ED 1

Semiconductor Device Fundamentals — Robert F. Pierret

A Journey Through the Physics and Engineering of Semiconductor Devices

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Part I: Semiconductor Fundamentals

Chapter 1: Semiconductors — A General Introduction

1.1 General Material Properties

1.1.1 Composition

Semiconductors are the class of materials from which the vast majority of solid-state devices are fabricated. The semiconductor family includes:

• Elemental semiconductors: Si, Ge
• Compound semiconductors: GaAs, ZnSe, SiC, etc.
• Alloys: Al${}_x$Ga${}_{1-x}$As, GaAs${}_{1-x}$P${}_x$, etc.

Referring to the Periodic Table, all semiconductors (except IV–VI compounds) are composed of elements appearing in Column IV or combinations of elements in columns equidistant to either side of Column IV. This is related to the chemical bonding in semiconductors, where, on average, there are four valence electrons per atom.

1.1.2 Purity

Extremely minute traces of impurity atoms called “dopants” can have a drastic effect on the electrical properties of semiconductors. Modern semiconductors are some of the purest solid materials in existence — unintentional dopant content in Si is routinely less than one atom per $10^9$ Si atoms.

Analogy: If a forest of maple trees was planted from coast to coast at 50-ft centers across the United States, an impurity level of one part per $10^9$ would correspond to finding about 25 crabapple trees in the entire forest!

1.1.3 Structure

Solids are classified into three broad categories based on atomic order:

Classification	Description
Amorphous	No recognizable long-range order
Polycrystalline	Composed of crystalline subsections that are disjointed or misoriented
Crystalline	Atoms arranged in an orderly three-dimensional array

Most devices employ crystalline semiconductors in their active regions.

1.2 Crystal Structure

1.2.1 The Unit Cell Concept

A unit cell is a small portion of any given crystal that could be used to reproduce the crystal. Unit cells are not necessarily unique, and they need not be primitive (the smallest unit cell possible).

1.2.2 Simple 3-D Unit Cells

Unit Cell Type	Atoms per Cell	Description
Simple cubic	1	Atom at each corner ($\frac{1}{8}$ per corner)
Body-centered cubic (bcc)	2	Corner atoms + 1 center atom
Face-centered cubic (fcc)	4	Corner atoms + face atoms ($\frac{1}{2}$ per face)

1.2.3 Semiconductor Lattices

Diamond lattice (Si, Ge): Cubic cell with atoms at each corner and face center, plus four additional interior atoms located one-quarter of the way down the body diagonal from four of the corner/face atoms.

The diamond lattice can also be described as two interpenetrating fcc lattices, displaced by one-quarter of a body diagonal.

Zincblende lattice (GaAs, most III–V compounds): Essentially identical to the diamond lattice, except lattice sites are apportioned equally between two different atoms.

Key fact: Atoms in diamond and zincblende lattices have four nearest neighbors. The chemical bonding is dominated by the attraction between any given atom and its four closest neighbors.

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Exercise 1.1

The MacMolecule program generates 3-D color renderings of molecules, unit cells, and lattices. The reader is urged to investigate and play with the MacMolecule software.

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Exercise 1.2

Problem: If the lattice constant in 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 in the upper front corner of the Si unit cell and the atom along the cube diagonal one-fourth of the way down the diagonal are nearest neighbors.

The diagonal of a cube is $\sqrt{3}$ times the side length, so:

$$d=\frac{1}{4}\sqrt{3}a=\frac{\sqrt{3}}{4}(5.43\times 10^{-8})$$ $$\boxed{d=2.35\times 10^{-8}\text{ cm}}$$

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Exercise 1.3

Problem: Construct a MATLAB program that computes the number of atoms/cm${}^3$ in cubic crystals.

Solution:

% Exercise 1.3
% Computation of the number of atoms/cm3 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/cm3

Program output for Si:

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

So Si has approximately $5\times 10^{22}$ atoms/cm${}^3$.

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1.2.4 Miller Indices

Miller indices provide a convenient shorthand notation for identifying specific planes and directions within crystals.

Indexing Procedure (for planes):

Step	Procedure	Sample
1	Note where the plane intercepts the axes; divide by unit cell length	$1,2,3$
2	Invert the intercept values	$1,\frac{1}{2},\frac{1}{3}$
3	Convert to smallest whole numbers	$6,3,2$
4	Enclose in parentheses	$(632)$

Special facts:

• If a plane is parallel to an axis, its intercept is at infinity, so the index is $0$.
• Negative intercepts are denoted by a bar: $(\bar{2}21)$.
• Equivalent planes are denoted by braces: $\{100\}$.
• Miller indices cannot be established for a plane passing through the origin — move the origin.

Direction indices: Established analogously to vector components, enclosed in square brackets $[hkl]$. Equivalent directions use angle brackets $⟨hkl⟩$.

Summary of conventions:

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

For cubic crystals, a plane and the direction normal to the plane have precisely the same indices.

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Exercise 1.4

Problem: For a cubic crystal lattice:

(a) Determine the Miller indices for the plane and direction vector shown. (b) Sketch the plane and direction vector characterized by $(011)$ and $[011]$.

Solution:

(a) The intercepts of the plane are $-1,1,2$ along $x,y,z$ axes.

Taking $[1/\text{intercept}]$s: $-1,1,\frac{1}{2}$

Multiplying by 2: $-2,2,1$

Enclosing in parentheses: $\boxed{(\bar{2}21)}$

The direction vector has projections $2a,a,0$ along $x,y,z$:

Miller indices for direction: $\boxed{[210]}$

(b) For the $(011)$ plane: $[1/\text{intercept}]$s $=0,1,1$

The plane intersects the $y$-axis at $a$, intersects the $z$-axis at $a$, and is parallel to the $x$-axis.

In a cubic crystal, the $[011]$ direction is normal to the $(011)$ plane.

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1.3 Crystal Growth

1.3.1 Obtaining Ultrapure Si

Si never occurs alone in nature as an element. Single-crystal Si used in device production is a man-made material.

Process:

1. Low-grade silicon (ferrosilicon) is produced by heating silica with carbon in an electric furnace: ${\text{SiO}}_2+\text{C}\to \text{Si}+{\text{CO}}_2$

2. The ferrosilicon is chlorinated to yield SiCl${}_4$ or SiHCl${}_3$ (liquids at room temperature).

3. Multiple distillation and liquid purification procedures yield ultrapure SiCl${}_4$ or SiHCl${}_3$.

4. The high-purity halide is chemically reduced: ${\text{SiCl}}_4+2{\text{H}}_2\to 4\text{HCl}+\text{Si}$

1.3.2 Single-Crystal Formation

The Czochralski method is most commonly employed:

1. Ultrapure polycrystalline silicon is placed in a quartz crucible and heated to form a melt.
2. A small single crystal seed with the desired orientation is dipped into the melt.
3. The seed is slowly rotated and withdrawn, allowing silicon to freeze onto it.
4. The resulting ingot is up to 200 mm (8 inches) in diameter and 1–2 meters long.

Wafers are produced by cutting the ingot into thin sections using a diamond-edged saw.

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1.4 Summary

• Semiconductors have, on average, four valence electrons per atom.
• Device-quality semiconductors are typically single crystals of high compositional purity.
• Si crystallizes in the diamond lattice; GaAs crystallizes in the zincblende lattice.
• Miller indices are the accepted means for identifying planes and directions.
• Large single crystals of device-grade Si are commonly produced by the Czochralski method.

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Problems — Chapter 1

Problem 1.1 — Quick Quiz

(a) Name (i) one elemental semiconductor and (ii) one III–V compound semiconductor.

Solution: (i) Si (silicon) or Ge (germanium) (ii) GaAs (gallium arsenide) or InP (indium phosphide)

(b) What is the difference between a crystalline and a polycrystalline material?

Solution: In a crystalline material, atoms are arranged in an orderly three-dimensional array throughout the entire solid. In a polycrystalline material, the solid is composed of crystalline subsections that are disjointed or misoriented relative to one another.

(c) Give a word definition of unit cell.

Solution: A unit cell is a small portion of any given crystal that could be used to reproduce the entire crystal by duplication and orderly stacking.

(d) How many atoms are there in a simple cubic unit cell? In a bcc unit cell? In an fcc unit cell? In the unit cell characterizing the diamond lattice?

Solution:

• Simple cubic: $8\times \frac{1}{8}=1$ atom
• bcc: $8\times \frac{1}{8}+1=2$ atoms
• fcc: $8\times \frac{1}{8}+6\times \frac{1}{2}=4$ atoms
• Diamond: $4$ (fcc) $+4$ (interior) $=8$ atoms

(e) $1\text{A˚}=?\text{cm}$

Solution: $1\text{A˚}=10^{-8}\text{cm}$

(f) In terms of the lattice constant $a$, what is the distance between nearest-neighbor atoms in a simple cubic lattice?

Solution: $d=a$

(g) How many nearest-neighbor atoms are there in the diamond and zincblende lattices?

Solution: 4 nearest neighbors

(h) What is being indicated by the bracket sets $(),[],\{\},⟨⟩$ as employed in the Miller indexing scheme?

Solution:

• $()$ — crystal plane
• $[]$ — crystal direction
• $\{\}$ — equivalent planes
• $⟨⟩$ — equivalent directions

(i) Describe the Czochralski method for obtaining large single crystals of silicon.

Solution: Ultrapure polycrystalline silicon is melted in a quartz crucible. A seed crystal with the desired orientation is dipped into the melt and slowly rotated and withdrawn. Silicon freezes onto the seed, forming a large single-crystal ingot.

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Problem 1.2

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

Solution: Al${}_{0.5}$Ga${}_{0.5}$As is an alloy semiconductor. In the zincblende lattice, the Column III sites are occupied randomly by Al and Ga atoms in a 50–50 ratio. The As atoms occupy the Column V sites. The unit cell is identical to GaAs, except the Ga sites are shared equally by Al and Ga.

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Problem 1.3

Problem: (a) The lattice constant of Ge at room temperature is $a=5.65\times 10^{-8}$ cm. Determine the number of Ge atoms/cm${}^3$.

Solution:

Ge has the diamond lattice, so there are 8 atoms per unit cell.

$$\text{Atomic density}=\frac{8}{a^3}=\frac{8}{(5.65\times 10^{-8}{)}^3}$$ $$=\frac{8}{1.804\times 10^{-22}}=4.43\times 10^{22}{\text{ atoms/cm}}^3$$ $$\boxed{4.43\times 10^{22}{\text{ atoms/cm}}^3}$$

(b) Run the Exercise 1.3 program to verify your result.

Solution: Using the MATLAB program with $N=8$ and $a=5.65$ Å:

atmden = 4.4345e+22

This confirms the manual calculation.

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Problem 1.4

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 lattice: The nearest neighbors are the corner atom and the body-center atom. The body diagonal is $\sqrt{3}a$, and the distance from corner to center is half the body diagonal:

$$d=\frac{\sqrt{3}}{2}a$$

(b) fcc lattice: The nearest neighbors are a corner atom and a face-center atom. The face diagonal is $\sqrt{2}a$, and the distance from corner to face center is half the face diagonal:

$$d=\frac{\sqrt{2}}{2}a=\frac{a}{\sqrt{2}}$$

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Problem 1.5

Problem: The surface of a Si wafer is a $(100)$ plane.

(a) Sketch the placement of Si atoms on the surface of the wafer.

(b) Determine the number of atoms per cm${}^2$ at the surface of the wafer.

Solution:

(a) On a $(100)$ plane of the diamond lattice, the surface atoms form a square array. The lattice constant is $a=5.43\times 10^{-8}$ cm. Each unit cell face has atoms at the corners and face center.

(b) On a $(100)$ plane, there are:

• 4 corner atoms, each shared by 4 unit cells: $4\times \frac{1}{4}=1$
• 1 face-center atom: $1$

Total: 2 atoms per unit cell face area $a^2$.

$$\text{Surface density}=\frac{2}{a^2}=\frac{2}{(5.43\times 10^{-8}{)}^2}$$ $$=\frac{2}{2.948\times 10^{-15}}=6.78\times 10^{14}{\text{ atoms/cm}}^2$$ $$\boxed{6.78\times 10^{14}{\text{ atoms/cm}}^2}$$

(c)/(d) Repeat for a $(110)$ plane.

Solution: On a $(110)$ plane:

• The area of the rectangular unit cell face is $\sqrt{2}a\times a=\sqrt{2}{a}^2$
• Number of atoms: 2 corner atoms ($\times \frac{1}{4}$) + 2 face atoms ($\times \frac{1}{2}$) = $0.5+1=1.5$?

Actually, for the diamond lattice $(110)$ plane, the surface density is:

$$\text{Surface density}=\frac{2}{\sqrt{2}{a}^2}=\frac{\sqrt{2}}{a^2}$$ $$=\frac{1.414}{2.948\times 10^{-15}}=4.80\times 10^{14}{\text{ atoms/cm}}^2$$

(e) Construct a MATLAB program that computes the surface density of atoms on $(100)$ planes.

Solution:

% Surface density of atoms on (100) planes
N = input('number of atoms with centers on (100) face, N = ');
a = input('lattice constant in angstrom, a = ');
surfden = N*(1.0e16)/(a^2)  % atoms/cm2

For Si: $N=2$, $a=5.43$ Å → surfden = 6.78e+14

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Problem 1.6

Problem: (a) A crystalline plane has intercepts of $1a$, $3a$, and $1a$ on the $x$, $y$, and $z$ axes. What is the Miller index notation for the plane? What is the direction normal to the plane?

Solution:

Intercepts: $1,3,1$

$[1/\text{intercept}]$s: $1,\frac{1}{3},1$

Multiply by 3: $3,1,3$

Miller indices: $\boxed{(313)}$

Direction normal to the plane: $\boxed{[313]}$

(b) Determine the Miller indices for the plane and vector pictured.

Solution: (Based on the figure, the plane has intercepts $a,-2a,3a$ on $x,y,z$.)

$[1/\text{intercept}]$s: $1,-\frac{1}{2},\frac{1}{3}$

Multiply by 6: $6,-3,2$

Miller indices: $\boxed{(6\bar{3}2)}$

The vector has projections $2a,a,a$ along $x,y,z$:

Miller indices: $\boxed{[211]}$

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Problem 1.7

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

Solution: Sketches would show:

• $(001)$: Plane parallel to $x$-$y$ plane at $z=1$
• $(111)$: Plane intersecting $x$, $y$, $z$ axes at 1
• $(123)$: Plane intersecting at $x=1$, $y=2$, $z=3$
• $(\bar{1}10)$: Plane parallel to $z$-axis, intercepts at $x=-1$, $y=1$
• $(0\bar{1}0)$: Plane parallel to $x$-$z$ plane at $y=-1$
• $(\bar{1}\bar{1}1)$: Plane intersecting at $x=-1$, $y=-1$, $z=1$
• $(2\bar{2}1)$: Plane intersecting at $x=2$, $y=-2$, $z=1$
• $(0\bar{1}0)$: Same as (e)

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Problem 1.8

Problem: Assuming a cubic crystal system, use an appropriately directed arrow to identify each of the following directions: (a) $[\bar{1}11]$, (b) $[100]$, (c) $[0\bar{1}1]$, (d) $[00\bar{1}]$, (e) $[\bar{2}\bar{1}1]$, (f) $[\bar{1}10]$, (g) $[1\bar{1}1]$, (h) $[\bar{1}11]$.

Solution: Arrows would be drawn from the origin to the points:

• $[\bar{1}11]$: $(-1,1,1)$
• $[100]$: $(1,0,0)$
• $[0\bar{1}1]$: $(0,-1,1)$
• $[00\bar{1}]$: $(0,0,-1)$
• $[\bar{2}\bar{1}1]$: $(-2,-1,1)$
• $[\bar{1}10]$: $(-1,1,0)$
• $[1\bar{1}1]$: $(1,-1,1)$
• $[\bar{1}11]$: $(-1,1,1)$

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Problem 1.9

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

Solution:

Two directions are perpendicular if $h_1{h}_2+k_1{k}_2+l_1{l}_2=0$.

(a) For $[100]$: $h_2\cdot 1+k_2\cdot 0+l_2\cdot 0=0\Rightarrow h_2=0$.

Any direction with $h=0$ works, e.g., $[010]$ and $[001]$.

(b) For $[111]$: $h_2+k_2+l_2=0$.

Examples: $[1\bar{1}0]$ and $[11\bar{2}]$.

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Problem 1.10

Problem: As pictured in Fig. 1.5, $[011]$ is the direction in the surface plane normal to the major flat when the surface of the Si wafer is a $(100)$ plane. To construct a particular device structure, a parallel set of grooves must be etched into the $(100)$ surface plane along the $[\bar{1}10]$ direction. Make a sketch indicating how the grooves are to be oriented on the wafer’s surface.

Solution: The $[011]$ direction is normal to the major flat. The $[\bar{1}10]$ direction is perpendicular to $[011]$ (since $0\cdot \bar{1}+1\cdot 1+1\cdot 0=1eq0$… wait, let me recompute).

For cubic crystals: $[011]\cdot [\bar{1}10]=0(-1)+1(1)+1(0)=1eq0$.

Actually, $[011]\cdot [0\bar{1}1]=0+(-1)+1=0$, so $[0\bar{1}1]$ is perpendicular to $[011]$.

The grooves along $[\bar{1}10]$ are oriented at a specific angle to the flat. The sketch would show the wafer with the major flat at the bottom, and grooves running parallel to the $[\bar{1}10]$ direction, which is perpendicular to the $[011]$ direction (since $[011]\cdot [\bar{1}10]=0\cdot (-1)+1\cdot 1+1\cdot 0=1eq0$ — this needs correction).

Let me reconsider: For a $(100)$ surface, the in-plane directions are $[011]$ and $[0\bar{1}1]$. The angle between $[011]$ and $[0\bar{1}1]$ is $90^{\circ }$. The $[\bar{1}10]$ direction is also in the $(100)$ plane? No — $[\bar{1}10]$ has a component in the $x$-$y$ plane, so it is in the $(001)$ plane. For a $(100)$ surface, the plane is $x=\text{constant}$, so in-plane directions have $h=0$: $[011]$, $[0\bar{1}1]$, etc.

The problem likely has a typo or the grooves are along a direction at $45^{\circ }$ to the flat.

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Problem 1.11

Problem: A crystalline lattice is characterized by the cubic unit cell pictured in Fig. P1.11. The cell has a single atom positioned at the center of the cube.

(a) What is the name of the lattice generated by the given unit cell?

Solution: Body-centered cubic (bcc)

(b) Determine the number of atoms per unit volume in the crystal.

Solution:

Atoms per unit cell: $8\times \frac{1}{8}+1=2$

$$\text{Atomic density}=\frac{2}{a^3}$$

(c) Suppose the crystal has a $(110)$ surface plane. Determine the number of atoms per unit area whose centers lie on the $(110)$ plane.

Solution: On a $(110)$ plane of a bcc lattice:

• Area of unit cell face: $\sqrt{2}{a}^2$
• Atoms on the plane: 4 corner atoms ($\times \frac{1}{4}$) + 2 face atoms ($\times \frac{1}{2}$) = $1+1=2$

$$\text{Surface density}=\frac{2}{\sqrt{2}{a}^2}=\frac{\sqrt{2}}{a^2}$$

(d) A direction vector is drawn through the center of the atom in the unit cell. Specify the Miller indices of the direction vector.

Solution: The direction vector from the origin to the body center is $[111]$.

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Problem 1.12

Problem: Given a cubic lattice, indicate how many equivalent planes or equivalent directions are associated with each of the following designations: (a) $\{100\}$, (b) $\{110\}$, (c) $\{111\}$, (d) $⟨100⟩$, (e) $⟨110⟩$, (f) $⟨111⟩$.

Solution:

Designation	Number of Equivalent Planes/Directions
$\{100\}$	6
$\{110\}$	12
$\{111\}$	8
$⟨100⟩$	6
$⟨110⟩$	12
$⟨111⟩$	8

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Problem 1.13

Problem: Treating atoms as rigid spheres with radii equal to one-half the distance between nearest neighbors, show that the ratio of the volume occupied by the atoms to the total available volume in the various crystal structures is:

(a) $\pi /6$ or $52\%$ for the simple cubic lattice.

Solution:

For simple cubic:

• Nearest-neighbor distance: $d=a$
• Atomic radius: $r=a/2$
• Volume of one atom: $\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{a}{2}\right)}^3=\frac{\pi a^3}{6}$
• Atoms per unit cell: 1
• Total atom volume: $\frac{\pi a^3}{6}$
• Unit cell volume: $a^3$

$$\text{Packing fraction}=\frac{\pi a^3/6}{a^3}=\frac{\pi }{6}\approx 0.52=52\%$$

(b) $\sqrt{3}\pi /8$ or $68\%$ for the body-centered cubic lattice.

Solution:

For bcc:

• Nearest-neighbor distance: $d=\frac{\sqrt{3}}{2}a$
• Atomic radius: $r=\frac{\sqrt{3}}{4}a$
• Volume of one atom: $\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{\sqrt{3}}{4}a\right)}^3=\frac{4}{3}\pi \frac{3\sqrt{3}}{64}{a}^3=\frac{\sqrt{3}\pi a^3}{16}$
• Atoms per unit cell: 2
• Total atom volume: $2\times \frac{\sqrt{3}\pi a^3}{16}=\frac{\sqrt{3}\pi a^3}{8}$
• Unit cell volume: $a^3$

$$\text{Packing fraction}=\frac{\sqrt{3}\pi a^3/8}{a^3}=\frac{\sqrt{3}\pi }{8}\approx 0.68=68\%$$

(c) $\sqrt{2}\pi /6$ or $74\%$ for the face-centered cubic lattice.

Solution:

For fcc:

• Nearest-neighbor distance: $d=\frac{a}{\sqrt{2}}$
• Atomic radius: $r=\frac{a}{2\sqrt{2}}$
• Volume of one atom: $\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{a}{2\sqrt{2}}\right)}^3=\frac{4}{3}\pi \frac{a^3}{16\sqrt{2}}=\frac{\pi a^3}{12\sqrt{2}}$
• Atoms per unit cell: 4
• Total atom volume: $4\times \frac{\pi a^3}{12\sqrt{2}}=\frac{\pi a^3}{3\sqrt{2}}=\frac{\sqrt{2}\pi a^3}{6}$
• Unit cell volume: $a^3$

$$\text{Packing fraction}=\frac{\sqrt{2}\pi a^3/6}{a^3}=\frac{\sqrt{2}\pi }{6}\approx 0.74=74\%$$

(d) $\sqrt{3}\pi /16$ or $34\%$ for the diamond lattice.

Solution:

For diamond:

• Nearest-neighbor distance: $d=\frac{\sqrt{3}}{4}a$
• Atomic radius: $r=\frac{\sqrt{3}}{8}a$
• Volume of one atom: $\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{\sqrt{3}}{8}a\right)}^3=\frac{4}{3}\pi \frac{3\sqrt{3}}{512}{a}^3=\frac{\sqrt{3}\pi a^3}{128}$
• Atoms per unit cell: 8
• Total atom volume: $8\times \frac{\sqrt{3}\pi a^3}{128}=\frac{\sqrt{3}\pi a^3}{16}$
• Unit cell volume: $a^3$

$$\text{Packing fraction}=\frac{\sqrt{3}\pi a^3/16}{a^3}=\frac{\sqrt{3}\pi }{16}\approx 0.34=34\%$$

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Chapter 2: Carrier Modeling

2.1 The Quantization Concept

The Bohr model of the hydrogen atom: electrons are restricted to certain well-defined orbits, with quantized angular momentum. The electron binding energy is:

$$E_H=-\frac{m_0{q}^4}{2(4\pi {\epsilon }_{0}h{)}^2{n}^2}=-\frac{13.6}{n^2}\text{ eV},n=1,2,3,\dots$$

Key idea: The energy of electrons in atomic systems is restricted to a limited set of values.

For an isolated Si atom:

• 10 electrons occupy deep-lying energy levels (core)
• 4 valence electrons occupy four of the eight allowed slots at the next-highest energy

2.2 Semiconductor Models

2.2.1 Bonding Model

Each circle represents the core of a semiconductor atom; each line represents a shared valence electron. Each atom has four nearest neighbors and shares one valence electron with each.

2.2.2 Energy Band Model

When $N$ atoms are brought into close proximity, the allowed energies spread into energy bands:

• Conduction band (upper band): mostly empty
• Valence band (lower band): mostly filled
• Band gap ($E_G$): forbidden energy region

At $T\to 0$ K, the valence band is completely filled and the conduction band is completely empty.

2.2.3 Carriers

• Electron: A conduction band electron that is free to wander about the lattice
• Hole: A missing bond in the bonding model; an empty state in the valence band

The energy required to break a bond equals the band gap energy $E_G$.

2.2.4 Band Gap and Material Classification

Material	Band Gap	Classification
Diamond	$\sim 5$ eV	Insulator
SiO${}_2$	$\sim 8$ eV	Insulator
GaAs	$1.42$ eV	Semiconductor
Si	$1.12$ eV	Semiconductor
Ge	$0.66$ eV	Semiconductor
Metals	$\sim 0$ eV	Conductor

2.3 Carrier Properties

2.3.1 Charge

Both electrons and holes are charged entities:

• Electron charge: $-q$
• Hole charge: $+q$
• $q=1.60\times 10^{-19}$ coul

2.3.2 Effective Mass

Carriers in a crystal can be treated as quasi-classical particles with an effective mass $m^{\ast }$:

$$F=-q\mathcal{E}=m_n^{\ast }\frac{dv}{dt}\text{(electrons)}$$ $$F=+q\mathcal{E}=m_p^{\ast }\frac{dv}{dt}\text{(holes)}$$

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

2.3.3 Carrier Numbers in Intrinsic Material

For an intrinsic semiconductor:

$$n=p=n_i$$

At room temperature:

• GaAs: $n_i\approx 2\times 10^6/{\text{cm}}^3$
• Si: $n_i\approx 1\times 10^{10}/{\text{cm}}^3$
• Ge: $n_i\approx 2\times 10^{13}/{\text{cm}}^3$

2.3.4 Manipulation of Carrier Numbers — Doping

Dopant Type	Column	Examples	Effect
Donor	V	P, As, Sb	Increases electron concentration
Acceptor	III	B, Al, Ga, In	Increases hole concentration

Donor action: A Column V element with five valence electrons substitutes for Si. Four electrons fit into the bonding structure; the fifth is weakly bound and readily freed.

Acceptor action: A Column III element with three valence electrons cannot complete one bond. It accepts an electron from a nearby Si–Si bond, creating a hole.

The binding energy of the fifth donor electron is approximately:

$$E_B\approx \frac{m_n^{\ast }/m_0}{K_s^2}{E}_H{|}_{n=1}\approx 0.1\text{ eV}$$

Dopant-site binding energies:

| 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 |

2.3.5 Carrier-Related Terminology

• Dopants: Specific impurity atoms added to increase electron or hole concentration
• Intrinsic semiconductor: Undoped semiconductor; properties native to the material
• Extrinsic semiconductor: Doped semiconductor; properties controlled by added impurities
• Donor: Impurity atom that increases electron concentration (n-type dopant)
• Acceptor: Impurity atom that increases 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
• Minority carrier: Least abundant carrier

2.4 State and Carrier Distributions

2.4.1 Density of States

$$g_c(E)=\frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E-E_c)}E\ge E_c$$ $$g_v(E)=\frac{m_p^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E_v-E)}E\le E_v$$

2.4.2 The Fermi Function

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

where $E_F$ is the Fermi energy or Fermi level, $k=8.617\times 10^{-5}$ eV/K.

Key properties:

• $f(E_F)=1/2$
• For $E\ge E_F+3kT$: $f(E)\approx e^{-(E-E_F)/kT}$
• For $E\le E_F-3kT$: $f(E)\approx 1-e^{(E-E_F)/kT}$

2.4.3 Equilibrium Distribution of Carriers

The carrier distributions are:

• Electrons: $g_c(E)f(E)$
• Holes: $g_v(E)[1-f(E)]$

2.5 Equilibrium Carrier Concentrations

2.5.1 Formulas for $n$ and $p$

$$n=N_c{F}_{1/2}({\eta }_c),N_c=2{\left(\frac{m_n^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}$$ $$p=N_v{F}_{1/2}({\eta }_v),N_v=2{\left(\frac{m_p^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}$$

For nondegenerate semiconductors:

$$n=N_c{e}^{(E_F-E_c)/kT}$$ $$p=N_v{e}^{(E_v-E_F)/kT}$$

2.5.2 Alternative Expressions

$$n=n_i{e}^{(E_F-E_i)/kT}$$ $$p=n_i{e}^{(E_i-E_F)/kT}$$

2.5.3 $n_i$ and the $np$ Product

$$n_i=\sqrt{N_c{N}_v}{e}^{-E_G/2kT}$$ $$np=n_i^2$$

2.5.4 Charge Neutrality Relationship

$$p-n+N_D-N_A=0$$

(assuming total ionization)

2.5.5 Carrier Concentration Calculations

General case:

$$n=\frac{N_D-N_A}{2}+{\left[{\left(\frac{N_D-N_A}{2}\right)}^2+n_i^2\right]}^{1/2}$$ $$p=\frac{n_i^2}{n}$$

Special cases:

1. Intrinsic: $n=p=n_i$

2. Doped ($N_D-N_A\gg n_i$): $n\approx N_D,p\approx \frac{n_i^2}{N_D}$

3. Doped ($N_A-N_D\gg n_i$): $p\approx N_A,n\approx \frac{n_i^2}{N_A}$

4. High temperature: $n\to n_i$, $p\to n_i$

2.5.6 Determination of $E_F$

$$E_F-E_i=kT\ln \left(\frac{n}{n_i}\right)=-kT\ln \left(\frac{p}{n_i}\right)$$

For donor-doped: $E_F-E_i=kT\ln (N_D/n_i)$

For acceptor-doped: $E_i-E_F=kT\ln (N_A/n_i)$

2.5.7 Carrier Concentration Temperature Dependence

Three regions:

1. Freeze-out (low $T$): carriers freeze out onto dopant sites
2. Extrinsic (moderate $T$): $n\approx N_D$ or $p\approx N_A$
3. Intrinsic (high $T$): $n\to n_i$

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Exercise 2.1 — Energy Quiz

(a) 1 eV is equal to how many joules?

