (16.15) with $\phi_s = \phi_F = 0.298\text{ V}$:

\begin{aligned} W &= \left[\frac{2K_S\varepsilon_0\phi_s}{qN_A}\right]^{1/2} \ &= \left[\frac{2(11.8)(8.85 \times 10^{-14}\text{ F/cm})(0.298\text{ V})}{(1.6 \times 10^{-19}\text{ C})(10^{15}\text{ cm}^{-3})}\right]^{1/2} \ &= \left[\frac{6.224 \times 10^{-13}}{1.6 \times 10^{-4}}\right]^{1/2} \ &= \left[3.89 \times 10^{-9}\text{ cm}^2\right]^{1/2} \ &= 6.24 \times 10^{-5}\text{ cm} = 0.624\ \mu\text{m} \end{aligned}

\boxed{W = 0.624\ \mu\text{m}}

**(c)** Surface electric field from (16.12) at $x = 0$:

\begin{aligned} \mathcal{E}_s &= \frac{qN_A}{K_S\varepsilon_0}W \ &= \frac{(1.6 \times 10^{-19}\text{ C})(10^{15}\text{ cm}^{-3})(6.24 \times 10^{-5}\text{ cm})}{(11.8)(8.85 \times 10^{-14}\text{ F/cm})} \ &= \frac{9.98 \times 10^{-9}}{1.044 \times 10^{-12}}\text{ V/cm} \ &\approx 9.56 \times 10^3\text{ V/cm} \end{aligned}

\boxed{\mathcal{E}_s \approx 9.56 \times 10^3\text{ V/cm}}

$$\ast \ast (d)\ast \ast Gatevoltagefrom(16.26):$$

\begin{aligned} V_G &= \phi_s + \frac{K_S}{K_O}\mathcal{E}_s x_o \ &= 0.298 + \left(\frac{11.8}{3.9}\right)(9.56 \times 10^3\text{ V/cm})(10^{-5}\text{ cm}) \ &= 0.298 + (3.026)(0.0956) \ &= 0.298 + 0.289 \ &\approx 0.587\text{ V} \end{aligned}

\boxed{V_G \approx 0.587\text{ V}}

$$Checkvia(16.28):$$

\mathcal{E}_s = \sqrt{\frac{2qN_A\phi_s}{K_S\varepsilon_0}} = \sqrt{\frac{2(1.6 \times 10^{-4})(0.298)}{1.044 \times 10^{-12}}} = 9.56 \times 10^3\text{ V/cm}

The calculation is completely consistent. **Comment.** These results are representative. In $kT/q$ units, $|\phi_F|$ typically lies between 9 and 18 at $300\text{ K}$, and for nondegenerate doping $|\phi_F|$ is less than half the Si band gap ($< 0.56\text{ V}$). The computed $W$ is less than $W_T$ for this doping, as expected (Fig. 16.9); indeed $W_T = \sqrt{2}\,W = 0.88\ \mu\text{m}$. The device parameters are identical to those of Fig. 16.10, and the computed $V_G$ matches the graphical value. --- ## 16.4 CapacitanceΓÇôVoltage 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 $C$ΓÇô$V$ 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 $C$ΓÇô$V$ meter. The meter superimposes a small a.c. signal (typically $\lesssim 15\text{ mV rms}$, commonly at $1\text{ MHz}$) 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 $C$ΓÇô$V$ curve, displayed on a monitor and printed. More sophisticated equipment (Fig. 16.11a) combines a high-frequency ($1\text{ MHz}$) $C$ΓÇô$V$ meter with low-frequency (quasistatic) capability. This section models the $C$ΓÇô$V$ 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 $N_D = 9.1 \times 10^{14}\text{ cm}^{-3}$ (100) Si with $x_o = 0.119\ \mu\text{m}$. Capacitance (normalized to $C_o$, vertical axis 0 to 1.0) is plotted against $V_G$ from about $-5\text{ V}$ to $+2\text{ V}$: it is flat near $1$ at positive $V_G$ (accumulation), falls through depletion at about $-1$ to $-3\text{ V}$, 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 OΓÇôS interface. The system can change very rapidly: majority carriers equilibrate with time constants of order $10^{-10}$ to $10^{-13}\text{ s}$. Thus at probing frequencies $\le 1\text{ MHz}$ 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_0 A_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_0 A_G}{x_o} \quad \text{(oxide capacitance)} \tag{16.30a}

C_s = \frac{K_S\varepsilon_0 A_G}{W} \quad \text{(semiconductor capacitance)} \tag{16.30b}

C(\text{depl}) = \frac{C_o C_s}{C_o + C_s} = \frac{C_o}{1 + \dfrac{K_O W}{K_S x_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 ($\omega \to 0$):** 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 \to 0 \tag{16.32}

- **Very high frequency ($\omega \to \infty$):** the sluggish generationΓÇôrecombination 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_o C_s}{C_o + C_s} = \frac{C_o}{1 + \dfrac{K_O W_T}{K_S x_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 $\omega \to 0$ 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 $C$ΓÇô$V$ characteristic (points aΓÇôg) and the block charge diagrams (1ΓÇô5) of Figure E16.4. For each biasing condition give the letter(s) of the bias point(s) on the $C$ΓÇô$V$ curve and the number of the block charge diagram. **Solution.** Reason from the polarity (n-type: positive $V_G$ = accumulation) and capacitance behavior: - **Accumulation:** large positive $V_G$, $C = C_o$, electrons accumulated at interface: point **g**, diagram **3**. - **Flat band:** $V_G = 0$, boundary between accumulation and depletion, zero charge: point **f**, diagram **2**. - **Depletion:** negative $V_G$, $C$ falling, depletion block of ionized donors: 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 inversion layer present: points **a / b / c** ($C_{min}$ 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 | --- #### Delta-Depletion Analysis In the delta-depletion formulation, accumulation and inversion layers are formally replaced by $\delta$-functions at the OΓÇôS interface:

C = C_o \quad \text{accumulation} \tag{16.34a}

C = \frac{C_o}{1 + \dfrac{K_O W}{K_S x_o}} \quad \text{depletion} \tag{16.34b}

C = C_o \quad \text{inversion } (\omega \to 0) \tag{16.34c}

C = \frac{C_o}{1 + \dfrac{K_O W_T}{K_S x_o}} \quad \text{inversion } (\omega \to \infty) \tag{16.34d}

To express depletion width $W$ as a function of $V_G$, invert (16.28) and substitute into (16.15):

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\text{)} \tag{16.36}

$$Substituting(16.35)into(16.34b)yields:$$

\begin{aligned} 1 + \frac{K_O W}{K_S x_o} &= 1 + \left[-1 + \sqrt{1 + \frac{V_G}{V_\delta}},\right] = \sqrt{1 + \frac{V_G}{V_\delta}} \end{aligned}

C = \frac{C_o}{\sqrt{1 + \dfrac{V_G}{V_\delta}}} \qquad \text{(depletion biases)} \tag{16.37}

A sample set of low- and high-frequency $C$ΓÇô$V$ curves from the delta-depletion analysis is shown in Fig. 16.13 ($x_o = 0.1\ \mu\text{m}$, $N_D = 10^{15}\text{ cm}^{-3}$, $T = 300\text{ K}$, $V_G$ from $-2$ to $+1\text{ V}$). --- ### 16.4.2 Computations and Observations #### Exact Computations The delta-depletion characteristics (Fig. 16.13) provide a first-order approximation but fail near transition points (accumulation to depletion, and depletion to inversion). Exact-charge formulations provide high accuracy (results in Appendix C). Observations from exact curves (Figs. 16.14ΓÇô16.16): - **Doping (Fig. 16.14):** 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. - **Oxide thickness (Fig. 16.15):** thicker oxide widens the depletion-bias region (a larger fraction of voltage drops across the oxide) and increases the minimum normalized capacitance $C/C_o$. - **Temperature (Fig. 16.16):** moderate sensitivity of the inversion-bias capacitance and near insensitivity of the depletion-bias capacitance to temperature. --- ### Exercise 16.5 **Problem.** (a) Using exact-charge relationships, write a MATLAB program that computes and plots low-frequency $C/C_o$ vs. $V_G$ for $U = \phi_s/(kT/q)$ stepped from $U = U_F - 21$ to $U_F + 21$. Set $N_A = 10^{15}\text{ cm}^{-3}$ and plot for $x_o = 0.1, 0.2, 0.3\ \mu\text{m}$. (b) Do the same for high-frequency $C/C_o$ with $x_o = 0.1\ \mu\text{m}$ and $N_A = 10^{14}, 10^{15}, 10^{16}\text{ cm}^{-3}$. **Method.** Using standard equilibrium exact-charge formulation for a p-substrate, with $U = \phi_s/(kT/q)$, $U_F = \phi_F/(kT/q) = \ln(N_A/n_i)$, and extrinsic Debye length $L_D = \sqrt{K_S\varepsilon_0 kT / (q^2 N_A)}$:

\begin{aligned} F(U, U_F) &= \sqrt{e^{-U} + U - 1 + e^{-2U_F}(e^U - U - 1)} \ \mathcal{E}_s &= \operatorname{sgn}(U)\frac{\sqrt{2},(kT/q)}{L_D}F \ V_G &= \phi_s + \frac{K_S}{K_O}x_o \mathcal{E}s \ C_s &= \frac{\sqrt{2},K_S\varepsilon_0}{L_D}\frac{\left|1 - e^{-U} + e^{-2U_F}(e^U - 1)\right|}{2F} \ \frac{C{LF}}{C_o} &= \frac{C_s}{C_o + C_s}, \quad C_o = \frac{K_O\varepsilon_0}{x_o} \end{aligned}

```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 (p-type, $N_A = 10^{15}\text{ cm}^{-3}$, $T = 300\text{ K}$):** | $x_o$ | Flat-band $C/C_o$ | LF exact dip: $\min C/C_o$ (at $V_G$) | Exact $V_T$ (at $\phi_s = 2\phi_F$) | | :---: | :---: | :---: | :---: | | $0.1\ \mu\text{m}$ | 0.70 | 0.28 (at $0.87\text{ V}$) | $1.005\text{ V}$ | | $0.2\ \mu\text{m}$ | 0.82 | 0.44 (at $1.24\text{ V}$) | $1.414\text{ V}$ | | $0.3\ \mu\text{m}$ | 0.87 | 0.54 (at $1.60\text{ V}$) | $1.823\text{ V}$ | Sample points for $x_o = 0.1\ \mu\text{m}$: - $U = -9.5 \implies V_G = -10.0\text{ V}$, $C/C_o = 0.995$ - $U = -0.5 \implies V_G = -0.045\text{ V}$, $C/C_o = 0.73$ - $U = +1.5 \implies V_G = 0.11\text{ V}$, $C/C_o = 0.60$ - $U = 11.5\ (\phi_F) \implies V_G = 0.57\text{ V}$, $C/C_o = 0.34$ - $U = 21.5 \implies V_G = 0.95\text{ V}$, $C/C_o = 0.31$ - $U = 23.5 \implies V_G = 1.03\text{ V}$, $C/C_o = 0.47$ - $U = 26.5 \implies V_G = 1.34\text{ V}$, $C/C_o = 0.88$ - $U = 32.5 \implies V_G = 10.6\text{ V}$, $C/C_o = 0.995$ For (b), the high-frequency inversion plateau is the depletion capacitance at $W_T$. Delta-depletion values for $x_o = 0.1\ \mu\text{m}$ from (16.33):

\begin{aligned} N_A = 10^{14}\text{ cm}^{-3}: \quad & W_T = 2.50\ \mu\text{m}, \quad C_{min}/C_o = \frac{1}{1 + \dfrac{3.9}{11.8}\dfrac{2.50}{0.1}} = 0.11 \ N_A = 10^{15}\text{ cm}^{-3}: \quad & W_T = 0.88\ \mu\text{m}, \quad C_{min}/C_o = \frac{1}{1 + \dfrac{3.9}{11.8}\dfrac{0.88}{0.1}} = 0.26 \ N_A = 10^{16}\text{ cm}^{-3}: \quad & W_T = 0.31\ \mu\text{m}, \quad C_{min}/C_o = \frac{1}{1 + \dfrac{3.9}{11.8}\dfrac{0.31}{0.1}} = 0.50 \end{aligned}

The exact $V_T$ values ($\phi_s = 2\phi_F$) are $0.59\text{ V}$, $1.01\text{ V}$, and $2.13\text{ V}$, respectively.

\boxed{\text{HF } C_{min}/C_o \approx 0.11, \ 0.26, \ 0.50 \quad \text{for } N_A = 10^{14}, 10^{15}, 10^{16}\text{ cm}

--- #### Practical Observations In modern MOS capacitors with long carrier lifetimes and low generation rates, even a $10\text{ Hz}$ test signal yields high-frequency-type characteristics. A genuine low-frequency curve requires indirect means such as the **quasistatic technique**: a slow (typically $10\text{--}100\text{ mV/s}$) linear voltage ramp is applied to the gate while monitoring displacement current into the gate, which is directly proportional to low-frequency capacitance. On the high side, standard measurements are performed at **$1\text{ MHz}$**. Frequencies $\gtrsim 1\text{ GHz}$ run into the response time of majority carriers and substrate resistance effects. High-frequency measurements also depend on ramp rate $R$ (Fig. 16.17). If ramped too quickly from accumulation into inversion, the inversion layer cannot form in time, driving the device into deep depletion. #### Deep Depletion In accumulation or depletion, majority carriers respond almost instantaneously. Entering inversion requires *minority* carriers, which must be thermally generated in the near-surface regionΓÇöa sluggish process. When swept rapidly, the depletion width expands beyond $W_T$ to balance the gate charge in the absence of minority carriers (Fig. 16.18a):

W > W_T

This nonequilibrium state is **deep depletion**. In the limit where no minority carriers are present (**total deep depletion**, Fig. 16.18b), the depletion capacitance relation (16.37) extends 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}

**Applications.** Deep depletion is central to **DRAMs** (deeply depleted MOS capacitors as storage elements) and **CCDs** (optically generated charge packets stored in potential wells). --- ### Exercise 16.6 **Problem.** The experimental $C$ΓÇô$V$ 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. **Point 1 to Point 2 (slow ramp):** The sweep rate is sufficiently slow for the semiconductor to maintain thermal equilibrium, tracing the standard high-frequency curve down to the equilibrium high-frequency minimum $C_{min}$ at the onset of inversion. 2. **Point 2 to Point 3 (fast ramp):** The sweep rate exceeds the thermal generation rate of minority carriers. The device is driven into **deep depletion** ($W > W_T$), causing capacitance to drop *below* the equilibrium minimum along the deep-depletion curve (16.38). 3. **Point 3 to Point 4 (sweep stopped):** With $V_G$ held constant, thermal generation gradually produces the required minority carriers. The inversion layer forms, the depletion width shrinks back to $W_T$, and the capacitance rises back to the equilibrium high-frequency value. --- ## 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: - **Accumulation:** pile-up of majority carriers at the oxideΓÇôsemiconductor interface. - **Flat band:** no bending of the semiconductor bands (zero net semiconductor charge). - **Depletion:** repulsion of majority carriers from the interface, leaving uncompensated impurity ions. - **Inversion:** pile-up of minority carriers at the oxideΓÇôsemiconductor interface. 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 $C$ΓÇô$V$ characteristics in the low- and high-frequency limits. Finally, 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 | 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 ΓÇö The Essentials > Constants used throughout: $T = 300\text{ K}$, $kT/q = 0.0259\text{ V}$, $n_i = 10^{10}\text{ cm}^{-3}$, $K_S = 11.8$, $K_O = 3.9$, $\varepsilon_0 = 8.85 \times 10^{-14}\text{ F/cm}$, $q = 1.6 \times 10^{-19}\text{ C}$. 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. We assume an **ideal MOS structure** (no oxide charge, no work-function difference, no interface traps) and examine the basic **long-channel enhancement-mode** device. Plan of the chapter: 1. Qualitative theory of operation. 2. Quantitative $I_D$ΓÇô$V_D$ relationships (square-law theory and bulk-charge theory). 3. a.c. response (small-signal circuits, cutoff frequency, characteristics). --- ## 17.1 Qualitative Theory of Operation ### The Structure A MOSFET is an **MOS capacitor with two pn junctions placed immediately next to the gated region** (Fig. 17.1). For an n-channel device on a p-type silicon substrate: - Two heavily doped $n^+$ regions form the **source (S)** and **drain (D)**. - Between them sits the **gate (G)**: metal or polysilicon over a thin $\text{SiO}_2$ layer. - The substrate (back, B) contacts the bulk. Source and back are grounded. Terminal conventions: - $V_G$ and $V_D$ are referenced to ground. - The drain current $I_D$ is the primary d.c. observable. - Electrons enter at the source, traverse the channel, and exit at the drain. - In normal operation the drain pn junction is reverse biased ($V_D \ge 0$). - $I_D$ is defined positive flowing from the external circuit into the drain terminal. ### Step 1: Gate Voltage Alone ($V_D = 0$) - For $V_G \le V_T$ (accumulation or depletion): The surface has no mobile electrons. An open circuit exists between source and drain (two back-to-back diodes). - For $V_G > V_T$ (inversion): Mobile electrons form a conductive surface inversion layer joining source and drain (Fig. 17.2a). Higher $V_G$ increases electron density and channel conductance. ### Step 2: Drain Voltage Application ($V_G > V_T$, Increasing $V_D$) The channel potential rises from $0$ at the source to $V_D$ at the drain. The local gate-to-channel potential drop decreases toward the drain ($V_G - V(y)$), reducing electron density near the drain: 1. **$V_D = 0$ (Fig. 17.2a):** Thermal equilibrium, $I_D = 0$. 2. **Small $V_D$ (Fig. 17.3, line O to A):** Linear ohmic behavior, $I_D \propto V_D$. 3. **Moderate $V_D$ (Fig. 17.2b):** Channel resistance increases near the drain; the $I_D$ΓÇô$V_D$ curve bends over. 4. **Pinch-off (Fig. 17.2c, point B):** At $V_D = V_{Dsat}$, the inversion density at the drain end drops to zero ($Q_N(L) = 0$). 5. **Post pinch-off, $V_D > V_{Dsat}$ (Fig. 17.2d):** A depleted region of length $\Delta L$ forms at the drain end, absorbing the extra voltage $V_D - V_{Dsat}$. For long channels ($\Delta L \ll L$), $I_D$ saturates at $I_{Dsat}$. ### Step 3: Family of Curves (Fig. 17.4) - $V_G \le V_T$: $I_D \approx 0$. - $V_G > V_T$: Output current $I_D$ increases with $V_G$, exhibiting linear and saturation regimes. --- ### 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.** - Fig. 17.1(a): p-type substrate becomes n-type substrate; $n^+$ source/drain become $p^+$ source/drain. - Fig. 17.1(b): Holes replace electrons as channel carriers; p-Si bulk becomes n-Si; terminal voltage condition $V_D \ge 0$ becomes $V_D \le 0$; current directions reverse such that positive $I_D$ flows from source to drain and out of the drain contact. - Fig. 17.2: Source and drain regions become $p^+$; polarity inequalities flip. - Fig. 17.3: $V_D \to -V_D$ and $V_{Dsat} \to -V_{Dsat}$ along the voltage axis. - Fig. 17.4: "$V_G$ increasing" becomes "$-V_G$ increasing", and $V_D \to -V_D$. Since $V_T < 0$ for a p-channel enhancement device, no current flows for $V_G > V_T$, and conduction occurs for $V_G < V_T$. --- ## 17.2 Quantitative $I_D$ΓÇô$V_D$ Relationships ### 17.2.1 Preliminary Considerations #### Threshold Voltage For an ideal structure, setting $\phi_s = 2\phi_F$ in (16.28):

V_T = 2\phi_F + \frac{K_S}{K_O}x_o\sqrt{\frac{4qN_A\phi_F}{K_S\varepsilon_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\varepsilon_0}} \qquad \text{ideal p-channel (n-bulk)} \tag{17.1b}

Using $C_o = K_O\varepsilon_0 / x_o$:

V_T = 2\phi_F + \frac{\sqrt{4qK_S\varepsilon_0 N_A\phi_F}}{C_o}

#### Effective Mobility Gate-induced transverse electric fields force carriers toward the interface, introducing surface scattering. The average mobility across the channel depth $x_c(y)$ is the **effective mobility** $\bar{\mu}_n$:

\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}

$$Theinversionlayerchargeperunitareais:$$

Q_N(y) = -q\int_0^{x_c(y)} n(x, y),dx \tag{17.3}

Expressing $\bar{\mu}_n$ in terms of $Q_N$:

\bar{\mu}_n = \frac{-q}{Q_N(y)}\int_0^{x_c(y)} \mu_n(x, y) n(x, y),dx \tag{17.4}

For long-channel devices, $\bar{\mu}_n$ is assumed independent of $y$ and $V_D$. Its gate-voltage dependence is empirically modelled as:

\bar{\mu}_n = \frac{\mu_0}{1 + \theta(V_G - V_T)} \tag{17.5}

--- ### 17.2.2 Square-Law Theory **Step 1: Current density.**

\mathbf{J}_N = q\mu_n n \boldsymbol{\mathcal{E}} + q D_N \nabla n \tag{17.6}

**Step 2: Drift simplification along the channel ($y$-direction).**

J_{Ny} = -q\mu_n n\frac{d\phi}{dy} \tag{17.7}

$$\ast \ast Step3:Totaldraincurrent.\ast \ast$$

I_D = -Z\int_0^{x_c(y)} J_{Ny},dx = Z\frac{d\phi}{dy}\left(q\int_0^{x_c} \mu_n n,dx\right) \tag{17.8a}

I_D = -Z\bar{\mu}_n Q_N(y)\frac{d\phi}{dy} \tag{17.9}

**Step 4: Integration along the channel length $L$.**

\int_0^L I_D,dy = -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}

$$\ast \ast Step5:Channelchargeapproximation.\ast \ast$$

\Delta Q_{gate} = -\Delta Q_{semi} = -Q_N \qquad (V_G \ge V_T) \tag{17.12}

Q_N = -C_o(V_G - V_T) \tag{17.14}

C_o = \frac{K_O\varepsilon_0}{x_o} \tag{17.15}

Accounting for the channel potential $\phi(y)$:

Q_N(y) = -C_o(V_G - V_T - \phi) \tag{17.16}

$$\ast \ast Step6:Integratetoobtainthecurrent-voltagerelationship.\ast \ast$$

\begin{aligned} I_D &= \frac{Z\bar{\mu}_n C_o}{L}\int_0^{V_D} (V_G - V_T - \phi),d\phi \ &= \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G - V_T)V_D - \frac{V_D^2}{2}\right] \end{aligned}

\boxed{I_D = \frac{Z\bar{\mu}n C_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}

$$Post-pinch-off:$$

I_D\big|{V_D > V{Dsat}} = I_D\big|{V_D = V{Dsat}} \equiv I_{Dsat} \tag{17.18}

I_{Dsat} = \frac{Z\bar{\mu}n C_o}{L}\left[(V_G - V_T)V{Dsat} - \frac{V_{Dsat}^2}{2}\right] \tag{17.19}

Pinch-off occurs when $Q_N(L) = 0$:

Q_N(L) = -C_o(V_G - V_T - V_{Dsat}) = 0 \tag{17.20}

V_{Dsat} = V_G - V_T \tag{17.21}

$$Substitutinginto(17.19):$$

\boxed{I_{Dsat} = \frac{Z\bar{\mu}_n C_o}{2L}(V_G - V_T)^2} \tag{17.22}

--- ### Exercise 17.2: The "Universal" Square-Law Characteristics **Problem.** Plot $I_D / (Z\bar{\mu}_n C_o / L)$ versus $V_D$ for $V_G - V_T = 1, 2, 3, 4\text{ V}$. **Solution.** Let $V_{GT} = V_G - V_T$. For $0 \le V_D \le V_{GT}$:

\frac{I_D}{Z\bar{\mu}n C_o / L} = V{GT}V_D - \frac{V_D

For $V_D > V_{GT}$:

\frac{I_{Dsat}}{Z\bar{\mu}n C_o / L} = \frac{V{GT}

Peak normalized values are $0.5, 2.0, 4.5, 8.0$ for $V_{GT} = 1, 2, 3, 4\text{ V}$. ```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 ``` --- ### 17.2.3 Bulk-Charge Theory **Step 1: Charge balance including depletion width variation $W(y)$.**

Q_N(y) = -C_o(V_G - V_T - \phi) + qN_A\big[W(y) - W_0\big] \tag{17.23}

$$\ast \ast Step2:Depletionwidths.\ast \ast$$

W(y) = \left[\frac{2K_S\varepsilon_0(2\phi_F + \phi)}{qN_A}\right]^{1/2} \tag{17.24}

W_0 = \left[\frac{2K_S\varepsilon_0(2\phi_F)}{qN_A}\right]^{1/2} \tag{17.25}

**Step 3: Define bulk-charge parameter $V_W$.**

V_W \equiv \frac{\sqrt{4qK_S\varepsilon_0 N_A\phi_F}}{C_o} \tag{17.26}

\begin{aligned} qN_A\big[W(y) - W_0\big] &= \sqrt{4qK_S\varepsilon_0 N_A\phi_F}\left[\sqrt{1 + \frac{\phi}{2\phi_F}} - 1\right] \ &= C_o V_W\left[\sqrt{1 + \frac{\phi}{2\phi_F}} - 1\right] \end{aligned}

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}

$$\ast \ast Step4:Integrate(17.27)in(17.11).\ast \ast$$

\begin{aligned} I_D &= \frac{Z\bar{\mu}_n C_o}{L}\int_0^{V_D}\left{(V_G - V_T + V_W) - \phi - V_W\sqrt{1 + \frac{\phi}{2\phi_F}}\right}d\phi \ &= \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G - V_T + V_W)V_D - \frac{V_D^2}{2} - \frac{2}{3}(2\phi_F)V_W\left{\left(1 + \frac{V_D}{2\phi_F}\right)^{3/2} - 1\right}\right] \end{aligned}

\boxed{I_D = \frac{Z\bar{\mu}_n C_o}{L}\left{(V_G - V_T + V_W)V_D - \frac{V_D^2}{2} - \frac{4\phi_F V_W}{3}\left[\left(1 + \frac{V_D}{2\phi_F}\right)^{3/2} - 1\right]\right}} \tag{17.28}

**Step 5: Saturation voltage $V_{Dsat}$.** Setting $Q_N(L) = 0$ at $\phi(L) = V_{Dsat}$ in (17.27), let $s = \sqrt{1 + V_{Dsat}/(2\phi_F)}$:

V_G - V_T - 2\phi_F(s^2 - 1) - V_W(s - 1) = 0

Since $V_T = 2\phi_F + V_W$:

2\phi_F s^2 + V_W s - V_G = 0

Solving for the positive root $s$:

s = \frac{-V_W + \sqrt{V_W^2 + 8\phi_F V_G}}{4\phi_F}

Since $V_{Dsat} = V_G - 2\phi_F - V_W s$:

\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_F V_G}} \tag{17.29}

--- ### Exercise 17.3: Diode-Connected MOSFET **Problem.** Gate and drain tied together. $x_o = 500\text{ \AA} = 5 \times 10^{-6}\text{ cm}$, $N_A = 10^{16}\text{ cm}^{-3}$, $Z/L = 10$, $\bar{\mu}_n = 625\text{ cm}^2/(\text{V}\cdot\text{s})$, $T = 300\text{ K}$. Using bulk-charge theory find $I_D$ for (a) $V_G = V_D = 1\text{ V}$; (b) $V_G = V_D = 3\text{ V}$. **Solution.** Compute bulk Fermi potential $\phi_F$:

\phi_F = 0.0259\ln\left(\frac{10^{16}}{10^{10}}\right) = 0.358\text{ V}

Compute threshold voltage $V_T$:

\begin{aligned} V_T &= 2\phi_F + \frac{K_S}{K_O}x_o\sqrt{\frac{4qN_A\phi_F}{K_S\varepsilon_0}} \ &= 0.716 + \left(\frac{11.8}{3.9}\right)(5 \times 10^{-6})\sqrt{\frac{4(1.6 \times 10^{-19})(10^{16})(0.358)}{(11.8)(8.85 \times 10^{-14})}} \ &= 0.716 + 0.708 = 1.42\text{ V} \end{aligned}

**(a)** For $V_G = V_D = 1\text{ V}$: Since $V_G = 1\text{ V} < V_T = 1.42\text{ V}$, the device is off:

\boxed{I_D = 0}

**(b)** For $V_G = V_D = 3\text{ V}$: Since $V_D = V_G > V_G - V_T > V_{Dsat}$, the transistor is in saturation ($I_D = I_{Dsat}$). Compute constants:

C_o = \frac{(3.9)(8.85 \times 10^{-14}\text{ F/cm})}{5 \times 10^{-6}\text{ cm}} = 6.90 \times 10^{-8}\text{ F/cm}

W_0 = \left[\frac{2(11.8)(8.85 \times 10^{-14})(0.716)}{(1.6 \times 10^{-19})(10^{16})}\right]^{1/2} = 3.06 \times 10^{-5}\text{ cm}

V_W = \frac{qN_A W_0}{C_o} = \frac{(1.6 \times 10^{-19})(10^{16})(3.06 \times 10^{-5})}{6.90 \times 10^{-8}} = 0.71\text{ V}

Compute $V_{Dsat}$ from (17.29) with $V_G = 3\text{ V}$:

V_{Dsat} \approx 1.15\text{ V}

$$Evaluateprefactorandcurrent:$$

\frac{Z\bar{\mu}_n C_o}{L} = 10(625)(6.90 \times 10^{-8}) = 4.31 \times 10^{-4}\text{ A/V}

Evaluating (17.28) at $V_D = 1.14\text{ V}$:

\begin{aligned} {\ldots} &= (3 - 1.42 + 0.71)(1.14) - \frac{(1.14)^2}{2} - \frac{4(0.358)(0.71)}{3}\left[\left(1 + \frac{1.14}{0.716}\right)^{3/2} - 1\right] \ &\approx 1.80 - 0.65 + 0.81 - 1.08 \approx 0.885\text{ V}^2 \end{aligned}

I_D = (4.31 \times 10^{-4}\text{ A/V}^2)(0.885\text{ V}^2) \approx 0.382\text{ mA}

\boxed{I_D \approx 0.382\text{ mA}}

--- ### 17.2.4 Charge-Sheet and Exact-Charge Theories Both square-law and bulk-charge theories have limitations: 1. $Q_N$ is assumed to be exactly zero for $V_G \le V_T$, failing to capture **subthreshold current**. 2. They do not self-saturate; saturation must be imposed explicitly post pinch-off. The **charge-sheet** and **exact-charge** theories remove both flaws, self-saturating and modelling subthreshold conduction accurately. --- ## 17.3 a.c. Response ### 17.3.1 Small-Signal Equivalent Circuits **Low frequencies.** Small-signal a.c. expansions:

i_d = I_D(V_D + v_d, , V_G + v_g) - I_D(V_D, V_G) \tag{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}

$$Conductances:$$

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)} \tag{17.33b}

i_d = g_d v_d + g_m v_g \tag{17.34}

Let $\xi \equiv Z\bar{\mu}_n C_o / L$. **Table 17.1: MOSFET small-signal parameters** | Theory | 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) | **High frequencies.** Include inter-electrode capacitances: - $C_{gd}$: overlap capacitance between gate and drain. - $C_{gs} \approx C_G = C_o Z L$: gate-to-source capacitance. ### 17.3.2 Cutoff Frequency Output short-circuit condition ($v_d = 0$):

i_{in} = j\omega(C_{gs} + C_{gd})v_g \approx j(2\pi f)C_G v_g \tag{17.35}

i_{out} = g_m v_g \tag{17.36}

Setting $|i_{out} / i_{in}| = 1$:

\frac{g_m}{2\pi f_{max} C_G} = 1 \implies f_{max} = \frac{g_m}{2\pi C_G}

Using $g_m = \xi V_D$ below pinch-off:

f_{max} = \frac{\bar{\mu}n V_D}{2\pi L^2} \qquad (V_D \le V{Dsat}) \tag{17.37}

--- ### 17.3.3 Small-Signal Characteristics **(a) $g_d$ versus $V_G$ at $V_D = 0$:**

g_d = \frac{Z\bar{\mu}_n C_o}{L}(V_G - V_T) \qquad (V_D = 0) \tag{17.38}

Extrapolating $g_d$ to zero gives $V_T$, and its slope provides $\bar{\mu}_n$. **(b) Gate capacitance $C_G$ versus $V_G$ at $V_D = 0$:** Resembles the low-frequency MOS-C curve because adjacent $n^+$ regions supply minority carriers laterally at high frequencies. --- ### Exercise 17.4: $g_m$ΓÇô$V_G$ Characteristics **Problem.** (a) Deduce $g_m$ vs. $V_G$ with $V_D$ fixed. Using square-law theory and $V_T = 2\text{ V}$, plot $g_m / (Z\bar{\mu}_n C_o / L)$ vs. $V_G$ for $V_D = 2, 4, 6\text{ V}$. (b) Compare with experiment. **Solution (a).** With $\xi = Z\bar{\mu}_n C_o / L$:

\frac{g_m}{\xi} = \begin{cases} 0, & V_G < V_T \ V_G - V_T, & 0 \le V_G - V_T \le V_D \quad \text{(saturation)} \ V_D, & V_G - V_T \ge V_D \quad \text{(below pinch-off)} \end{cases}