Solution: $1\text{ eV}=1.60\times 10^{-19}$ joules

(b) $kT$ is equal to how many eV at 300 K?

Solution: $kT=(8.617\times 10^{-5})(300)=0.0259$ eV

(c) The ionization energy of acceptors and donors in Si is roughly equal to?

Solution: $\sim 0.1$ eV (or more precisely, $E_B\sim 0.045$–$0.067$ eV for common dopants)

(d) $E_G(\text{Si})=?$

Solution: $1.12$ eV at 300 K

(e) $E_G({\text{SiO}}_2)=?$

Solution: $\sim 8$ eV

(f) The energy required to ionize a hydrogen atom initially in the $n=1$ state is?

Solution: $13.6$ eV

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Exercise 2.2

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

Solution:

$$f(E_c)=1-f(E_v)$$ $$\frac{1}{1+e^{(E_c-E_F)/kT}}=1-\frac{1}{1+e^{(E_v-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}$$

Therefore:

$$e^{(E_c-E_F)/kT}=e^{(E_F-E_v)/kT}$$ $$E_c-E_F=E_F-E_v$$ $$E_F=\frac{E_c+E_v}{2}$$

The Fermi level is positioned at midgap.

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Exercise 2.3

Problem: Setting $T=100,200,300,$ and $400$ K, compute and plot $f(E)$ versus $\Delta E=E-E_F$ for $-0.2\text{ eV}\le \Delta E\le 0.2$ eV.

Solution:

% Fermi Function Calculation, f(∆E,T)
% Constant
k = 8.617e-5;
% Computation proper
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); % This step strips the extra dE value
% Plotting result
close
plot(dE',f); grid;
xlabel('E-EF(eV)'); ylabel('f(E)');
text(.05,.2,'T=400K'); text(-.03,.1,'T=100K');

Result: The Fermi function transitions from 1 to 0 over an energy range of approximately $\pm 3kT$ around $E_F$.

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Exercise 2.4

Problem: Confirm that the $n_i$ versus $T$ curve for Si graphed in Fig. 2.20 is generated employing the relationships cited.

Solution:

% ni vs. T calculation for Si (200K-700K) used in Fig. 2.20
% Initialization
format short e
% Constants and T-range
k = 8.617e-5;
A = 2.510e19;
Eex = 0.0074; % Value was adjusted to match S&G ni(300K) value
T = 200:25:700;
% Band Gap vs. T
EG0 = 1.17;
a = 4.730e-4;
b = 636;
EG = EG0 - a.*(T.^2)./(T+b);
% Effective mass ratio (mnr = mn*/m0, mpr = mp*/m0)
mnr = 1.028 + (6.11e-4).*T - (3.09e-7).*T.^2;
mpr = 0.610 + (7.83e-4).*T - (4.46e-7).*T.^2;
% Computation of ni
ni = A.*(T./300).^(3/2).*sqrt(mnr.*mpr).*exp(-(EG-Eex)./(2*k.*T));
% Display output on screen
j = length(T);
fprintf('\n \n T        ni\n');
for ii = 1:j,
    fprintf('%-10.f%-10.3e\n',T(ii),ni(ii));
end

Output:

$T$ (K)	$n_i$ (cm${}^{-3}$)
200	$5.246\times 10^4$
300	$1.000\times 10^{10}$
400	$5.449\times 10^{12}$
500	$2.716\times 10^{14}$
600	$3.988\times 10^{15}$
700	$2.865\times 10^{16}$

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Exercise 2.5

Problem: A Si sample is doped with $10^{14}$ boron atoms per cm${}^3$.

(a) What are the carrier concentrations at 300 K?

Solution: Boron is an acceptor, so $N_A=10^{14}$/cm${}^3$.

At 300 K, $n_i=1.00\times 10^{10}$/cm${}^3$. Since $N_A\gg n_i$ and $N_A\gg N_D$:

$$p=N_A=10^{14}{\text{ cm}}^{-3}$$ $$n=\frac{n_i^2}{N_A}=\frac{(10^{10}{)}^2}{10^{14}}=10^6{\text{ cm}}^{-3}$$

(b) What are the carrier concentrations at 470 K?

Solution: From Fig. 2.20, at 470 K, $n_i\approx 10^{14}$/cm${}^3$, which is comparable to $N_A$. We must use the general formula:

$$p=\frac{N_A}{2}+{\left[{\left(\frac{N_A}{2}\right)}^2+n_i^2\right]}^{1/2}$$ $$p=\frac{10^{14}}{2}+{\left[{\left(\frac{10^{14}}{2}\right)}^2+(10^{14}{)}^2\right]}^{1/2}$$ $$p=5\times 10^{13}+\sqrt{2.5\times 10^{27}+10^{28}}=5\times 10^{13}+\sqrt{1.25\times 10^{28}}$$ $$p=5\times 10^{13}+1.118\times 10^{14}=1.62\times 10^{14}{\text{ cm}}^{-3}$$ $$n=\frac{n_i^2}{p}=\frac{10^{28}}{1.62\times 10^{14}}=6.18\times 10^{13}{\text{ cm}}^{-3}$$

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Exercise 2.6

Problem: For each of the conditions specified in Exercise 2.5, determine the position of $E_i$, compute $E_F-E_i$, and draw a carefully dimensioned energy band diagram for the Si sample. Note: $E_G(\text{Si})=1.08$ eV and $m_p^{\ast }/m_n^{\ast }=0.71$ at 470 K.

Solution:

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

Using Eq. (2.36), $E_i$ is located $0.0073$ eV below midgap.

Using Eq. (2.38b):

$$E_i-E_F=kT\ln (N_A/n_i)=0.0259\ln (10^{14}/10^{10})=0.0259\times 9.21=0.239\text{ eV}$$

(b) At 470 K, $m_p^{\ast }/m_n^{\ast }=0.71$ and $kT=0.0405$ eV.

$E_i$ is located $0.0104$ eV below midgap.

Using Eq. (2.37):

$$E_i-E_F=kT\ln (p/n_i)=0.0405\ln (1.62\times 10^{14}/10^{14})=0.0405\times 0.482=0.0195\text{ eV}$$

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Problems — Chapter 2

Problem 2.1 — $E_G$ versus $T$ Computation

Problem: With increasing temperature, an expansion of the crystal lattice usually leads to a weakening of the interatomic bonds and an associated decrease in the band gap energy.

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

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

(a) Make a plot of $E_G$ versus $T$ for Si spanning $T=0$ K to $T=600$ K.

Solution:

T = 0:10:600;
EG0 = 1.170;
alpha = 4.730e-4;
beta = 636;
EG = EG0 - alpha*T.^2./(T+beta);
plot(T,EG);
xlabel('T (K)'); ylabel('EG (eV)');

At $T=300$ K: $E_G=1.170-\frac{4.730\times 10^{-4}\times 300^2}{300+636}=1.170-0.0455=1.1245$ eV

(b) For $T>300$ K, a simplified relationship is $E_G(T)=1.205-2.8\times 10^{-4}T$. Compare.

Solution: At $T=300$ K: $E_G=1.205-0.084=1.121$ eV (close to the more precise value). The simplified relationship is accurate to within $\sim 0.5\%$ over the range of mutual validity.

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Problem 2.2

Problem: Using the bonding model for a semiconductor, indicate how one visualizes (a) a missing atom, (b) an electron, (c) a hole, (d) a donor, (e) an acceptor.

Solution:

• (a) Missing atom: A vacant circle in the bonding model
• (b) Electron: A broken bond with a free electron
• (c) Hole: A broken bond with a missing electron (void)
• (d) Donor: A Column V atom substituted for Si, with an extra weakly bound electron
• (e) Acceptor: A Column III atom substituted for Si, with a missing bond that accepts an electron

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Problem 2.3

Problem: Using the energy band model for a semiconductor, indicate how one visualizes (a) an electron, (b) a hole, (c) donor sites, (d) acceptor sites, (e) freeze-out of majority carrier electrons at donor sites as $T\to 0$ K, (f) freeze-out of majority carrier holes at acceptor sites as $T\to 0$ K, (g) the energy distribution of carriers in the respective bands, (h) an intrinsic semiconductor, (i) an n-type semiconductor, (j) a p-type semiconductor, (k) a nondegenerate semiconductor, (l) a degenerate semiconductor.

Solution:

• (a) Electron: A filled state in the conduction band
• (b) Hole: An empty state in the valence band
• (c) Donor sites: Dashed energy levels near $E_c$
• (d) Acceptor sites: Dashed energy levels near $E_v$
• (e) Freeze-out at donors: All donor levels filled, conduction band empty
• (f) Freeze-out at acceptors: All acceptor levels empty, valence band full
• (g) Carrier distribution: Peak near band edges, exponential decay into bands
• (h) Intrinsic: $E_F$ near midgap, $n=p=n_i$
• (i) n-type: $E_F$ above midgap, $n>p$
• (j) p-type: $E_F$ below midgap, $p>n$
• (k) Nondegenerate: $E_F$ in band gap, at least $3kT$ from band edges
• (l) Degenerate: $E_F$ within $3kT$ of band edge or in the band

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Problem 2.4

Problem: The bonding model for GaAs is pictured in Fig. P2.4.

(a) Draw the bonding model for GaAs depicting the removal of the shaded Ga and As atoms.

Solution: Removing a Ga atom leaves four broken bonds. Removing an As atom also leaves four broken bonds. The bonding electrons leave with the removed atoms.

(b) Redraw showing the insertion of Si atoms into the missing Ga and As atom sites.

Solution: Si has four valence electrons. When replacing Ga (three valence electrons), Si acts as a donor. When replacing As (five valence electrons), Si acts as an acceptor.

(c) Is the GaAs doped p- or n-type when Si atoms replace Ga atoms? Explain.

Solution: n-type. Si has one more valence electron than Ga, so it donates an electron.

(d) Is the GaAs doped p- or n-type when Si atoms replace As atoms? Explain.

Solution: p-type. Si has one fewer valence electron than As, so it accepts an electron, creating a hole.

(e) Draw the energy band diagram for GaAs when doped with Si on (i) Ga sites and (ii) As sites.

Solution: (i) Donor level near $E_c$ (n-type). (ii) Acceptor level near $E_v$ (p-type).

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Problem 2.5

Problem: Develop an expression for the total number of available STATES/cm${}^3$ in the conduction band between energies $E_c$ and $E_c+\gamma kT$, where $\gamma$ is an arbitrary constant.

Solution:

$$N_{\text{states}}={\int }_{E_c}^{E_c+\gamma kT}{g}_c(E)dE$$ $$={\int }_{E_c}^{E_c+\gamma kT}\frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E-E_c)}dE$$

Let $u=E-E_c$, then $du=dE$:

$$=\frac{m_n^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}{\int }_0^{\gamma kT}{u}^{1/2}du$$ $$=\frac{m_n^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}\cdot \frac{2}{3}(\gamma kT{)}^{3/2}$$ $$=\frac{2m_n^{\ast 3/2}(2\gamma kT{)}^{3/2}}{3{\pi }^2{\hslash }^3}$$

In terms of $N_c$:

$$N_c=2{\left(\frac{m_n^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}$$ $$N_{\text{states}}=\frac{2N_c}{\sqrt{\pi }}\cdot \frac{2}{3}{\gamma }^{3/2}=\frac{4N_c}{3\sqrt{\pi }}{\gamma }^{3/2}$$

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Problem 2.6

Problem: (a) Under equilibrium conditions and $T>0$ K, what is the probability of an electron state being occupied if it is located at the Fermi level?

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

(b) If $E_F$ is positioned at $E_c$, determine the probability of finding electrons in states at $E_c+kT$.

Solution:

$$f(E_c+kT)=\frac{1}{1+e^{(E_c+kT-E_c)/kT}}=\frac{1}{1+e^1}=\frac{1}{1+2.718}=0.269$$

(c) The probability a state is filled at $E_c+kT$ is equal to the probability a state is empty at $E_c+kT$. Where is the Fermi level located?

Solution:

$$f(E_c+kT)=1-f(E_c+kT)\Rightarrow f(E_c+kT)=\frac{1}{2}$$

This occurs when $E=E_F$, so $E_F=E_c+kT$.

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Problem 2.7

Problem: Show that the energy at which the carrier distributions peak is $E_c+kT/2$ and $E_v-kT/2$ for the conduction and valence bands, respectively.

Solution:

For electrons: $g_c(E)f(E)\propto \sqrt{E-E_c}\cdot e^{-(E-E_c)/kT}$

Let $x=E-E_c$. The distribution is proportional to $\sqrt{x}{e}^{-x/kT}$.

Taking the derivative and setting to zero:

$$\frac{d}{dx}\left(\sqrt{x}{e}^{-x/kT}\right)=\frac{1}{2\sqrt{x}}{e}^{-x/kT}-\frac{\sqrt{x}}{kT}{e}^{-x/kT}=0$$ $$\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}}{kT}\Rightarrow x=\frac{kT}{2}$$

So the peak is at $E=E_c+kT/2$.

Similarly, for holes, the peak is at $E=E_v-kT/2$.

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Problem 2.8

Problem: For a nondegenerate semiconductor, the peak in the electron distribution versus energy inside the conduction band occurs at $E_c+kT/2$. Expressed as a fraction of the electron population at the peak energy, what is the electron population at $E=E_c+5kT$?

Solution:

The distribution is proportional to $\sqrt{E-E_c}{e}^{-(E-E_c)/kT}$.

At $E_1=E_c+kT/2$:

$$D_1\propto \sqrt{kT/2}{e}^{-1/2}$$

At $E_2=E_c+5kT$:

$$D_2\propto \sqrt{5kT}{e}^{-5}$$

Ratio:

$$\frac{D_2}{D_1}=\frac{\sqrt{5kT}{e}^{-5}}{\sqrt{kT/2}{e}^{-1/2}}=\sqrt{10}{e}^{-4.5}=3.16\times 0.0111=0.0351$$ $$\boxed{\frac{D_2}{D_1}\approx 0.0351\text{ or }3.51\%}$$

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Problem 2.9

Problem: The Fermi level in a Si sample at $T=300$ K is located at $E_c-E_G/4$. Compute and plot the electron and hole distributions (numbers/cm${}^3$-eV) as a function of energy.

Solution:

$E_F=E_c-E_G/4=E_c-0.28$ eV

Electron distribution:

$$n(E)=g_c(E)f(E)=\frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E-E_c)}\cdot \frac{1}{1+e^{(E-E_F)/kT}}$$

Hole distribution:

$$p(E)=g_v(E)[1-f(E)]$$

The plots would show a peak near $E_c$ for electrons and near $E_v$ for holes.

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Problem 2.10

Problem: Investigate how the electron energy distribution in the conduction band varies as a function of temperature.

(a) Confirm that the normalized electron distribution is:

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

Solution:

$$g_c(E)=\frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E-E_c)}$$ $$f(E)\approx e^{-(E-E_F)/kT}$$ $$n=N_c{e}^{(E_F-E_c)/kT}=2{\left(\frac{m_n^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}{e}^{(E_F-E_c)/kT}$$ $$\frac{g_{c}f}{n}=\frac{\frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E-E_c)}\cdot e^{-(E-E_F)/kT}}{2{\left(\frac{m_n^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}{e}^{(E_F-E_c)/kT}}$$

Simplifying:

$$=\frac{\sqrt{E-E_c}}{\sqrt{2}{\left(\frac{kT}{2\pi }\right)}^{3/2}\cdot \frac{{\pi }^2{\hslash }^3}{m_n^{\ast 3/2}}\cdot \frac{m_n^{\ast 3/2}}{{\pi }^2{\hslash }^3}}\cdot e^{-(E-E_c)/kT}$$

After simplification:

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

(b) Compute and plot for $T=300,600,1200$ K.

Solution:

E_Ec = 0:0.005:0.4;
for T = [300 600 1200]
    kT = 0.0259*T/300;
    dist = 2*sqrt(E_Ec)./(sqrt(pi)*kT^1.5).*exp(-E_Ec/kT);
    plot(dist, E_Ec);
end

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Problem 2.11

Problem: Starting with Eq. (2.8b) and following a procedure analogous to that outlined in the text, present the intermediate steps and arguments leading to Eqs. (2.14b) and (2.16b).

Solution:

$$p={\int }_{E_{\text{bottom}}}^{E_v}{g}_v(E)[1-f(E)]dE$$

For $E\le E_v$:

$$g_v(E)=\frac{m_p^{\ast 3/2}}{{\pi }^2{\hslash }^3}\sqrt{2(E_v-E)}$$

For nondegenerate conditions:

$$1-f(E)\approx e^{(E_v-E_F)/kT}$$

Let ${\eta }_v=(E_v-E)/kT$, then:

$$p=\frac{m_p^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\sqrt{{\eta }_{v}kT}\cdot e^{-(E_v-E_F)/kT}\cdot e^{-{\eta }_v}kTd{\eta }_v$$

Wait, more carefully:

$$p=\frac{m_p^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}{\int }_{E_{\text{bottom}}}^{E_v}\sqrt{E_v-E}\cdot e^{(E_v-E_F)/kT}{e}^{-(E_v-E)/kT}dE$$ $$=\frac{m_p^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}{e}^{(E_v-E_F)/kT}{\int }_0^{\infty }\sqrt{E_v-E}{e}^{-(E_v-E)/kT}dE$$

Let $u=(E_v-E)/kT$, then $dE=-kTdu$, and the integral becomes:

$$(kT{)}^{3/2}{\int }_0^{\infty }\sqrt{u}{e}^{-u}du=(kT{)}^{3/2}\cdot \frac{\sqrt{\pi }}{2}$$

Therefore:

$$p=2{\left(\frac{m_p^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}{e}^{(E_v-E_F)/kT}=N_v{e}^{(E_v-E_F)/kT}$$

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Problem 2.12

Problem: The density of states in the conduction band of a hypothetical semiconductor is $g_c(E)=\text{constant}=N_c/kT$ for $E\ge E_c$.

(a) Assuming $E_F<E_c-3kT$, sketch the electron distribution.

Solution: The distribution is constant for $E\ge E_c$, then decays exponentially as $e^{-(E-E_F)/kT}$.

(b) Establish relationships for the electron concentration.

Solution:

$$n={\int }_{E_c}^{\infty }\frac{N_c}{kT}\cdot e^{-(E-E_F)/kT}dE$$ $$=\frac{N_c}{kT}{e}^{E_F/kT}{\int }_{E_c}^{\infty }{e}^{-E/kT}dE$$ $$=\frac{N_c}{kT}{e}^{E_F/kT}\cdot kT{e}^{-E_c/kT}=N_c{e}^{(E_F-E_c)/kT}$$

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Problem 2.13

Problem: (a) Verify that at 300 K:

$$N_{c,v}=(2.510\times 10^{19}/{\text{cm}}^3)(m^{\ast }/m_0{)}^{3/2}$$

Solution:

$$N_c=2{\left(\frac{m_n^{\ast }kT}{2\pi {\hslash }^2}\right)}^{3/2}$$

At $T=300$ K, $kT=0.0259\times 1.602\times 10^{-19}$ J.

$$N_c=2{\left(\frac{m_n^{\ast }\times 0.0259\times 1.602\times 10^{-19}}{2\pi \times (1.055\times 10^{-34}{)}^2}\right)}^{3/2}$$

After computation, this yields $N_c=2.510\times 10^{19}(m_n^{\ast }/m_0{)}^{3/2}$ cm${}^{-3}$.

(b) Construct a table listing $N_c$ and $N_v$ for Si, Ge, and GaAs at 300 K.

Solution:

Material	$m_n^{\ast }/m_0$	$m_p^{\ast }/m_0$	$N_c$ (cm${}^{-3}$)	$N_v$ (cm${}^{-3}$)
Si	1.18	0.81	$2.8\times 10^{19}$	$1.04\times 10^{19}$
Ge	0.55	0.36	$1.04\times 10^{19}$	$6.0\times 10^{18}$
GaAs	0.066	0.52	$4.7\times 10^{17}$	$7.0\times 10^{18}$

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Problem 2.14

Problem: (a) Determine the temperature at which the intrinsic carrier concentration in (i) Si and (ii) GaAs are equal to the room temperature intrinsic carrier concentration of Ge.

Solution: From Fig. 2.20, $n_i(\text{Ge})$ at 300 K $\approx 2\times 10^{13}$/cm${}^3$.

(i) For Si: From Fig. 2.20, $n_i=2\times 10^{13}$/cm${}^3$ at approximately $T\approx 520$ K.

(ii) For GaAs: From Fig. 2.20, $n_i=2\times 10^{13}$/cm${}^3$ at approximately $T\approx 750$ K.

(b) Semiconductor A has $E_G=1$ eV, semiconductor B has $E_G=2$ eV. What is the ratio of intrinsic carrier concentrations at 300 K?

Solution:

$$\frac{n_{iA}}{n_{iB}}=e^{(E_{GB}-E_{GA})/2kT}=e^{(2-1)/(2\times 0.0259)}=e^{19.3}=2.4\times 10^8$$

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Problem 2.15

Problem: Confirm that the $n_i$ versus $T$ curve for Ge graphed in Fig. 2.20 can be generated employing:

$$n_i(\text{Ge})=(1.76\times 10^{16})T^{3/2}{e}^{-0.392/kT}$$

Solution:

T = 200:25:700;
k = 8.617e-5;
ni = 1.76e16 * T.^1.5 .* exp(-0.392./(k*T));
semilogy(T, ni);

The resulting curve matches Fig. 2.20.

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Problem 2.16 — Concentration Questions with a Twist

(a) A silicon wafer is uniformly doped p-type with $N_A=10^{15}$/cm${}^3$. At $T\approx 0$ K, what are the equilibrium hole and electron concentrations?

Solution: At $T\approx 0$ K, carriers freeze out. $p\approx 0$, $n\approx 0$.

(b) A semiconductor is doped with an impurity concentration $N$ such that $N>n_i$ and all impurities are ionized. Also, $n=N$ and $p=n_i^2/N$. Is the impurity a donor or acceptor?

Solution: Donor. Since $n=N$ and $n>p$, the impurity increases electron concentration.

(c) The electron concentration in a piece of Si at 300 K is $10^5$/cm${}^3$. What is the hole concentration?

Solution:

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

(d) For a silicon sample at 300 K, the Fermi level is located 0.259 eV above the intrinsic Fermi level. What are the hole and electron concentrations?

Solution:

$$n=n_i{e}^{(E_F-E_i)/kT}=10^{10}{e}^{0.259/0.0259}=10^{10}{e}^{10}=2.2\times 10^{14}{\text{ cm}}^{-3}$$ $$p=\frac{n_i^2}{n}=\frac{10^{20}}{2.2\times 10^{14}}=4.5\times 10^5{\text{ cm}}^{-3}$$

(e) In a nondegenerate germanium sample at room temperature, $n=10^{13}$/cm${}^3$, $n_i=2p$, and $N_A=0$. Determine $n_i$ and $N_D$.

Solution: $n_i=2p$, and $np=n_i^2=4p^2$, so $p=n_i/2$.

Also, $n+N_A=p+N_D$ with $N_A=0$: $N_D=n-p$.

$np=n_i^2=4p^2\Rightarrow n=4p\Rightarrow p=n/4=2.5\times 10^{12}$ cm${}^{-3}$

$n_i=2p=5\times 10^{12}$ cm${}^{-3}$

$N_D=n-p=10^{13}-2.5\times 10^{12}=7.5\times 10^{12}$ cm${}^{-3}$

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Problem 2.17

Problem: Determine the equilibrium electron and hole concentrations in a uniformly doped Si sample.

(a) $T=300$ K, $N_A\le N_D$, $N_D=10^{15}$/cm${}^3$.

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

(b) $T=300$ K, $N_A=10^{16}$/cm${}^3$, $N_D\le N_A$.

Solution: $p\approx N_A=10^{16}$/cm${}^3$, $n=10^{20}/10^{16}=10^4$ cm${}^{-3}$.

(c) $T=300$ K, $N_A=9\times 10^{15}$/cm${}^3$, $N_D=10^{16}$/cm${}^3$.

Solution: $N_D-N_A=10^{15}$/cm${}^3$ (n-type). $n\approx 10^{15}$/cm${}^3$, $p\approx 10^5$ cm${}^{-3}$.

(d) $T=450$ K, $N_A=0$, $N_D=10^{14}$/cm${}^3$.

Solution: From Fig. 2.20, $n_i(450\text{ K})\approx 10^{13}$/cm${}^3$. Since $n_i$ is comparable to $N_D$, use the general formula:

$$n=\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}=5\times 10^{13}+\sqrt{2.5\times 10^{27}+10^{26}}$$ $$=5\times 10^{13}+5.1\times 10^{13}=1.01\times 10^{14}{\text{ cm}}^{-3}$$ $$p=\frac{n_i^2}{n}=\frac{10^{26}}{1.01\times 10^{14}}=9.9\times 10^{11}{\text{ cm}}^{-3}$$

(e) $T=650$ K, $N_A=0$, $N_D=10^{14}$/cm${}^3$.

Solution: At 650 K, $n_i\gg N_D$, so the sample is intrinsic: $n\approx p\approx n_i\approx 10^{16}$/cm${}^3$ (from Fig. 2.20).

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Problem 2.18

Problem: For each of the conditions specified in Problem 2.17, determine the position of $E_i$, compute $E_F-E_i$, and draw a carefully dimensioned energy band diagram. Note: $E_G(\text{Si})=1.08$ eV at 450 K and $1.015$ eV at 650 K.

Solution:

(a) $n=10^{15}$/cm${}^3$:

$$E_F-E_i=kT\ln (n/n_i)=0.0259\ln (10^{15}/10^{10})=0.0259\times 11.51=0.298\text{ eV}$$

$E_F$ is 0.298 eV above $E_i$.

(b) $p=10^{16}$/cm${}^3$:

$$E_i-E_F=kT\ln (p/n_i)=0.0259\ln (10^{16}/10^{10})=0.0259\times 13.82=0.358\text{ eV}$$

$E_F$ is 0.358 eV below $E_i$.

(c) $n=10^{15}$/cm${}^3$: Same as (a), $E_F-E_i=0.298$ eV.

(d) $n=1.01\times 10^{14}$/cm${}^3$ at 450 K:

$kT=0.0259\times 450/300=0.03885$ eV

$$E_F-E_i=0.03885\ln (1.01\times 10^{14}/10^{13})=0.03885\times 2.31=0.0898\text{ eV}$$

(e) At 650 K, intrinsic: $E_F\approx E_i$.

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Problem 2.19

Problem: Assuming a nondegenerate Si sample and total impurity atom ionization, construct a MATLAB program that computes $n$, $p$, and $E_F-E_i$ given $T$, $N_D$, and $N_A$.

Solution:

% Problem 2.19: Carrier concentrations and Fermi level
T = input('Temperature in Kelvin, T = ');
ND = input('Donor concentration in cm^-3, ND = ');
NA = input('Acceptor concentration in cm^-3, NA = ');
 
% Constants
k = 8.617e-5;
q = 1.602e-19;
EG0 = 1.17;
alpha = 4.730e-4;
beta = 636;
 
% Band gap at T
EG = EG0 - alpha*T^2/(T+beta);
 
% Intrinsic carrier concentration (using Exercise 2.4 fit)
Eex = 0.0074;
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 = 2.510e19 * (T/300)^1.5 * sqrt(mnr*mpr) * exp(-(EG-Eex)/(2*k*T));
 
% Carrier concentrations
ND_NA = ND - NA;
if abs(ND_NA) > 10*ni
    if ND_NA > 0
        n = ND_NA;
        p = ni^2/n;
    else
        p = abs(ND_NA);
        n = ni^2/p;
    end
else
    n = ND_NA/2 + sqrt((ND_NA/2)^2 + ni^2);
    p = ni^2/n;
end
 
% Fermi level
EF_Ei = k*T*log(n/ni);
 
fprintf('ni = %.4e cm^-3\n', ni);
fprintf('n = %.4e cm^-3\n', n);
fprintf('p = %.4e cm^-3\n', p);
fprintf('EF - Ei = %.4f eV\n', EF_Ei);

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Problem 2.20

Problem: Verify that the maximum nondegenerate donor and acceptor doping concentrations in Si at room temperature are $N_D\approx 1.6\times 10^{18}$/cm${}^3$ and $N_A\approx 9.1\times 10^{17}$/cm${}^3$.

Solution:

The nondegenerate limit is $E_F=E_c-3kT$ for donors.

$$n=N_D=N_c{e}^{(E_F-E_c)/kT}=N_c{e}^{-3}=2.8\times 10^{19}\times 0.0498=1.39\times 10^{18}{\text{ cm}}^{-3}$$

Using more precise values: $N_c=2.8\times 10^{19}$ cm${}^{-3}$, $e^{-3}=0.0498$:

$$N_D=2.8\times 10^{19}\times 0.0498=1.39\times 10^{18}{\text{ cm}}^{-3}$$

Hmm, this gives $1.39\times 10^{18}$ rather than $1.6\times 10^{18}$. The text may use slightly different values. With $N_c=3.2\times 10^{19}$: $N_D=1.6\times 10^{18}$ cm${}^{-3}$.

For acceptors: $N_A=N_v{e}^{-3}=1.04\times 10^{19}\times 0.0498=5.2\times 10^{17}$ cm${}^{-3}$.

Hmm, this gives $5.2\times 10^{17}$ rather than $9.1\times 10^{17}$. Using $N_v=1.83\times 10^{19}$: $N_A=9.1\times 10^{17}$ cm${}^{-3}$.

---

Problem 2.21

Problem: Construct a computer program to produce a plot of $E_F-E_i$ versus $N_A$ or $N_D$ similar to Fig. 2.21.

Solution:

NA = logspace(14, 19, 100);
kT = 0.0259;
ni = 1e10;
EF_Ei = -kT*log(NA/ni); % for p-type
ND = logspace(14, 19, 100);
EF_Ei_n = kT*log(ND/ni); % for n-type
semilogx(NA, EF_Ei, ND, EF_Ei_n);
xlabel('NA or ND (cm^-3)');
ylabel('EF - Ei (eV)');
legend('p-type', 'n-type');

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Problem 2.22 — GaAs Considerations

(a) Make a sketch similar to Fig. 2.14 specifically appropriate for GaAs, taking into account that $m_n^{\ast }eq{m}_p^{\ast }$ in GaAs.

Solution: In GaAs, $m_n^{\ast }=0.066m_0$ and $m_p^{\ast }=0.52m_0$. The density of states $g_c(E)$ is much smaller than $g_v(E)$ because of the smaller electron effective mass. The peak of the electron distribution is lower and the peak of the hole distribution is higher.

(b) Based on (a), would you expect $E_i$ in GaAs to lie above or below midgap?

Solution: Since $m_p^{\ast }>m_n^{\ast }$, $N_v>N_c$. From $E_i=(E_c+E_v)/2+(3kT/4)\ln (m_p^{\ast }/m_n^{\ast })$, since $m_p^{\ast }/m_n^{\ast }>1$, $E_i$ lies above midgap.

(c) Determine the position of the intrinsic Fermi level in GaAs at 300 K.

Solution:

$$E_i=\frac{E_c+E_v}{2}+\frac{3kT}{4}\ln \left(\frac{m_p^{\ast }}{m_n^{\ast }}\right)$$ $$=\text{midgap}+\frac{3\times 0.0259}{4}\ln \left(\frac{0.52}{0.066}\right)=\text{midgap}+0.0194\times 2.06=\text{midgap}+0.040\text{ eV}$$

$E_i$ is 0.040 eV above midgap.

(d) Determine the maximum nondegenerate donor and acceptor doping concentrations in GaAs at room temperature.

Solution:

$N_c=4.7\times 10^{17}$ cm${}^{-3}$, $N_v=7.0\times 10^{18}$ cm${}^{-3}$

$$N_D=N_c{e}^{-3}=4.7\times 10^{17}\times 0.0498=2.3\times 10^{16}{\text{ cm}}^{-3}$$ $$N_A=N_v{e}^{-3}=7.0\times 10^{18}\times 0.0498=3.5\times 10^{17}{\text{ cm}}^{-3}$$

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Problem 2.23

Problem: Given an $N_A=10^{14}$/cm${}^3$ doped Si sample:

(a) Present a qualitative argument that leads to the approximate positioning of the Fermi level as $T\to 0$ K.

Solution: At $T\to 0$ K, carriers freeze out. The Fermi level approaches the acceptor level $E_A$, which is slightly above $E_v$.

(b) Construct a MATLAB program to calculate and plot $E_F$ versus $T$ for $200\text{ K}\le T\le 500$ K.

Solution:

T = 200:10:500;
NA = 1e14;
k = 8.617e-5;
EG0 = 1.17; alpha = 4.730e-4; beta = 636;
for i = 1:length(T)
    EG = EG0 - alpha*T(i)^2/(T(i)+beta);
    ni = 2.510e19*(T(i)/300)^1.5*sqrt((1.028+6.11e-4*T(i)-3.09e-7*T(i)^2)*(0.610+7.83e-4*T(i)-4.46e-7*T(i)^2))*exp(-(EG-0.0074)/(2*k*T(i)));
    if NA > 10*ni
        EF_Ei(i) = -k*T(i)*log(NA/ni);
    else
        p = NA/2 + sqrt((NA/2)^2 + ni^2);
        EF_Ei(i) = -k*T(i)*log(p/ni);
    end
end
plot(T, EF_Ei);
xlabel('T (K)'); ylabel('EF - Ei (eV)');

(c) What do you conclude relative to the general behavior of the Fermi level positioning as a function of temperature?

Solution: As temperature increases, $E_F$ moves toward midgap (intrinsic behavior). At low temperatures, $E_F$ is near the acceptor level.

(d) Run the program to determine what happens if $N_A$ is progressively increased from $10^{14}$ to $10^{18}$ cm${}^{-3}$.

Solution: Increasing $N_A$ pushes $E_F$ further below $E_i$ at low temperatures.

(e) How would your answers be modified if the Si sample was doped with donors instead of acceptors?

Solution: The Fermi level would move above $E_i$ instead of below.

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Chapter 3: Carrier Action

3.1 Drift

3.1.1 Definition-Visualization

Drift is charged-particle motion in response to an applied electric field. Carriers accelerate in the field direction (holes) or opposite to it (electrons), but collisions with impurities and lattice atoms interrupt the acceleration.

On a macroscopic scale, drift is visualized as all carriers of a given type moving at a constant drift velocity $v_d$.

3.1.2 Drift Current

For holes:

$$J_{p|drift}=qp{\mu }_p\mathcal{E}$$

For electrons:

$$J_{n|drift}=qn{\mu }_n\mathcal{E}$$

where ${\mu }_n$ and ${\mu }_p$ are the electron and hole mobilities.

3.1.3 Mobility

Standard units: cm${}^2$/V-sec

Sample values at 300 K:

• Si: ${\mu }_n\approx 1360$, ${\mu }_p\approx 460$ cm${}^2$/V-sec (at $N_D=10^{14}$/cm${}^3$)
• GaAs: ${\mu }_n\approx 8000$, ${\mu }_p\approx 400$ cm${}^2$/V-sec

Relationship to scattering:

$$\mu =\frac{q\tau }{m^{\ast }}$$

Doping dependence: Mobilities decrease with increasing doping above $\sim 10^{15}$ cm${}^{-3}$.

Temperature dependence: $\mu \propto T^{-2.3}$ for electrons, $\mu \propto T^{-2.2}$ for holes (at low doping).

3.1.4 Resistivity

$$\rho =\frac{1}{q({\mu }_{n}n+{\mu }_{p}p)}$$

For n-type:

$$\rho \approx \frac{1}{q{\mu }_n{N}_D}$$

For p-type:

$$\rho \approx \frac{1}{q{\mu }_p{N}_A}$$

3.1.5 Band Bending

The electric field is related to the slope of the energy bands:

$$\mathcal{E}=\frac{1}{q}\frac{d{E}_c}{dx}=\frac{1}{q}\frac{d{E}_v}{dx}=\frac{1}{q}\frac{d{E}_i}{dx}$$

The electrostatic potential is:

$$V=-\frac{1}{q}(E_c-E_{\text{ref}})$$

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Exercise 3.1

Problem: Construct a log-log plot of ${\mu }_n$ and ${\mu }_p$ versus $N_A$ or $N_D$ for Si at 300 K.

Solution:

% Mobility versus Dopant Concentration (Si 300K)
% Fit Parameters
NDref = 1.3e17; NAref = 2.35e17;
munmin = 92; mupmin = 54.3;
mun0 = 1268; mup0 = 406.9;
an = 0.91; ap = 0.88;
% Mobility Calculation
N = logspace(14,19);
mun = munmin + mun0./(1+(N/NDref).^an);
mup = mupmin + mup0./(1+(N/NAref).^ap);
% Plotting results
loglog(N,mun,N,mup); grid;
axis([1.0e14 1.0e19 1.0e1 1.0e4]);
xlabel('NA or ND (cm-3)');
ylabel('Mobility (cm2/V-sec)');
text(1.0e15,1500,'Electrons');
text(1.0e15,500,'Holes');
text(1.0e18,2000,'Si, 300K');

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Exercise 3.2

Problem: Consider the energy band diagram with $E_c-E_F=E_G/4$ at $x=\pm L$ and $E_F-E_v=E_G/4$ at $x=0$. Take $E_F$ as the energy reference.

(a) Sketch the electrostatic potential $V(x)$.

Solution: $V(x)$ has the same shape as the “upside-down” of $E_c(x)$.

(b) Sketch the electric field $\mathcal{E}(x)$.

Solution: $\mathcal{E}(x)$ is proportional to the slope of the bands. It is zero at $x=\pm L$ and $x=0$, maximum in magnitude between these points.

(c) Ascertain the K.E. and P.E. of the electrons and holes pictured.

Solution:

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

(d) Determine the resistivity of the $x>L$ portion.

Solution: In the $x>L$ region, $E_c-E_F=E_G/4=0.28$ eV.

$$N_A=p=n_i{e}^{(E_i-E_F)/kT}=10^{10}{e}^{0.28/0.0259}=4.96\times 10^{14}{\text{ cm}}^{-3}$$

From Fig. 3.5(a), ${\mu }_p=459$ cm${}^2$/V-sec.

$$\rho =\frac{1}{q{\mu }_{p}p}=\frac{1}{(1.6\times 10^{-19})(459)(4.96\times 10^{14})}=27.5\text{ ohm-cm}$$

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3.2 Diffusion

3.2.1 Definition-Visualization

Diffusion is the process whereby particles tend to spread out or redistribute as a result of random thermal motion, migrating from regions of high concentration to regions of low concentration.

3.2.2 Hot-Point Probe Measurement

The hot-point probe determines whether a semiconductor is n- or p-type. The heated probe creates energetic carriers that diffuse away, creating a charge imbalance detected by a center-zero milliammeter.

3.2.3 Diffusion and Total Currents

Fick’s law:

$$F=-D\nabla \eta$$

Diffusion current densities:

$$J_{p|diff}=-q{D}_p\nabla p$$ $$J_{n|diff}=q{D}_n\nabla n$$

Total currents:

$$J_p=q{\mu }_{p}p\mathcal{E}-q{D}_p\nabla p$$ $$J_n=q{\mu }_{n}n\mathcal{E}+q{D}_n\nabla n$$ $$J=J_n+J_p$$

3.2.4 Relating Diffusion Coefficients/Mobilities

Einstein relationship:

$$\frac{D_n}{{\mu }_n}=\frac{D_p}{{\mu }_p}=\frac{kT}{q}$$

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Exercise 3.3

The “DiffDemo” MATLAB file simulates diffusion in a one-dimensional system.

% Simulation of Diffusion (DiffDemo)
% One-dimensional system, right and left jumps equally probable
% Initialization
close
x = [0.5 1.5 2.5 3.5 4.5];
y = [1.0e6 0 0 0 0]; % NOTE: initial position can be changed
[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; % NOTE: increase N for extended run
% Computations and Plotting
for ii = 1:N,
    % Diffusion step calculation
    bin(1) = round(y(1)/2 + y(2)/2);
    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);
    y = bin;
    % Plotting the result
    [xp,yp] = bar(x,y);
    axis(axis);
    plot(xp,yp); text(0.5,1.1e6,['t=',num2str(ii)]);
    axis([0,5,0,1.2e6]);
    pause(0.5)
end

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Exercise 3.4

Problem: (a) The semiconductor is in equilibrium. How does one deduce this from the energy band diagram?

Solution: The Fermi level is invariant as a function of position.

(b) What is the electron current density and hole current density at $x=\pm L/2$?

Solution: $J_n=0$ and $J_p=0$ everywhere under equilibrium conditions.

(c) Roughly sketch $n$ and $p$ versus $x$.

Solution: $n$ is highest at $x=0$ and decreases symmetrically. $p$ is lowest at $x=0$ and increases symmetrically.

(d) Is there an electron diffusion current at $x=\pm L/2$?

Solution: Yes. At $x=-L/2$, $dn/dx>0$, so $J_{n|diff}$ flows in the $+x$ direction. At $x=+L/2$, $dn/dx<0$, so $J_{n|diff}$ flows in the $-x$ direction.

(e) Is there an electron drift current at $x=\pm L/2$?

Solution: Yes. $J_{n|drift}$ is in the $-x$ direction at $x=-L/2$ and in the $+x$ direction at $x=+L/2$, canceling the diffusion component.

(f) What is the hole diffusion coefficient in the $x>L$ region?

Solution: From part (d) of Exercise 3.2, $N_A\approx 5\times 10^{14}$/cm${}^3$ and ${\mu }_p=459$ cm${}^2$/V-sec.

$$D_p=\frac{kT}{q}{\mu }_p=0.0259\times 459=11.9{\text{ cm}}^2/\text{sec}$$

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3.3 Recombination-Generation

3.3.1 Definition-Visualization

• Recombination: A process whereby electrons and holes are annihilated or destroyed
• Generation: A process whereby electrons and holes are created

Processes:

• Band-to-band recombination
• R-G center recombination (indirect)
• Auger recombination
• Band-to-band generation
• R-G center generation
• Impact ionization

3.3.2 Momentum Considerations

Direct semiconductors: Conduction band minimum and valence band maximum at $k=0$ (e.g., GaAs)

Indirect semiconductors: Conduction band minimum at $keq0$ (e.g., Si, Ge)

Band-to-band recombination is much more efficient in direct semiconductors.

3.3.3 R-G Statistics

For low-level injection in an n-type material:

$${\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}=-\frac{\Delta p}{{\tau }_p}$$

For low-level injection in a p-type material:

$${\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=-\frac{\Delta n}{{\tau }_n}$$

General case:

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

3.3.4 Minority Carrier Lifetimes

The minority carrier lifetimes ${\tau }_n$ and ${\tau }_p$ are the average times an excess minority carrier lives in a sea of majority carriers.

Lifetime measurement: Photoconductivity decay

$$\Delta p(t)=\Delta p_0{e}^{-t/{\tau }_p}$$

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Sample Problem No. 1

Problem: A uniformly donor-doped silicon wafer at room temperature is suddenly illuminated with light at $t=0$. $N_D=10^{15}$/cm${}^3$, ${\tau }_p=10^{-6}$ sec, $G_L=10^{17}$ electron-hole pairs/cm${}^3$-sec. Determine $\Delta p_n(t)$ for $t>0$.

Solution:

Step 1: Given: Si, $T=300$ K, $N_D=10^{15}$/cm${}^3$, $G_L=10^{17}$/cm${}^3$-sec.

Step 2: Equilibrium: $n_0=N_D=10^{15}$/cm${}^3$, $p_0=n_i^2/N_D=10^5$ cm${}^{-3}$.

Step 3: Qualitative analysis: $\Delta p_n(t)$ starts at zero, builds up, and eventually reaches a steady state $G_L{\tau }_p$.

Step 4: The minority carrier diffusion equation with no position dependence:

$$\frac{d\Delta p_n}{dt}=-\frac{\Delta p_n}{{\tau }_p}+G_L$$

General solution:

$$\Delta p_n(t)=G_L{\tau }_p+A{e}^{-t/{\tau }_p}$$

Boundary condition: $\Delta p_n(0)=0\Rightarrow A=-G_L{\tau }_p$

$$\Delta p_n(t)=G_L{\tau }_p(1-e^{-t/{\tau }_p})$$

Step 5: Check: At $t=0$, $\Delta p_n=0$. As $t\to \infty$, $\Delta p_n\to G_L{\tau }_p=10^{17}\times 10^{-6}=10^{11}$ cm${}^{-3}$.

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Sample Problem No. 2

Problem: The $x=0$ end of a uniformly doped semi-infinite bar of silicon with $N_D=10^{15}$/cm${}^3$ is illuminated so as to create $\Delta p_{n0}=10^{10}$/cm${}^3$ excess holes at $x=0$. The wavelength of the illumination is such that no light penetrates into the interior ($x>0$). Determine $\Delta p_n(x)$.

Solution:

Steady state: $\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{L_p^2}=0$ for $x>0$

General solution: $\Delta p_n(x)=A{e}^{-x/L_p}+B{e}^{x/L_p}$

Boundary conditions:

• $\Delta p_n(0)=\Delta p_{n0}$
• $\Delta p_n(\infty )=0\Rightarrow B=0$

Therefore:

$$\Delta p_n(x)=\Delta p_{n0}{e}^{-x/L_p}$$

where $L_p=\sqrt{D_p{\tau }_p}$ is the hole diffusion length.

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3.4 Equations of State

3.4.1 Continuity Equations

$$\frac{\partial n}{\partial t}=\frac{1}{q}\nabla \cdot J_n+{\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial n}{\partial t}|}_{\text{other processes}}$$ $$\frac{\partial p}{\partial t}=-\frac{1}{q}\nabla \cdot J_p+{\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}+{\frac{\partial p}{\partial t}|}_{\text{other processes}}$$

3.4.2 Minority Carrier Diffusion Equations

$$\frac{\partial \Delta n_p}{\partial t}=D_n\frac{{\partial }^2\Delta n_p}{\partial x^2}-\frac{\Delta n_p}{{\tau }_n}+G_L$$ $$\frac{\partial \Delta p_n}{\partial t}=D_p\frac{{\partial }^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{{\tau }_p}+G_L$$

3.4.3 Simplifications and Solutions

Common simplifications:

Simplification	Effect
Steady state	$\partial \Delta /\partial t\to 0$
No concentration gradient	${\partial }^2\Delta /\partial x^2\to 0$
No drift current	$\mathcal{E}=0$
No thermal R-G	$\Delta /\tau \to 0$
No light	$G_L=0$

Common solutions:

Solution	Given	Simplified Equation	Solution
1	Steady state, no light	$0=D_n\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n}$	$\Delta n_p(x)=A{e}^{-x/L_n}+B{e}^{x/L_n}$
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	$0=D_n\frac{d^2\Delta n_p}{d{x}^2}$	$\Delta n_p(x)=A+Bx$

3.5 Supplemental Concepts

3.5.1 Diffusion Lengths

$$L_p=\sqrt{D_p{\tau }_p}\text{(holes in n-type)}$$ $$L_n=\sqrt{D_n{\tau }_n}\text{(electrons in p-type)}$$

Physically, $L_p$ and $L_n$ represent the average distance minority carriers can diffuse into a sea of majority carriers before being annihilated.

3.5.2 Quasi-Fermi Levels

Under nonequilibrium conditions:

$$n=n_i{e}^{(F_N-E_i)/kT}$$ $$p=n_i{e}^{(E_i-F_P)/kT}$$

where $F_N$ and $F_P$ are the quasi-Fermi levels for electrons and holes.

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Exercise 3.5

Problem: In Sample Problem No. 2, nonpenetrating illumination causes $\Delta p_n(x)=\Delta p_{n0}\exp (-x/L_p)$. Establish relationships for $F_N$ and $F_P$ in the illuminated bar.

Solution:

Since $n\approx n_0$, $F_N\approx E_F$.

$$F_P=E_i-kT\ln (p/n_i)=E_i-kT\ln [p_{n0}/n_i+(\Delta p_{n0}/n_i)e^{-x/L_p}]$$

For $\Delta p_n(x)\gg p_{n0}$:

$$F_P\approx E_i-kT\ln [(\Delta p_{n0}/n_i)e^{-x/L_p}]=E_i-kT\ln (\Delta p_{n0}/n_i)+kT(x/L_p)$$

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Problems — Chapter 3

Problem 3.1

Problem: Using the energy band diagram, indicate how one visualizes (a) the existence of an electric field, (b) an electron with K.E. = 0, (c) a hole with K.E. = $E_G/4$, (d) photogeneration, (e) direct thermal generation, (f) band-to-band recombination, (g) recombination via R-G centers, (h) generation via R-G centers.

Solution:

• (a) Band bending (slope of $E_c$ or $E_v$)
• (b) Electron at the conduction band edge
• (c) Hole at $E_v+E_G/4$ (i.e., $E=E_v-E_G/4$ from the hole’s perspective)
• (d) Electron excited from valence band to conduction band by photon
• (e) Electron excited from valence band to conduction band by thermal energy
• (f) Electron in conduction band annihilates with hole in valence band
• (g) Electron trapped at R-G center, then annihilates with hole
• (h) Electron and hole generated via R-G center

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Problem 3.2 — Short Answer

(a) A 1-cm-long semiconductor bar has a cross-sectional area of $10^{-2}$ cm${}^2$ and a resistance of 100 ohms. What is the hole mobility inside the bar?

Solution: Assuming p-type with $N_A=10^{15}$/cm${}^3$:

$$R=\frac{\rho L}{A}\Rightarrow \rho =\frac{RA}{L}=\frac{100\times 10^{-2}}{1}=1\text{ ohm-cm}$$ $$\rho =\frac{1}{q{\mu }_{p}p}\Rightarrow {\mu }_p=\frac{1}{q\rho p}=\frac{1}{(1.6\times 10^{-19})(1)(10^{15})}=6250{\text{ cm}}^2/\text{V-sec}$$

(b) Name the two dominant carrier scattering mechanisms in nondegenerately doped semiconductors of device quality.

Solution: Lattice scattering and ionized impurity scattering.

(c) For a given semiconductor, the carrier mobilities in intrinsic material are (choose one: higher than, lower than, the same as) those in heavily doped material. Briefly explain.

Solution: Higher than. In heavily doped material, ionized impurity scattering is significant, reducing mobility.

(d) Two GaAs wafers, one n-type and one p-type, are uniformly doped such that $N_D(\text{wafer 1})=N_A(\text{wafer 2})>n_i$. Which wafer will exhibit the larger resistivity? Explain.

Solution: The p-type wafer (wafer 2) will have larger resistivity because ${\mu }_p<{\mu }_n$ in GaAs.

(e) The electron mobility in a silicon sample is determined to be 1300 cm${}^2$/V-sec at room temperature. What is the electron diffusion coefficient?

Solution: $D_n=(kT/q){\mu }_n=0.0259\times 1300=33.7$ cm${}^2$/sec.

(f) What is the algebraic statement of low-level injection?

Solution: $\Delta n\ll p_0$ (in p-type) or $\Delta p\ll n_0$ (in n-type).

(g) Light is used to create excess carriers in silicon. These excess carriers will predominantly recombine via (choose one: band-to-band, R-G center, or photo) recombination.

Solution: R-G center recombination (Si is an indirect semiconductor).

(h) Prior to processing, a silicon sample contains $N_D=10^{14}$/cm${}^3$ donors and $N_T=10^{15}$/cm${}^3$ R-G centers. After processing, the sample contains $N_D=10^{17}$/cm${}^3$ donors and $N_T=10^{10}$/cm${}^3$ R-G centers. Did processing increase or decrease the minority carrier lifetime? Explain.

Solution: Processing decreased $N_T$ by a factor of $10^5$, which increases the lifetime by the same factor (since $\tau \propto 1/N_T$).

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Problem 3.3

Problem: Complete part (b) of Exercise 3.1.

Solution: Part (b) requires constructing log-log plots of ${\mu }_n$ and ${\mu }_p$ versus $T$ for various dopings. The results should match Figs. 3.7(a) and (b).

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Problem 3.4

Problem: The measurement described in the reference is used to determine the mobility in an $N_D<10^{15}$/cm${}^3$ Si sample as a function of temperature. Use the MATLAB polyfit function to determine the best fit value of the power factor $b$.

Solution:

T = [290 280 270 260 250 240 230 220 210 200 190 180 170 160 150];
mun = [1501 1646 1805 1985 2185 2415 2675 2978 3306 3743 4209 4619 5216 5910 6757];
lnT = log(T);
lnmun = log(mun);
p = polyfit(lnT, lnmun, 1);
b = -p(1);
% b ≈ 2.3

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Problem 3.5 — Intrinsic and Maximum Resistivity

Problem: (a) Determine the resistivity of intrinsic Ge, Si, and GaAs at 300 K.

Solution:

For intrinsic material: $\rho =\frac{1}{q{n}_i({\mu }_n+{\mu }_p)}$

Material	$n_i$ (cm${}^{-3}$)	${\mu }_n$ (cm${}^2$/V-sec)	${\mu }_p$ (cm${}^2$/V-sec)	$\rho$ (ohm-cm)
Ge	$2\times 10^{13}$	3900	1900	47
Si	$10^{10}$	1350	480	$2.3\times 10^5$
GaAs	$2\times 10^6$	8500	400	$3.5\times 10^8$

(b) Determine the maximum possible resistivity of Ge, Si, and GaAs at 300 K.

Solution: The maximum resistivity occurs for intrinsic material (minimum carrier concentration). The values are the same as part (a).

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Problem 3.6 — More Resistivity Questions

Problem: (a) A silicon sample at room temperature is uniformly doped with $10^{17}$/cm${}^3$ donors. Calculate the resistivity.

Solution:

$n\approx N_D=10^{17}$/cm${}^3$. From Fig. 3.5(a), ${\mu }_n\approx 800$ cm${}^2$/V-sec.

$$\rho =\frac{1}{q{\mu }_n{N}_D}=\frac{1}{(1.6\times 10^{-19})(800)(10^{17})}=0.078\text{ ohm-cm}$$

(b) The sample is compensated by adding $N_A=10^{16}$/cm${}^3$ acceptors. Calculate the resistivity.

Solution: $N_D-N_A=9\times 10^{16}$/cm${}^3$. From Fig. 3.5(a), ${\mu }_n\approx 850$ cm${}^2$/V-sec.

$$\rho =\frac{1}{(1.6\times 10^{-19})(850)(9\times 10^{16})}=0.082\text{ ohm-cm}$$

(c) Compute the resistivity of intrinsic silicon at room temperature.

Solution:

$$\rho =\frac{1}{q{n}_i({\mu }_n+{\mu }_p)}=\frac{1}{(1.6\times 10^{-19})(10^{10})(1350+480)}=3.4\times 10^5\text{ ohm-cm}$$

(d) A 500-ohm resistor is to be made from a bar-shaped piece of n-type Si. The bar has a cross-sectional area of $10^{-2}$ cm${}^2$ and a current-carrying length of 1 cm. Determine the doping required.

Solution:

$$\rho =\frac{RA}{L}=\frac{500\times 10^{-2}}{1}=5\text{ ohm-cm}$$

From Fig. 3.8(a), $\rho =5$ ohm-cm corresponds to $N_D\approx 10^{15}$/cm${}^3$.

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Problem 3.7

Problem: Construct a plot of the silicon resistivity versus impurity concentration at $T=300$ K.

Solution:

N = logspace(14, 20, 100);
% n-type
mun = 92 + 1268./(1 + (N/1.3e17).^0.91);
rho_n = 1./(1.6e-19 * mun .* N);
% p-type
mup = 54.3 + 406.9./(1 + (N/2.35e17).^0.88);
rho_p = 1./(1.6e-19 * mup .* N);
loglog(N, rho_n, N, rho_p);
xlabel('NA or ND (cm^-3)');
ylabel('Resistivity (ohm-cm)');
legend('n-type', 'p-type');

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Problem 3.8

Problem: Resistors in ICs are sometimes thin semiconductor layers near the surface of a wafer.

(a) Given a bar-shaped layer of width $W$, length $L$, and depth $d$, and assuming an arbitrary $N_D(x)$ variation with depth $x$ from the wafer surface, show that:

$$R=\frac{L}{W}\frac{1}{q{\int }_0^d{\mu }_n(x)N_D(x)dx}$$

Solution:

The conductance of a differential layer at depth $x$ is:

$$dG=\frac{q{\mu }_n(x)N_D(x)Wdx}{L}$$

Integrating:

$$G=\frac{W}{L}q{\int }_0^d{\mu }_n(x)N_D(x)dx$$ $$R=\frac{1}{G}=\frac{L}{W}\frac{1}{q{\int }_0^d{\mu }_n(x)N_D(x)dx}$$

(b) Taking $N_D(x)=N_{D0}\exp (-ax)+N_{DB}$, compute and plot $R$ versus $N_{D0}$.

Solution:

ND0 = logspace(14, 18, 100);
NDB = 1e14; d = 5e-4; a = 1e4; L = 1; W = 1;
for i = 1:length(ND0)
    x = linspace(0, d, 100);
    ND = ND0(i)*exp(-a*x) + NDB;
    mun = 92 + 1268./(1 + (ND/1.3e17).^0.91);
    integrand = mun.*ND;
    R(i) = L/W / (1.6e-19 * trapz(x, integrand));
end
semilogx(ND0, R);
xlabel('ND0 (cm^-3)'); ylabel('R (ohms)');

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Problem 3.9

Problem: Modify your mobility versus $T$ program to compute and construct semilog plots of $\rho$ versus $T$ for $200\text{ K}\le T\le 500$ K and $N_D$ or $N_A$ stepped in decade values from $10^{14}$/cm${}^3$ to $10^{18}$/cm${}^3$.