```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)** In experiment, $g_m$ rolls off for $V_G > V_D + V_T$ due to mobility degradation from surface scattering ($\bar{\mu}_n(V_G)$, Eq. 17.5). --- ## 17.4 Summary - Gate-induced inversion creates a channel between source and drain. - The drain voltage induces pinch-off near the drain, leading to current saturation. - Square-law theory provides simple, useful analytical trends; bulk-charge theory accounts for the increase in depletion charge along the channel. - Cutoff frequency scales as $1/L^2$, making channel length scaling critical for high-speed operation. --- # Chapter 17 Problems | 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 | 2ΓÇô3 | 10 | n to p in Subsection 17.2.2 | | 17.8 | 17.2.2 | 1 | 5 | Plot $I_D$ΓÇô$V_D$, $\bar{\mu}_n$ per Eq. (17.5) | | 17.9 | 17.2.2 | 3 | 10 | $V_D = V_G - V_B$ | | 17.10 | 17.2.2 | 3ΓÇô4 | 10 | $I_D$ versus $V_G - V_T$ | | 17.11 | 17.2.2 | 3ΓÇô4 | 10 | Circular MOSFET | | 17.12 | 17.2.2 | 3 | 12 | $I_D$ temperature dependence | | 17.13 | 17.2.2 | 3ΓÇô4 | 12 | Include $R_S$, $R_D$ in theory | | 17.14 | 17.2.3 | 3ΓÇô4 | 10 | Derive Eq. (17.29) | | 17.15 | 17.2.3 | 3 | 10 | Plot bulk-charge $I_D$ΓÇô$V_D$ | | 17.16 | 17.2.4 | 4 | 25 | Plot sheet, exact $I_D$ΓÇô$V_D$ | | 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 | 2ΓÇô3 | 15 | General MOSFET review | | 17.21 | 17.3.3 | 3 | 12 | Small-signal characteristics | | 17.22 | 17.4 | 2 | 12 | Compare FETs | Constants: $T = 300\text{ K}$, $kT/q = 0.0259\text{ V}$, $n_i = 10^{10}\text{ cm}^{-3}$, $K_S = 11.8$, $K_O = 3.9$, $\varepsilon_0 = 8.85 \times 10^{-14}\text{ F/cm}$, $q = 1.6 \times 10^{-19}\text{ C}$. Let $\xi \equiv Z\bar{\mu}_n C_o / L$. --- ## 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 $I_D$ΓÇô$V_D$ 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?** 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 SiΓÇôSiO$_2$ 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}_n C_o}{2L}(V_G - V_T)

(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 axis), 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 $C$ΓÇô$V_G$ ($V_D=0$) a low-frequency curve even at $1\text{ MHz}$?** 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, $T=300\text{ K}$, $Z=50\ \mu\text{m}$, $L=5\ \mu\text{m}$, $x_o=0.05\ \mu\text{m}$, $N_A=10^{16}\text{ cm}^{-3}$, $\bar{\mu}_n=800\text{ cm}^2/\text{V}\cdot\text{s}$ (independent of $V_G$). Find: (a) $V_T$; (b) $I_{D\text{sat}}$ (square law) if $V_G-V_T=2\text{ V}$; (c) $I_{D\text{sat}}$ (bulk charge) if $V_G-V_T=2\text{ V}$; (d) $g_d$ if $V_G-V_T=2\text{ V}$, $V_D=0$; (e) $g_m$ (square law) if $V_G-V_T=2\text{ V}$, $V_D=2\text{ V}$; (f) $g_m$ (bulk charge), same bias; (g) $f_{\max}$ if $V_G-V_T=2\text{ V}$, $V_D=1\text{ V}$. **Solution.** Common quantities: 1. Built-in bulk Fermi potential:

\begin{aligned} \phi_F &= \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right) \ &= 0.0259\ln\left(\frac{10^{16}}{10^{10}}\right) \ &= 0.358\text{ V} \end{aligned}

$$2.Oxidecapacitanceperunitarea:$$

\begin{aligned} C_o &= \frac{K_O\epsilon_0}{x_o} \ &= \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})}{5\times10^{-6}\text{ cm}} \ &= 6.90\times10^{-8}\text{ F/cm}^2 \end{aligned}

3. Process transconductance parameter $\xi$:

\begin{aligned} \xi &= \frac{Z}{L}\bar{\mu}_n C_o \ &= \left(\frac{50\ \mu\text{m}}{5\ \mu\text{m}}\right)(800\text{ cm}^2/\text{V}\cdot\text{s})(6.90\times10^{-8}\text{ F/cm}^2) \ &= (10)(800)(6.90\times10^{-8}) \ &= 5.52\times10^{-4}\text{ A/V}^2 \end{aligned}

**(a)** Calculate the bulk-charge parameter $V_W$:

\begin{aligned} V_W &= \frac{\sqrt{4qK_S\epsilon_0N_A\phi_F}}{C_o} \ &= \frac{\sqrt{4(1.6\times10^{-19}\text{ C})(11.8)(8.85\times10^{-14}\text{ F/cm})(10^{16}\text{ cm}^{-3})(0.358\text{ V})}}{6.90\times10^{-8}\text{ F/cm}^2} \end{aligned}

$$Evaluatingthenumerator:$$

\sqrt{2.39\times10^{-15}}\approx 4.88\times10^{-8}\text{ C/cm}

$$Thus:$$

V_W = \frac{4.88\times10^{-8}}{6.90\times10^{-8}} = 0.708\text{ V}

$$Thethresholdvoltageisthen:$$

\begin{aligned} V_T &= 2\phi_F + V_W \ &= 2(0.358) + 0.708 \ &= 0.716 + 0.708 \ &= \boxed{1.42\text{ V}} \end{aligned}

$$\ast \ast (b)\ast \ast Squarelaw,(17.22):$$

\begin{aligned} I_{D\text{sat}} &= \frac{\xi}{2}(V_G-V_T)^2 \ &= \frac{5.52\times10^{-4}\text{ A/V}^2}{2}(2\text{ V})^2 \ &= (2.76\times10^{-4})(4) \ &= 1.10\times10^{-3}\text{ A} \ &= \boxed{1.10\text{ mA}} \end{aligned}

**(c)** Bulk charge theory: With $V_G - V_T = 2\text{ V}$, the applied gate voltage is:

V_G = V_T + 2 = 1.42 + 2 = 3.42\text{ V}

$$From(17.29),computethediscriminantterm:$$

\begin{aligned} \sqrt{V_W^2+8\phi_F V_G} &= \sqrt{(0.708)^2 + 8(0.358)(3.42)} \ &= \sqrt{0.501 + 9.795} \ &= \sqrt{10.296} \ &= 3.209 \end{aligned}

Compute the parameter $s$:

\begin{aligned} s &= \frac{-V_W + \sqrt{V_W^2+8\phi_F V_G}}{4\phi_F} \ &= \frac{-0.708 + 3.209}{4(0.358)} \ &= \frac{2.501}{1.432} \ &= 1.747 \end{aligned}

$$Thesaturationdrainvoltageis:$$

\begin{aligned} V_{D\text{sat}} &= V_G - 2\phi_F - V_W s \ &= 3.42 - 0.716 - (0.708)(1.747) \ &= 3.42 - 0.716 - 1.237 \ &= 1.47\text{ V} \end{aligned}

From (17.28) evaluated at $V_D=V_{D\text{sat}}=1.47\text{ V}$:

\begin{aligned} {\ldots} &= (V_G - V_T + V_W)V_D - \frac{V_D^2}{2} - \frac{4\phi_F V_W}{3}\left[\left(1+\frac{V_D}{2\phi_F}\right)^{3/2}-1\right] \ &= (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) \ &= 2.901 - 1.467 \ &\approx 1.435\text{ V}^2 \end{aligned}

$$Thesaturationdraincurrentis:$$

\begin{aligned} I_{D\text{sat}} &= \xi{\ldots} \ &= (5.52\times10^{-4}\text{ A/V}^2)(1.435\text{ V}^2) \ &= 0.792\times10^{-3}\text{ A} \ &= \boxed{0.79\text{ mA}} \end{aligned}

(The bulk-charge value is about $28\%$ below the square-law value.) **(d)** At $V_D=0$ (Table 17.1 or (17.38)):

\begin{aligned} g_d &= \xi(V_G-V_T) \ &= (5.52\times10^{-4}\text{ A/V}^2)(2\text{ V}) \ &= 1.10\times10^{-3}\text{ S} \ &= \boxed{1.10\text{ mS}} \end{aligned}

$$\ast \ast (e)\ast \ast Squarelaw:$$

V_{D\text{sat}} = V_G - V_T = 2\text{ V} = V_D

$$Sincethedevicesitsexactlyatthepinch-offedge,bothbelow-andabove-pinch-offformulasagree:$$

\begin{aligned} g_m &= \xi V_D \ &= (5.52\times10^{-4}\text{ A/V}^2)(2\text{ V}) \ &= 1.10\times10^{-3}\text{ S} \ &= \boxed{1.10\text{ mS}} \end{aligned}

$$\ast \ast (f)\ast \ast Bulkchargetheory:$$

V_{D\text{sat}} = 1.47\text{ V} < V_D = 2\text{ V}

The device is saturated, so $g_m=\xi V_{D\text{sat}}$:

\begin{aligned} g_m &= (5.52\times10^{-4}\text{ A/V}^2)(1.47\text{ V}) \ &= 0.811\times10^{-3}\text{ S} \ &= \boxed{0.81\text{ mS}} \end{aligned}

**(g)** Below pinch-off ($V_D=1\text{ V} < V_{D\text{sat}}$ in either theory):

\begin{aligned} f_{\max} &= \frac{\bar{\mu}_n V_D}{2\pi L^2} \ &= \frac{(800\text{ cm}^2/\text{V}\cdot\text{s})(1\text{ V})}{2\pi(5\times10^{-4}\text{ cm})^2} \ &= \frac{800}{2\pi(2.5\times10^{-7})} \ &= \frac{800}{1.571\times10^{-6}} \ &= 5.09\times10^{8}\text{ Hz} \ &\approx \boxed{509\text{ MHz}} \end{aligned}

--- ## 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 occurs at $V_G=V_T<0$. - 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 + \frac{\sqrt{4qK_S\epsilon_0N_A\phi_F},x_o}{K_O\epsilon_0}

$$with:$$

\phi_F = \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right)

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 $N_A=10^{16}\text{ cm}^{-3}$: - $x_o=0$: $V_T=0.72\text{ V}$ - $x_o=0.02\ \mu\text{m}$: $V_T=1.00\text{ V}$ - $x_o=0.05\ \mu\text{m}$: $V_T=1.42\text{ V}$ - $x_o=0.1\ \mu\text{m}$: $V_T=2.13\text{ V}$ The $x_o=0$ curve runs from about $0.48\text{ V}$ at $10^{14}\text{ cm}^{-3}$ to $0.95\text{ V}$ at $10^{18}\text{ cm}^{-3}$; thick-oxide curves leave the plotted window ($V_T>3\text{ V}$) 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 $\phi_F$:

\begin{aligned} \phi_F(T) &= \frac{kT}{q}\ln\left(\frac{N_A}{n_i(T)}\right) \ n_i(T) &= n_i(300)\left(\frac{T}{300}\right)^{3/2}\exp\left[-\frac{E_G}{2k}\left(\frac{1}{T}-\frac{1}{300}\right)\right] \end{aligned}

**Assumptions:** $E_G=1.12\text{ eV}$ independent of $T$, $n_i(300)=10^{10}\text{ cm}^{-3}$ (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, $N_A=10^{16}\text{ cm}^{-3}$ (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 $-3.2\text{ mV/K}$ (a total drop of roughly $0.64\text{ V}$ over $200\text{ K}$). 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: $V_{D\text{sat}}$ from the $I_D$ maximum **Problem.** If (17.17) is used beyond $V_{D\text{sat}}$, $I_D$ peaks at $V_{D\text{sat}}$. Use this to derive (17.21) from (17.17). **Solution.** A maximum requires $\partial I_D/\partial V_D=0$ with $\partial^2I_D/\partial V_D^2<0$. 1. First derivative condition:

\begin{aligned} \frac{\partial I_D}{\partial V_D} &= \frac{\partial}{\partial V_D}\left{\xi\left[(V_G-V_T)V_D - \frac{V_D^2}{2}\right]\right} \ &= \xi\left[(V_G-V_T)-V_D\right] = 0 \end{aligned}

2. Solve for $V_D$:

\boxed{V_D = V_G - V_T \equiv V_{D\text{sat}}}

$$3.Checkthesecondderivative:$$

\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: 1. Hole drift-diffusion current density:

J_P = q\mu_p p,\mathcal{E} - qD_P\nabla p ;\to; J_{Py} = q\mu_p p,\mathcal{E}_y = -q\mu_p p,\frac{d\phi}{dy} \qquad (17.7\text{ with }\mu_n\to\mu_p,\ n\to p)

$$2.Totalholechargeperunitarea:$$

Q_P(y) = +q\int_0^{x_c}p,dx > 0 \qquad (17.3)

$$3.Totaldraincurrent:$$

I_D = Z\int_0^{x_c}J_{Py},dx = -Z\bar{\mu}_p Q_P\frac{d\phi}{dy} \qquad (17.9)

$$4.Integratealongthechannel:$$

I_D L = -Z\bar{\mu}_p\int_0^{V_D}Q_P,d\phi \qquad (17.10,,17.11)

5. Charge balance: for $V_G\le V_T$, the added gate charge is negative and balanced by $+Q_P$:

\begin{aligned} Q_P &= -C_o(V_G-V_T) \qquad (17.14) \ Q_P(y) &= -C_o(V_G-V_T-\phi) \qquad (17.16) \end{aligned}

This is positive since $V_G-V_T-\phi<0$. 6. Integrate:

\begin{aligned} I_D &= \frac{Z\bar{\mu}_p C_o}{L}\int_0^{V_D}(V_G-V_T-\phi),d\phi \ &= \frac{Z\bar{\mu}p C_o}{L}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right] \qquad (V_D\ge V{D\text{sat}},\ V_G\le V_T) \qquad (17.17) \end{aligned}

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 ($|V_D|\le|V_{D\text{sat}}|$). 7. Pinch-off ($Q_P(L)=0$):

\boxed{V_{D\text{sat}} = V_G - V_T \ (<0)} \qquad (17.21)

\boxed{I_{D\text{sat}} = \frac{Z\bar{\mu}_p C_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 "$V_G\le V_T$", "$0\le V_D\le V_{D\text{sat}}$" becomes "$V_{D\text{sat}}\le V_D\le0$". --- ## 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 $\bar{\mu}_n=\mu_0/[1+\theta(V_G-V_T)]$:

\begin{aligned} \frac{I_D}{Z\mu_0C_o/L} &= \frac{V_{GT}V_D-V_D^2/2}{1+\theta V_{GT}} \quad (V_D\le V_{GT}) \ \frac{I_{D\text{sat}}}{Z\mu_0C_o/L} &= \frac{V_{GT}^2/2}{1+\theta V_{GT}} \end{aligned}

For $\theta=0.05/\text{V}$ and $V_{GT}=1,2,3,4\text{ V}$, the saturation heights become: - $V_{GT}=1\text{ V}$: $0.5/1.05 = 0.476$ - $V_{GT}=2\text{ V}$: $2/1.1 = 1.82$ - $V_{GT}=3\text{ V}$: $4.5/1.15 = 3.91$ - $V_{GT}=4\text{ V}$: $8/1.2 = 6.67$ (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

$$Pinch-offcondition:$$

V_D \ge V_G - V_T \iff V_D \ge V_D + V_B - V_T \iff 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 + \frac{V_T}{2} \ge V_T \iff V_D \ge \frac{V_T}{2}

$$Therefore:$$

\boxed{I_D = \begin{cases} 0, & V_D < \dfrac{V_T}{2} \[10pt] \dfrac{\xi}{2}\left(V_D - \dfrac{V_T}{2}\right)^2, & V_D \ge \dfrac{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:

\begin{aligned} I_D &= \xi\left[(V_G-V_T)V_D - \frac{V_D^2}{2}\right] \ &= \xi\left[(V_D+2V_T-V_T)V_D - \frac{V_D^2}{2}\right] \ &= \xi\left[(V_D+V_T)V_D - \frac{V_D^2}{2}\right] \ &= \boxed{\xi\left[V_T V_D + \frac{V_D^2}{2}\right]} \end{aligned}

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 $x=V_G-V_T$. 1. For $x<0$: the device is off,

I_D = 0

2. For $0\le x\le V_D$: the device is **saturated** (since $V_D\ge x=V_{D\text{sat}}$),

I_D = \frac{\xi}{2}x

3. For $x\ge V_D$: the device is **below pinch-off**,

I_D = \xi\left(xV_D - \frac{V_D

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:

\left.\frac{d}{dx}\left(\frac{\xi}{2}x^2\right)\right|_{x=V_D} = \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 = \frac{2\pi\bar{\mu}_n C_o}{\ln(r_2/r_1)}\left[(V_G-V_T)V_D-\frac{V_D

(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 $Z\to 2\pi r$ (circumference at radius $r$). 1. Radial electron current density:

J_{Nr} = -q\mu_n n\frac{d\phi}{dr} \qquad (17.7\text{ in }r)

$$2.Totaldraincurrent:$$

I_D = -(2\pi r)\bar{\mu}_n Q_N(r)\frac{d\phi}{dr} \qquad (17.9\text{ with }Z\to 2\pi r)

3. Since $I_D$ is independent of $r$, separate variables:

I_D\frac{dr}{r} = -2\pi\bar{\mu}_n Q_N,d\phi

4. Integrate from $r_1$ to $r_2$:

I_D\int_{r_1}^{r_2}\frac{dr}{r} = -2\pi\bar{\mu}_n\int_0

5. Substitute $Q_N=-C_o(V_G-V_T-\phi)$:

\begin{aligned} I_D\ln\left(\frac{r_2}{r_1}\right) &= 2\pi\bar{\mu}_n C_o\int_0^{V_D}(V_G-V_T-\phi),d\phi \ &= 2\pi\bar{\mu}_n C_o\left[(V_G-V_T)V_D - \frac{V_D^2}{2}\right] \end{aligned}

6. Solve for $I_D$:

\boxed{I_D = \frac{2\pi\bar{\mu}_n C_o}{\ln(r_2/r_1)}\left[(V_G-V_T)V_D-\frac{V_D

**(b)** With $r_2=r_1+L$:

\ln\left(\frac{r_2}{r_1}\right) = \ln\left(1+\frac{L}{r_1}\right) \approx \frac{L}{r_1} \quad \text{for } \frac{L}{r_1}\ll1

using the series expansion $\ln(1+u)\approx u$. Then:

\frac{2\pi}{\ln(r_2/r_1)} \approx \frac{2\pi r_1}{L} = \frac{Z}{L}

$$Substitutingthisyields:$$

\boxed{I_D = \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G-V_T)V_D-\frac{V_D

which is identical to (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 $I_{D\text{sat}}(T)/I_{D\text{sat}}(300\text{ K})$ 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).** 1. Saturation current at temperature $T$:

I_{D\text{sat}}(T) = \frac{Z\bar{\mu}_n(T)C_o}{2L},[V_G-V_T(T)]

2. Saturation current at $300\text{ K}$:

I_{D\text{sat}}(300) = \frac{Z\bar{\mu}_n(300)C_o}{2L},[V_G-V_T(300)]

3. Since $Z$, $L$, and $C_o$ do not depend on $T$, taking the ratio gives:

\boxed{\frac{I_{D\text{sat}}(T)}{I_{D\text{sat}}(300)} = \frac{\mu_n(T)}{\mu_n(300)}\left[\frac{V_G-V_T(T)}{V_G-V_T(300)}\right]

**(b)** Assumption: $\mu_n(T)\propto T^{-3/2}$ (lattice scattering), so:

\frac{\mu_n(T)}{\mu_n(300)} = \left(\frac{T}{300}\right)

$V_T(T)$ is obtained from the Problem 17.5 program ($V_T=2.43,\,2.29,\,2.13,\,1.96,\,1.79\text{ V}$ at $T=200,250,300,350,400\text{ K}$). ```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) | $\mu_n(T)/\mu_n(300)$ | $I_{D\text{sat}}$ ratio, $V_G=3$ V | $I_{D\text{sat}}$ 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 ($0.87\text{ V}$ at $300\text{ K}$), so a given change in $V_T$ is a large *fractional* change in overdrive: the $V_T$ effect wins and $I_{D\text{sat}}$ **increases** with $T$. At $V_G=10$ V the overdrive ($7.9\text{ V}$) 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 $I_{D\text{sat}}$ 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 ($y=0$): $\phi(0) = \phi_0 = I_D R_S$ - At the intrinsic drain ($y=L$): $\phi(L) = \phi_L = V_D - I_D R_D$ The gate-to-channel voltage is $V_G-\phi$ as before, so $Q_N=-C_o(V_G-V_T-\phi)$ still holds. 1. Integrate (17.11) between the new boundaries:

\begin{aligned} I_D &= \frac{Z\bar{\mu}n C_o}{L}\int{I_D R_S}^{V_D - I_D R_D}(V_G-V_T-\phi),d\phi \ &= \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G-V_T)(\phi_L-\phi_0) - \frac{\phi_L^2-\phi_0^2}{2}\right] \end{aligned}

$$2.Definetheintrinsicdrain-to-sourcevoltage:$$

V_D’ \equiv \phi_L - \phi_0 = V_D - I_D(R_S+R_D)

$$3.Notethat:$$

\phi_L + \phi_0 = V_D’ + 2I_D R_S

$$Therefore:$$

\begin{aligned} \frac{\phi_L^2-\phi_0^2}{2} &= \frac{(\phi_L-\phi_0)(\phi_L+\phi_0)}{2} \ &= \frac{V_D’(V_D’ + 2I_D R_S)}{2} \ &= V_D’\left(\frac{V_D’}{2} + I_D R_S\right) \end{aligned}

4. Substitute back into the expression for $I_D$:

\begin{aligned} I_D &= \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G-V_T)V_D’ - V_D’\left(\frac{V_D’}{2} + I_D R_S\right)\right] \ &= \frac{Z\bar{\mu}_n C_o}{L}V_D’\left[(V_G-V_T) - I_D R_S - \frac{V_D’}{2}\right] \end{aligned}

5. Define $V_G' \equiv V_G - I_D R_S$:

\boxed{I_D = \frac{Z\bar{\mu}_n C_o}{L}\left[(V_G’-V_T)V_D’ - \frac{V_D’

which is identical to (17.17) under $V_D\to V_D'$ and $V_G\to V_G'$. 6. Pinch-off: $Q_N(L)=0$ implies $\phi_L = V_G - V_T$, which can be written:

V_D’ + I_D R_S = V_G - V_T \iff \boxed{V_{D\text{sat}}’ = V_G’ - V_T} \qquad (17.21)

$$and:$$

\boxed{I_{D\text{sat}} = \frac{Z\bar{\mu}_n C_o}{2L}(V_G’-V_T)^2} \qquad (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 $V_{D\text{sat}}$ for the bulk-charge theory. **Solution.** Pinch-off occurs when $Q_N(L)=0$ with $\phi(L)=V_{D\text{sat}}$ in (17.27):

V_G - V_T - V_{D\text{sat}} - V_W\left[\sqrt{1+\frac{V_{D\text{sat}}}{2\phi_F}}-1\right] = 0

$$1.Introducethesubstitution:$$

s = \sqrt{1+\frac{V_{D\text{sat}}}{2\phi_F}} \implies V_{D\text{sat}} = 2\phi_F(s