Solution:

T = 200:10:500;
ND = [1e14 1e15 1e16 1e17 1e18];
for j = 1:length(ND)
    for i = 1:length(T)
        kT = 0.0259*T(i)/300;
        mun = (92 + 1268./(1+(ND(j)/1.3e17).^0.91)) * (T(i)/300)^(-2.3);
        rho(j,i) = 1/(1.6e-19 * mun * ND(j));
    end
end
semilogy(T, rho);
xlabel('T (K)'); ylabel('Resistivity (ohm-cm)');

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Problem 3.10 — Temperature Sensor

Problem: Design a temperature sensor for measuring ambient outside temperature in W. Lafayette, IN ($-30^{\circ }$C to $40^{\circ }$C) based on the change in resistance of a bar-shaped piece of Si.

Solution:

(a) Doping and dimensions: Choose $N_D=10^{16}$/cm${}^3$ (moderate doping for reasonable sensitivity). For $R$ between 1 and 1000 ohms, with $L/W=10$ and $d=0.1$ cm, $W=0.01$ cm, $L=0.1$ cm:

$$R=\frac{\rho L}{Wd}=\frac{\rho \times 0.1}{0.01\times 0.1}=100\rho$$

At $T=300$ K, $\rho \approx 0.5$ ohm-cm, so $R\approx 50$ ohms.

(b) Sensitivity: $dR/dT=100d\rho /dT$. Since $\rho \propto {\mu }^{-1}\propto T^{2.3}$, $d\rho /dT>0$. Lower doping gives higher sensitivity.

(c) Upper limit: intrinsic temperature. Lower limit: freeze-out.

(d) Plot $R$ versus $T$.

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Problem 3.11

Problem: Considering a nondegenerate semiconductor at room temperature, compute the average speed of an electron corresponding to the peak of the electron distribution.

Solution:

The peak of the electron distribution is at $E=E_c+kT/2$. The kinetic energy is $kT/2$.

$$\frac{1}{2}{m}_n^{\ast }{v}^2=\frac{kT}{2}\Rightarrow v=\sqrt{\frac{kT}{m_n^{\ast }}}$$

For Si: $m_n^{\ast }=1.18m_0=1.18\times 9.109\times 10^{-31}$ kg

$$v=\sqrt{\frac{(1.38\times 10^{-23})(300)}{1.18\times 9.109\times 10^{-31}}}=\sqrt{\frac{4.14\times 10^{-21}}{1.075\times 10^{-30}}}=\sqrt{3.85\times 10^9}=6.2\times 10^4\text{ m/sec}$$

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Problem 3.12 — Interpretation of Energy Band Diagrams

Problem: Six different silicon samples at 300 K are characterized by the energy band diagrams in Fig. P3.12.

Solution: (Answers will depend on the specific diagram chosen.)

For a sample with $E_F$ constant: equilibrium prevails. $V(x)$ is the upside-down of $E_c(x)$. $\mathcal{E}(x)$ is the slope of $E_c(x)$. Carrier concentrations are determined from $E_c-E_F$ and $E_F-E_v$.

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Problem 3.13

Problem: The nonuniform doping in the central region of a BJT creates a built-in field. Suppose the BJT is Si at room temperature with a central region of length $L$. The acceptor doping is:

$$p(x)\equiv N_A(x)=n_i{e}^{(a-x)/b}0\le x\le L$$

where $a=1.8$ μm, $b=0.1$ μm, $L=0.8$ μm.

(a) Draw the energy band diagram.

Solution:

$$N_A(x)=n_i{e}^{(a-x)/b}$$

At $x=0$: $N_A=10^{10}{e}^{18}=6.6\times 10^{17}$ cm${}^{-3}$

At $x=L=0.8$ μm: $N_A=10^{10}{e}^{10}=2.2\times 10^{14}$ cm${}^{-3}$

$E_i-E_F=kT\ln (N_A/n_i)=(a-x)kT/b$

At $x=0$: $E_i-E_F=1.8\times 0.0259/0.1=0.466$ eV

At $x=L$: $E_i-E_F=(1.8-0.8)\times 0.0259/0.1=0.259$ eV

(b) Make a sketch of the $\mathcal{E}$-field and compute the value at $x=L/2$.

Solution:

$$\mathcal{E}(x)=\frac{kT}{q}\frac{d{N}_A/dx}{N_A}=\frac{kT}{q}\left(-\frac{1}{b}\right)=-\frac{0.0259}{0.1\times 10^{-4}}=-2590\text{ V/cm}$$

(c) Is the built-in electric field such as to aid the motion of minority carrier electrons in going from $x=0$ to $x=L$?

Solution: Yes. The field points in the $-x$ direction, aiding electron motion in the $+x$ direction.

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Problem 3.14

Problem: (a) Roughly sketch the expected variation of $D_n$ and $D_p$ versus doping for Si at $T=300$ K.

Solution: $D=(kT/q)\mu$. Since mobility decreases with doping, $D$ decreases with doping.

(b) Construct a plot of $D_n$ and $D_p$ versus $N_A$ or $N_D$ for Si at 300 K.

N = logspace(14, 18, 100);
Dn = 0.0259 * (92 + 1268./(1+(N/1.3e17).^0.91));
Dp = 0.0259 * (54.3 + 406.9./(1+(N/2.35e17).^0.88));
loglog(N, Dn, N, Dp);
xlabel('NA or ND (cm^-3)'); ylabel('Diffusion coefficient (cm^2/sec)');

(c) Why was the upper doping limit taken to be $10^{18}$/cm${}^3$?

Solution: Because above this doping, the semiconductor becomes degenerate and the Einstein relationship (which assumes nondegeneracy) is no longer valid.

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Problem 3.15

Problem: Show that the general-case R-G relationship of Eq. (3.35) reduces to the special-case relationships of Eqs. (3.34) under low-level injection conditions.

Solution:

For an n-type material under low-level injection:

• $n\approx n_0\gg p$
• $p=p_0+\Delta p$
• $n_1=n_i{e}^{(E_T-E_i)/kT}$, $p_1=n_i{e}^{(E_i-E_T)/kT}$

Eq. (3.35):

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

With $n\approx n_0$, $np-n_i^2=n_0(p_0+\Delta p)-n_i^2=n_0\Delta p$ (since $n_0{p}_0=n_i^2$).

Denominator: ${\tau }_p(n_0+n_1)+{\tau }_n(p_0+\Delta p+p_1)$

Since $n_0\gg n_1,p_0,p_1,\Delta p$:

$$\approx {\tau }_p{n}_0$$

Therefore:

$${\frac{\partial p}{\partial t}|}_{\text{R-G}}=-\frac{n_0\Delta p}{{\tau }_p{n}_0}=-\frac{\Delta p}{{\tau }_p}$$

which is Eq. (3.34a).

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Problem 3.16

Problem: Explain why the minority carrier diffusion equation is called a diffusion equation, a minority carrier equation, and is valid only under low-level injection.

Solution:

• Diffusion equation: It contains a second-order spatial derivative term ($D{\partial }^2\Delta /\partial x^2$) that describes diffusion.
• Minority carrier equation: It applies to minority carriers only — electrons in p-type material and holes in n-type material.
• Low-level injection: The derivation neglects the drift component of the current, which is valid only when the minority carrier concentration is much smaller than the majority carrier concentration.

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Problem 3.17

Problem: Show that under steady state conditions, $\Delta p_n(x)=\Delta p_{n0}(1-x/L)$ for $0\le x\le L$ is the special-case solution if R-G is negligible and $\Delta p_n(0)=\Delta p_{n0}$, $\Delta p_n(L)=0$.

Solution:

With R-G negligible, the diffusion equation reduces to:

$$\frac{d^2\Delta p_n}{d{x}^2}=0$$

General solution: $\Delta p_n(x)=A+Bx$

Boundary conditions:

• $\Delta p_n(0)=A=\Delta p_{n0}$
• $\Delta p_n(L)=A+BL=0\Rightarrow B=-\Delta p_{n0}/L$

Therefore:

$$\Delta p_n(x)=\Delta p_{n0}(1-x/L)$$

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Problem 3.18

Problem: The earth is hit by a mysterious ray that momentarily wipes out all minority carriers. A uniformly doped silicon wafer with $N_A=10^{16}$/cm${}^3$, ${\tau }_n=10^{-6}$ sec, $T=300$ K is struck at $t=0$.

(a) Determine $\Delta n_p(t)$ for $t>0$.

Solution:

The minority carrier diffusion equation:

$$\frac{d\Delta n_p}{dt}=-\frac{\Delta n_p}{{\tau }_n}$$

With $\Delta n_p(0)=-n_{p0}$ (all minority carriers wiped out):

$$\Delta n_p(t)=-n_{p0}{e}^{-t/{\tau }_n}$$

(b) Does generation or recombination dominate at $t=0^{+}$?

Solution: Generation dominates because $\Delta n_p<0$ (deficit of minority carriers).

(c) Do low-level injection conditions exist at $t=0$?

Solution: Yes, because $|\Delta n_p|=n_{p0}\ll p_{p0}=N_A$.

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Problem 3.19

Problem: A silicon wafer ($N_A=10^{14}$/cm${}^3$, ${\tau }_n=1$ μsec) is first illuminated for $t>{\tau }_n$ with $G_{L0}=10^{16}$ cm${}^{-3}$-sec${}^{-1}$. At $t=0$ the light intensity is reduced to $G_L=G_{L0}/2$. Determine $\Delta n_p(t)$ for $t\ge 0$.

Solution:

For $t<0$ (steady state): $\Delta n_p=G_{L0}{\tau }_n=10^{16}\times 10^{-6}=10^{10}$ cm${}^{-3}$.

For $t\ge 0$:

$$\frac{d\Delta n_p}{dt}=-\frac{\Delta n_p}{{\tau }_n}+G_{L0}/2$$

General solution:

$$\Delta n_p(t)=G_{L0}{\tau }_n/2+A{e}^{-t/{\tau }_n}$$

At $t=0$: $\Delta n_p(0)=G_{L0}{\tau }_n=10^{10}$ cm${}^{-3}$

$$10^{10}=5\times 10^9+A\Rightarrow A=5\times 10^9$$ $$\Delta n_p(t)=5\times 10^9(1+e^{-t/{\tau }_n}){\text{ cm}}^{-3}$$

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Problem 3.20

Problem: A semi-infinite p-type bar is illuminated with light generating $G_L$ electron-hole pairs/cm${}^3$-sec uniformly. Carriers are extracted at $x=0$ making $\Delta n_p=0$ at $x=0$. Determine $\Delta n_p(x)$.

Solution:

Steady state diffusion equation:

$$D_n\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n}+G_L=0$$

Boundary conditions:

• $\Delta n_p(0)=0$
• $\Delta n_p(\infty )=G_L{\tau }_n$

Solution:

$$\Delta n_p(x)=G_L{\tau }_n(1-e^{-x/L_n})$$

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Problem 3.21

Problem: The two ends of a uniformly doped n-type Si bar of length $L$ are illuminated so as to create $\gamma N_D$ excess holes at both $x=0$ and $x=L$. No light penetrates into the interior.

(a) Sketch the expected general form of $\Delta p_n(x)$.

Solution: Symmetric distribution with maxima at both ends and a minimum in the middle.

(b) Do low-level injection conditions prevail?

Solution: Yes, since $\gamma \ll 1$ and $\Delta p_n\le \gamma N_D\ll N_D$.

(c) Write the differential equation to solve.

Solution:

$$D_p\frac{d^2\Delta p_n}{d{x}^2}-\frac{\Delta p_n}{{\tau }_p}=0\text{for }0<x<L$$

(d) Write the general solution and boundary conditions.

Solution:

$$\Delta p_n(x)=A{e}^{-x/L_p}+B{e}^{x/L_p}$$

Boundary conditions: $\Delta p_n(0)=\Delta p_n(L)=\gamma N_D$

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Problem 3.22

Problem: The $x>0$ portion of an infinite semiconductor is illuminated with light generating $G_L=10^{15}$ cm${}^{-3}$-sec${}^{-1}$ uniformly. $G_L=0$ for $x<0$. Si, $N_D=10^{15}$/cm${}^3$, ${\tau }_p=10^{-6}$ sec, $T=300$ K.

(a) What is the hole concentration at $x=-\infty$?

Solution: $p=p_{n0}=n_i^2/N_D=10^5$ cm${}^{-3}$.

(b) What is the hole concentration at $x=+\infty$?

Solution: $p=p_{n0}+G_L{\tau }_p=10^5+10^{15}\times 10^{-6}=10^9$ cm${}^{-3}$.

(c) Do low-level injection conditions prevail?

Solution: Yes, since $\Delta p=10^9\ll N_D=10^{15}$.

(d) Determine $\Delta p_n(x)$ for all $x$.

Solution:

For $x>0$: $\Delta p_n(x)=G_L{\tau }_p+A{e}^{-x/L_p}$

For $x<0$: $\Delta p_n(x)=B{e}^{x/L_p}$

Continuity at $x=0$: $G_L{\tau }_p+A=B$

Continuity of derivative at $x=0$: $-A/L_p=B/L_p\Rightarrow -A=B$

Therefore: $G_L{\tau }_p+A=-A\Rightarrow A=-G_L{\tau }_p/2$, $B=G_L{\tau }_p/2$

For $x>0$: $\Delta p_n(x)=G_L{\tau }_p(1-\frac{1}{2}{e}^{-x/L_p})$

For $x<0$: $\Delta p_n(x)=\frac{G_L{\tau }_p}{2}{e}^{x/L_p}$

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Problem 3.23 — CdS Photoconductor

Problem: (a) Speculate why the conducting film has a snake-like pattern.

Solution: To increase the resistance path length and achieve a high dark resistance in a compact area.

(b) The VT333 resistor pattern is approximately 0.3 mm wide and 3 cm long. Estimating the deposited CdS film to be 5 μm thick, $N_D=10^{13}$/cm${}^3$, and ${\mu }_n=100$ cm${}^2$/V-sec, compute the dark resistance.

Solution:

$$R=\frac{L}{W}\frac{1}{q{\mu }_n{N}_{D}d}=\frac{3}{0.03}\frac{1}{(1.6\times 10^{-19})(100)(10^{13})(5\times 10^{-4})}$$ $$=100\times \frac{1}{8\times 10^{-8}}=1.25\times 10^9\text{ ohms}$$

(c) The VT333 exhibited a 250 ohm resistance when illuminated with a microscope light. Can the usual relationships be used to determine $G_L$?

Solution: Not directly, because high-level injection conditions likely prevail, and the simple relationships assume low-level injection.

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Problem 3.24

Problem: The equilibrium and steady state conditions before and after illumination of a semiconductor are characterized by the energy band diagrams in Fig. P3.24.

(a) Determine $n_0$ and $p_0$.

Solution: From the before-illumination diagram: $E_F-E_i=0.3$ eV.

$$n_0=n_i{e}^{0.3/0.0259}=10^{10}{e}^{11.58}=1.07\times 10^{15}{\text{ cm}}^{-3}$$ $$p_0=n_i^2/n_0=10^{20}/1.07\times 10^{15}=9.35\times 10^4{\text{ cm}}^{-3}$$

(b) Determine $n$ and $p$ under steady state illumination.

Solution: From the after-illumination diagram: $F_N-E_i=0.3$ eV, $E_i-F_P=0.318$ eV.

$$n=n_i{e}^{0.3/0.0259}=1.07\times 10^{15}{\text{ cm}}^{-3}$$ $$p=n_i{e}^{0.318/0.0259}=10^{10}{e}^{12.28}=2.15\times 10^{15}{\text{ cm}}^{-3}$$

(c) Determine $N_D$.

Solution: $N_D\approx n_0=1.07\times 10^{15}$ cm${}^{-3}$.

(d) Do we have low-level injection when illuminated?

Solution: No, because $p=2.15\times 10^{15}>n_0=1.07\times 10^{15}$.

(e) What is the resistivity before and after illumination?

Solution:

Before: $\rho =1/(q{\mu }_n{n}_0)=1/[(1.6\times 10^{-19})(1345)(1.07\times 10^{15})]=4.34$ ohm-cm

After: The conductivity is $\sigma =q({\mu }_{n}n+{\mu }_{p}p)$. With ${\mu }_n=1345$, ${\mu }_p=458$:

$$\sigma =(1.6\times 10^{-19})[(1345)(1.07\times 10^{15})+(458)(2.15\times 10^{15})]$$ $$=(1.6\times 10^{-19})[1.44\times 10^{18}+9.85\times 10^{17}]=0.388{\text{ (ohm-cm)}}^{-1}$$ $$\rho =2.58\text{ ohm-cm}$$

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Problem 3.25

Problem: A portion ($0\le x\le L$) of a Si sample, uniformly doped with $N_D=10^{15}$/cm${}^3$, is subject to a steady state perturbation such that:

$$n=N_D,p=n_i(1-x/L)+n_i^2/N_D$$

Sketch the energy band diagram for the perturbed region including $E_c$, $E_v$, $E_i$, $F_N$, and $F_P$.

Solution:

$n=N_D$, so $F_N=E_F$ (constant).

$$p=n_i(1-x/L)+n_i^2/N_D$$ $$F_P=E_i-kT\ln (p/n_i)=E_i-kT\ln [(1-x/L)+n_i/N_D]$$

At $x=0$: $p=n_i+n_i^2/N_D\approx n_i$, so $F_P\approx E_i$.

At $x=L$: $p=n_i^2/N_D=p_0$, so $F_P=E_F$.

---

Problem 3.26

Problem: Write a MATLAB program that automatically plots $F_N$ and $F_P$ versus $x$ for the experimental situation in Sample Problem No. 2.

Solution:

ND = input('Donor concentration in cm^-3, ND = ');
Dpn0 = input('Excess hole concentration at x=0, Dpn0 = ');
ni = 1e10; kT = 0.0259; taup = 1e-6;
Dp = 11.9; Lp = sqrt(Dp*taup);
x = 0:0.01:10*Lp;
p = ni^2/ND + Dpn0*exp(-x/Lp);
n = ni^2./p;
FN = ni*exp(0); % Actually FN is constant
FP = -(kT)*log(p/ni);
plot(x/Lp, (FP)/kT);
xlabel('x/Lp'); ylabel('(FP - Ei)/kT');

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Chapter 4: Basics of Device Fabrication (Read Only)

4.1 Fabrication Processes

4.1.1 Oxidation

$$\text{Si}+{\text{O}}_2\to {\text{SiO}}_2$$ $$\text{Si}+2{\text{H}}_2\text{O}\to {\text{SiO}}_2+2{\text{H}}_2$$

Dry oxidation: Used for critical insulator regions (e.g., gate oxide). Superior Si–SiO${}_2$ interface properties.

Wet oxidation: Faster rate, used for thicker barrier oxides.

4.1.2 Diffusion

Dopant atoms diffuse into the semiconductor at elevated temperatures.

Predeposition:

$$N_1(x,t_1)=N_0\text{erfc}\left(\frac{x}{2\sqrt{D_1{t}_1}}\right)$$

Drive-in:

$$N_2(x,t_2)=\frac{N_0}{\sqrt{\pi D_2{t}_2}}{e}^{-(x/2\sqrt{D_2{t}_2}{)}^2}$$

4.1.3 Ion Implantation

Impurity ions are accelerated to high energies and shot into the semiconductor.

Gaussian distribution:

$$N(x)=\frac{\varphi }{\Delta R_p\sqrt{2\pi }}{e}^{-(1/2)[(x-R_p)/\Delta R_p{]}^2}$$

where $R_p$ is the projected range and $\Delta R_p$ is the straggle.

4.1.4 Lithography

Process for selectively removing thin films:

1. Apply photoresist
2. Expose through mask
3. Develop
4. Etch
5. Remove resist

4.1.5 Thin-Film Deposition

• Evaporation: Source material heated in vacuum, deposits on substrate
• Sputtering: Ar${}^{+}$ ions bombard source, ejected atoms deposit on substrate
• CVD: Chemical reaction of gas-phase components forms film

4.1.6 Epitaxy

Deposition of a crystalline layer that is an extension of the underlying semiconductor lattice.

4.2 Device Fabrication Examples

4.2.1 pn Junction Diode Fabrication

1. Start with p-type Si wafer
2. Thermal oxidation
3. Lithography to open diffusion holes
4. Phosphorus diffusion
5. Metallization (Al)
6. Lithography to pattern metal

4.2.2 Computer CPU Process Flow

The Intel CHMOS process flow includes:

1. Starting silicon and well definition
2. Field threshold implant and field oxidation
3. Gate oxidation
4. Transistor threshold ion implant
5. Polysilicon-to-diffusion contacts
6. Polycide deposition, doping, patterning, oxidation
7. Source/drain implant and poly re-ox
8. Doped glass deposition
9. Contact mask and etch
10. Metal deposition and definition
11. Passivation

4.3 Summary

Common component processes: oxidation, diffusion, ion implantation, lithography, evaporation, sputtering, CVD, and epitaxy.

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Part IIA: pn Junction Diodes

Chapter 5: pn Junction Electrostatics

5.1 Preliminaries

5.1.1 Junction Terminology/Idealized Profiles

The metallurgical junction occurs at the plane where $N_D-N_A=0$.

The doping profile is a plot of net doping concentration versus position.

Idealized profiles:

• Step junction: Abrupt change in doping at the junction
• Linearly graded junction: Linear variation of doping across the junction

5.1.2 Poisson’s Equation

$$\nabla \cdot \mathcal{E}=\frac{\rho }{K_s{\epsilon }_0}$$

In one dimension:

$$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_s{\epsilon }_0}$$

where:

$$\rho =q(p-n+N_D-N_A)$$

5.1.3 Qualitative Solution

The energy band diagram for a pn junction under equilibrium is constructed by:

1. Drawing the isolated p and n energy band diagrams
2. Aligning them to the position-independent Fermi level
3. Connecting the band edges

The electrostatic variables are then deduced:

• $V(x)$: upside-down of $E_c(x)$
• $\mathcal{E}(x)$: slope of $E_c(x)$
• $\rho (x)$: slope of $\mathcal{E}(x)$

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

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

For nondegenerately doped step junctions.

5.1.5 The Depletion Approximation

Assumptions:

1. Carriers are negligible compared to net doping in the depletion region
2. Charge density is zero outside the depletion region

5.2 Quantitative Electrostatic Relationships

5.2.2 Step Junction with $V_A=0$

Charge density:

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

Electric field:

$$\mathcal{E}(x)=\{\begin{array}{ll}-\frac{q{N}_A}{\epsilon }(x+x_p)& -x_p\le x\le 0\\ \frac{q{N}_D}{\epsilon }(x_n-x)& 0\le x\le x_n\end{array}$$

Potential:

$$V(x)=\{\begin{array}{ll}\frac{q{N}_A}{2\epsilon }(x+x_p{)}^2& -x_p\le x\le 0\\ V_{bi}-\frac{q{N}_D}{2\epsilon }(x_n-x{)}^2& 0\le x\le x_n\end{array}$$

Depletion widths:

$$x_n={\left[\frac{2K_s{\epsilon }_0}{q}\frac{N_A}{N_D(N_A+N_D)}{V}_{bi}\right]}^{1/2}$$ $$x_p={\left[\frac{2K_s{\epsilon }_0}{q}\frac{N_D}{N_A(N_A+N_D)}{V}_{bi}\right]}^{1/2}$$ $$W=x_n+x_p={\left[\frac{2K_s{\epsilon }_0}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)V_{bi}\right]}^{1/2}$$

5.2.3 Step Junction with $V_A

eq 0$

Replace $V_{bi}$ by $V_{bi}-V_A$ in all expressions.

5.2.5 Linearly Graded Junctions

$$N_D-N_A=ax$$ $$W={\left[\frac{12K_s{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}$$

5.3 Summary

The pn junction electrostatics provides the foundation for diode analysis. The depletion approximation is the most important simplifying approximation in device modeling.

---

Exercise 5.1

Problem: Compute and plot $V_{bi}$ as a function of doping on the lightly doped side of Si p${}^{+}$-n and n${}^{+}$-p step junctions at 300 K.

Solution:

% Vbi Computation (p+/n and n+/p junctions)
EG = 1.12;
kT = 0.0259;
ni = 1.0e10;
ND = logspace(14,17);
Vbi = EG/2 + kT.*log(ND./ni);
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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Exercise 5.2

Problem: Perform a sample computation to gauge the size of $W$ and $|\mathcal{E}{|}_{\max }$ under equilibrium conditions. Si step junction at 300 K with $N_A=10^{17}$/cm${}^3$ and $N_D=10^{14}$/cm${}^3$.

Solution:

$$V_{bi}=0.0259\ln \left(\frac{10^{17}\times 10^{14}}{10^{20}}\right)=0.0259\ln (10^{11})=0.656\text{ V}$$ $$x_n={\left[\frac{2\times 11.8\times 8.85\times 10^{-14}\times 0.656}{1.6\times 10^{-19}\times 10^{14}}\right]}^{1/2}=2.93\times 10^{-4}\text{ cm}=2.93\text{ μm}$$ $$x_p=\frac{N_D}{N_A}{x}_n=2.93\times 10^{-7}\text{ cm}$$ $$W=x_n+x_p\approx 2.93\text{ μm}$$ $$|\mathcal{E}{|}_{\max }=\frac{q{N}_D{x}_n}{K_s{\epsilon }_0}=\frac{(1.6\times 10^{-19})(10^{14})(2.93\times 10^{-4})}{(11.8)(8.85\times 10^{-14})}=4.49\times 10^3\text{ V/cm}$$

---

Exercise 5.3

Problem: Construct a log-log plot of the depletion width versus impurity concentration on the lightly doped side of Si p${}^{+}$-n and n${}^{+}$-p step junctions at 300 K for $V_A=0.5$ V, 0 V, and $-10$ V.

Solution:

T = 300;
k = 8.617e-5;
e0 = 8.85e-14;
q = 1.602e-19;
KS = 11.8;
ni = 1e10;
EG = 1.12;
NB = logspace(14,17);
VA = [0.5 0 -10];
Vbi = EG/2 + k*T*log(NB./ni);
for i = 1:3
    W(i,:) = 1.0e4*sqrt(2*KS*e0/q*(Vbi-VA(i))./NB);
end
loglog(NB, W); grid;
axis([1e14 1e17 1e-1 1e1]);
xlabel('NA or ND (cm-3)'); ylabel('W (micrometers)');

---

Exercise 5.4

Problem: Construct a fully dimensioned energy band diagram using the “Diagram Generator” program.

Solution:

% Equilibrium Energy Band Diagram Generator
T = 300;
k = 8.617e-5;
e0 = 8.85e-14;
q = 1.602e-19;
KS = 11.8;
ni = 1.0e10;
EG = 1.12;
xleft = -3.5e-4;
xright = -xleft;
NA = input('Please enter p-side doping (cm^-3), NA = ');
ND = input('Please enter n-side doping (cm^-3), ND = ');
Vbi = k*T*log((NA*ND)/ni^2);
xN = sqrt(2*KS*e0/q*NA*Vbi/(ND*(NA+ND)));
xP = sqrt(2*KS*e0/q*ND*Vbi/(NA*(NA+ND)));
x = linspace(xleft, xright, 200);
Vx1 = 0.5*q*ND*(xN-x).^2/(KS*e0).*(x>=0 & x<=xN);
Vx2 = 0.5*q*NA*(xP+x).^2/(KS*e0).*(x>=-xP & x<0);
Vx = Vx1 + Vx2;
VMAX = 3;
EF = Vx(1) + VMAX/2 - k*T*log(NA/ni);
plot(x, -Vx+EG/2+VMAX/2);
axis([xleft xright 0 VMAX]);
axis('off'); hold on
plot(x, -Vx-EG/2+VMAX/2);
plot(x, -Vx+VMAX/2, 'w:');
plot([xleft xright], [EF EF], 'w');
plot([0 0], [0.15*VMAX 0.5], 'w-');

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Problems — Chapter 5

Problem 5.1 — True or False

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

Solution: True.

(b) The $\rho$ that appears in Poisson’s equation is the charge density and has units of coul/cm${}^3$.

Solution: 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.

Solution: False. It is due to the depletion of mobile carriers, leaving behind ionized dopant sites.

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

Solution: True.

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

Solution: True.

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

Solution: False. Ohmic contacts have negligible voltage drop; the built-in voltage is unchanged.

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

Solution: True.

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

Solution: True.

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

Solution: False. It decreases with forward biasing.

(j) The depletion width in a linearly graded junction varies as $(V_{bi}-V_A{)}^{1/3}$.

Solution: True.

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Problem 5.2

Problem: The energy band diagram of a Si step junction at room temperature has $E_i-E_F=E_G/4$ on the p-side and $E_F-E_i=E_G/4$ on the n-side.

(a) Draw the equilibrium energy band diagram.

Solution: The Fermi level is constant. On the p-side, $E_F$ is $E_G/4$ below $E_i$; on the n-side, $E_F$ is $E_G/4$ above $E_i$.

(b) Determine $V_{bi}$.

Solution:

$$V_{bi}=\frac{1}{q}[(E_i-E_F{)}_{p\text{-side}}+(E_F-E_i{)}_{n\text{-side}}]=\frac{E_G}{4q}+\frac{E_G}{4q}=\frac{E_G}{2q}=0.56\text{ V}$$

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Problem 5.3

Problem: Consider the p${}_1$-p${}_2$ “isotype” step junction.

(a) Draw the equilibrium energy band diagram.

Solution: The Fermi level is constant. Band bending occurs at the junction.

(b) Derive an expression for the built-in voltage.

Solution:

$$V_{bi}=\frac{kT}{q}\ln \left(\frac{N_{A1}}{N_{A2}}\right)$$

(c) Make rough sketches of potential, electric field, and charge density.

Solution: $\rho$: negative step from $q{N}_{A1}$ to $q{N}_{A2}$. $\mathcal{E}$: slope. $V$: quadratic.

(d) Briefly describe the depletion approximation.

Solution: Carriers are neglected in the depletion region; $\rho =q(N_D-N_A)$.

(e) Can the depletion approximation be invoked?

Solution: Yes, for the isotype junction as well.

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Problem 5.4

Problem: A Si step junction at room temperature has $N_A=2\times 10^{17}$/cm${}^3$ and $N_D=10^{17}$/cm${}^3$. Compute (a) $V_{bi}$, (b) $x_p$, $x_n$, $W$, (c) $\mathcal{E}$ at $x=0$, (d) $V$ at $x=0$, (e) make sketches.

Solution:

(a) $V_{bi}=0.0259\ln \left(\frac{2\times 10^{17}\times 10^{17}}{10^{20}}\right)=0.0259\ln (2\times 10^{14})=0.859$ V

(b) $x_n={\left[\frac{2\times 11.8\times 8.85\times 10^{-14}\times 0.859}{1.6\times 10^{-19}\times 10^{17}}\right]}^{1/2}=1.06\times 10^{-5}$ cm

$x_p=\frac{N_D}{N_A}{x}_n=5.3\times 10^{-6}$ cm

$W=x_n+x_p=1.59\times 10^{-5}$ cm

(c) $\mathcal{E}(0)=-\frac{q{N}_A{x}_p}{K_s{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(2\times 10^{17})(5.3\times 10^{-6})}{(11.8)(8.85\times 10^{-14})}=-1.62\times 10^5$ V/cm

(d) $V(0)=\frac{q{N}_A{x}_p^2}{2K_s{\epsilon }_0}=\frac{(1.6\times 10^{-19})(2\times 10^{17})(5.3\times 10^{-6}{)}^2}{2\times 11.8\times 8.85\times 10^{-14}}=0.43$ V

---

Problem 5.5

Problem: Repeat Problem 5.4 with $N_A=10^{17}$/cm${}^3$.

Solution: Similar calculations with symmetric doping. $V_{bi}=0.756$ V, $x_n=x_p=7.2\times 10^{-6}$ cm, $W=1.44\times 10^{-5}$ cm, $\mathcal{E}(0)=-1.10\times 10^5$ V/cm, $V(0)=0.378$ V.

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Problem 5.6

Problem: Compute using the MATLAB diary function.

Solution: Use the program from Exercise 5.4 with instructor-supplied values.

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Problem 5.7

Problem: Compute and present coordinated plots of $\mathcal{E}$ and $V$ inside the junction for $N_A=10^{16}$/cm${}^3$, $N_D=2\times 10^{14}$/cm${}^3$, and $V_A=-20$ V.

Solution:

NA = 1e16; ND = 2e14; VA = -20;
Vbi = 0.0259*log(NA*ND/1e20);
xN = sqrt(2*11.8*8.85e-14/1.6e-19*NA*(Vbi-VA)/(ND*(NA+ND)));
xP = sqrt(2*11.8*8.85e-14/1.6e-19*ND*(Vbi-VA)/(NA*(NA+ND)));
x = linspace(-xP, xN, 100);
E = -1.6e-19*NA*(x+xP)/(11.8*8.85e-14).*(x<0) + ...
    1.6e-19*ND*(xN-x)/(11.8*8.85e-14).*(x>=0);
V = 1.6e-19*NA*(x+xP).^2/(2*11.8*8.85e-14).*(x<0) + ...
    (Vbi-VA - 1.6e-19*ND*(xN-x).^2/(2*11.8*8.85e-14)).*(x>=0);
subplot(2,1,1); plot(x,E); ylabel('E (V/cm)');
subplot(2,1,2); plot(x,V); ylabel('V (V)');

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Problem 5.8

Problem: Sketch the expected charge density, electric field, and electrostatic potential based on the depletion approximation for the special doping profile.

Solution: The charge density follows the doping profile in the depletion region. The electric field is the integral of charge density. The potential is the negative integral of electric field.

---

Problem 5.9

Problem: A pn junction diode has $N_D-N_A=N_0[1-\exp (-x)]$.

(a) Give a concise statement of the depletion approximation.

Solution: Carriers are neglected in the depletion region; $\rho =q(N_D-N_A)$ in the depletion region and zero outside.

(b) Sketch the charge density.

Solution: $\rho$ follows the doping profile in the depletion region.

(c) Establish an expression for the electric field.

Solution:

$$\mathcal{E}(x)=\frac{q{N}_0}{K_s{\epsilon }_0}\left[x+\frac{1}{\alpha }(e^{-\alpha x}-1)\right]+C$$

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Problem 5.10

Problem: A pn junction diode has a doping profile as sketched.

(a) What is the built-in voltage?

Solution: $V_{bi}=\frac{kT}{q}\ln \left[\frac{n(x_n)}{n(-x_p)}\right]$

(b) Invoking the depletion approximation, sketch the charge density.

Solution: $\rho$ follows the doping profile in the depletion region.

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Problem 5.11 — p-i-n Diode

Problem: The p-i-n diode has a middle intrinsic region.

(a) Roughly sketch charge density, electric field, and electrostatic potential.

Solution: $\rho$: positive in n-region, negative in p-region, zero in i-region. $\mathcal{E}$: constant in i-region. $V$: linear in i-region.

(b) What is the built-in voltage drop?

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

(c) Establish quantitative relationships.

Solution: The depletion region extends into the i-region. The electric field is approximately constant in the i-region.

---

Problem 5.12

Problem: The electrostatic potential in the depletion region is:

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

(a) Establish an expression for the electric field.

Solution:

$$\mathcal{E}(x)=-\frac{dV}{dx}=-\frac{V_{bi}\pi }{2W}\cos \left(\frac{\pi x}{W}\right)$$

(b) Establish an expression for the charge density.

Solution:

$$\rho (x)=-K_s{\epsilon }_0\frac{d\mathcal{E}}{dx}=\frac{K_s{\epsilon }_0{V}_{bi}{\pi }^2}{2W^2}\sin \left(\frac{\pi x}{W}\right)$$

(c) Determine $N_D-N_A$ versus $x$.

Solution:

$$N_D-N_A=\frac{\rho }{q}=\frac{K_s{\epsilon }_0{V}_{bi}{\pi }^2}{2q{W}^2}\sin \left(\frac{\pi x}{W}\right)$$

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Problem 5.13

Problem: Compute and present coordinated plots of $\mathcal{E}$ and $V$ for a linearly graded junction.

Solution: Use the relationships from Subsection 5.2.5.

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Problem 5.14

Problem: Obtain a closed-form solution for the electrostatic variables inside a pn step junction without invoking the depletion approximation.

Solution: The exact solution involves the Fermi-Dirac integral:

$$\mathcal{E}=\pm \frac{kT}{q{L}_D}F(U,U_F)$$

where:

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

The position is obtained from:

$$x=\mp L_D{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime },U_F)}$$

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Chapter 6: pn Junction Diode — I-V Characteristics

6.1 The Ideal Diode Equation

6.1.1 Qualitative Derivation

Under forward bias, the potential hill is lowered, allowing more carriers to diffuse across the junction. The forward current increases exponentially with voltage.

Under reverse bias, the potential hill is increased, cutting off majority carrier injection and leaving only a small minority carrier current.

The ideal diode equation:

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

6.1.2 Quantitative Solution Strategy

Assumptions:

1. Steady state
2. Nondegenerately doped step junction
3. One-dimensional
4. Low-level injection
5. No other processes (G${}_L$ = 0)

Solution approach:

1. Solve minority carrier diffusion equations in quasineutral regions
2. Evaluate current densities at depletion region edges
3. Sum edge currents and multiply by area

Boundary conditions:

At ohmic contacts:

$$\Delta n_p(-\infty )=0,\Delta p_n(+\infty )=0$$

At depletion region edges (law of the junction):

$$\Delta n_p(-x_p)=\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)$$ $$\Delta p_n(x_n)=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$$

6.1.3 Derivation Proper

For holes on the n-side:

$$\Delta p_n(x^{\prime })=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)e^{-x^{\prime }/L_p}$$ $$J_p(x^{\prime })=-q{D}_p\frac{d\Delta p_n}{d{x}^{\prime }}=q\frac{D_p}{L_p}\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)e^{-x^{\prime }/L_p}$$

For electrons on the p-side:

$$\Delta n_p(x^{\prime \prime })=\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)e^{-x^{\prime \prime }/L_n}$$ $$J_n(x^{\prime \prime })=q{D}_n\frac{d\Delta n_p}{d{x}^{\prime \prime }}=q\frac{D_n}{L_n}\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)e^{-x^{\prime \prime }/L_n}$$

Total current:

$$I=I_0(e^{q{V}_A/kT}-1)$$ $$I_0=qA\left(\frac{D_p}{L_p}\frac{n_i^2}{N_D}+\frac{D_n}{L_n}\frac{n_i^2}{N_A}\right)$$

6.1.4 Examination of Results

Ideal I-V:

• Reverse bias: $I\to -I_0$ for $|V_A|>$ few $kT/q$
• Forward bias: $I\to I_0{e}^{q{V}_A/kT}$

Semilog plot: Slope = $q/kT$, intercept = $\ln (I_0)$.

6.2 Deviations from the Ideal

6.2.1 Ideal Theory vs. Experiment

Major deviations:

1. Reverse-bias breakdown
2. Excess current at small forward biases and all reverse biases (R-G current)
3. Slope-over at high forward biases (series resistance or high-level injection)

6.2.2 Reverse-Bias Breakdown

Avalanching: Impact ionization creates carrier multiplication.

Zener process: Tunneling in heavily doped junctions.

Breakdown voltage:

$$V_{BR}\propto \frac{1}{N_B^{0.75}}$$

6.2.3 The R-G Current

For reverse bias $>$ few $kT/q$:

$$I_{R-G}=-\frac{qA{n}_{i}W}{2{\tau }_0}$$

where:

$${\tau }_0=\frac{2}{\frac{1}{{\tau }_n}+\frac{1}{{\tau }_p}}$$

For forward bias:

$$I_{R-G}\propto e^{q{V}_A/2kT}$$

6.2.4 $V_A\to V_{bi}$ High-Current Phenomena

Series resistance:

$$V=V_A-I{R}_s$$

High-level injection: Current varies as $e^{q{V}_A/2kT}$.

6.3 Special Considerations

6.3.1 Charge Control Approach

$$\frac{d{Q}_p}{dt}=i_{DIFF}-\frac{Q_p}{{\tau }_p}$$

6.3.2 Narrow-Base Diode

For a p${}^{+}$-n diode with base width $x_c$:

$$\Delta p_n(x^{\prime })=\Delta p_n(0)\frac{\sinh [(x_c-x^{\prime })/L_p]}{\sinh (x_c/L_p)}$$ $$I_{DIFF}=I_0^{\prime }(e^{q{V}_A/kT}-1)$$ $$I_0^{\prime }=\frac{qA{D}_p{n}_i^2}{L_p{N}_D}\coth \left(\frac{x_c}{L_p}\right)$$

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Exercise 6.1

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

Solution:

$$I_0\propto \frac{1}{N_D}$$ $$\frac{I_{01}}{I_{02}}\approx \frac{N_{D2}}{N_{D1}}=10$$

Diode 1 has approximately 10 times larger saturation current.

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Exercise 6.2

Problem: An ideal Si p${}^{+}$-n step junction diode at 300 K has $A=10^{-4}$ cm${}^2$ and ${\tau }_p=10^{-6}$ sec. Write a MATLAB program that computes and plots the ideal I-V characteristic.

Solution:

k = 8.617e-5; q = 1.6e-19;
A = 1.0e-4; ni = 1.0e10; taup = 1.0e-6;
ND = input('Input the n-side doping concentration, ND = ');
T = 300;
NAref = 2.35e17; mupmin = 54.3; mup0 = 406.9; ap = 0.88;
mup = mupmin + mup0./(1+(ND./NAref).^ap);
VA = linspace(-1, 0.2);
DP = k*T.*mup;
LP = sqrt(DP.*taup);
I0 = q.*A.*(DP./LP).*(ni^2/ND);
I = I0.'*(exp(VA./(k.*T))-1);
plot(VA, I); grid;
xlabel('VA (volts)'); ylabel('I (amps)');

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Exercise 6.3

Problem: Figure E6.3 is a dimensioned plot of the steady state carrier concentrations inside a pn junction diode.

(a) Is the diode forward or reverse biased?

Solution: Forward biased. There is a pile-up of minority carriers at the depletion region edges.

(b) Do low-level injection conditions prevail?

Solution: Yes. $\Delta p_n\ll n_n$ and $\Delta n_p\ll p_p$.

(c) Determine the applied voltage $V_A$.

Solution:

$$V_A=\frac{kT}{q}\ln \left[\frac{n(x_n)p(x_n)}{n_i^2}\right]=0.0259\ln \left(\frac{10^{15}\times 10^{15}}{10^{20}}\right)=0.0259\ln (10^{10})=0.596\text{ V}$$

(d) Determine the hole diffusion length $L_p$.

Solution:

$$L_p=\frac{x^{\prime }}{\ln [p_n(0)/p_n(x^{\prime })]}=\frac{1.6\times 10^{-2}}{\ln (10^{15}/10^8)}=\frac{1.6\times 10^{-2}}{\ln (10^7)}=\frac{1.6\times 10^{-2}}{16.12}=9.9\times 10^{-4}\text{ cm}$$

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Exercise 6.4

Problem: Derive an expression for the reverse-bias saturation current based on approximate squared-off distributions.

Solution:

The generation rate in the quasineutral regions is:

$${\frac{\partial n}{\partial t}|}_{\text{thermal R-G}}=\frac{n_i^2}{N_A{\tau }_n}\text{for }-L_n-x_p\le x\le -x_p$$ $${\frac{\partial p}{\partial t}|}_{\text{thermal R-G}}=\frac{n_i^2}{N_D{\tau }_p}\text{for }{x}_n\le x\le x_n+L_p$$

Generation is taking place in volumes $A{L}_n$ and $A{L}_p$:

$$I_0=qA\left(\frac{L_n{n}_i^2}{{\tau }_n{N}_A}+\frac{L_p{n}_i^2}{{\tau }_p{N}_D}\right)$$

One obtains the usual $I_0$ expression.

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Exercise 6.5

Problem: Compute and plot the multiplication factor versus $|V_A|/V_{BR}$ when $m=3$ and $m=6$.

Solution:

x = linspace(0, 0.99);
M3 = 1./(1-x.^3);
M6 = 1./(1-x.^6);
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');

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Exercise 6.6

Problem: Compute $I_{01}$ and $I_{02}$ for a Si p${}^{+}$-n step junction diode at 300 K with $A=10^{-4}$ cm${}^2$, $N_D=10^{16}$/cm${}^3$, and ${\tau }_n={\tau }_p={\tau }_0=10^{-6}$ sec.

Solution:

$V_{bi}=(E_G/2q)+(kT/q)\ln (N_D/n_i)=0.56+0.0259\ln (10^6)=0.56+0.358=0.918$ V

$W=\sqrt{2K_s{\epsilon }_0{V}_{bi}/(q{N}_D)}=0.344$ μm

$I_{01}=5.38\times 10^{-16}$ A

$I_{02}=2.08\times 10^{-13}$ A

For Ge: $I_{01}=6.23\times 10^{-9}$ A, $I_{02}=8.34\times 10^{-10}$ A

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Exercise 6.7

Problem: Determine the $I_{01}$, $I_{02}$, $n_1$, and $n_2$ that provide an optimal match to the forward bias I-V data in Fig. 6.10(a).

Solution:

From the semilog plot:

• $I_{01}\approx 10^{-14}$ A, $n_1\approx 1.1$
• $I_{02}\approx 10^{-12}$ A, $n_2\approx 1.5$

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Exercise 6.8

Problem: Crudely estimate the series resistance of the diode in Fig. 6.10(a).

Solution:

At $I=80$ mA, $V_A=0.85$ V. The ideal diode curve would give 80 mA at $V=0.78$ V.

$$\Delta V=0.85-0.78=0.07\text{ V}$$ $$R_s=\frac{\Delta V}{I}=\frac{0.07}{0.08}\approx 1\text{ ohm}$$

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Exercise 6.9

Problem: Identify the cause of each nonideal I-V feature.

Solution:

• A: Avalanching and/or Zener process (3)
• B: Thermal generation in the depletion region (6)
• C: Thermal recombination in the depletion region (2)
• D: High-level injection (10)
• E: Series resistance (8)

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Exercise 6.10

Problem: A planar Si p${}^{+}$-n step junction diode at room temperature has $N_D=10^{16}$/cm${}^3$ and n-side thickness $x_c=2$ μm. Determine the punch-through voltage.

Solution:

At punch-through, $x_n=x_c=2$ μm.

$$x_n=\sqrt{\frac{2K_s{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)}$$ $$V_{bi}-V_A=\frac{q{N}_D{x}_n^2}{2K_s{\epsilon }_0}=\frac{(1.6\times 10^{-19})(10^{16})(2\times 10^{-4}{)}^2}{2\times 11.8\times 8.85\times 10^{-14}}=30.6\text{ V}$$

$V_{bi}=0.92$ V, so $V_A=0.92-30.6=-29.7$ V.

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Problems — Chapter 6

Problem 6.1 — Ideal Diode Quiz

(a) The forward-bias current is associated with what type of carrier activity?

Solution: Majority carrier injection across the junction.

(b) The reverse-bias current is associated with what type of carrier activity?

Solution: Minority carriers wandering into the depletion region and being swept to the other side.

(c) Why is the reverse-bias current expected to be small in magnitude and to saturate at a small reverse voltage?

Solution: It is associated with minority carriers, which are few in number. Once the majority carrier diffusion current is cut off (at a few $kT/q$), the current saturates.

(d) Under reverse biasing, what processes occur in the quasineutral regions adjacent to the depletion region edges?

Solution: Generation and diffusion.

(e) Why can’t the minority carrier diffusion equations be used to determine the minority carrier concentrations and current in the depletion region?

Solution: Because the depletion approximation assumes the electric field is non-zero in the depletion region, and the diffusion equations assume $\mathcal{E}\approx 0$.

(f) What a priori justification is there for assuming recombination-generation is negligible throughout the depletion region?

Solution: None — it is an assumption that leads to a simple solution. Its validity is checked by comparison with experiment.

(g) What exactly is a “wide-base” diode?

Solution: A diode whose contacts are several minority carrier diffusion lengths or more from the edges of the depletion region.

(h) What is the “law of the junction”?

Solution: $np=n_i^2{e}^{q{V}_A/kT}$ at the depletion region edges.

(i) Given a semilog plot of the forward bias I-V characteristic, how does one determine $I_0$?

Solution: Extrapolate the linear region to $V_A=0$; the intercept is $\ln (I_0)$.

(j) True or false: The reverse-bias saturation current may be alternatively viewed as arising from minority carrier generation in the quasineutral regions.

Solution: True.

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Problem 6.2

Problem: Compare the I-V characteristics of an actual Si pn junction diode at room temperature and the theoretical ideal diode characteristic.

Solution:

Feature	Ideal	Actual
Reverse bias	Saturates at $-I_0$	Increases with reverse bias (R-G current)
Small forward bias	$I\propto e^{q{V}_A/kT}$	$I\propto e^{q{V}_A/2kT}$ (R-G current dominates)
Large forward bias	$I\propto e^{q{V}_A/kT}$	Slope-over (series resistance)
Breakdown	None	Occurs at $V_{BR}$

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Problem 6.3

Problem: Sketch the energy band diagram for an ideal p${}^{+}$-n step junction diode showing carrier activity when (a) $V_A=0$, (b) $V_A>0$, (c) $V_A<0$.

Solution:

• (a) Equilibrium: constant $E_F$, potential hill $=q{V}_{bi}$
• (b) Forward bias: potential hill reduced by $q{V}_A$, $E_{Fp}-E_{Fn}=q{V}_A$
• (c) Reverse bias: potential hill increased by $q|V_A|$, $E_{Fn}-E_{Fp}=q|V_A|$

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Problem 6.4

Problem: Construct a composite energy-band/circuit diagram showing the overall view of carrier activity inside a forward-biased pn junction diode.

Solution: Similar to Fig. 6.2, but with the forward-bias polarity. Electrons are injected from the n-side to the p-side; holes from the p-side to the n-side. Recombination occurs in the quasineutral regions, and carriers are resupplied from the contacts.

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Problem 6.5

Problem: Show that the Eq. (6.15) boundary condition can be derived by continuing to assume $J_N=0$ when $V_{A}eq0$.

Solution:

Under equilibrium, $J_N=0$ gives:

$$\mathcal{E}=\frac{kT}{q}\frac{dn/dx}{n}$$

Integrating across the depletion region:

$$V_{bi}-V_A=\frac{kT}{q}\ln \left[\frac{n(x_n)}{n(-x_p)}\right]$$ $$n(-x_p)=n(x_n)e^{-(V_{bi}-V_A)/(kT/q)}$$

With $n(x_n)=N_D$ and $V_{bi}=(kT/q)\ln (N_A{N}_D/n_i^2)$:

$$n(-x_p)=N_D\frac{n_i^2}{N_A{N}_D}{e}^{q{V}_A/kT}=\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}$$

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Problem 6.6

Problem: Considering an n${}^{+}$-p silicon step junction to be ideal, perform calculations to determine typical current levels.

Solution:

Use the ideal diode equation with $A=10^{-3}$ cm${}^2$, $N_A=10^{17}$/cm${}^3$, ${\tau }_n=10^{-6}$ sec.

(a) At 300 K:

• $V_A=-50$ V: $I\approx -I_0$
• $V_A=-0.1$ V: $I\approx -I_0$
• $V_A=0.1$ V: $I\approx I_0(e^{3.86}-1)\approx 47I_0$
• $V_A=0.5$ V: $I\approx I_0(e^{19.3}-1)\approx 2.4\times 10^8{I}_0$

(b) At 500 K: $n_i$ is much larger, so $I_0$ is much larger.

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Problem 6.7

Problem: Explore how the ideal diode I-V characteristic varies as a function of temperature.

Solution:

T = 300; k = 8.617e-5; q = 1.6e-19;
A = 1e-4; ND = 1e16;
taup = 1e-6; Vbi = 1.12/2 + 0.0259*log(ND/1e10);
W = sqrt(2*11.8*8.85e-14*Vbi/(q*ND));
VA = linspace(-1, 0.2);
for T = [295 300 305]
    kT = k*T;
    ni = 2.510e19*(T/300)^1.5*exp(-(1.12-0.0074)/(2*kT));
    mup = 54.3 + 406.9/(1+(ND/2.35e17)^0.88);
    DP = kT*mup; LP = sqrt(DP*taup);
    I0 = q*A*(DP/LP)*(ni^2/ND);
    I = I0*(exp(VA/kT)-1);
    semilogy(VA, I); hold on;
end

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Problem 6.8 — Photodiode/Solar Cell

Problem: A pn junction photodiode has uniform photogeneration $G_L$.

(a) What is the excess minority carrier concentration on the n-side at $x\to \infty$?

Solution: $\Delta p_n(\infty )=G_L{\tau }_p$

(b) Derive the I-V characteristic.

Solution:

$$I=I_0(e^{q{V}_A/kT}-1)-qA(L_n+W+L_p)G_L$$

(c) Sketch the characteristics for different $G_L$ values.

Solution: The characteristics are translated downward along the current axis with increasing $G_L$.

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Problem 6.9

Problem: Construct a plot illustrating the solar cell characteristics.

Solution:

VA = linspace(-0.5, 0.1);
kT = 0.0259; I0 = 1e-10;
for IL = [0 -1 -2 -4]
    I = I0*(exp(VA/kT)-1) + IL;
    plot(VA, I); hold on;
end
xlabel('VA (volts)'); ylabel('I (amps)');

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Problem 6.10

Problem: Figure P6.10 is a dimensioned plot of steady state carrier concentrations.

(a) Is the diode forward or reverse biased?

Solution: Forward biased (pile-up of minority carriers).

(b) Do low-level injection conditions prevail?

Solution: Yes.

(c) What are the p-side and n-side doping concentrations?

Solution: $p_p=10^{17}$/cm${}^3$ → $N_A=10^{17}$/cm${}^3$. $n_n=10^{16}$/cm${}^3$ → $N_D=10^{16}$/cm${}^3$.

(d) Determine $V_A$.

Solution:

$$V_A=0.0259\ln \left(\frac{10^{14}\times 10^{14}}{10^{20}}\right)=0.0259\ln (10^8)=0.477\text{ V}$$

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Problem 6.11

Problem: A voltage $V_A=23.03(kT/q)$ is applied to a step junction diode with $N_A=10^{17}$/cm${}^3$ and $N_D=10^{16}$/cm${}^3$.

Solution: $V_A=23.03\times 0.0259=0.596$ V.

$n_p=n_i^2/N_A{e}^{q{V}_A/kT}=10^{20}/10^{17}{e}^{23.03}=10^3\times 10^{10}=10^{13}$ cm${}^{-3}$

$p_n=n_i^2/N_D{e}^{q{V}_A/kT}=10^{20}/10^{16}{e}^{23.03}=10^4\times 10^{10}=10^{14}$ cm${}^{-3}$

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Problem 6.12 — Temperature Sensor

Problem: A silicon diode is used as a temperature sensor with constant forward current.

(a) Compute and plot $V_A$ versus $T$ for $0\le T{(}^{\circ }\text{C})\le 100^{\circ }\text{C}$.

Solution:

I = 1e-4; k = 8.617e-5;
T = 273:373;
for i = 1:length(T)
    ni = 2.510e19*(T(i)/300)^1.5*exp(-1.12/(2*k*T(i)));
    I0 = 1e-10*(ni/1e10)^2;
    VA(i) = k*T(i)*log(I/I0 + 1);
end
plot(T, VA);
xlabel('T (K)'); ylabel('VA (volts)');

(b) Would it be preferable to monitor using $I=10^{-4}$ A or $I=10^{-6}$ A?

Solution: $I=10^{-6}$ A gives higher sensitivity ($d{V}_A/dT$ is larger) because the diode is further from the high-injection regime.

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Problem 6.13

Problem: Given a planar p${}^{+}$-n Si step junction diode with $N_D=10^{15}$/cm${}^3$ and $T=300$ K, determine (a) $V_{BR}$, (b) depletion width at breakdown, (c) maximum electric field at breakdown.

Solution:

(a) From Fig. 6.11, $V_{BR}\approx 200$ V for $N_D=10^{15}$/cm${}^3$.

(b) At breakdown, $V_A=-V_{BR}$:

$$W=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{15}}(0.92+200)}=1.62\times 10^{-3}\text{ cm}$$

(c) ${\mathcal{E}}_{\max }=\frac{q{N}_{D}W}{K_s{\epsilon }_0}=2.78\times 10^5$ V/cm

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Problem 6.14

Problem: Model carrier multiplication and avalanche breakdown.

Solution:

NB = 2e16; VBR = 60*(NB/1e16)^(-0.75);
VA = linspace(-VBR, 0);
M = 1./(1-(-VA/VBR).^6);
I_I0 = M;
plot(VA, I_I0);
xlabel('VA (volts)'); ylabel('I/I0');

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Problem 6.15

Problem: Repeat Problem 6.14 replacing $I_0$ by the reverse bias $I_{R-G}$.

Solution: Normalize the current axis to the magnitude at $V_A=-V_{BR}/2$.

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Problem 6.16

Problem: Determine the temperature where $I_{DIFF}=I_{R-G}$ at a reverse bias of $V_{bi}-V_A=V_{BR}/2$.

Solution: This requires solving:

$$\frac{qA{n}_i^2}{N_D}\sqrt{\frac{D_p}{{\tau }_p}}=\frac{qA{n}_{i}W}{2{\tau }_0}$$

Since $n_i^2$ increases faster with temperature than $n_i$, the diffusion component dominates at high temperatures.

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Problem 6.17

Problem: Derive an expression for the R-G current from a special region of width $d$ with three times as many R-G centers.

Solution:

$$I_{R-G}=qA\left(\frac{n_{i}d}{2{\tau }_0}\right)\times 3+qA\left(\frac{n_i(W-d)}{2{\tau }_0}\right)=\frac{qA{n}_i}{2{\tau }_0}(W+2d)$$

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Problem 6.18

Problem: A Si diode has $I_0=10^{-12}$ A, $R_s=10$ ohms.

(a) Determine the forward-bias current at which the applied voltage differs from the ideal by 10%.

Solution:

$$I{R}_s=0.1V_A\Rightarrow I=\frac{0.1V_A}{10}=0.01V_A$$

Solving iteratively with $I=10^{-12}{e}^{V_A/0.0259}$ gives $I\approx 10^{-3}$ A.

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Problem 6.19

Problem: Determine $I_{01}$, $I_{02}$, $n_1$, and $n_2$ from the 1N757 diode I-V data.

Solution:

Using least squares fit:

VA = [0.02:0.02:0.46];
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];
y = log(I);
p = polyfit(VA, y, 1);
n1 = 1/(0.0259*p(1));
I01 = exp(p(2));

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Problem 6.20

Problem: Determine the $R_s$ value of the 1N757 diode.

Solution: Following the procedure in Fig. 6.16, plot $\Delta V$ versus $I$ and find the slope.

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Problem 6.21

Problem: The model shown in Fig. P6.21 represents real pn junction diodes.

(a) Confirm the total current relationship.

Solution:

$$I=I_{01}(e^{q{V}_j/n_{1}kT}-1)+I_{02}(e^{q{V}_j/n_{2}kT}-1)+\frac{V_j}{R_{SH}}$$

with $V_A=V_j+I{R}_s$.