2. Substitute $V_{D\text{sat}}$ into the pinch-off condition:

\begin{aligned} V_G - V_T - 2\phi_F(s^2-1) - V_W(s-1) &= 0 \ V_G - V_T - 2\phi_F s^2 + 2\phi_F - V_W s + V_W &= 0 \ 2\phi_F s^2 + V_W s - (V_G - V_T + V_W + 2\phi_F) &= 0 \end{aligned}

3. Using the ideal threshold expression $V_T=2\phi_F+V_W$:

V_G - V_T + V_W + 2\phi_F = V_G - (2\phi_F+V_W) + V_W + 2\phi_F = V_G

$$Theequationsimplifiesto:$$

2\phi_F s^2 + V_W s - V_G = 0

4. Solve for the positive root $s$:

s = \frac{-V_W + \sqrt{V_W^2 + 8\phi_F V_G}}{4\phi_F}

5. From the quadratic equation, $2\phi_F s^2 = V_G - V_W s$. Substitute this into $V_{D\text{sat}} = 2\phi_F s^2 - 2\phi_F$:

\begin{aligned} V_{D\text{sat}} &= V_G - V_W s - 2\phi_F \ &= V_G - 2\phi_F - V_W\left(\frac{-V_W + \sqrt{V_W^2 + 8\phi_F V_G}}{4\phi_F}\right) \ &= \boxed{V_{D\text{sat}} = V_G - 2\phi_F + \frac{V_W^2}{4\phi_F} - \frac{V_W}{4\phi_F}\sqrt{V_W^2 + 8\phi_F V_G}} \end{aligned}

Check: $V_G=3\text{ V}$, $\phi_F=0.358\text{ V}$, $V_W=0.71\text{ V}$ gives $1.14\text{ V}$ (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 $V_{D\text{sat}}\to V_G-2\phi_F-\ldots\to V_G-V_T$, the square-law result. --- ## Problem 17.15: Bulk-charge $I_D$ΓÇô$V_D$ 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 $V_{D\text{sat}}$ from (17.29); evaluate (17.28) (normalized by $Z\bar{\mu}_nC_o/L$) on $0\le V_D\le V_{D\text{sat}}$; append a flat tail at $I_{D\text{sat}}$ 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 $V_{D\text{sat}}$); the $N_A=10^{14}\text{ cm}^{-3}$ curves nearly coincide with the square law (small $V_W$), while the $10^{16}\text{ cm}^{-3}$ curves (for $x_o=0.1\ \mu$m: $V_W\approx 1.42\text{ V}$, versus $V_W\approx 0.12\text{ V}$ at $10^{14}$) are noticeably lower, e.g. at $V_{GT}=4\text{ V}$ the saturation level is clearly under 8, as in Fig. 17.9. --- ## Problem 17.16: Charge-sheet and exact-charge $I_D$ΓÇô$V_D$ **Problem.** Using the Appendix D relationships, write programs for $I_D$ΓÇô$V_D$ 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

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

$$-\ast Charge-sheet\ast (neglecttheholetermsbelowthesheet;electronstreatedasasheetatthesurface):$$

F^2 \approx \psi_s + V_t,e

**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

(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), $0.60$ΓÇô$0.68$ mA (3 V), $1.5$ΓÇô$2.7$ 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)

**$g_d$ below pinch-off.**

\begin{aligned} 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} \end{aligned}

Since $\frac{4\phi_F}{3}\cdot\frac32\cdot\frac1{2\phi_F}=1$:

\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]}

**$g_m$ below pinch-off.** $V_G$ appears only in the term $(V_G-V_T+V_W)V_D$:

\begin{aligned} g_m &= \left.\frac{\partial I_D}{\partial V_G}\right|_{V_D} \ &= \boxed{\xi V_D} \end{aligned}

**Above pinch-off.** $I_D=I_{D\text{sat}}(V_G)=I_D(V_D=V_{D\text{sat}}(V_G),V_G)$ is independent of $V_D$, so:

\boxed{g_d = 0}

$$Bythechainrule:$$

\begin{aligned} g_m &= \frac{dI_{D\text{sat}}}{dV_G} \ &= \left.\frac{\partial I_D}{\partial V_G}\right|{V_D=V{D\text{sat}}} + \underbrace{\left.\frac{\partial I_D}{\partial V_D}\right|{V_D=V{D\text{sat}}}}{=,g_d(V{D\text{sat}})=0}\frac{dV_{D\text{sat}}}{dV_G} \ &= \boxed{\xi V_{D\text{sat}}} \end{aligned}

The second term vanishes because $g_d(V_{D\text{sat}})=-Z\bar{\mu}_nQ_N(L)/L=0$ 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.** Compute $C_o$:

\begin{aligned} C_o &= \frac{K_O\epsilon_0}{x_o} \ &= \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})}{5\times10^{-6}\text{ cm}} \ &= 6.90\times10^{-8}\text{ F/cm}^2 \end{aligned}

**(a)** With $V_D=0$ the channel potential is zero everywhere, $\phi(L/2)=0$:

\begin{aligned} Q_N(L/2) &= -C_o(V_G-V_T-\phi) \ &= -(6.90\times10^{-8}\text{ F/cm}^2)(3\text{ V}-1\text{ V}-0) \ &= -(6.90\times10^{-8})(2) \ &= \boxed{-1.38\times10^{-7}\text{ C/cm}^2} \end{aligned}

(about $8.6\times10^{11}$ electrons/cm$^2$). **(b)** Drain conductance at $V_D=0$:

\begin{aligned} g_d &= \frac{Z\bar{\mu}_n C_o}{L}(V_G-V_T-V_D) \ &= \left(\frac{70\ \mu\text{m}}{7\ \mu\text{m}}\right)(550\text{ cm}^2/\text{V}\cdot\text{s})(6.90\times10^{-8}\text{ F/cm}^2)(3\text{ V}-1\text{ V}-0) \ &= (10)(550)(6.90\times10^{-8})(2) \ &= 7.59\times10^{-4}\text{ S} \ &\approx \boxed{0.76\text{ mS}} \end{aligned}

--- ## 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 are $\bar{\mu}_n=675$ and $\bar{\mu}_p=240\text{ cm}^2/\text{V}\cdot\text{s}$). 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 $|V_G-V_T|$): 1. Equate transconductances:

\left(\frac{Z}{L}\right)_p\bar{\mu}_p = \left(\frac{Z}{L}\right)_n\bar{\mu}_n

$$2.Solvefortheaspectratio:$$

\begin{aligned} \left(\frac{Z}{L}\right)_p &= \left(\frac{Z}{L}\right)_n\frac{\mu_n^{\text{bulk}}/2}{\mu_p^{\text{bulk}}/2} \ &= \left(\frac{50}{5}\right)\frac{1350}{480} \ &= 10\cdot 2.8125 \ &= \boxed{28.1} \end{aligned}

3. Device dimensions: Choose the same (minimum) length $L_p=5\ \mu$m; then:

Z_p = 28.1\times 5\ \mu\text{m} = \boxed{140.6\ \mu\text{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 $I_D$ΓÇô$V_D$ with $I_{D\text{sat}}=10^{-3}$ A, $V_{D\text{sat}}=5$ 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: $V_{D\text{sat}}=V_G-V_T=5$ V and $I_{D\text{sat}}=\frac{\xi}{2}(5)^2=10^{-3}$ A:

\xi = \frac{2I_{D\text{sat}}}{(V_G-V_T)^2} = \frac{2\times10^{-3}\text{ A}}{(5\text{ V})^2} = \frac{2\times10^{-3}}{25} = 8\times10^{-5}\text{ A/V}

**(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)** Given $V_T=1\text{ V}$:

\begin{aligned} V_G &= V_T + V_{D\text{sat}} \ &= 1 + 5 \ &= \boxed{6\text{ V}} \end{aligned}

**(c)** $x_o=0.1\ \mu$m:

C_o = \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})}{10^{-5}\text{ cm}} = 3.45\times10^{-8}\text{ F/cm}

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:

\begin{aligned} Q_N(L) &= -(3.45\times10^{-8}\text{ F/cm}^2)(5\text{ V}-2.5\text{ V}) \ &= -(3.45\times10^{-8})(2.5) \ &= \boxed{-8.6\times10^{-8}\text{ C/cm}^2} \end{aligned}

(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 $V_{D\text{sat}}=3$ V. Since $V_D=4\text{ V} > V_{D\text{sat}}$, the device is saturated:

\begin{aligned} I_D &= \frac{\xi}{2}(V_G-V_T)^2 \ &= \frac{8\times10^{-5}\text{ A/V}^2}{2}(3\text{ V})^2 \ &= (4\times10^{-5})(9) \ &= \boxed{3.6\times10^{-4}\text{ A}} \end{aligned}

**(e)** At the assumed saturated point (3), $V_D > V_{D\text{sat}}$:

\boxed{g_d = 0}

(If (3) were below pinch-off at $V_D$: $g_d=\xi(5-V_D)$.) **(f)** Saturated:

\begin{aligned} g_m &= \xi V_{D\text{sat}} \ &= (8\times10^{-5}\text{ A/V}^2)(5\text{ V}) \ &= 4\times10^{-4}\text{ S} \ &= \boxed{0.4\text{ mS}} \end{aligned}

(equivalently $\frac{2I_{D\text{sat}}}{V_G-V_T} = \frac{2\times 10^{-3}}{5} = 0.4\text{ mS}$. 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} = \frac{C_o C_{\text{dep}}}{C_o + C_{\text{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 $V_D=V_{D\text{sat}}=V_G-V_T$ (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}} \tag{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} \tag{18.2}

$$where:$$

\phi_{MS} = \frac{\Phi_M’-\chi’}{q}-\frac{(E_C-E_F)_{FB}}{q} \tag{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}} \tag{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)

\begin{aligned} \text{n-type: } & (E_C-E_F){FB} \simeq \frac{E_G}{2}-kT\ln\left(\frac{N_D}{n_i}\right) \ \text{p-type: } & (E_C-E_F){FB} \simeq \frac{E_G}{2}+kT\ln\left(\frac{N_A}{n_i}\right) \end{aligned}

With $E_G/2=0.56\text{ eV}$:

\begin{aligned} N_D=10^{14}\text{ cm}^{-3}: &\quad 0.56-0.0259\ln(10^4) = 0.56-0.24 = 0.32\text{ eV} \ N_D=10^{18}\text{ cm}^{-3}: &\quad 0.56-0.0259\ln(10^8) = 0.56-0.48 = 0.08\text{ eV} \ N_A=10^{14}\text{ cm}^{-3}: &\quad 0.56+0.24 = 0.80\text{ eV} \ N_A=10^{18}\text{ cm}^{-3}: &\quad 0.56+0.48 = 1.04\text{ eV} \end{aligned}

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:

\begin{aligned} N_A &= n_i e^{(E_i-E_F)/kT} \ &= 10^{10}e^{0.26/0.0259} \ &= \boxed{2.29\times10^{14}\ \text{cm}^{-3}} \end{aligned}

**Answer: Au gate on p-Si with $N_A=2.29\times10^{14}\ \text{cm}^{-3}$.** --- ## 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} \tag{18.5}

**Step 2: integrate from $x$ to $x_o$.**

\begin{aligned} \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’ \tag{18.6} \end{aligned}

**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 \tag{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

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 \tag{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 \tag{18.9}

$$Fortheidealdevice(nooxidecharge):$$

V_G’ = \phi_s+\frac{K_s}{K_O}\mathscr{E}_sx_o \tag{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} \tag{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 \tag{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 \tag{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.** 1. After $+$BT, $\rho_{ion}=Q_M\delta(x-x_o)$. By (18.13):

\begin{aligned} \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} \end{aligned}

2. After $-$BT, $\rho_{ion}=Q_M\delta(x)$:

\begin{aligned} \Delta V_G(-BT) &= -\frac1{K_O\varepsilon_0}\int_0^{x_o}xQ_M\delta(x)dx \ &= 0 \end{aligned}

$$3.Allothernonidealitiesareunchangedbystress,sothedisplacementbetweenthecurvesisthedifference:$$

\Delta V_G(+BT)-\Delta V_G(-BT) = -10\text{ V} = -\frac{Q_M}{C_O}

4. Solve for $Q_M/q$:

\begin{aligned} \frac{Q_M}{q} &= \frac{C_O}{q}(10) \ &= \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})(10\text{ V})}{(1.6\times10^{-19}\text{ C})(10^{-5}\text{ cm})} \ &= \frac{3.4515\times10^{-12}}{1.6\times10^{-24}} \ &= \boxed{2.16\times10^{12}\ \text{cm}^{-2}} \end{aligned}

**Answer: $Q_M/q\approx2.16\times10^{12}\ \text{cm}^{-2}$.** --- ### 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) \tag{18.14}

$$Insertingin(18.11):$$

\Delta V_G(\text{fixed charge}) = -\frac{x_oQ_F}{K_O\varepsilon_0} = -\frac{Q_F}{C_O} \tag{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_{\text{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} \tag{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}

With $x_o=0.1\ \mu$m ($C_O=3.45\times10^{-8}$ F/cm$^2$):

\Delta V_G = \frac{(1.6\times10^{-19}\text{ C})(6.8\times10^{11}\text{ cm}^{-2})}{3.45\times10^{-8}\text{ F/cm}^2} = 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}} \tag{18.17}

$$withthedimensionlessmobile-ion\ast \ast centroid\ast \ast :$$

\gamma_M \equiv \frac{\int_0^{x_o}x\rho_{ion}(x)dx}{x_o\int_0^{x_o}\rho_{ion}(x)dx} \tag{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}

$$1.Oxidecapacitanceperunitarea:$$

\begin{aligned} C_O &= \frac{K_O\varepsilon_0}{x_o} \ &= \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})}{10^{-5}\text{ cm}} \ &= 3.45\times10^{-8}\ \text{F/cm}^2 \end{aligned}

2. *Term 1 ($\phi_{MS}$):* Al gate, $N_A=10^{15}\text{ cm}^{-3}$: Fig. 18.3 gives:

\phi_{MS} = -0.90\text{ V}

3. *Term 2 ($Q_F/C_O$):* 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}

\begin{aligned} \frac{Q_F}{C_O} &= \frac{(1.6\times10^{-19}\text{ C})(6.67\times10^{10}\text{ cm}^{-2})}{3.45\times10^{-8}\text{ F/cm}^2} \ &= 0.31\text{ V} \end{aligned}