(b) Construct a computer program to investigate the effect of $R_s$, $R_{SH}$, and $I_{01}$, $I_{02}$.

Solution:

I01 = 1e-13; I02 = 1e-9; n1 = 1; n2 = 2;
Rs = 1; RSH = 1e12; T = 300; kT = 0.0259;
Vj = linspace(0, 0.8, 1000);
for i = 1:length(Vj)
    I(i) = I01*(exp(Vj(i)/n1/kT)-1) + I02*(exp(Vj(i)/n2/kT)-1) + Vj(i)/RSH;
    VA(i) = Vj(i) + I(i)*Rs;
end
semilogy(VA, I);
xlabel('VA (volts)'); ylabel('I (amps)');

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Problem 6.22

Problem: Consider the special silicon p${}^{+}$-n step junction diode with ${\tau }_p=\infty$ for $0\le x\le x_b$ and ${\tau }_p=0$ for $x_b\le x\le x_c$. Develop an expression for the I-V characteristic.

Solution:

For $0\le x\le x_b$: $\Delta p_n(x)=\Delta p_n(0)(1-x/x_b)$ (linear, no recombination)

For $x_b\le x\le x_c$: $\Delta p_n(x)=0$ (immediate recombination)

$$I=qA\frac{D_p}{x_b}\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$$

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Problem 6.23

Problem: Reconsider the special diode with ${\tau }_p$ non-zero but small.

Solution:

$$I=qA\frac{D_p}{L_p}\frac{n_i^2}{N_D}\coth \left(\frac{x_b}{L_p}\right)(e^{q{V}_A/kT}-1)+\text{contribution from }{x}_b\le x\le x_c$$

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Problem 6.24

Problem: A planar Si p${}^{+}$-n step junction diode at 300 K has $N_D=10^{16}$/cm${}^3$. Denuding has created a reduced R-G center concentration.

(a) What is $V_{bi}$?

Solution: $V_{bi}=0.0259\ln (10^{16}\times 10^{16}/10^{20})=0.92$ V

(b) What is $V_{BR}$?

Solution: From Fig. 6.11, $V_{BR}\approx 55$ V.

(c) What applied voltage is necessary to expand the n-side depletion region edge to $x=x_b=2$ μm?

Solution:

$$V_{bi}-V_A=\frac{q{N}_D{x}_b^2}{2K_s{\epsilon }_0}=30.6\text{ V}\Rightarrow V_A=-29.7\text{ V}$$

(d) If the minority carrier lifetimes are ${\tau }_1$ and ${\tau }_2$, what is the ${\tau }_{p1}/{\tau }_{p2}$ ratio?

Solution: ${\tau }_{p1}/{\tau }_{p2}=N_{T2}/N_{T1}=10$

(e) Establish expressions for the diffusion current.

Solution:

$$I_{DIFF}=qA\frac{D_p}{L_{p1}}\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)\text{(near junction)}$$

(f) Establish expressions for the R-G current.

Solution:

$$I_{R-G}=\frac{qA{n}_i{x}_b}{2{\tau }_{01}}+\frac{qA{n}_i(W-x_b)}{2{\tau }_{02}}$$

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Chapter 7: pn Junction Diode — Small-Signal Admittance

7.1 Introduction

The small-signal admittance of a pn junction diode is:

$$Y=G+j\omega C$$

7.2 Reverse-Bias Junction Capacitance

7.2.1 General Information

$$C_J=\frac{K_s{\epsilon }_{0}A}{W}$$

7.2.2 C-V Relationships

For a one-sided power-law profile $N_B(x)=b{x}^m$:

$$W={\left[\frac{(m+2)K_s{\epsilon }_0}{qb}(V_{bi}-V_A)\right]}^{1/(m+2)}$$ $$C_J=\frac{K_s{\epsilon }_{0}A}{W}$$

For a step junction ($m=0$, $b=N_B$):

$$C_J=\frac{C_{J0}}{(1-V_A/V_{bi}{)}^{1/2}}$$

7.2.3 Parameter Extraction/Profiling

$$N_B(x)=\frac{2}{q{K}_s{\epsilon }_0{A}^2\left|\frac{d(1/C_J^2)}{d{V}_A}\right|}$$ $$x=\frac{K_s{\epsilon }_{0}A}{C_J}$$

7.2.4 Reverse-Bias Conductance

$$G_0=\frac{dI}{d{V}_A}$$

7.3 Forward-Bias Diffusion Admittance

7.3.1 General Information

The diffusion admittance arises from minority carrier charge oscillations.

7.3.2 Admittance Relationships

For a p${}^{+}$-n diode:

$$Y_D=G_0\sqrt{1+j\omega {\tau }_p}$$ $$G_D=\frac{G_0}{\sqrt{2}}{\left(\sqrt{1+(\omega {\tau }_p{)}^2}+1\right)}^{1/2}$$ $$C_D=\frac{G_0}{\omega \sqrt{2}}{\left(\sqrt{1+(\omega {\tau }_p{)}^2}-1\right)}^{1/2}$$

At low frequencies ($\omega {\tau }_p\ll 1$):

$$G_D\to G_0,C_D\to \frac{G_0{\tau }_p}{2}$$

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Exercise 7.1

Problem: Compute and plot normalized $C_J/C_{J0}$ versus $V_A/V_{bi}$ for linearly graded, step, and $m=-1$ hyperabrupt junctions.

Solution:

m = [1 0 -1];
s = 1./(m+2);
x = linspace(-25, 0);
y = [];
for i = 1:3
    y = [y; 1./(1-x).^s(i)];
end
plot(x, y); grid;
axis([-25 0 0 1]);
xlabel('VA/Vbi'); ylabel('CJ/CJ0');
text(-20, 0.42, 'linear (m=1)');
text(-20, 0.27, 'step (m=0)');
text(-20, 0.10, 'hyperabrupt (m=-1)');

---

Exercise 7.2

Problem: Analyze the 1N5472A C-V data to confirm the abrupt nature of the junction and determine the p-side doping.

Solution:

% Subtract stray capacitance
C = [150 125 100 75 50 25 0]; % pF
C = C - 2; % pF
VA = [-25 -20 -15 -10 -5 0 5]; % volts
invC2 = 1./(C*1e-12).^2;
p = polyfit(VA, invC2, 1);
slope = p(1); intercept = p(2);
NA = 2/(1.6e-19*11.8*8.85e-14*(3.72e-4)^2*abs(slope));
Vbi = -intercept/slope;

Results: $N_A\approx 8.84\times 10^{15}$/cm${}^3$, $V_{bi}\approx 0.70$ V.

---

Exercise 7.3

Problem: Confirm that $R_s$ and $G_0$ may be neglected in modeling the admittance.

Solution:

$G_0=\Delta I/\Delta V=40$ pA / 18 V $=2.22\times 10^{-12}$ S

$\omega C=2\pi \times 10^5\times 63\times 10^{-12}=3.96\times 10^{-5}$ S

Since $\omega C\gg G_0$, $Y\approx j\omega C$.

$|Z|=1/(\omega C)=2.52\times 10^4$ ohms $\gg R_s=1$ ohm.

---

Exercise 7.4

Problem: Compute and plot $C_J$, $C_D$, and $C_J+C_D$ versus $V_A$ for the 1N5472A diode.

Solution:

CJ0 = 120e-12; Vbi = 0.7; kT = 0.0259;
taun = 5e-7; I0 = 8e-13; n1 = 1.22;
wt = input('input the angular-frequency*lifetime product');
VA = linspace(0, 0.65);
CJ = CJ0./sqrt(1-VA./Vbi);
G0 = I0/kT*exp(VA./(n1*kT));
CD = taun.*G0./(sqrt(2)*wt)*sqrt(sqrt(1+(wt)^2)-1);
semilogy(VA, CJ, '-r'); hold on;
semilogy(VA, CD, '-g');
semilogy(VA, CJ+CD);
xlabel('VA (volts)'); ylabel('C (farads)');

---

Problems — Chapter 7

Problem 7.1 — Quick Quiz

(a) What is the physical origin of the junction capacitance?

Solution: The in-and-out movement of majority carriers about the steady state depletion width in response to the a.c. signal.

(b) Sketch the $m=-1$ hyperabrupt profile of an n${}^{+}$-p junction.

Solution: The doping decreases as one moves away from the junction.

(c) Define quasistatically.

Solution: The device follows the a.c. signal as if it were a d.c. bias.

(d) Define varactor.

Solution: A variable capacitor diode.

(e) Define profiling.

Solution: Determining the doping concentration as a function of position from C-V data.

(f) Make a sketch of the low-frequency conductance of an ideal diode.

Solution: $G_0=0$ for reverse bias $>$ few $kT/q$; $G_0\propto e^{q{V}_A/kT}$ for forward bias.

(g) What is the physical origin of the diffusion admittance?

Solution: Minority carrier charge oscillations in the quasineutral regions.

(h) Why does the diffusion conductance increase with increasing $\omega {\tau }_p$ when $\omega {\tau }_p\gg 1$?

Solution: Because the minority carriers have difficulty following the a.c. signal, leading to out-of-phase oscillations that enhance the conductance.

---

Problem 7.2

Problem: Derive Eq. (7.6) for the one-sided power-law profile.

Solution: Following the procedure in Chapter 5, with $N_B(x)=b{x}^m$:

$$W={\left[\frac{(m+2)K_s{\epsilon }_0}{qb}(V_{bi}-V_A)\right]}^{1/(m+2)}$$

---

Problem 7.3

Problem: Construct a computer program that yields fully dimensioned reverse-bias C-V curves.

Solution:

A = input('Diode area (cm^2), A = ');
NB = input('Lightly doped side concentration (cm^-3), NB = ');
Vmax = input('Largest reverse-bias voltage (V), Vmax = ');
T = 300; KS = 11.8; e0 = 8.85e-14; q = 1.6e-19;
Vbi = 0.0259*log(NB/1e10) + 0.56;
VA = linspace(-Vmax, 0);
W = sqrt(2*KS*e0/q*(Vbi-VA)./NB);
C = KS*e0*A./W;
plot(VA, C);
xlabel('VA (volts)'); ylabel('C (farads)');

---

Problem 7.4

Problem: Analyze the 1N4002 C-V data.

Solution:

Subtract 3 pF stray capacitance. Plot $1/C^2$ versus $V_A$ and fit a straight line.

VA = [0 -0.2 -0.4 -0.6 -0.8 -1.0 -1.4 -1.8 -2.2 -2.6 -3.0 ...
      -4.0 -5.0 -6.0 -7.0 -8.0 -9.0 -10.0 -11.0 -12.0 -13.0 ...
      -14.0 -15.0];
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];
C = (C - 3)*1e-12;
invC2 = 1./C.^2;
p = polyfit(VA, invC2, 1);

---

Problem 7.5

Problem: Derive Eqs. (7.12) and (7.13).

Solution:

From the C-V relationship:

$$C_J=\frac{K_s{\epsilon }_{0}A}{W}$$ $$\frac{d(1/C_J^2)}{d{V}_A}=\frac{2}{q{K}_s{\epsilon }_0{A}^2{N}_B(x)}$$

Therefore:

$$N_B(x)=\frac{2}{q{K}_s{\epsilon }_0{A}^2\left|\frac{d(1/C_J^2)}{d{V}_A}\right|}$$ $$x=\frac{K_s{\epsilon }_{0}A}{C_J}$$

---

Problem 7.6

Problem: Construct a computer program that accepts C-V data and outputs a plot of $N_B(x)$ versus $x$.

Solution: Use the relationships from Problem 7.5 with numerical differentiation.

---

Problem 7.7

Problem: Verify the accuracy of Fig. 7.10.

Solution: Compute and plot $C_D/G_0$ and $G_D/G_0$ versus $\omega {\tau }_p$.

---

Problem 7.8

Problem: What is the relative size of the capacitive and conductive components of the diffusion admittance?

Solution:

For $\omega {\tau }_p\ll 1$: $\omega C_D/G_D\to \omega {\tau }_p/2$

For $\omega {\tau }_p\gg 1$: $\omega C_D/G_D\to 1$

---

Problem 7.9

Problem: Determine the apparent ${\tau }_n$ of the 1N5472A diode.

Solution:

Given $C_D/G_D={\tau }_n/2$ for $\omega {\tau }_n\ll 1$:

$V_A$ (V)	$G_D$ (S)	${\tau }_n$ (sec)
0.50	$2.00\times 10^{-4}$	—
0.52	$3.90\times 10^{-4}$	—
0.54	$7.15\times 10^{-4}$	—
0.56	$1.33\times 10^{-3}$	—
0.58	$2.28\times 10^{-3}$	—

Average ${\tau }_n\approx 5\times 10^{-7}$ sec.

---

Chapter 8: pn Junction Diode — Transient Response

8.1 Turn-Off Transient

8.1.1 Introduction

The storage delay time $t_s$ is the period during which the reverse current remains constant.

The reverse recovery time $t_{rr}$ is the total time for the reverse current to decay to 10% of its maximum magnitude.

8.1.2 Qualitative Analysis

The root cause of the delay is the stored excess minority carriers in the quasineutral regions. These carriers must be removed before the diode can switch to the off-state.

Two removal mechanisms:

1. Recombination
2. Reverse current flow

8.1.3 The Storage Delay Time

Charge control analysis:

$$\frac{d{Q}_p}{dt}=i_{DIFF}-\frac{Q_p}{{\tau }_p}$$

For $0^{+}\le t\le t_s$ with $i=-I_R$:

$$t_s={\tau }_p\ln \left(1+\frac{I_F}{I_R}\right)$$

More precise analysis:

$$\text{erf}\left(\sqrt{\frac{t_s}{{\tau }_p}}\right)=\frac{1}{1+I_R/I_F}$$

8.1.4 General Information

• $t_s$ increases with $I_F$
• $t_s$ decreases with $I_R$
• $t_s\propto {\tau }_p$
• Step-recovery diodes have very short $t_s$

8.2 Turn-On Transient

For a current pulse from $i=0$ to $i=I_F$:

$$Q_p(t)=I_F{\tau }_p(1-e^{-t/{\tau }_p})$$ $$v_A(t)=\frac{kT}{q}\ln \left[1+\frac{I_F}{I_0}(1-e^{-t/{\tau }_p})\right]$$

---

Exercise 8.1

Problem: Predict how key factors affect the i-t transient.

Solution:

• (a) Increasing $I_F$ increases stored charge, so $t_s$ increases.
• (b) Increasing $I_R$ increases removal rate, so $t_s$ decreases.
• (c) Decreasing ${\tau }_p$ increases recombination rate, so $t_s$ decreases.

---

Exercise 8.2

The CPG program visualizes the turn-off and turn-on transients.

% Exercise 8.2 -- Turn-off/Turn-on Concentration Plot Generator
S = menu('Choose the desired plot', 'OFF-Linear', 'OFF-Semilog', ...
          'ON-Linear', 'ON-Semilog');
if S <= 2
    Iratio = input('Please input the IR/IF ratio: IR/IF = ');
    if Iratio == 0
        TS = 1;
    else
        TS = (erfinv(1/(1+Iratio)))^2;
    end
else
    TS = 0;
end
if S == 1 || S == 2
    X = 0:0.03:3;
    T = TS/10:TS/10:TS;
else
    X = 0:0.03:3;
    T = [0.1:0.1:2];
end
y0 = exp(-X);
% ... (plotting commands)

---

Problems — Chapter 8

Problem 8.1 — Quick Quiz

(a) Define storage delay time.

Solution: The time period during which the reverse current remains constant after switching.

(b) Define recovery time.

Solution: The total time for the reverse current to decay to 10% of its maximum magnitude.

(c) True or false: $t_s$ increases with increasing $I_F$ and decreases with increasing $I_R$. Explain.

Solution: True. Larger $I_F$ means more stored charge; larger $I_R$ means faster removal.

(d) What is the root cause of the delay?

Solution: Stored excess minority carriers in the quasineutral regions.

(e) Name the two mechanisms that remove excess stored charge.

Solution: Recombination and reverse current flow.

(f) True or false: If $\Delta p_n(x,t)>0$, $v_A>0$.

Solution: True.

(g) True or false: If $i<0$, the slope of the linear $p_n(x,t)$ versus $x$ plot must be positive at $x=x_n$.

Solution: True.

(h) What is special about a step-recovery diode?

Solution: Very short $t_s$ (~1 nsec), used as pulse generators and harmonic generators.

(i) True or false: Increasing both $I_F$ and $I_R$ by a factor of 2 will have no effect on $t_s$.

Solution: True, since $t_s$ depends on the ratio $I_R/I_F$.

(j) True or false: Recombination actually acts to retard the build-up of stored carriers during the turn-on transient.

Solution: True.

---

Problem 8.2

Problem: The hole concentration on the n-side of a pn step junction diode is as pictured.

(a) Is the junction forward or reverse biased?

Solution: Forward biased (excess holes).

(b) If $p_{n0}=10^4$ cm${}^{-3}$ and $T=300$ K, determine $v_A$.

Solution: From the plot, $p_n(0)=10^8$ cm${}^{-3}$.

$$v_A=\frac{kT}{q}\ln \left(\frac{p_n(0)}{p_{n0}}\right)=0.0259\ln \left(\frac{10^8}{10^4}\right)=0.0259\ln (10^4)=0.239\text{ V}$$

(c) Is there a forward or reverse current flowing?

Solution: Forward current (since the junction is forward biased).

---

Problem 8.3

Problem: Determine whether the revised $t_s$ expression is a significant improvement.

Solution:

$$Q_p(t_s)=\frac{I_R{\tau }_p}{1+I_R/I_F}$$ $$t_s={\tau }_p\ln \left[\frac{1+I_F/I_R}{1+\frac{I_R/I_F}{1+I_R/I_F}}\right]$$

This provides a modest improvement over the simpler expression.

---

Problem 8.4

Problem: An ideal p${}^{+}$-n step junction diode carrying forward current $I_F$ is suddenly open-circuited at $t=0$.

(a) Sketch the expected variation of $p_n(x,t)$ versus $x$.

Solution: The stored charge decays by recombination only.

(b) Derive an expression for the stored hole charge.

Solution:

$$Q_p(t)=Q_p(0)e^{-t/{\tau }_p}$$

where $Q_p(0)=I_F{\tau }_p$.

(c) Assuming quasistatic decay, derive an expression for $v_A(t)$.

Solution:

$$v_A(t)=\frac{kT}{q}\ln \left[\frac{Q_p(t)}{Q_p(0)}{e}^{q{V}_{ON}/kT}\right]=V_{ON}-\frac{kT}{q}\frac{t}{{\tau }_p}$$

(d) Does this suggest anything?

Solution: The open-circuit voltage decay is linear with time, providing a method to measure ${\tau }_p$.

---

Problem 8.5

Problem: An ideal p${}^{+}$-n step junction diode is switched with a current pulse from $I=0$ to $I=1$ mA at $t=0$. Calculate the time for the diode voltage to reach 90% and 95% of its final value.

Solution:

$v_A(t)=\frac{kT}{q}\ln \left[1+\frac{I_F}{I_0}(1-e^{-t/{\tau }_p})\right]$

$V_{ON}=\frac{kT}{q}\ln (I_F/I_0)$

For 90%: $0.9V_{ON}=\frac{kT}{q}\ln \left[1+\frac{I_F}{I_0}(1-e^{-t/{\tau }_p})\right]$

With $I_F/I_0=10^{12}$ and $V_{ON}=0.0259\ln (10^{12})=0.716$ V:

$0.644=0.0259\ln [1+10^{12}(1-e^{-t/{\tau }_p})]$

$e^{24.86}=1+10^{12}(1-e^{-t/{\tau }_p})$

$6.25\times 10^{10}=10^{12}(1-e^{-t/{\tau }_p})$

$1-e^{-t/{\tau }_p}=0.0625$

$t=-{\tau }_p\ln (0.9375)=0.0645{\tau }_p$

For 95%: $t=0.142{\tau }_p$

---

Problem 8.6

Problem: An ideal pn step junction diode initially forward biased at $I_{F1}$ is pulsed to $I_2>I_{F1}$ at $t=0$.

(a) Sketch the expected variation of $p_n(x,t)$.

Solution: The stored charge increases from its initial value to a new steady state.

(b) Assuming quasistatic build-up, derive an expression for $v_A(t)$.

Solution:

$$Q_p(t)=I_2{\tau }_p+(I_{F1}-I_2){\tau }_p{e}^{-t/{\tau }_p}$$ $$v_A(t)=\frac{kT}{q}\ln \left[\frac{Q_p(t)}{I_0{\tau }_p}+1\right]$$

---

Problem 8.7

Problem: A diode is forward biased at 1 mA, then pulsed to $-1$ mA at $t=0$, then back to 1 mA after 1 μsec.

(a) Sketch $i(t)$.

Solution: Forward at 1 mA, then reverse at $-1$ mA for 1 μsec, then forward at 1 mA again.

(b) Establish an expression for $v_A(t)$ for $t>1$ μsec.

Solution: The stored charge builds up from its value at $t=1$ μsec.

---

Problem 8.8

Problem: Compare the exact turn-on $v_A(t)$ with the quasistatic Eq. (8.16) solution.

Solution:

VON = 0.5; T = 300; kT = 0.0259;
t_taup = 0.1:0.1:2;
for i = 1:length(t_taup)
    vA_exact(i) = kT*log(exp(VON/kT)*(1 - exp(-t_taup(i))));
    vA_quasi(i) = kT*log(1 + exp(VON/kT)*(1 - exp(-t_taup(i))));
end
plot(t_taup, vA_exact, t_taup, vA_quasi);
xlabel('t/taup'); ylabel('vA (volts)');
legend('Exact', 'Quasistatic');

---

Chapter 9: Optoelectronic Diodes (Read Only)

9.1 Introduction

Three functional categories of photodevices:

1. Photodetectors: Convert photo-energy to electrical energy for detection
2. Solar cells: Convert photo-energy to electrical power
3. LEDs and laser diodes: Convert electrical energy to photo-energy

9.2 Photodiodes

9.2.1 pn Junction Photodiodes

$$I=I_{dark}+I_L$$ $$I_L=-qA(L_n+W+L_p)G_L$$

Spectral response: Limited by band gap at long wavelengths, surface recombination at short wavelengths.

9.2.2 p-i-n and Avalanche Photodiodes

p-i-n photodiodes: The intrinsic region is fully depleted, providing fast response.

$$f_{\max }=\frac{1}{W_i/v_{sat}}$$

Avalanche photodiodes: Operated near breakdown for internal gain.

9.3 Solar Cells

9.3.1 Solar Cell Basics

Parameters:

• $V_{oc}$: open circuit voltage
• $I_{sc}$: short circuit current
• Fill factor: $FF=\frac{V_m{I}_m}{V_{oc}{I}_{sc}}$

9.3.2 Efficiency Considerations

$$\eta =\frac{FF\cdot I_{sc}\cdot V_{oc}}{P_{in}}$$

Loss mechanisms:

• Photons with $h\nu <E_G$ are not absorbed
• Photons with $h\nu >E_G$ lose excess energy as heat
• Reflection
• Recombination

9.3.3 Solar Cell Technology

Three categories:

1. Thin-film (amorphous Si, CdTe, CuInSe${}_2$)
2. Single-crystal (Si, GaAs)
3. Concentrator

9.4 LEDs

9.4.1 General Overview

Requirements for visible LEDs:

1. Direct semiconductor
2. Band gap between 1.77 eV and 3.10 eV
3. Amenable to pn junction formation

9.4.2 Commercial LEDs

Material	Color	Peak λ (μm)	η (%)
GaAs${}_{0.6}$P${}_{0.4}$	Red	0.650	0.2
GaAs${}_{0.35}$P${}_{0.65}$:N	Orange-Red	0.630	0.7
GaAs${}_{0.14}$P${}_{0.86}$:N	Yellow	0.585	0.2
GaP:N	Green	0.565	0.4
GaP:Zn-O	Red	0.700	2
AlGaAs	Red	0.650	4-16
AlInGaP	Orange	0.620	6
AlInGaP	Yellow	0.585	5
AlInGaP	Green	0.570	1
SiC	Blue	0.470	0.02
GaN	Blue	0.450	2

9.4.3 LED Packaging and Photon Extraction

Total internal reflection limits light extraction. The epoxy dome and reflective cup help.

---

Part IIB: BJTs and Other Junction Devices

Chapter 10: BJT Fundamentals

10.1 Terminology

The BJT contains three adjoining, alternately doped regions:

• Emitter: Heavily doped outer region
• Base: Narrow central region
• Collector: Moderately doped outer region

Terminal currents and voltages:

$$I_E=I_B+I_C$$ $$V_{EB}+V_{BC}+V_{CE}=0$$

Biasing modes:

Mode	E-B Junction	C-B Junction
Saturation	Forward	Forward
Active	Forward	Reverse
Inverted	Reverse	Forward
Cutoff	Reverse	Reverse

10.2 Fabrication

Double-diffused pnp transistor and integrated circuit npn transistor.

10.3 Electrostatics

Under equilibrium and normal operating conditions, the BJT can be viewed as two independent pn junctions.

10.4 Introductory Operational Considerations

Under active mode biasing:

• Emitter injects carriers into base
• Base is narrow, so most carriers diffuse through to collector
• Collector collects the carriers

10.5 Performance Parameters

Emitter efficiency:

$$\gamma =\frac{I_{Ep}}{I_E}$$

Base transport factor:

$${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}$$

Common base d.c. current gain:

$${\alpha }_{dc}=\gamma {\alpha }_T$$

Common emitter d.c. current gain:

$${\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}$$

---

Exercise 10.1

Problem: Draw the energy band diagram for (a) an npn transistor under equilibrium, (b) a pnp transistor under active mode biasing.

Solution:

(a) For npn: Flip the pnp diagram upside down.

(b) For pnp active mode: E-B junction forward biased (potential hill reduced), C-B junction reverse biased (potential hill increased).

---

Exercise 10.2

Problem: Construct a MATLAB program that draws the equilibrium energy band diagram for a BJT.

Solution: (The BJT_Eband program is provided in Appendix M.)

---

Problems — Chapter 10

Problem 10.1 — Quick Quiz

(a) What is the name of the circuit configuration where $I_B$ is the input current and $I_C$ is the output current?

Solution: Common emitter.

(b) What is the name of the circuit configuration where $I_C$ is the output current and $V_{EC}$ is the output voltage?

Solution: Common emitter.

(c) Name the four biasing modes.

Solution: Active, inverted, saturation, cutoff.

(d) What is the purpose of the “buried layer” found in integrated circuit BJTs?

Solution: To provide a low-resistance path between the active collector region and the top-side collector contact.

(e) In a standard pnp BJT, what are the relative sizes of $N_{AE}$, $N_{DB}$, and $N_{AC}$?

Solution: $N_{AE}>N_{DB}>N_{AC}$

(f) What is the standard symbol for the width of the depletion region in a pn junction diode? What is the standard symbol for the quasineutral width of the base in a BJT?

Solution: Diode: $W$; BJT base: $W$

(g) The base region in a BJT is narrow. What is the precise definition of “narrow”?

Solution: The base width is much less than the minority carrier diffusion length in the base.

(h) Why is it necessary for the base region to be narrow?

Solution: So that most injected carriers diffuse through to the collector without recombining.

(i) Define in words what is meant by “emitter efficiency.”

Solution: The fraction of the emitter current that is useful (carriers injected into the base).

(j) Define in words what is meant by “base transport factor.”

Solution: The fraction of injected carriers that successfully diffuse across the base and enter the collector.

---

Problem 10.2

Problem: Complete the tables indicating the polarity of input and output voltages.

Solution:

(a) pnp:

Mode	$V_{EB}$	$V_{CB}$
Active	+	-
Inverted	-	+
Saturation	+	+
Cutoff	-	-

(b) npn:

Mode	$V_{BE}$	$V_{BC}$
Active	+	-
Inverted	-	+
Saturation	+	+
Cutoff	-	-

---

Problem 10.3

Problem: A Si pnp BJT with $N_{AE}=5\times 10^{17}$/cm${}^3$, $N_{DB}=10^{15}$/cm${}^3$, $N_{AC}=10^{14}$/cm${}^3$, and $W_B=3$ μm is at equilibrium.

(a) Sketch the energy band diagram.

Solution: Constant $E_F$. Band bending at both junctions.

(b) Sketch potential, electric field, and charge density.

Solution: $\rho$: negative in p-regions, positive in n-region. $\mathcal{E}$: slope. $V$: quadratic.

(c) Calculate the net potential difference between collector and emitter.

Solution:

$$V_{CE}=V_{bi,EB}-V_{bi,CB}=\frac{kT}{q}\ln \left(\frac{N_{AE}{N}_{DB}}{n_i^2}\right)-\frac{kT}{q}\ln \left(\frac{N_{AC}{N}_{DB}}{n_i^2}\right)=\frac{kT}{q}\ln \left(\frac{N_{AE}}{N_{AC}}\right)$$ $$=0.0259\ln \left(\frac{5\times 10^{17}}{10^{14}}\right)=0.0259\ln (5000)=0.220\text{ V}$$

(d) Determine the quasineutral width of the base.

Solution: $W=W_B-x_{nEB}-x_{nCB}$

$x_{nEB}=\sqrt{\frac{2K_s{\epsilon }_0}{q}\frac{N_{AE}}{N_{DB}(N_{AE}+N_{DB})}{V}_{bi,EB}}$

$V_{bi,EB}=0.0259\ln (5\times 10^{17}\times 10^{15}/10^{20})=0.0259\ln (5\times 10^{12})=0.756$ V

$x_{nEB}=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}}\frac{5\times 10^{17}}{10^{15}(5\times 10^{17}+10^{15})}(0.756)}=9.94\times 10^{-5}$ cm

$x_{nCB}=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}}\frac{10^{14}}{10^{15}(10^{14}+10^{15})}(0.656)}=2.93\times 10^{-4}$ cm

$W=3\times 10^{-4}-9.94\times 10^{-5}-2.93\times 10^{-4}<0$ — the base is punched through!

---

Problem 10.4

Problem: A Si npn with $N_{DE}=10^{18}$/cm${}^3$, $N_{AB}=10^{16}$/cm${}^3$, $N_{DC}=10^{15}$/cm${}^3$, $W_B=2$ μm at equilibrium.

Solution: Similar to Problem 10.3 with n and p interchanged.

---

Problem 10.5

Problem: Biases of $V_{EB}=0.5$ V and $V_{CB}=2$ V are applied to the Problem 10.3 BJT.

Solution: E-B forward biased, C-B reverse biased (active mode).

---

Problem 10.6

Problem: For a typically doped Si npn transistor, sketch the energy band diagram, potential, electric field, and charge density under active mode biasing.

Solution: E-B junction forward biased, C-B junction reverse biased. The Fermi level is split into quasi-Fermi levels.

---

Problem 10.7

Problem: A pnp BJT is saturation biased with $I_C>0$. Construct figures similar to Figs. 10.8 and 10.9.

Solution: Both junctions forward biased. Carrier injection at both junctions.

---

Problem 10.8

Problem: An npn BJT is biased into cutoff. Construct figures showing carrier activity and diffusion currents.

Solution: Both junctions reverse biased. Only small leakage currents flow.

---

Problem 10.9

Problem: Given a pnp BJT where $I_{Ep}=1$ mA, $I_{En}=0.01$ mA, $I_{Cp}=0.98$ mA, and $I_{Cn}=0.1$ μA, calculate: (a) ${\alpha }_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) ${\alpha }_{dc}$, ${\beta }_{dc}$, (e) $I_{CBO}$, $I_{CEO}$, (f) effect of $I_{Cp}$ increase, (g) effect of $I_{En}$ increase.

Solution:

(a) ${\alpha }_T=I_{Cp}/I_{Ep}=0.98/1=0.98$

(b) $\gamma =I_{Ep}/(I_{Ep}+I_{En})=1/1.01=0.990$

(c) $I_E=1.01$ mA, $I_C=0.98$ mA, $I_B=I_E-I_C=0.03$ mA

(d) ${\alpha }_{dc}=\gamma {\alpha }_T=0.990\times 0.98=0.970$

${\beta }_{dc}={\alpha }_{dc}/(1-{\alpha }_{dc})=0.970/0.030=32.3$

(e) $I_{CBO}=I_{Cn}=0.1$ μA

$I_{CEO}=I_{CBO}/(1-{\alpha }_{dc})=0.1/0.030=3.33$ μA

(f) Increasing $I_{Cp}$ increases ${\alpha }_T$ and ${\beta }_{dc}$.

(g) Increasing $I_{En}$ decreases $\gamma$ and ${\beta }_{dc}$.

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Problem 10.10

Problem: Given an npn BJT where $I_{En}=100$ μA, $I_{Ep}=1$ μA, $I_{Cn}=99$ μA, and $I_{Cp}=0.1$ μA, calculate: (a) ${\alpha }_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) ${\alpha }_{dc}$, ${\beta }_{dc}$, (e) $I_{CBO}$, $I_{CEO}$, (f) effect of $I_{Cn}$ increase, (g) effect of $I_{Ep}$ increase.

Solution:

(a) ${\alpha }_T=I_{Cn}/I_{En}=99/100=0.99$

(b) $\gamma =I_{En}/(I_{En}+I_{Ep})=100/101=0.990$

(c) $I_E=101$ μA, $I_C=99.1$ μA, $I_B=1.9$ μA

(d) ${\alpha }_{dc}=0.990\times 0.99=0.980$

${\beta }_{dc}=0.980/0.020=49$

(e) $I_{CBO}=I_{Cp}=0.1$ μA

$I_{CEO}=0.1/0.020=5$ μA

(f) Increasing $I_{Cn}$ increases ${\beta }_{dc}$.

(g) Increasing $I_{Ep}$ decreases ${\beta }_{dc}$.

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Problem 10.11

Problem: Explain why $I_{CBO}$ given by Eq. (10.12) is devoid of a hole component.

Solution: $I_{CBO}$ is the collector current when $I_E=0$ (emitter open). Under this condition, there is no hole injection from the emitter. The only current is the reverse saturation current of the C-B junction, which is dominated by minority carriers (electrons in the p-collector for a pnp transistor). The hole component would come from the p-collector side, but since the collector is lightly doped, this contribution is negligible.

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Chapter 11: BJT Static Characteristics

11.1 Ideal Transistor Analysis

11.1.1 Solution Strategy

Basic assumptions:

1. pnp BJT with nondegenerate uniformly doped regions
2. Steady state
3. One-dimensional
4. Low-level injection
5. No other processes
6. R-G negligible in depletion regions
7. Emitter and collector much wider than diffusion lengths

Boundary conditions:

Emitter:

$$\Delta n_E(x^{\prime \prime }\to \infty )=0,\Delta n_E(x^{\prime \prime }=0)=n_{E0}(e^{q{V}_{EB}/kT}-1)$$

Base:

$$\Delta p_B(0)=p_{B0}(e^{q{V}_{EB}/kT}-1),\Delta p_B(W)=p_{B0}(e^{q{V}_{CB}/kT}-1)$$

Collector:

$$\Delta n_C(x^{\prime }\to \infty )=0,\Delta n_C(x^{\prime }=0)=n_{C0}(e^{q{V}_{CB}/kT}-1)$$

11.1.2 General Solution ($W$ Arbitrary)

Base solution:

$$\Delta p_B(x)=\Delta p_B(0)\frac{\sinh [(W-x)/L_B]}{\sinh (W/L_B)}+\Delta p_B(W)\frac{\sinh (x/L_B)}{\sinh (W/L_B)}$$

Current components:

$$I_{Ep}=qA\frac{D_B}{L_B}{p}_{B0}\left[\frac{\cosh (W/L_B)}{\sinh (W/L_B)}(e^{q{V}_{EB}/kT}-1)-\frac{1}{\sinh (W/L_B)}(e^{q{V}_{CB}/kT}-1)\right]$$ $$I_{Cp}=qA\frac{D_B}{L_B}{p}_{B0}\left[\frac{1}{\sinh (W/L_B)}(e^{q{V}_{EB}/kT}-1)-\frac{\cosh (W/L_B)}{\sinh (W/L_B)}(e^{q{V}_{CB}/kT}-1)\right]$$

11.1.3 Simplified Relationships ($W\ll L_B$)

$$\Delta p_B(x)=\Delta p_B(0)+[\Delta p_B(W)-\Delta p_B(0)]\frac{x}{W}$$ $$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}$$ $${\alpha }_T=\frac{1}{1+\frac{1}{2}(W/L_B{)}^2}$$ $${\alpha }_{dc}=\gamma {\alpha }_T$$ $${\beta }_{dc}=\frac{1}{\frac{1}{2}(W/L_B{)}^2+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}$$

11.1.4 Ebers-Moll Equations and Model

$$I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)-{\alpha }_R{I}_{R0}(e^{q{V}_{CB}/kT}-1)$$ $$I_C={\alpha }_F{I}_{F0}(e^{q{V}_{EB}/kT}-1)-I_{R0}(e^{q{V}_{CB}/kT}-1)$$

with ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$.

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Exercise 11.1

Problem: (a) Show that the expressions for the emitter and collector terminal currents reduce to the ideal diode equation in the limit where $W\gg L_B$.

Solution:

For $W\gg L_B$: $\cosh (W/L_B)/\sinh (W/L_B)\to 1$, $1/\sinh (W/L_B)\to 0$.

$$I_E=qA\left(\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\right)(e^{q{V}_{EB}/kT}-1)$$ $$I_C=-qA\left(\frac{D_C}{L_C}{n}_{C0}+\frac{D_B}{L_B}{p}_{B0}\right)(e^{q{V}_{CB}/kT}-1)$$

These are ideal diode equations.

(b) Show that setting $V_{CB}=0$ in the emitter current expression leads to the narrow-base diode result.

Solution: With $V_{CB}=0$, the $e^{q{V}_{CB}/kT}-1$ term vanishes. The emitter current becomes:

$$I_E=qA\left(\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right)(e^{q{V}_{EB}/kT}-1)$$

This is the narrow-base diode result.

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Exercise 11.2

Problem: Show that Eq. (11.40) is the solution to the minority carrier diffusion equation if thermal recombination-generation is negligible.

Solution:

With R-G negligible:

$$\frac{d^2\Delta p_B}{d{x}^2}=0$$

General solution: $\Delta p_B(x)=A_1+A_{2}x$

Applying boundary conditions gives:

$$\Delta p_B(x)=\Delta p_B(0)+[\Delta p_B(W)-\Delta p_B(0)]\frac{x}{W}$$

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Exercise 11.3

Problem: Consider a pnp transistor ($W\ll L_B$) biased into saturation. Sketch the perturbed minority carrier distribution.

Solution:

In saturation, both junctions are forward biased. The normalized carrier excess is:

Emitter: $\Delta n_E/n_{E0}=(e^{q{V}_{EB}/kT}-1)e^{-x^{\prime \prime }/L_E}$

Base: Linear from $\Delta p_B(0)/p_{B0}=(e^{q{V}_{EB}/kT}-1)$ to $\Delta p_B(W)/p_{B0}=(e^{q{V}_{CB}/kT}-1)$

Collector: $\Delta n_C/n_{C0}=(e^{q{V}_{CB}/kT}-1)e^{-x^{\prime }/L_C}$

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Exercise 11.4

Problem: Work out an expression for the base transport factor using the linear approximation for $\Delta p_B(x)$.

Solution:

Since $\Delta p_B(x)$ is linear, $d\Delta p_B/dx$ is constant. Therefore $I_{Cp}=I_{Ep}$ and:

$${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=1$$

This is consistent with the assumption of negligible recombination in the base.

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Exercise 11.5

Problem: Show that the forward current gain in the Ebers-Moll formulation equals the common base current gain.

Solution:

Dividing Eq. (11.46) by Eq. (11.45a):

$${\alpha }_F=\frac{qA\frac{D_B}{L_B}{p}_{B0}/\sinh (W/L_B)}{qA\left(\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right)}$$

This equals ${\alpha }_{dc}$ from Eq. (11.33).

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Exercise 11.6

Problem: Show that the Ebers-Moll equations can be manipulated to obtain $I_C={\alpha }_{dc}{I}_E+I_{CBO}$.

Solution:

Under active mode, the Ebers-Moll equations reduce to:

$$I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)+{\alpha }_R{I}_{R0}$$ $$I_C={\alpha }_F{I}_{F0}(e^{q{V}_{EB}/kT}-1)+I_{R0}$$

Solving the $I_E$ equation for the voltage factor and substituting:

$$I_C={\alpha }_F{I}_E+(1-{\alpha }_F{\alpha }_R)I_{R0}$$

Identifying $I_{CBO}=(1-{\alpha }_F{\alpha }_R)I_{R0}$ gives the desired result.

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Exercise 11.7

Problem: Construct a MATLAB program that calculates and plots the input and output characteristics of a pnp BJT.

Solution: (The BJT program is provided in Appendix M.)

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Exercise 11.8

Problem: For the BJT with parameters in Table 11.1, determine the C-B junction voltage that will lead to punch-through.

Solution:

At punch-through, $W=0$:

$$W_B=x_{nEB}+x_{nCB}$$

Solving for $V_{CB}$:

$$V_{CB}=V_{bi,CB}-\frac{(W_B-x_{nEB}{)}^2}{\frac{2K_s{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}}$$

With the parameters from Table 11.1, $V_{CB}\approx -235$ V.

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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.

(a) Which transistor has the larger punch-through voltage?

Solution: Transistor A (base is more heavily doped than collector, so the C-B depletion region extends mostly into the collector).

(b) If limited by avalanche breakdown, which has the larger $V_{CBO}$?

Solution: Transistor A (lightly doped collector side determines breakdown).

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Exercise 11.10

Problem: Modify the BJT program to include base width modulation and carrier multiplication.

Solution: (The BJTplus program is provided in Appendix M.)

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Problems — Chapter 11

Problem 11.1 — Quick Quiz

(a) In the ideal transistor analysis, what assumptions are made about the regional E, B, C doping concentrations and quasineutral widths?

Solution: $N_E\gg N_B>N_C$, and the quasineutral widths of the emitter and collector are much greater than the minority carrier diffusion lengths. The base width is arbitrary.

(b) Roughly sketch the minority carrier distribution in the base of a pnp BJT under active mode biasing when $W/L_B\ll 1$ and $W/L_B\gg 1$.

Solution: For $W/L_B\ll 1$: linear. For $W/L_B\gg 1$: exponential decay.

(c) A pnp BJT with $W/L_B\ll 1$ is active mode biased. Given $\Delta p_{B\max }=10p_{B0}$, make sketches of $\Delta p_B(x)$, $\Delta p_B(x)/p_{B0}$, and $p_B(x)$.

Solution: Linear distribution from $10p_{B0}$ at $x=0$ to $\sim 0$ at $x=W$.

(d) What are typical sizes for $\gamma$, $\alpha$, ${\alpha }_{dc}$, and ${\beta }_{dc}$?

Solution: $\gamma \approx 0.99$, ${\alpha }_T\approx 0.99$, ${\alpha }_{dc}\approx 0.98$, ${\beta }_{dc}\approx 50-300$.

(e) Establish a general expression for $I_B=I_B(V_{EB},V_{CB})$ in terms of the Ebers-Moll parameters.

Solution:

$$I_B=I_E-I_C=(1-{\alpha }_F)I_{F0}(e^{q{V}_{EB}/kT}-1)+(1-{\alpha }_R)I_{R0}(e^{q{V}_{CB}/kT}-1)$$

(f) Referring to Fig. 11.5(d), what is the cause of the quasilinear slope at lower $V_{EC}$? What is the cause of the sharp upward curvature as $V_{EC}\to V_{CEO}$?

Solution: Quasilinear slope: base width modulation. Sharp upward curvature: carrier multiplication and avalanche breakdown.

(g) It is established that $V_{CBO}$ of a BJT is limited by punch-through. Nevertheless, someone claims $V_{CEO}<V_{CBO}$. Is this possible?

Solution: Yes, because carrier multiplication and feedback can cause $V_{CEO}<V_{CBO}$ even if punch-through limits $V_{CBO}$.

(h) What is the cause of “current crowding”?

Solution: The lateral voltage drop in the base from the center to the edges of the emitter, causing higher current density at the emitter periphery.

(i) Cite the beneficial effects associated with a graded base.

Solution: Built-in electric field assists minority carrier transport across the base, reducing transit time and recombination, increasing gain and high-frequency response.

(j) What is a Gummel plot?

Solution: A semilog plot of $I_B$ and $I_C$ versus $V_{EB}$ with $V_{CB}$ constant.

(k) What is the cause of the decrease in ${\beta }_{dc}$ at low and high $I_C$ values?

Solution: Low $I_C$: R-G current. High $I_C$: high-level injection, current crowding, series resistance.

(l) Would there be any advantage to constructing a Si BJT where the entire emitter is made of polysilicon?

Solution: Yes, polysilicon emitters improve gain by reducing the back-injection of carriers into the emitter.

(m) Define “heterojunction.”

Solution: A junction between two dissimilar semiconductors.

(n) What are the differences between a BJT and an HBT?

Solution: In an HBT, the emitter is made from a wider band gap semiconductor than the base, allowing heavy base doping without sacrificing gain.

(o) What is the cause of the decrease in ${\beta }_{dc}$ at low $I_C$ values as exemplified by Fig. 11.16?

Solution: The R-G current from the E-B depletion region becomes a significant fraction of the base current at low current levels.

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Problem 11.2

Problem: Determine $\gamma$, ${\alpha }_T$, ${\beta }_{dc}$, $I_B$, and whether R-G current is negligible.

Solution:

From the figure: $I_{Ep}=0.999$ mA, $I_{En}=0.001$ mA, $I_{Cp}=0.998$ mA, $I_{Cn}=0.001$ μA.

$\gamma =0.999/1=0.999$

${\alpha }_T=0.998/0.999=0.999$

${\beta }_{dc}=\gamma {\alpha }_T/(1-\gamma {\alpha }_T)=0.998/0.002=499$

$I_B=I_E-I_C=1-0.998=0.002$ mA = 2 μA

The R-G current is negligible because $I_{Cn}$ is much smaller than $I_{Cp}$.

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Problem 11.3

Problem: Indicate the required revisions for an npn BJT.

Solution: Interchange n and p, reverse voltage polarities, interchange electron and hole currents.

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Problem 11.4

Problem: Sample computations for a pnp BJT with $V_{EB}=0.7$ V, $V_{CB}=-5$ V, $W=W_B$, $A=10^{-4}$ cm${}^2$, and material parameters from Table 11.1.

Solution:

(a) $W/L_B=2\times 10^{-4}/3.36\times 10^{-3}=0.0595$

$\cosh (W/L_B)=1.00177$, $\sinh (W/L_B)=0.0595$

(b) $I_{Ep}=qA\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}(e^{q{V}_{EB}/kT}-1)=1.60\times 10^{-3}$ A

$I_{En}=qA\frac{D_E}{L_E}{n}_{E0}(e^{q{V}_{EB}/kT}-1)=1.60\times 10^{-5}$ A

$I_{Cp}=qA\frac{D_B}{L_B}{p}_{B0}\frac{1}{\sinh (W/L_B)}(e^{q{V}_{EB}/kT}-1)=1.59\times 10^{-3}$ A

$I_{Cn}=qA\frac{D_C}{L_C}{n}_{C0}=1.60\times 10^{-14}$ A

(c) $I_E=1.62\times 10^{-3}$ A, $I_C=1.59\times 10^{-3}$ A, $I_B=2.6\times 10^{-5}$ A

(d) $I_{B1}=I_{En}=1.60\times 10^{-5}$ A, $I_{B2}=I_{Ep}-I_{Cp}=1.0\times 10^{-5}$ A, $I_{B3}=I_{Cn}=1.6\times 10^{-14}$ A

(e) $\gamma =0.990$, ${\alpha }_T=0.994$, ${\alpha }_{dc}=0.984$, ${\beta }_{dc}=61.5$

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Problem 11.5

Problem: Examine how the minority carrier distribution in the base varies with $W/L_B$.

Solution:

W_LB = [10 5 1 0.5 0.1];
x = linspace(0, 1, 100);
for i = 1:length(W_LB)
    y = sinh(W_LB(i)*(1-x))./sinh(W_LB(i));
    plot(x, y); hold on;
end
xlabel('x/W'); ylabel('Delta pB(x)/Delta pB(0)');

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Problem 11.6

Problem: Graphically illustrate how the minority carrier distribution in the base varies with biasing mode.

Solution: Four plots for active, inverted, saturation, and cutoff.

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Problem 11.7

Problem: Sketch the majority and minority carrier distributions under (a) active, (b) inverted, (c) saturation, (d) cutoff biasing.

Solution:

• (a) Active: minority carriers pile up at E-B edge, decrease to zero at C-B edge.
• (b) Inverted: minority carriers pile up at C-B edge, decrease to zero at E-B edge.
• (c) Saturation: minority carriers pile up at both edges.
• (d) Cutoff: minority carriers depleted at both edges.

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Problem 11.8

Problem: Determine the polarity of $V_{EB}$, $V_{CB}$, magnitude of $V_{CB}$, and biasing mode.

Solution:

From the figure: $\Delta n_E/n_{E0}=-1$ (reverse biased E-B), $\Delta p_B/p_{B0}=10$ (forward biased C-B).

$V_{CB}=0.0259\ln (11)=0.062$ V

Biasing mode: inverted.

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Problem 11.9

Problem: Complete the table indicating the effect of changes on $\gamma$, ${\alpha }_T$, and ${\beta }_{dc}$.

Solution:

Change	Effect on $\gamma$	Effect on ${\alpha }_T$	Effect on ${\beta }_{dc}$
Increase $W_B$	No effect	Decrease	Decrease
Increase ${\tau }_B$	No effect	Increase	Increase
Increase $N_B$	Decrease	No effect	Decrease
Increase ${\tau }_E$	Increase	No effect	Increase
Increase $N_E$	Increase	No effect	Increase

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Problem 11.10

Problem: Construct four log-log plots of ${\beta }_{dc}$ versus ${\tau }_B/{\tau }_{B0}$, $N_B/N_{B0}$, ${\tau }_E/{\tau }_{E0}$, and $N_E/N_{E0}$.

Solution: Use the mobility fit relationships from Exercise 3.1.

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Problem 11.11

Problem: Examine the effect of a shallow emitter on the performance parameters.

Solution:

(a) Revised $I_{En}$: $I_{En}=qA\frac{D_E}{W_E}{n}_{E0}(e^{q{V}_{EB}/kT}-1)$

(b) Revised performance parameters:

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}}$$ $${\alpha }_T=\frac{1}{1+\frac{1}{2}(W/L_B{)}^2}$$

(c) For $W\ll L_B$ and $W_E\ll L_E$:

$${\beta }_{dc}=\frac{1}{\frac{1}{2}(W/L_B{)}^2+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}}$$

(d) As $W_E$ decreases, $\gamma$ decreases and ${\beta }_{dc}$ decreases.

(e) Sketch showing linear distribution in the shallow emitter.

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Problem 11.12

Problem: Derive the I-V relationship for various “diode” connections of a BJT.

Solution: (Six possible connections, using Ebers-Moll equations.)

For example, with B open ($I_B=0$):

$$I_C=\frac{{\alpha }_F}{1-{\alpha }_R}{I}_E+\frac{1-{\alpha }_F{\alpha }_R}{1-{\alpha }_R}{I}_{R0}$$

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Problem 11.13

Problem: Draw the simplified large-signal equivalent circuit at the active/saturation boundary.

Solution: Both junctions forward biased. Use the Ebers-Moll model with $V_{CB}=0$.

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Problem 11.14

Problem: Perform the algebraic manipulations to derive the computational equations for the BJT characteristics.

Solution:

(a) Common base output: $I_C={\alpha }_F{I}_E-(1-{\alpha }_F{\alpha }_R)I_{R0}(e^{q{V}_{CB}/kT}-1)$

(b) Common emitter input: Derive from $I_B=I_E-I_C$.

(c) Common emitter output: Substitute $I_E=I_C+I_B$ into the Ebers-Moll equations.

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Problem 11.15

Problem: Show that the common emitter input characteristics are relatively insensitive to base width modulation when the gain is limited by emitter efficiency.

Solution: When $\frac{D_E{N}_{B}W}{D_B{N}_E{L}_E}\gg \frac{1}{2}(W/L_B{)}^2$, the W-dependent term in the denominator of ${\beta }_{dc}$ is negligible.

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Problem 11.16

Problem: How is the Exercise 11.8 result modified if $V_{EB}=0.7$ V?

Solution: $x_{nEB}$ decreases slightly because $V_{bi,EB}-V_{EB}$ is smaller. The punch-through voltage increases slightly.

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Problem 11.17

Problem: Two pnp BJTs are identical except the emitter and collector region doping are interchanged.

(a) Which transistor has the greater emitter efficiency?

Solution: Transistor A (higher emitter doping).

(b) Which transistor will exhibit greater sensitivity to base width modulation under active mode biasing?

Solution: Transistor B (collector is more heavily doped, so C-B depletion region extends more into the base).

(c) If limited by avalanche breakdown, which has the larger $V_{CBO}$?

Solution: Transistor A (collector is more lightly doped).

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Problem 11.18

Problem: Determine the Early voltage from the characteristics.

Solution: From the slope of the linear region: $V_A=I_C/\text{slope}$.

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Problem 11.19

Problem: Generate plots of the I-V characteristics for diode-like operation.

Solution: Use the BJTplus program with appropriate modifications.

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Problem 11.20

Problem: Construct a theoretical Gummel plot.

Solution:

VEB = linspace(0.3, 0.8);
% Ideal currents
IC_ideal = IF0*(exp(VEB/kT)-1);
IB_ideal = (1-aF)*IF0*(exp(VEB/kT)-1);
% Corrected for R-G and high-level injection
IRG = 1e-14*(exp(VEB/(1.5*kT))-1);
IB_corrected = IB_ideal + IRG;
gamma = 1./(1+exp((VEB-0.75)/(2*kT)));
IC_corrected = IC_ideal.*gamma;
semilogy(VEB, IC_ideal, '--', VEB, IB_ideal, '--', ...
         VEB, IC_corrected, VEB, IB_corrected);
xlabel('VEB (volts)'); ylabel('Current (A)');

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Problem 11.21

Problem: Explain the $V_{EC}$ dependence of ${\beta }_{dc}$.

Solution: Increasing $V_{EC}$ increases the reverse bias across the C-B junction, which decreases $W$ (base width modulation), increasing ${\beta }_{dc}$.

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Chapter 12: BJT Dynamic Response Modeling

12.1 Small-Signal Equivalent Circuits

12.1.1 Generalized Two-Port Model

$$i_b=g_{11}{v}_{be}+g_{12}{v}_{ce}$$ $$i_c=g_{21}{v}_{be}+g_{22}{v}_{ce}$$

12.1.2 Hybrid-Pi Models

Simplified model:

$$g_m=\frac{q{I}_C}{kT},r_{\pi }=\frac{{\beta }_{dc}}{g_m}$$

Complete model:

$$g_m=g_{21}-g_{12},r_0=\frac{1}{g_{22}+g_{12}},r_{\pi }=\frac{1}{g_{11}+g_{12}},r_{\mu }=-\frac{1}{g_{12}}$$

High-frequency model: Includes $C_{cb}$, $C_{eb}$, $r_b$, $r_c$, $r_e$.

12.2 Transient (Switching) Response

12.2.1 Qualitative Observations

Turn-on: stored charge builds up in the base.

Turn-off: stored charge is removed.

12.2.2 Charge Control Relationships

$$\frac{d{Q}_B}{dt}=i_B-\frac{Q_B}{{\tau }_B}$$ $$i_C=\frac{Q_B}{{\tau }_t}$$

where ${\tau }_t=W^2/2D_B$ is the base transit time.

12.2.3 Quantitative Analysis

Turn-on transient:

$$Q_B(t)=I_{BB}{\tau }_B(1-e^{-t/{\tau }_B})$$ $$t_r={\tau }_B\ln \left(\frac{I_{BB}{\tau }_B}{I_{BB}{\tau }_B-I_{CC}{\tau }_t}\right)$$

Turn-off transient:

$$Q_B(t)=\left[\frac{I_{CC}{\tau }_t}{1+\xi }+(1+\xi )I_{BB}{\tau }_B\right]e^{-t/{\tau }_B}-\frac{I_{BB}{\tau }_B}{1+\xi }$$ $$t_{sd}={\tau }_B\ln \left[\frac{1+\xi }{I_{CC}{\tau }_t/(I_{BB}{\tau }_B)+\xi }\right]$$

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Problems — Chapter 12

Problem 12.1 — Quick Quiz

(a) In words, what is the quasistatic assumption?

Solution: The device can follow the a.c. signal as if it were a d.c. bias.

(b) Why are there separate definitions for the npn and pnp $g_{ij}$ conductances?

Solution: Because the current directions and voltage polarities are reversed.

(c) What is the origin of the “Hybrid-Pi” name?

Solution: The circuit uses a combination of hybrid (h) parameters and a pi-shaped arrangement.

(d) State the names of the Hybrid-Pi parameters.

Solution: $g_m$ = transconductance, $r_0$ = output resistance, $r_{\pi }$ = input resistance, $r_{\mu }$ = feedthrough resistance.

(e) What is the justification for adding capacitors to the low-frequency Hybrid-Pi model?

Solution: To account for the junction capacitances at high frequencies.

(f) Describe what happens to $i_B$, $i_C$, and the minority carrier concentration once the transistor enters saturation.

Solution: $i_C$ saturates, $i_B$ remains constant, and the minority carrier concentration continues to build up.

(g) Give the word and mathematical definitions of the “base transit time.”

Solution: The average time for minority carriers to diffuse across the quasineutral base. ${\tau }_t=W^2/2D_B$.

(h) How are ${\tau }_B$ and ${\tau }_t$ related to ${\beta }_{dc}$?

Solution: ${\beta }_{dc}={\tau }_B/{\tau }_t$.

(i) Why does an $i_B<0$ during turn-off give rise to a reduction in both storage-delay and fall times?

Solution: It aids the withdrawal of stored charge from the base.

(j) What is a Schottky diode clamp?

Solution: A Schottky diode connected between the collector and base to prevent saturation.

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Problem 12.2

Problem: Establish a small-signal equivalent circuit for the common base configuration.

Solution: Use the two-port model with appropriate terminal identifications.

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Problem 12.3

Problem: Determine $g_m$ and $r_{\pi }$ for the 2N2605 pnp BJT with $I_B=5$ μA and $V_{EC}=10$ V.

Solution:

From Fig. 11.5(d), at $I_B=5$ μA, $I_C\approx 1$ mA.

$$g_m=\frac{q{I}_C}{kT}=\frac{1\times 10^{-3}}{0.0259}=3.86\times 10^{-2}\text{ S}$$ $${\beta }_{dc}=I_C/I_B=1\text{ mA}/5\text{ μA}=200$$ $$r_{\pi }=\frac{{\beta }_{dc}}{g_m}=\frac{200}{0.0386}=5180\text{ ohms}$$

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Problem 12.4

Problem: Perform the circuit analysis to confirm Eqs. (12.10).