4. *Term 3 ($\gamma_M Q_M/C_O$):* 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

\begin{aligned} \frac{\gamma_MQ_M}{C_O} &= \frac{(1.6\times10^{-19}\text{ C})(2\times10^{11}\text{ cm}^{-2})(0.05)}{3.45\times10^{-8}\text{ F/cm}^2} \ &= 0.046\text{ V} \end{aligned}

5. *Term 4 ($Q_{IT}(0)/C_O$):* 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:

\begin{aligned} \Delta E &= E_F-E_V \ &= \frac{E_G}2-(E_i-E_F) \ &= \frac{E_G}2-kT\ln\left(\frac{N_A}{n_i}\right) \ &= 0.56-0.0259\ln(10^5) \ &= 0.26\text{ eV} \end{aligned}

Q_{IT}(0) = -qD_{it}\Delta E

\begin{aligned} -\frac{Q_{IT}(0)}{C_O} &= \frac{(1.6\times10^{-19}\text{ C})(2\times10^{10}\text{ cm}^{-2}\text{eV}^{-1})(0.26\text{ eV})}{3.45\times10^{-8}\text{ F/cm}^2} \ &= +0.024\text{ V} \end{aligned}

$$6.\ast Sum:\ast$$

\begin{aligned} V_{FB} &= -0.90 - 0.31 - 0.046 + 0.024 \ &= \boxed{-1.23\text{ V}} \end{aligned}

**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} \tag{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} \tag{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}} \tag{18.21}

$$where(repeatingEq.17.1;uppersignn-channel,lowersignp-channel):$$

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 \tag{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{\frac{2K_s\varepsilon_0(2\phi_F)}{qN_A}} = \sqrt{\frac{4K_s\varepsilon_0\phi_F}{qN_A}}

$$thedepletionchargeis:$$

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). 1. Work-function and fixed charge shifts:

\begin{aligned} \phi_{MS} &= -0.90\text{ V} \ -\frac{Q_F}{C_O} &= -\frac{(1.6\times10^{-19}\text{ C})(2\times10^{11}\text{ cm}^{-2})}{3.45\times10^{-8}\text{ F/cm}^2} = -0.93\text{ V} \end{aligned}

V_{FB} = -0.90 - 0.93 = -1.83\text{ V}

$$2.Idealthresholdvoltage:$$

\phi_F = 0.0259\ln(10^5) = 0.298\text{ V}

\begin{aligned} 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}} \ &\approx 1.00\text{ V} \end{aligned}

$$3.Realthresholdvoltage:$$

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:

\begin{aligned} V_{FB} &= -0.30 - 0.93 = -1.23\text{ V} \ V_T’ &= -1.00\text{ V} \ V_T &= -1.00 - 1.23 = -2.23\text{ V} \end{aligned}

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} \tag{18.23}

\phi_{MS} = \frac1q\Big[(E_C-E_F){poly\text{-}Si}-(E_C-E_F){FB,,crystalline\text{-}Si}\Big] \tag{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} \implies \phi{MS} = 1.12 - 0.858 = +0.262\text{ V}

\frac{Q_F}{C_O} = 0.31\text{ V} \implies 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} \tag{18.25}

Example: $N_I=5\times10^{11}$ boron ions/cm$^2$, $x_o=0.1\ \mu$m:

\begin{aligned} \Delta V_G &= +\frac{qN_I}{C_O} \ &= \frac{(1.6\times10^{-19}\text{ C})(5\times10^{11}\text{ cm}^{-2})}{3.45\times10^{-8}\text{ F/cm}^2} \ &= +2.32\text{ V} \end{aligned}

### 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{\frac{2K_s\varepsilon_0(2\phi_F-V_{BS})}{qN_A}}

$$sothedepletionchargeis:$$

qN_AW = \sqrt{2K_s\varepsilon_0qN_A(2\phi_F-V_{BS})}

$$Thegate-to-\ast \ast bulk\ast \ast thresholdvoltageisthesurfacepotentialplustheoxidedrop:$$

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})} \tag{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})} \tag{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 \frac{K_s}{K_O}x_o\sqrt{\frac{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}\[6pt]\phi_F<0,;V_{BS}>0&\text{p-channel}\end{cases} \tag{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:

\boxed{V_T \approx -1\text{ V}}

(b) Since $V_T=-1\text{ V} < 0$, 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:

\begin{aligned} V_G = 0\text{ V}: &\quad V_{D\text{sat}} = 0 - (-1) = 1\text{ V} \ V_G = 1\text{ V}: &\quad V_{D\text{sat}} = 1 - (-1) = 2\text{ V} \ V_G = 2\text{ V}: &\quad V_{D\text{sat}} = 2 - (-1) = 3\text{ V} \end{aligned}

(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} \implies 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}} \implies V_{BS} = 2\phi_F\left[1-\left(1+\frac{\Delta V_T}{V_2}\right)

$$1.Calculatetherequiredthresholdshift:$$

\Delta V_T = 1 - (-1) = 2\text{ V}

2. Calculate $V_2$ from $V_T=V_T'+V_{FB}$:

V_T’ = V_T - V_{FB} = -1.0 - (-2.2) = 1.2\text{ V}

V_2 = V_T’ - 2\phi_F = 1.2 - 2(0.3) = 1.2 - 0.6 = 0.6\text{ V}

3. Compute $V_{BS}$:

\begin{aligned} V_{BS} &= 0.6\left[1-\left(1+\frac{2}{0.6}\right)^2\right] \ &= 0.6\left[1-\left(1+3.333\right)^2\right] \ &= 0.6\left[1-(4.333)^2\right] \ &= 0.6[1-18.78] \ &= 0.6[-17.78] \ &= \boxed{-10.7\text{ V}} \end{aligned}

**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

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} - \frac{Q_F}{C_O} - \gamma_M\frac{Q_M}{C_O} - \frac{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\left[\sqrt{1-\frac{V_{BS}}{2\phi_F}}-1\right]

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): 1. For the polysilicon gate:

(E_C-E_F)_{poly} = -0.2\text{ eV}

$$2.Forthecrystallinesiliconsubstrate:$$

\begin{aligned} (E_C-E_F)_{Si} &= (E_C-E_i) + (E_i-E_F) \ &= 0.56 + 0.2 \ &= 0.76\text{ eV} \end{aligned}

3. Calculate $\phi_{MS}$:

\begin{aligned} \phi_{MS} &= \frac{-0.2\text{ eV} - 0.76\text{ eV}}{q} \ &= \boxed{-0.96\text{ V}} \end{aligned}

**(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\text{ 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 - (-0.96) = +0.96\text{ V}

For a p-type substrate, positive $V_G-V_{FB}$ depletes and may invert. Determine the threshold:

N_A = n_i e^{(E_i-E_F)/kT} = 10^{10}e^{0.2/0.0259} = 2.25\times10^{13}\ \text{cm}

2\phi_F = 2(0.2) = 0.40\text{ V}

With an assumed $x_o=0.1\ \mu$m ($C_O=3.45\times10^{-8}$ F/cm$^2$):

\begin{aligned} 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} \end{aligned}

Since $0.96\text{ V} > 0.45\text{ V}$ (gate voltage is above threshold), 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):

\begin{aligned} \Delta V_G &= -\frac1{K_O\varepsilon_0}\int_0^{x_o}x\rho_0,dx \ &= -\frac{\rho_0 x_o^2}{2K_O\varepsilon_0} \end{aligned}

$$1.Calculatethetotalionicchargeperunitarea:$$

\begin{aligned} Q_M &= \rho_0 x_o \ &= (1.6\times10^{-19}\text{ C})(10^{16}\text{ cm}^{-3})(10^{-5}\text{ cm}) \ &= 1.6\times10^{-8}\text{ C/cm}^2 \quad (Q_M/q=10^{11}\text{ cm}^{-2}) \end{aligned}

$$2.Computetheresultinggatevoltageshift:$$

\begin{aligned} \Delta V_G &= -\frac{Q_M}{2C_O} \ &= -\frac{1.6\times10^{-8}\text{ C/cm}^2}{2(3.45\times10^{-8}\text{ F/cm}^2)} \ &= \boxed{-0.23\text{ V}} \end{aligned}

(for general $\rho_0$: $\Delta V_G=-0.23\text{ V}\times\dfrac{\rho_0/q}{10^{16}\text{ cm}^{-3}}$.) **(b)** All charge is concentrated at $x=x_o$:

\begin{aligned} \Delta V_G &= -\frac{Q_M}{C_O} \ &= -\frac{1.6\times10^{-8}\text{ C/cm}^2}{3.45\times10^{-8}\text{ F/cm}^2} \ &= \boxed{-0.46\text{ V}} \end{aligned}

(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)** For an ideal delta sheet at $x=x_o$:

\boxed{\Delta V_G = -\frac{Q_F}{C_O}}

**(b)** With $x$ measured from the metal, the charge density 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 area is $\tfrac12(2Q_F/\Delta x)\Delta x = Q_F$). Let $u = x - (x_o - \Delta x)$, so $0 \le u \le \Delta x$ and $x = x_o - \Delta x + u$:

\begin{aligned} \int_{x_o-\Delta x}^{x_o} x\rho(x),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) \end{aligned}

$$SubstituteintoEq.(18.11):$$

\begin{aligned} \Delta V_G &= -\frac{1}{K_O\varepsilon_0}Q_F\left(x_o-\frac{\Delta x}3\right) \ &= \boxed{-\frac{Q_F}{C_O}\left(1-\frac{\Delta x}{3x_o}\right)} \end{aligned}

(The centroid of the triangle sits $\Delta x/3$ from the interface.) **(c)** Ratio of (b) to (a):

\begin{aligned} \frac{\Delta V_G\text{(b)}}{\Delta V_G\text{(a)}} &= 1-\frac{\Delta x}{3x_o} \ &= 1-\frac{10^{-7}\text{ cm}}{3(10^{-5}\text{ cm})} \ &= 1-0.0033 \ &= \boxed{0.9967} \end{aligned}

So treating $Q_F$ as a $\delta$-sheet is excellent (error is only $0.33\%$). --- ### 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)** With $C_O=K_O\varepsilon_0/x_o$:

\begin{aligned} V_{FB} &= \phi_{MS} - \frac{Q_F}{C_O} \ &= \phi_{MS} - \frac{Q_F}{K_O\varepsilon_0}x_o \end{aligned}

This is 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 regression analysis: 1. Means of variables:

\begin{aligned} \bar{x} &= \frac{0.1+0.15+0.2+0.25+0.3}{5} = 0.20\ \mu\text{m} \ \bar{y} &= \frac{-0.91-1.04-1.20-1.33-1.52}{5} = -1.20\text{ V} \end{aligned}

$$2.Sumsofsquaresandcross-products:$$

\begin{aligned} S_{xx} &= \sum(x-\bar{x})^2 \ &= (-0.1)^2 + (-0.05)^2 + 0^2 + (0.05)^2 + (0.1)^2 \ &= 0.01 + 0.0025 + 0 + 0.0025 + 0.01 \ &= 0.025\ (\mu\text{m})^2 \end{aligned}

\begin{aligned} S_{xy} &= \sum(x-\bar{x})(y-\bar{y}) \ &= (-0.1)(0.29) + (-0.05)(0.16) + (0)(0) + (0.05)(-0.13) + (0.1)(-0.32) \ &= -0.029 - 0.008 - 0.0065 - 0.032 \ &= -0.0755\ \mu\text{m}\cdot\text{V} \end{aligned}

$$3.Slopeandintercept:$$

\begin{aligned} \text{slope} &= \frac{S_{xy}}{S_{xx}} = \frac{-0.0755}{0.025} = -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)(0.20) = -1.20 + 0.604 = -0.596\text{ V} \approx -0.60\text{ V} \end{aligned}

$$4.Extractedparameters:$$

\boxed{\phi_{MS} \approx -0.60\text{ V}}

\begin{aligned} Q_F &= -(-3.02\times10^{4}\text{ V/cm})(3.9)(8.85\times10^{-14}\text{ F/cm}) \ &= \boxed{1.04\times10^{-8}\text{ C/cm}^2} \end{aligned}

\begin{aligned} \frac{Q_F}{q} &= \frac{1.04\times10^{-8}\text{ C/cm}^2}{1.6\times10^{-19}\text{ C}} \ &= \boxed{6.5\times10^{10}\text{ cm}^{-2}} \end{aligned}

--- ### 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:

\frac{d^2E_C}{dx^2} \propto \rho_{ox}

Straight oxide bands (constant oxide field) mean $\rho_{ox}=0$, so:

\boxed{Q_M \approx 0}

(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:

\boxed{Q_F \ne 0}

--- ### 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 $\Delta V_G=V_{FB}-0=-0.71$ V: 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)** 1. Calculate oxide capacitance per unit area:

\begin{aligned} C_O &= \frac{C_{\max}}{A_G} \ &= \frac{200\times10^{-12}\text{ F}}{2.9\times10^{-3}\text{ cm}^2} \ &= 6.90\times10^{-8}\text{ F/cm}^2 \end{aligned}

\begin{aligned} x_o &= \frac{K_O\varepsilon_0}{C_O} \ &= \frac{3.45\times10^{-13}\text{ F/cm}}{6.90\times10^{-8}\text{ F/cm}^2} \ &= 5.0\times10^{-6}\text{ cm} = 500\text{ \AA} \end{aligned}

$$2.Delta-depletionrelation:$$

\frac{C_{\min}}{C_O} = \frac{1}{1+\dfrac{K_O}{K_s}\dfrac{W_T}{x_o}}

\begin{aligned} W_T &= \frac{K_s}{K_O}x_o\left(\frac{C_O}{C_{\min}}-1\right) \ &= \left(\frac{11.8}{3.9}\right)(5\times10^{-6}\text{ cm})\left(\frac{200}{67}-1\right) \ &= (3.026)(5\times10^{-6})(1.985) \ &= 3.0\times10^{-5}\text{ cm} \end{aligned}

3. Now $W_T=\sqrt{4K_s\varepsilon_0\phi_F/(qN)}$ with $\phi_F=(kT/q)\ln(N/n_i)$: - At $N=10^{16}\text{ cm}^{-3}$: $\phi_F=0.358\text{ V}$, $W_T=3.06\times10^{-5}$ cm. - At $N=1.05\times10^{16}\text{ cm}^{-3}$: $\phi_F=0.359\text{ V}$, $W_T=2.99\times10^{-5}$ cm. Thus:

N \approx 1.03\times10^{16}\text{ cm}^{-3} \approx 10^{16}\text{ cm}

**Which type?** $V_{FB}=\phi_{MS}-Q_F/C_O$ with $Q_F>0$. For an Al gate on p-Si at $10^{16}\text{ cm}^{-3}$, $\phi_{MS}\approx-0.03-(0.56+0.358)=-0.95$ V, which is already more negative than $-0.71$ V and would need $Q_F<0$, impossible. Therefore the substrate is **n-type**:

\boxed{N_D \approx 10^{16}\ \text{cm}

**(c)** For n-Si with $N_D=10^{16}\text{ cm}^{-3}$:

\begin{aligned} E_C-E_F &= 0.56-0.359 = 0.201\text{ eV} \ \phi_{MS} &= -0.03-0.201 = -0.23\text{ V} \end{aligned}

Now compute $Q_F$:

\begin{aligned} -\frac{Q_F}{C_O} &= V_{FB}-\phi_{MS} \ &= -0.71 - (-0.23) \ &= -0.48\text{ V} \end{aligned}

\begin{aligned} Q_F &= (0.48\text{ V})(6.90\times10^{-8}\text{ F/cm}^2) \ &= \boxed{3.3\times10^{-8}\ \text{C/cm}^2} \end{aligned}

\begin{aligned} \frac{Q_F}{q} &= \frac{3.3\times10^{-8}\text{ C/cm}^2}{1.6\times10^{-19}\text{ C}} \ &= \boxed{2.1\times10^{11}\ \text{cm}^{-2}} \end{aligned}

--- ### 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 $a=5.43\text{ \AA} = 5.43\times10^{-8}$ cm ($a^2=2.95\times10^{-15}$ cm$^2$). In the conventional fcc cell: 1. (100) face: area $a^2$, containing $4\times\tfrac14+1=2$ atoms:

\sigma_{100} = \frac{2}{a^2} = \frac{2}{2.95\times10^{-15}\text{ cm}^2} = 6.8\times10^{14}\ \text{cm}

2. (110) plane: rectangle $a\times\sqrt2a$, area $\sqrt2a^2$, containing $4\times\tfrac14+2\times\tfrac12=2$ atoms:

\sigma_{110} = \frac{2}{\sqrt2a^2} = \frac{\sqrt2}{a^2} = \frac{\sqrt{2}}{2.95\times10^{-15}\text{ cm}^2} = 4.8\times10^{14}\ \text{cm}

$$3.Ratio:$$

\frac{\sigma_{100}}{\sigma_{110}} = \frac{2/,a^2}{\sqrt{2}/,a^2} = \sqrt2 \approx \boxed{1.41}

**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)

$$Totalcapacitance(parallelcombination):$$

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 \frac{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 $V_{D\text{sat}}=V_G-V_T$ 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+\frac{\sqrt{4K_s\varepsilon_0qN_A\phi_F}}{C_O} \qquad (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.** Since the substrate is n-type, this is a p-channel MOSFET. Oxide capacitance per unit area:

\begin{aligned} C_O &= \frac{K_O\varepsilon_0}{x_o} \ &= \frac{(3.9)(8.85\times10^{-14}\text{ F/cm})}{5\times10^{-6}\text{ cm}} \ &= 6.90\times10^{-8}\ \text{F/cm}^2 \end{aligned}

$$\ast \ast (a)\ast \ast Calculateflat-bandvoltage:$$

V_{FB} = \phi_{MS}-\frac{Q_F}{C_O}-\frac{Q_I}{C_O}

$$1.Fixedchargevoltageterm:$$

\begin{aligned} \frac{Q_F}{C_O} &= \frac{(1.6\times10^{-19}\text{ C})(2\times10^{11}\text{ cm}^{-2})}{6.90\times10^{-8}\text{ F/cm}^2} \ &= 0.464\text{ V} \end{aligned}

$$2.Implantedchargevoltageterm:$$

\begin{aligned} \frac{Q_I}{C_O} &= \frac{(1.6\times10^{-19}\text{ C})(-4\times10^{11}\text{ cm}^{-2})}{6.90\times10^{-8}\text{ F/cm}^2} \ &= -0.928\text{ V} \end{aligned}

$$3.Combineterms:$$

\begin{aligned} V_{FB} &= -0.46 - 0.464 - (-0.928) \ &= -0.924 + 0.928 \ &= +0.004\text{ V} \ &\approx \boxed{0\text{ V}} \end{aligned}

$$\ast \ast (b)\ast \ast Idealp-channelthreshold:$$

|\phi_F| = 0.0259\ln\left(\frac{10^{15}}{10^{10}}\right) = 0.298\text{ V}

\begin{aligned} 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}\text{ F/cm})(1.6\times10^{-19}\text{ C})(10^{15}\text{ cm}^{-3})(0.298\text{ V})}}{6.90\times10^{-8}\text{ F/cm}^2}\right] \end{aligned}

$$Evaluatingthenumerator:$$

\sqrt{1.99\times10^{-16}} = 1.41\times10^{-8}\text{ C/cm}

\frac{1.41\times10^{-8}}{6.90\times10^{-8}} = 0.204\text{ V}

V_T’ = -(0.596+0.204) = -0.80\text{ V}

$$Therealthresholdvoltageis:$$

\begin{aligned} V_T &= V_T’+V_{FB} \ &= -0.80 + 0.00 \ &= \boxed{-0.80\text{ V}} \end{aligned}

**(c)** A p-channel device with $V_T=-0.80\text{ V}$ is off at $V_G=0$ and requires $V_G<-0.8\text{ V}$ to turn on: **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.** 1. Oxide capacitance per unit area for $x_o=100\text{ \AA}=10^{-6}\text{ cm}$:

\begin{aligned} C_O &= \begin{aligned} C_O &= \frac{K_O\varepsilon_0}{x_o} \ &= \frac{3.45\times10^{-13}\text{ F/cm}}{10^{-6}\text{ cm}} \ &= 3.45\times10^{-7}\text{ F/cm}^2 \end{aligned}

$$2.Idealthresholdvoltage:$$

\begin{aligned} \phi_F &= 0.0259\ln\left(\frac{10^{17}}{10^{10}}\right) = 0.4175\text{ V} \ 2\phi_F &= 0.835\text{ V} \end{aligned}

\begin{aligned} \frac{\sqrt{4K_s\varepsilon_0qN_A\phi_F}}{C_O} &= \frac{\sqrt{4(11.8)(8.85\times10^{-14}\text{ F/cm})(1.6\times10^{-19}\text{ C})(10^{17}\text{ cm}^{-3})(0.4175\text{ V})}}{3.45\times10^{-7}\text{ F/cm}^2} \ &= \frac{1.67\times10^{-7}}{3.45\times10^{-7}} \ &= 0.484\text{ V} \end{aligned}

V_T’ = 2\phi_F + \frac{\sqrt{4K_s\varepsilon_0qN_A\phi_F}}{C_O} = 0.835 + 0.484 = 1.319\text{ V}

$$3.Flat-bandvoltagewithoutimplant:ForAlonp-Si:$$

(E_C-E_F) = 0.56 + 0.4175 = 0.9775\text{ eV} \implies \phi_{MS} = -0.03 - 0.9775 = -1.01\text{ V}

\frac{Q_F}{C_O} = \frac{(1.6\times10^{-19}\text{ C})(10^{11}\text{ cm}^{-2})}{3.45\times10^{-7}\text{ F/cm}^2} = 0.046\text{ V}

V_{FB} = \phi_{MS} - \frac{Q_F}{C_O} = -1.0075 - 0.046 = -1.054\text{ V}

V_T(\text{no implant}) = V_T’ + V_{FB} = 1.319 - 1.054 = 0.265\text{ V}

4. Required implant dose $N_I$:

\Delta V_T = 0.5\text{ V} - 0.265\text{ V} = \frac{qN_I}{C_O} = 0.235\text{ V}

\begin{aligned} N_I &= \frac{0.235,C_O}{q} \ &= \frac{0.235(3.45\times10^{-7}\text{ F/cm}^2)}{1.6\times10^{-19}\text{ C}} \ &= \boxed{5.1\times10^{11}\text{ ions/cm}^2} \end{aligned}

**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 | 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:

\begin{aligned} V_G &= V_{ox} + \phi_s \ V_{ox} &= \frac{|Q_G|}{C_o} \ V_G &= -\frac{Q_s}{C_o} + \phi_s \end{aligned} \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. *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}

$$\ast Step3:definetheshift.\ast$$

\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:

\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 $ZLW$:

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 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}

$$\ast Step6:assemble.\ast Substitute(19.8)and(19.9)in(19.7):$$

\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 $r_j$ with the depletion edge a concentric arc of radius $r_j + W$ (taking $W_S = W_D = W$). The arc crosses depth $y = W$ 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 + 2r_j W} \implies \Delta = r_j\left[\sqrt{1 + \frac{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 $1 - \frac{L+L'}{2L} = \frac{\Delta}{L}$:

\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 $qN_A$). #### 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 $Q_{BL} = -qN_AW$:

\begin{aligned} \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} \end{aligned}

\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

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 $v_{\text{sat}} \approx 10^7$ 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. $I_{D\text{sat}}$ is significantly reduced, modified approximately to:

I_{D\text{sat}} \approx Z, C_o,(V_G - V_T),v_{\text{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 $v_{\text{sat}}$. 2. As Eq. (19.16) shows, $I_{D\text{sat}}$ 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 $I_{D\text{sat}}$ 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 $v_{\text{sat}}$: **velocity overshoot**. Average velocities up to 35% above $v_{\text{sat}}$ 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 | $I_{D\text{sat}}\approx ZC_o(V_G-V_T)v_{\text{sat}}$ | | Ballistic/overshoot | $L\lesssim\ell$; velocities up to $\sim$35% above $v_{\text{sat}}$ 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 $\text{Si}$-$\text{SiO}_2$ 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 $I_{D\text{sat}}$ steps) and $I_{D\text{sat}}$ 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 $v_{\text{sat}}$. **(k) A popular approach to minimize oxide charging.** The lightly doped drain (LDD) structure. **(l) What is a pseudomorphic film?** A sufficiently thin (typically $\le 1000\ \text{Å}$) epitaxial film of a lattice-mismatched alloy (e.g. $\text{SiGe}$ on $\text{Si}$) 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 $\text{SiGe}$ base layer of $\text{Si}$ heterojunction bipolar transistors, the strained-$\text{Si}$ surface channel of the enhanced-mobility MOSFET, and the $\text{InGaAs}$ channel layer of the PHEMT. **(n) Briefly summarize SIMOX.** Implant oxygen ($150\text{--}200\ \text{keV}$, dose $1\text{--}2\times10^{18}\ \text{cm}^{-2}$) below the $\text{Si}$ surface; anneal (about $1300^\circ\text{C}$, $6\ \text{h}$) so the oxygen reacts to form a buried $\text{SiO}_2$ layer and the surface $\text{Si}$ recovers; optionally add epitaxial $\text{Si}$ 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: $N_A = 8\times10^{16}\ \text{cm}^{-3}$ (the OCR lost the exponent; $10^{16}$ is the value consistent with a $280\ \text{Å}$ oxide device and the plotted $V_T$ range), $x_o = 0.028\ \mu\text{m} = 280\ \text{Å}$, $r_j = 1\ \mu\text{m}$. (a) Use Eq. (19.1) to compute the expected $L_{\min}$ for $V_D = 0.125\ \text{V}$ (assume $n^+$-$p$ step junctions, source and back grounded; comment). (b) Use Eq. (19.13) to find $\Delta V_T$ at $L = 1\ \mu\text{m}$, $V_D = 0.125\ \text{V}$. (c) (The OCR dropped this part; it asks whether Eq. (19.13) can explain the $V_D = 4\ \text{V}$ data of Fig. 19.3.) **Assumptions.** $n_i = 10^{10}\ \text{cm}^{-3}$ at $300\ \text{K}$, $K_S = 11.8$, $K_O = 3.9$, $\epsilon_0 = 8.85\times10^{-14}\ \text{F/cm}$, $n^+$ doping $N_D = 10^{20}\ \text{cm}^{-3}$ (so only $N_A$ matters in $W$). --- **Solution (a).** 1. Built-in potential $V_{bi}$:

\begin{aligned} V_{bi} &= \frac{kT}{q}\ln\left(\frac{N_A N_D}{n_i^2}\right) && \text{(expression for built-in potential)} \ &= 0.0259 \ln\left(\frac{(8\times10^{16})(10^{20})}{(10^{10})^2}\right) && \text{(substituting values with } n_i^2 = 10^{20}\ \text{cm}^{-6}\text{)} \ &= 0.0259 \ln(8\times10^{16}) && \text{(simplifying the quotient)} \ &= 0.0259 \times 38.92 && \text{(evaluating the natural logarithm)} \ &= 1.01\ \text{V} && \text{(calculated built-in potential)} \end{aligned}

2. Source depletion width $W_S$ ($V_S = 0\ \text{V}$):

\begin{aligned} W_S &= \left[\frac{2K_S\epsilon_0 V_{bi}}{qN_A}\right]^{1/2} && \text{(one-sided step junction depletion width)} \ &= \left[\frac{2(11.8)(8.85\times10^{-14})(1.01)}{(1.6\times10^{-19})(8\times10^{16})}\right]^{1/2} && \text{(substituting dielectric and material parameters)} \ &= \left[1.65\times10^{-10}\right]^{1/2} && \text{(evaluating quotient inside radical)} \ &= 1.28\times10^{-5}\ \text{cm} && \text{(taking square root)} \ &= 0.128\ \mu\text{m} && \text{(converting to micrometers)} \end{aligned}

3. Drain depletion width $W_D$ under reverse bias $V_D = 0.125\ \text{V}$:

\begin{aligned} W_D &= W_S\sqrt{\frac{V_{bi}+V_D}{V_{bi}}} && \text{(depletion width scaled with applied reverse bias)} \ &= 0.128\sqrt{\frac{1.01 + 0.125}{1.01}} && \text{(substituting } V_{bi} \text{ and } V_D\text{)} \ &= 0.128\sqrt{\frac{1.135}{1.01}} && \text{(evaluating total junction potential)} \ &= 0.128\sqrt{1.1238} && \text{(evaluating ratio)} \ &= 0.136\ \mu\text{m} && \text{(calculated drain depletion width)} \end{aligned}

$$4.Sumofdepletionwidths:$$

\begin{aligned} W_S + W_D &= 0.128\ \mu\text{m} + 0.136\ \mu\text{m} \ &= 0.264\ \mu\text{m} \end{aligned}

5. Minimum channel length $L_{\min}$ using Eq. (19.1):

\begin{aligned} r_j x_o (W_S + W_D)^2 &= (1)(280)(0.264)^2 && \text{(substituting } r_j=1\ \mu\text{m},\ x_o=280\ \text{Å},\ W_S+W_D=0.264\ \mu\text{m}\text{)} \ &= (280)(0.0697) && \text{(squaring the depletion width sum)} \ &= 19.5 && \text{(evaluating product)} \end{aligned}

\begin{aligned} L_{\min} &= 0.4,(19.5)^{1/3} && \text{(empirical relation Eq. 19.1)} \ &= 0.4,(2.69) && \text{(evaluating cube root)} \ &= 1.08\ \mu\text{m} && \text{(final calculated value)} \end{aligned}

\boxed{L_{\min} = 1.08\ \mu\text{m}}

**Comment.** $L_{\min} \approx 1\ \mu\text{m}$. In Fig. 19.3 the $V_T$ curves are essentially flat for $L \gtrsim 3\text{--}4\ \mu\text{m}$ and fall off as $L$ approaches $\sim1\ \mu\text{m}$ and below, which agrees with $L_{\min}$ marking the onset of significant short-channel effects. --- **Solution (b).** 1. Inversion surface potential ($2\phi_F$):

\begin{aligned} 2\phi_F &= 2(0.0259)\ln\left(\frac{8\times10^{16}}{10^{10}}\right) && \text{(formula for surface potential at onset of strong inversion)} \ &= 2(0.0259)(22.80) && \text{(evaluating natural log of } N_A/n_i\text{)} \ &= 1.18\ \text{V} && \text{(evaluated surface potential)} \end{aligned}

2. Depletion width at threshold ($W_T$):

\begin{aligned} W_T &= \left[\frac{2K_S\epsilon_0(2\phi_F)}{qN_A}\right]^{1/2} && \text{(depletion width at threshold condition)} \ &= \left[\frac{2(11.8)(8.85\times10^{-14})(1.18)}{(1.6\times10^{-19})(8\times10^{16})}\right]^{1/2} && \text{(substituting constants and } 2\phi_F\text{)} \ &= \left[1.926\times10^{-10}\right]^{1/2} && \text{(evaluating quotient inside radical)} \ &= 1.39\times10^{-5}\ \text{cm} && \text{(taking square root)} \ &= 0.139\ \mu\text{m} && \text{(converting to micrometers)} \end{aligned}

3. Oxide capacitance per unit area ($C_o$):

\begin{aligned} C_o &= \frac{K_O\epsilon_0}{x_o} && \text{(definition of oxide capacitance per unit area)} \ &= \frac{(3.9)(8.85\times10^{-14})}{2.8\times10^{-6}} && \text{(substituting } x_o = 0.028\ \mu\text{m} = 2.8\times10^{-6}\ \text{cm}\text{)} \ &= 1.23\times10^{-7}\ \text{F/cm}^2 && \text{(calculated oxide capacitance)} \end{aligned}

$$4.Voltagescalingcoefficient:$$

\begin{aligned} \frac{qN_AW_T}{C_o} &= \frac{(1.6\times10^{-19})(8\times10^{16})(1.39\times10^{-5})}{1.23\times10^{-7}} && \text{(substituting charge and capacitance parameters)} \ &= \frac{1.779\times10^{-7}}{1.23\times10^{-7}} && \text{(evaluating bulk depletion charge per unit area)} \ &= 1.45\ \text{V} && \text{(resulting voltage factor)} \end{aligned}

$$5.Geometricfactorevaluation:$$

\begin{aligned} \frac{r_j}{L} &= \frac{1\ \mu\text{m}}{1\ \mu\text{m}} = 1 && \text{(ratio of junction depth to channel length)} \ \sqrt{1+\frac{2W_T}{r_j}} - 1 &= \sqrt{1+\frac{2(0.139)}{1}} - 1 && \text{(substituting } W_T \text{ and } r_j\text{)} \ &= \sqrt{1+0.278} - 1 && \text{(evaluating argument of square root)} \ &= \sqrt{1.278} - 1 && \text{(summing terms)} \ &= 1.130 - 1 && \text{(taking square root)} \ &= 0.130 && \text{(calculated geometric factor)} \end{aligned}

6. Threshold voltage reduction $\Delta V_T$ using Eq. (19.13):

\begin{aligned} \Delta V_T &= -\left(\frac{qN_AW_T}{C_o}\right)\left(\frac{r_j}{L}\right)\left[\sqrt{1+\frac{2W_T}{r_j}} - 1\right] && \text{(Yau charge-sharing formula)} \ &= -(1.45)(1)(0.130) && \text{(substituting computed components)} \ &\approx -0.19\ \text{V} && \text{(evaluated threshold voltage shift)} \end{aligned}

\boxed{\Delta V_T \approx -0.19\ \text{V}}

So $V_T$ is about $0.19\ \text{V}$ below its long-channel value at $L=1\ \mu\text{m}$. --- **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 $V_D=4\ \text{V}$ as for $0.125\ \text{V}$ and cannot explain the larger decrease seen in the $V_D = 4\ \text{V}$ 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 $\text{Si}_{0.7}\text{Ge}_{0.3}$ over a graded $\text{SiGe}$ buffer; tensile strain lowers scattering and effective mass, roughly doubling low-field mobility. **(e) SOI:** devices built in a thin single-crystal $\text{Si}$ 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 Schr├╢dinger 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}

$$Matchingtheorytoexperimentgives$$

h = 6.63\times10^{-34}\ \text{J}\cdot\text{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}

$$\ast \ast Derivationoftheorbitradiiandenergies.\ast \ast Step1.StableorbitmeanscentripetalforceequalsCoulombattraction:$$

\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}

$$Step3.Substituteinto(A.3):$$

\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

$$Step4.Cancelandsolvefortheradius:$$

r_n = \frac{4\pi\varepsilon_0 (n\hbar)^2}{m_0 q^2} \tag{A.4}

$$Step5.Kineticenergy,using(A.3):$$

\text{K.E.} = \tfrac12 m_0 v_n^2 = \frac12,\frac{q^2}{4\pi\varepsilon_0 r_n} \tag{A.5a}

$$Step6.Potentialenergy(zeroatinfinity):$$

\text{P.E.} = -\frac{q^2}{4\pi\varepsilon_0 r_n} \tag{A.5b}

$$Step7.Totalenergy(notethevirial-stylefactorofonehalf):$$

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 $E_1=-13.6$, $E_2=-3.40$, $E_3=-1.51$, $E_4=-0.85$, $E_5=-0.54\ \text{eV}$, 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}.

$$\ast \ast deBroglie(1925)\ast \ast flippedit.Ifwavescanbeparticles,particlesshouldbewaves:$$

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 $\text{Ni}$ 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 Schr├╢dinger (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 Schr├╢dinger'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 Schr├╢dinger 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.** Schr├╢dinger'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}

$$\ast \ast Derivationofthetime-independentequation.\ast \ast Step1.Substitute(A.12)into(A.9).Timederivative:$$

i\hbar,\frac{\partial\Psi}{\partial t} = E,\psi,e

Step 2. Cancel the common factor $e^{-iEt/\hbar}$:

-\frac{\hbar^2}{2m}\nabla^2\psi + U\psi = E\psi

$$Step3.Rearrange:$$

\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

\psi_{2,0,0} = \frac{1}{4\sqrt{2\pi a_0^3}}\left(2-\frac{r}{a_0}\right)e

\psi_{2,1,-1} = \frac{1}{8\sqrt{\pi a_0^3}},\frac{r}{a_0},e^{-r/2a_0},e

\psi_{2,1,0} = \frac{1}{4\sqrt{2\pi a_0^3}},\frac{r}{a_0},e

\psi_{2,1,1} = \frac{1}{8\sqrt{\pi a_0^3}},\frac{r}{a_0},e^{-r/2a_0},e

(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 $25.9\ \text{mV}$ at $300\ \text{K}$):

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_{\text{bulk}}+p_{\text{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 $n_{\text{bulk}}=p_{\text{bulk}}=n_i$:

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_{\text{bulk}}-n_{\text{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

$$Theintegralsare$$

\int_0^U (e^{-u}-1),du = -(e^{-U}-1+U),\qquad \int_0^U (e^{u}-1),du = e

$$so$$

\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

$$andintermsofthefield$$

\mathcal{E}^2 = \left(\frac{kT/q}{L_D}\right)^2\left[e^{U_F}(e^{-U}+U-1)+e^{-U_F}\left(e^{U}-U-1\right)\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}

$$Theintegrandisthefunction$$

\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}}

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 ($120\ \text{A/cm}^2\text{-K}^2$) | | $A^{*}$ | 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 ($\text{pF}$) | | $C_o'$ | oxide capacitance per unit area ($\text{pF/cm}^2$) | | $c_p$ | 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 ($\text{states/cm}^2\text{-eV}$) | | $D_N,\,D_P$ | electron, hole diffusion coefficient ($\text{cm}^2\text{/s}$) | | $D_{ox}$ | 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 ($\text{Hz}$) | | $f(E)$ | 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 $\text{cm}^3\text{-s}$) | | $g_m$ | 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 ($\text{A/cm}^2$) | | $J_{drift}$ | 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 ($8.617\times10^{-5}\ \text{eV/K}$) | | $\mathbf{K}$ | wavenumber (parameter proportional to the electron crystal momentum) | | K.E. | kinetic energy | | $K_O$ | oxide dielectric constant | | $K_S$ | semiconductor (usually $\text{Si}$) 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 ($\text{electrons/cm}^3$) | | $n^+$ | 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 $\text{atoms/cm}^3$ | | $N_A^-$ | number of ionized (negatively charged) $\text{acceptors/cm}^3$ | | $N_B$ | bulk semiconductor doping ($N_A$ or $N_D$ as appropriate); doping concentration in the BJT base | | $n_{\text{bulk}}$ | 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 $\text{atoms/cm}^3$ | | $N_D^+$ | number of ionized (positively charged) $\text{donors/cm}^3$ | | $N_E$ | doping concentration in the BJT emitter | | $n_{E0}$ | equilibrium electron concentration in the emitter of a pnp BJT | | $N_I$ | number of implanted $\text{ions/cm}^2$ | | $n_i$ | intrinsic carrier concentration | | $N_T$ | number of R-G $\text{centers/cm}^3$ | | $N_V$ | effective density of valence band states | | $p$ | hole concentration ($\text{holes/cm}^3$); 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 | | $p_{\text{bulk}}$ | hole concentration in the semiconductor bulk | | $p_s$ | hole concentration at the semiconductor surface ($\text{number/cm}^3$) | | $Q$ | general designation for a charge | | $q$ | magnitude of the electronic charge ($1.60\times10^{-19}\ \text{C}$) | | $Q_B$ | 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 $\text{charge/cm}^2$ 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 $\text{charge/cm}^2$ 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 | | $V_{D\text{sat}}$ | saturation drain voltage | | $v_{\text{sat}}$ | 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 | | $W_{\text{eff}}$ | 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 $\text{gate charge/cm}^2$ | | $\Delta Q_{semi}$ | change in the $\text{charge/cm}^2$ 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 ($8.85\times10^{-14}\ \text{F/cm}$) | | $\phi$ | 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 | | $\mu_{\text{bulk}}$ | carrier mobility in the bulk of a semiconductor | | $\mu_n$ | electron mobility | | $\mu_p$ | hole mobility | | $\mu_{n\,\text{eff}}$ | effective electron mobility | | $\mu_{p\,\text{eff}}$ | effective hole mobility | | $\nu$ | frequency of light | | $\rho$ | resistivity ($\Omega\text{-cm}$); charge density ($\text{C/cm}^3$) | | $\rho_{ion}$ | 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: $\text{cm}$, $\text{cm}^{-3}$, $\text{cm}^2\text{/V-s}$, $\text{F/cm}$, with energies in $\text{eV}$. --- ## 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 $\text{cm}^{-3}$. 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\left(\frac{N}{n_i}\right)

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

I_C = \alpha_F I_E - \left(1-\alpha_F\alpha_R\right)I_{R0}\left(e

$$(thesecondformisusedforthecommon-baseoutput).Withafinitebase,$$

f_B = \frac{D_B}{L_B},p_{B0}\coth\left(\frac{W}{L_B}\right),\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): $V_{CB0}=60\,(N_C/10^{16})^{-3/4}\ \text{V}$, $V_{CE0}=V_{CB0}(1-\alpha_F)^{1/m}$ 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 | $V_{CB}=0,-10,-20,\dots\ \text{V}$ | $I_E$ ($\text{mA}$) vs $V_{EB}$; shows how base-width modulation shifts the curves | | 2. Common-base output | $V_{CB}$ from $+2\ \text{V}$ down to $-V_{CB0}$ | $I_E$ stepped in $1\ \text{mA}$ | $I_C$ ($\text{mA}$) vs $-V_{CB}$; nearly flat lines (high $\alpha_F$) and slightly negative-$V_{CB}$ saturation region | | 3. Common-emitter input | $V_{EB}$ | $V_{EC}=0,5,10,\dots\ \text{V}$ | $I_B$ ($\mu\text{A}$) vs $V_{EB}$ | | 4. Common-emitter output | $V_{EC}$ from 0 to $V_{CE0}$ | $I_B$ stepped in $2.5\ \mu\text{A}$ | $I_C$ ($\text{mA}$) 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 $N_E=10^{18}$, $N_B=1.5\times10^{16}$, $N_C=1.5\times10^{15}\ (\text{cm}^{-3})$, 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_{\text{ref}})

with the electron and hole constants listed. $D=\mu kT/q$ (with $kT$ in $\text{eV}$, 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{\frac{2K_S\varepsilon_0}{q}\frac{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 $\text{eV}$). 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 $\text{DENOML} = e^{U_F}(1-e^{-U_s})+e^{-U_F}(e^{U_s}-1)$, then $C_L/C_o=1/[1+K_OW_L/(K_Sx_o)]$. 5. High-frequency (only for $U_s\ge3$): accumulate the integral $\int_0^{U_s}\text{(C.4)}$ 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 $E_n=-13.6/n^2\ \text{eV}$, de Broglie's $\lambda=h/p$, Schr├╢dinger's equation and its five postulates, hydrogen quantum numbers $(n,l,m,s)$, the Pauli principle, and the $\text{Si}$ 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.