Solution: Comparing the two-port model and the Hybrid-Pi model:

$$g_m=g_{21}-g_{12}$$ $$r_0=\frac{1}{g_{22}+g_{12}}$$ $$r_{\pi }=\frac{1}{g_{11}+g_{12}}$$ $$r_{\mu }=-\frac{1}{g_{12}}$$

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Problem 12.5

Problem: Write a MATLAB program that calculates the Hybrid-Pi parameters.

Solution: Use the BJTplus program and numerically differentiate.

---

Problem 12.6

Problem: Derive an expression for $f_T$, the unity beta frequency.

Solution:

$$f_T=\frac{g_m}{2\pi (C_{\pi }+C_{\mu })}$$

---

Problem 12.7

Problem: Show that the diffusion component of $C_{eb}$ is given by the expression.

Solution: From the narrow-base diode analysis:

$$C_{eb|diff}=\frac{2}{3}\frac{(W^2/2D_B)}{kT/q}{I}_C=\frac{2}{3}\frac{{\tau }_t}{kT/q}{I}_C=\frac{2}{3}{g}_m{\tau }_t$$

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Problem 12.8

Problem: Estimate $I_{CC}{\tau }_t/I_{BB}{\tau }_B$ for switching between the off and on points.

Solution: From Fig. 12.3(b): $I_{CC}=V_{CC}/R_L=5$ mA, $I_{BB}=V_S/R_s=30$ μA.

$$\frac{I_{CC}{\tau }_t}{I_{BB}{\tau }_B}=\frac{5\times 10^{-3}}{30\times 10^{-6}}\frac{{\tau }_t}{{\tau }_B}=167\frac{{\tau }_t}{{\tau }_B}$$

If ${\beta }_{dc}={\tau }_B/{\tau }_t=100$, then the ratio is 1.67.

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Problem 12.9

Problem: Rise and storage-delay times.

Solution:

(a) $t_r/{\tau }_B$ versus $x=I_{CC}{\tau }_t/(I_{BB}{\tau }_B)$:

$$\frac{t_r}{{\tau }_B}=\ln \left(\frac{1}{1-x}\right)$$

(b) $t_{sd}/{\tau }_B$ versus $x$ for $\xi =0$ and $\xi =1$.

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Problem 12.10

Problem: Fall time.

Solution:

(a) Using the measurement-based definition (10% to 90%):

$$t_f={\tau }_B\ln \left[\frac{I_{CC}{\tau }_t/(I_{BB}{\tau }_B)+\xi }{0.1I_{CC}{\tau }_t/(I_{BB}{\tau }_B)+\xi }\right]$$

(b) Plot $t_f/{\tau }_B$ versus $x$ for $\xi =0$ and $\xi =1$.

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Chapter 13: PNPN Devices (Read Only)

13.1 Silicon Controlled Rectifier (SCR)

The SCR contains four alternately doped regions (P1-N2-P3-N4) and three junctions (J12, J23, J34).

13.2 SCR Operational Theory

The two-transistor model: The SCR is conceptually subdivided into a pnp transistor (P1-N2-P3) and an npn transistor (N2-P3-N4).

$$I_{AK}=\frac{I_{R01}+I_{R02}}{1-({\alpha }_1+{\alpha }_2)}$$

Switching occurs when ${\alpha }_1+{\alpha }_2\to 1$.

13.3 Practical Turn-On/Turn-Off Considerations

• di/dt burnout: Excessive gate current rise rate causes localized heating.
• du/dt effect: Rapid voltage changes can trigger premature conduction.
• Shorted-cathode configuration: Improves reproducibility.
• Triggering time: $t_{ON}\approx \sqrt{W_2^2{W}_3^2/(2D_p{D}_n)}$

13.4 Other PNPN Devices

• SCS: Silicon controlled switch (dual-gate SCR)
• DIAC: Bilateral diode
• TRIAC: Bilateral SCR
• PUT: Programmable unijunction transistor

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Chapter 14: MS Contacts and Schottky Diodes

14.1 Ideal MS Contacts

Semiconductor Type	Condition	Contact Nature
n-type	${\Phi }_M>{\Phi }_s$	Rectifying
n-type	${\Phi }_M<{\Phi }_s$	Ohmic
p-type	${\Phi }_M<{\Phi }_s$	Rectifying
p-type	${\Phi }_M>{\Phi }_s$	Ohmic

Barrier height:

$${\Phi }_B={\Phi }_M-\chi \text{(n-type)}$$ $${\Phi }_B=E_G+\chi -{\Phi }_M\text{(p-type)}$$

14.2 Schottky Diode

14.2.1 Electrostatics

$$V_{bi}=\frac{1}{q}[{\Phi }_B-(E_c-E_F{)}_{FB}]$$ $$W=\sqrt{\frac{2K_s{\epsilon }_0}{q{N}_D}(V_{bi}-V_A)}$$

14.2.2 I-V Characteristics

$$I=I_S(e^{q{V}_A/kT}-1)$$ $$I_S=A{A}^{\ast }{T}^2{e}^{-{\Phi }_B/kT}$$

where $A^{\ast }=120$ A/cm${}^2$-K${}^2$ for free electrons.

14.2.3 a.c. Response

$$C=\frac{K_s{\epsilon }_{0}A}{W}$$

The Schottky diode does not exhibit a diffusion capacitance.

14.2.4 Transient Response

Very fast switching due to negligible minority carrier storage.

14.3 Practical Contact Considerations

• Fermi-level pinning: ${\Phi }_B$ is often independent of ${\Phi }_M$.
• Ohmic contacts: Formed by heavily doping the semiconductor beneath the contact.
• Spiking: Al penetration through shallow junctions.
• Silicides: TiSi${}_2$, etc., for high-temperature stability.

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Exercise 14.1

Problem: (a) Construct the equilibrium energy band diagram for an ideal p-type semiconductor to metal contact where ${\Phi }_M<{\Phi }_s$.

Solution: Rectifying contact for p-type.

(b) Repeat when ${\Phi }_M>{\Phi }_s$.

Solution: Ohmic contact for p-type.

(c) Verify that an ideal MS contact on p-type is rectifying if ${\Phi }_M<{\Phi }_s$ and ohmic if ${\Phi }_M>{\Phi }_s$.

Solution: The barrier to hole flow from the semiconductor to the metal is ${\Phi }_B=E_G+\chi -{\Phi }_M$. If ${\Phi }_M<{\Phi }_s$, this barrier is large, and the contact is rectifying.

(d) Establish an expression for the barrier height.

Solution:

$${\Phi }_B=E_G+\chi -{\Phi }_M$$

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Exercise 14.2

Problem: Copper on n-type Si. ${\Phi }_M=4.65$ eV, $\chi =4.03$ eV, $N_D=10^{16}$/cm${}^3$, $T=300$ K.

Solution:

(a) ${\Phi }_B={\Phi }_M-\chi =4.65-4.03=0.62$ eV

(b) $V_{bi}={\Phi }_B-(E_c-E_F{)}_{FB}=0.62-0.20=0.42$ V

(c) $W=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{16}}(0.42)}=0.234$ μm

(d) $|\mathcal{E}{|}_{\max }=\frac{q{N}_{D}W}{K_s{\epsilon }_0}=3.59\times 10^4$ V/cm

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Exercise 14.3

Problem: Perform a least squares fit to determine $I_S$ and $n_1$.

Solution:

I = [4.047e-7 2.792e-6 1.890e-5 1.263e-4 8.084e-4 4.487e-3];
VA = [0.10 0.15 0.20 0.25 0.30 0.35];
y = log(I);
c = polyfit(VA, y, 1);
n1 = 1/(0.0259*c(1));
Is = exp(c(2));
% n1 = 1.03, Is = 1.02e-8 A

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Exercise 14.4

Problem: Estimate the barrier height ${\Phi }_B$.

Solution:

From the $1/C^2$ versus $V_A$ plot, $V_{bi}\approx 0.6$ V.

Slope $\approx -5.5\times 10^{20}$ F${}^{-2}$-V${}^{-1}$

$$N_D=\frac{2}{q{K}_s{\epsilon }_0{A}^2|\text{slope}|}=9.7\times 10^{15}{\text{ cm}}^{-3}$$ $$(E_c-E_F{)}_{FB}=\frac{kT}{q}\ln (N_D/n_i)=0.20\text{ eV}$$ $${\Phi }_B=V_{bi}+(E_c-E_F{)}_{FB}=0.6+0.2=0.8\text{ eV}$$

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Problems — Chapter 14

Problem 14.1 — Quick Quiz

(a) What are the differences between an MS diode, a Schottky diode, and a hot carrier diode?

Solution: All are the same — a rectifying metal-semiconductor contact. “Hot carrier diode” emphasizes the majority carrier nature.

(b) An ideal MS contact is formed with ${\Phi }_M$. Under what conditions is it ohmic? Rectifying?

Solution: n-type: ohmic if ${\Phi }_M<{\Phi }_s$, rectifying if ${\Phi }_M>{\Phi }_s$. p-type: opposite.

(c) Are there any significant differences in the electrostatic formulations between a pn junction and an MS diode?

Solution: Yes — in the MS diode, the negative charge is a $\delta$-function at the metal surface, while in the pn junction it is distributed over the depletion region.

(d) Name the dominant current component in MS diodes.

Solution: Thermionic emission (majority carrier injection over the barrier).

(e) How was the $I_{Mo\to S}$ component determined without a detailed mathematical analysis?

Solution: By requiring the currents to balance at equilibrium.

(f) Explain why MS diodes do not exhibit a diffusion capacitance.

Solution: Because there is negligible minority carrier injection and storage.

(g) Explain why MS diodes switch very rapidly.

Solution: Because there is little stored charge to remove.

(h) What is a “Schottky diode clamp”?

Solution: A Schottky diode connected between the base and collector of a BJT to prevent saturation.

(i) Describe the usual procedure for forming practical ohmic contacts.

Solution: Heavily dope the semiconductor beneath the contact to enable tunneling.

(j) What is “spiking”?

Solution: Nonuniform penetration of Al through shallow junctions.

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Problem 14.2

Problem: For each MS parameter combination, draw the equilibrium energy band diagram and confirm the ohmic or rectifying nature.

Solution:

• A: ${\Phi }_M=4.75$, $\chi =4.00$, n-Ge, $N_D=10^{16}$: Rectifying (Φ_M > Φ_s)
• B: Same but p-Ge, $N_A=10^{17}$: Ohmic (Φ_M > Φ_s)
• C: ${\Phi }_M=4.00$, $\chi =4.03$, n-Si, $N_D=10^{15}$: Ohmic (Φ_M < Φ_s)
• D: ${\Phi }_M=4.25$, $\chi =4.03$, p-Si, $N_A=10^{16}$: Ohmic (Φ_M > Φ_s? No, p-type ohmic requires Φ_M > Φ_s, and Φ_s for p-type is χ + E_G - (E_c - E_F) ≈ 4.03 + 1.12 - 0.2 = 4.95, so Φ_M < Φ_s → Rectifying)
• E: ${\Phi }_M=4.75$, $\chi =4.07$, n-GaAs, $N_D=10^{16}$: Rectifying
• F: Same but p-GaAs, $N_A=10^{17}$: Ohmic

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Problem 14.3

Problem: Gold on n-type Si. ${\Phi }_M=5.10$ eV, $\chi =4.03$ eV, $N_D=10^{15}$/cm${}^3$.

Solution:

(a) ${\Phi }_B=5.10-4.03=1.07$ eV

(b) $V_{bi}=1.07-0.26=0.81$ V

(c) $W=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{15}}(0.81)}=1.03$ μm

(d) $|\mathcal{E}{|}_{\max }=1.57\times 10^4$ V/cm

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Problem 14.4

Problem: Construct a plot of the equilibrium depletion width versus $N_D$ for Si MS diodes.

Solution:

ND = logspace(14, 17);
for PHIB = [0.5 0.6 0.7]
    Vbi = PHIB - 0.0259*log(ND/1e10);
    W = sqrt(2*11.8*8.85e-14./(1.6e-19*ND).*Vbi);
    loglog(ND, W); hold on;
end
xlabel('ND (cm^-3)'); ylabel('W (micrometers)');

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Problem 14.5

Problem: Construct a MATLAB program that draws the equilibrium energy band diagram for an MS diode.

Solution: Similar to the MOS diagram generator, but with appropriate boundary conditions.

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Problem 14.6

Problem: Fill in the missing steps in the derivation of Eq. (14.18).

Solution:

Starting from Eq. (14.16):

$$I_{So\to M}=-qA{\int }_{-\infty }^{-v_{\min }}{v}_{x}n(v_x)d{v}_x$$

Substituting Eq. (14.17) and integrating:

$$I_{So\to M}=A{A}^{\ast }{T}^2{e}^{-{\Phi }_B/kT}{e}^{q{V}_A/kT}$$

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Problem 14.7 — MS Photodiode

Problem: An n-type Si MS diode is illuminated.

(a) Short-circuited: The photogenerated holes are swept to the metal, electrons to the semiconductor. Short-circuit photocurrent flows from metal to semiconductor.

(b) Open-circuited: The bands shift to maintain zero current.

(c) Derive the photocurrent expression.

Solution:

$$I_L=-qA(L_n+W+L_p)G_L$$

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Problem 14.8

Problem: Graphically illustrate the effect of series resistance on the I-V characteristic.

Solution:

Is = 1e-10; Rs = [0 0.1 1 10];
VA = linspace(0, 0.6);
for i = 1:length(Rs)
    Vj = VA;
    for j = 1:length(VA)
        I(j) = Is*(exp(Vj(j)/0.0259)-1);
        Vj(j) = VA(j) - I(j)*Rs(i);
    end
    semilogy(VA, I); hold on;
end

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Problem 14.9

Problem: Estimate the minority carrier injection ratio.

Solution:

$$\frac{I_{DIFF}}{I_{TE}}=\frac{qA\frac{D_p}{L_p}\frac{n_i^2}{N_D}{e}^{q{V}_A/kT}}{A{A}^{\ast }{T}^2{e}^{-{\Phi }_B/kT}{e}^{q{V}_A/kT}}=\frac{q{D}_p{n}_i^2}{L_p{N}_D{A}^{\ast }{T}^2{e}^{-{\Phi }_B/kT}}$$

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Problem 14.10

Problem: Examine the magnitude and effect of Schottky barrier lowering.

Solution:

(a) $\Delta {\Phi }_B=\sqrt{\frac{q{\mathcal{E}}_s}{4\pi K_s{\epsilon }_0}}$

(b) $I(V_A)/I(V_A=0)=e^{\Delta {\Phi }_B/kT}$

(c) The reverse current increases with reverse bias due to barrier lowering.

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Problem 14.11

Problem: Analyze the $1/C^2$ versus $V_A$ data.

Solution:

VA = [-1.09 -2.08 -3.07 -4.06 -5.05 -6.04 -7.03 -8.02 -9.01 -10.00];
invC2 = [0.953 1.494 2.035 2.579 3.125 3.673 4.217 4.763 5.320 5.890]*1e20;
p = polyfit(VA, invC2, 1);
Vbi = -p(2)/p(1);
ND = 2/(1.6e-19*11.8*8.85e-14*(1.5e-3)^2*abs(p(1)));

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Problem 14.12

Problem: Linearly graded MS diode.

(a) Derive solutions for $\rho$, $\mathcal{E}$, $V$, and $W$.

Solution:

$$\rho (x)=qax$$ $$\mathcal{E}(x)=\frac{qa}{2K_s{\epsilon }_0}(W^2-x^2)$$ $$V(x)=V_{bi}-V_A-\frac{qa}{2K_s{\epsilon }_0}\left(W^{2}x-\frac{x^3}{3}\right)$$ $$W={\left[\frac{12K_s{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}$$

(b) Determine $V_{bi}$.

Solution: $V_{bi}=\frac{2kT}{q}\ln \left(\frac{a{W}_0}{2n_i}\right)$

(c) Establish an expression for the junction capacitance.

Solution:

$$C=\frac{K_s{\epsilon }_{0}A}{W}$$

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Part III: Field Effect Devices

Chapter 15: Field Effect Introduction — The J-FET and MESFET

15.1 General Introduction

The field effect is the modulation of the conductivity of a semiconductor by an electric field applied normal to the surface.

Key devices:

• J-FET: Junction field effect transistor (Shockley, 1952)
• MESFET: Metal-semiconductor field effect transistor
• MOSFET: Metal-oxide-semiconductor field effect transistor (Kahng and Atalla, 1960)
• MODFET/HEMT: Modulation-doped field effect transistor

15.2 J-FET

15.2.1 Introduction

The J-FET uses pn junction depletion regions to modulate the channel conductance.

15.2.2 Qualitative Theory of Operation

As $V_D$ increases, the depletion regions widen, narrowing the channel. Pinch-off occurs when the depletion regions touch.

15.2.3 Quantitative $I_D$-$V_D$ Relationships

Gradual channel approximation:

$$I_D=\frac{2qZ{\mu }_n{N}_{D}a}{L}\left[V_D-\frac{2}{3}\frac{(V_D+V_{bi}-V_G{)}^{3/2}-(V_{bi}-V_G{)}^{3/2}}{\sqrt{V_{bi}-V_p}}\right]$$

for $0\le V_D\le V_{Dsat}$ and $V_p\le V_G\le 0$.

Pinch-off voltage:

$$V_{Dsat}=V_p-V_G$$

Saturation current:

$$I_{Dsat}=\frac{2qZ{\mu }_n{N}_{D}a}{L}\left[\frac{(V_{bi}-V_p{)}^{3/2}-(V_{bi}-V_G{)}^{3/2}}{3\sqrt{V_{bi}-V_p}}\right]$$

Square-law relationship:

$$I_{Dsat}=I_{D0}{\left(1-\frac{V_G}{V_p}\right)}^2$$

15.2.4 a.c. Response

Small-signal parameters:

$$g_m=\frac{\partial I_D}{\partial V_G}{|}_{V_D},g_d=\frac{\partial I_D}{\partial V_D}{|}_{V_G}$$

15.3 MESFET

15.3.1 General Information

Similar to J-FET but uses a Schottky barrier gate.

15.3.2 Short-Channel Considerations

Variable mobility model:

$$I_D=\frac{I_D(\text{long-channel})}{1+\frac{{\mu }_0{V}_D}{v_{sat}L}}$$

Saturated velocity model:

$$I_{Dsat}=q{v}_{sat}{N}_{D}Z[a-W(0)]$$

Two-region model: Combines gradual channel approximation with saturated velocity model.

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Exercise 15.1

Problem: An n-channel Si J-FET with $N_D=10^{16}$/cm${}^3$. Determine the maximum junction-to-junction half-width $a$.

Solution:

For saturating characteristics, $a\le W_{BR}$.

From Fig. 6.11, $V_{BR}\approx 55$ V. $V_{bi}=0.92$ V.

$$a_{\max }=W_{BR}=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{16}}(0.92+55)}=2.70\text{ μm}$$

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Exercise 15.2

Problem: Compare $I_{Dsat}/I_{D0}$ versus $V_G/V_p$ from Eq. (15.13) and the square-law.

Solution:

z = -0.4; x = linspace(0, 1);
Num = -x + (2/3)*(1-z)*(1-((x-z)/(1-z)).^1.5);
Den = -1 - (2/3)*(z-1)*(1-(z/(z-1))^1.5);
yP = Num./Den;
yS = (1-x).^2;
plot(x, yP, x, yS, '--');
xlabel('VG/VP'); ylabel('IDsat/ID0');

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Exercise 15.3

Problem: Show that source and drain resistances are taken into account by replacing $V_D$ with $V_D-I_D(R_S+R_D)$ and $V_G$ with $V_G-I_D{R}_S$.

Solution: The channel voltages at $y=0$ and $y=L$ are $V(0)=I_D{R}_S$ and $V(L)=V_D-I_D{R}_D$. Substituting into the integral and simplifying yields the modified equations.

---

Exercise 15.4

Problem: Show that source and drain resistances give rise to a reduced effective transconductance.

Solution:

$$g_m^{\prime }=\frac{g_m}{1+g_m{R}_S}$$

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Problems — Chapter 15

Problem 15.1 — Quick Quiz

(a) Define “field effect.”

Solution: The modulation of semiconductor conductivity by an electric field applied normal to the surface.

(b) For a p-channel J-FET, does the drain current flow into or out of the drain contact?

Solution: Out of the drain contact (holes flow from source to drain).

(c) What is meant by “pinch-off”?

Solution: The condition where the depletion regions touch, completely depleting the channel.

(d) What is the mathematical definition of drain conductance and transconductance?

Solution: $g_d=\partial I_D/\partial V_D{|}_{V_G}$, $g_m=\partial I_D/\partial V_G{|}_{V_D}$

(e) What do MESFET, D-MESFET, and E-MESFET stand for?

Solution: Metal-Semiconductor FET, Depletion-mode MESFET, Enhancement-mode MESFET.

(f) Why is the magnitude of the electric field in the channel of concern in modeling short-channel MESFETs?

Solution: Because at high fields, the carrier drift velocity saturates and is no longer proportional to the electric field.

(g) What is the primary difference between the long-channel and two-region short-channel theories?

Solution: The two-region model includes a region where the carrier velocity is saturated, while the long-channel theory assumes $v=\mu \mathcal{E}$ throughout.

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Problem 15.2 — Composite Transistor

Problem: Analyze the composite transistor structure.

Solution: (a) With D-B tied together and $d\le L_p$, the device operates as a J-FET. (b) With E-C tied together, the device operates as back-to-back diodes. (c) With E-C tied together and $d<2W_{BR}$, the device operates as a punch-through device.

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Problem 15.3

Problem: Derive an expression for the channel voltage position.

Solution:

$$y=L\frac{V(y)-\frac{2}{3}\frac{(V(y)+V_{bi}-V_G{)}^{3/2}-(V_{bi}-V_G{)}^{3/2}}{\sqrt{V_{bi}-V_p}}}{V_D-\frac{2}{3}\frac{(V_D+V_{bi}-V_G{)}^{3/2}-(V_{bi}-V_G{)}^{3/2}}{\sqrt{V_{bi}-V_p}}}$$

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Problem 15.4

Problem: Derive Eq. (15.12) from the condition that $I_D$ is maximum at $V_D=V_{Dsat}$.

Solution: Setting $\partial I_D/\partial V_D=0$ in Eq. (15.9) yields $V_{Dsat}=V_G-V_p$.

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Problem 15.5

Problem: J-FET with bottom gate tied to source.

Solution: (a) The depletion region extends further from the top gate. (b) $V_p^{\prime }=V_p-V_{bi}$. (c) Similar. (d) $V_{Dsat}$ from the condition that the top gate pinches off. (e) $V_{Dsat}$ is larger for $V_{GB}=0$. (f) Modify the analysis with asymmetric gates.

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Problem 15.6

Problem: J-FET with linearly graded junctions.

Solution: (a) $W(V)={\left[\frac{12K_s{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}$ (b) Modify the analysis with the new $W$ expression.

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Problem 15.7

Problem: Graphically illustrate the effect of $R_S$ and $R_D$ on the $I_D$-$V_D$ characteristics.

Solution: Use the modified equations from Exercise 15.3.

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Problem 15.8 — Pinch Resistor

Problem: (a) Establish expressions for the d.c. and a.c. conductance.

Solution: $G=I/V$, $g=dI/dV$

(b) Compute the values.

Solution: Use the J-FET equations with $V_G=0$.

---

Problem 15.9

Problem: Derive the cutoff frequency expression.

Solution: $f_{\max }=\frac{g_m}{2\pi C_G}$

---

Problem 15.10

Problem: Temperature dependence of the J-FET transconductance.

Solution: Compute $g_m(T)/g_m(300)$ versus $T$ and determine the power law.

---

Problem 15.11

Problem: Back-gated MESFET.

Solution: Derive the $I_D$-$V_D$ relationship with both top and bottom gates.

---

Problem 15.12

Problem: Derive Eq. (15.22).

Solution: Substitute the field-dependent mobility into the long-channel derivation.

---

Problem 15.13

Problem: Derive Eq. (15.26).

Solution: Differentiate the channel voltage expression with respect to position and set $v=v_{sat}$.

---

Problem 15.14

Problem: Construct a MATLAB program for the two-region model.

Solution: Implement the two-region equations and plot.

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Chapter 16: MOS Fundamentals

16.1 Ideal Structure Definition

The ideal MOS structure:

1. Metal gate is equipotential
2. Oxide is perfect insulator
3. No charge centers in oxide or at interface
4. Semiconductor is uniformly doped
5. Semiconductor is sufficiently thick
6. Ohmic back contact
7. One-dimensional
8. ${\Phi }_M={\Phi }_s$

16.2 Electrostatics — Mostly Qualitative

16.2.1 Visualization Aids

Energy band diagram: Constructed by aligning Fermi levels and inserting the oxide.

Block charge diagrams: Show the approximate charge distribution.

16.2.2 Effect of an Applied Bias

Bias	n-type	p-type
Accumulation	$V_G>0$	$V_G<0$
Flat band	$V_G=0$	$V_G=0$
Depletion	$V_T<V_G<0$	$0<V_G<V_T$
Inversion	$V_G<V_T$	$V_G>V_T$

16.3 Electrostatics — Quantitative Formulation

16.3.1 Semiconductor Electrostatics

Doping parameter:

$${\varphi }_F=\frac{kT}{q}\ln \left(\frac{N_A}{n_i}\right)\text{(p-type)}$$

Depletion width:

$$W=\sqrt{\frac{2K_s{\epsilon }_0}{q{N}_A}{\varphi }_s}$$

Surface potential at inversion:

$${\varphi }_s=2{\varphi }_F$$

Maximum depletion width:

$$W_T=\sqrt{\frac{2K_s{\epsilon }_0}{q{N}_A}(2{\varphi }_F)}$$

16.3.2 Gate Voltage Relationship

$$V_G={\varphi }_s+\frac{K_s}{K_O}{x}_0{\mathcal{E}}_s$$

For depletion:

$$V_G={\varphi }_s+\frac{K_s}{K_O}{x}_0\sqrt{\frac{2q{N}_A{\varphi }_s}{K_s{\epsilon }_0}}$$

16.4 Capacitance-Voltage Characteristics

16.4.1 Theory and Analysis

Accumulation: $C\approx C_0=\frac{K_O{\epsilon }_{0}A}{x_0}$

Depletion:

$$C=\frac{C_0{C}_s}{C_0+C_s}$$

where $C_s=\frac{K_s{\epsilon }_{0}A}{W}$

Inversion (low frequency): $C\approx C_0$

Inversion (high frequency):

$$C=\frac{C_0{C}_{s\min }}{C_0+C_{s\min }}$$

16.4.2 Computations and Observations

Delta-depletion analysis:

$$C=\frac{C_0}{\sqrt{1+\frac{2K_s^2{\epsilon }_0{V}_G}{q{N}_A{K}_O^2{x}_0^2}}}$$

Deep depletion: Occurs when the MOS-C is not allowed to equilibrate under inversion biasing.

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Exercise 16.1

Problem: Construct line plots identifying the voltage ranges for accumulation, depletion, and inversion.

Solution:

For n-type:

• Inversion: $V_G<V_T$
• Depletion: $V_T<V_G<0$
• Accumulation: $V_G>0$

For p-type:

• Accumulation: $V_G<0$
• Depletion: $0<V_G<V_T$
• Inversion: $V_G>V_T$

---

Exercise 16.2

Problem: For each ${\varphi }_F$, ${\varphi }_s$ set, indicate the doping type and biasing condition.

Solution:

(i) ${\varphi }_F=12$, ${\varphi }_s=12$ (in $kT/q$ units): p-type, mid-depletion (ii) ${\varphi }_F=-9$, ${\varphi }_s=3$: n-type, accumulation (iii) ${\varphi }_F=-9$, ${\varphi }_s=-18$: n-type, inversion (iv) ${\varphi }_F=15$, ${\varphi }_s=36$: p-type, inversion (v) ${\varphi }_F=-15$, ${\varphi }_s=0$: n-type, flat band

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Exercise 16.3

Problem: MOS-C at 300 K, $x_0=0.1$ μm, $N_A=10^{15}$/cm${}^3$. Compute (a) ${\varphi }_F$, (b) $W$ when ${\varphi }_s={\varphi }_F$, (c) ${\mathcal{E}}_s$ when ${\varphi }_s={\varphi }_F$, (d) $V_G$ when ${\varphi }_s={\varphi }_F$.

Solution:

(a) ${\varphi }_F=0.0259\ln (10^{15}/10^{10})=0.298$ V

(b) $W=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{15}}(0.298)}=0.624$ μm

(c) ${\mathcal{E}}_s=\frac{q{N}_{A}W}{K_s{\epsilon }_0}=9.56\times 10^3$ V/cm

(d) $V_G=0.298+\frac{11.8\times 10^{-5}\times 9.56\times 10^3}{3.9}=0.587$ V

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Exercise 16.4

Problem: Complete the table identifying capacitance and charge diagram for each biasing condition.

Solution:

Bias Condition	Capacitance	Charge Diagram
Accumulation	g	3
Depletion	e	1
Inversion	a/b/c	4
Flat band	f	2
Depl/inv transition	d	5

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Exercise 16.5

Problem: Write a MATLAB program to calculate and plot low-frequency C-V characteristics.

Solution: (The MOS_CV program is provided in Appendix M.)

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Exercise 16.6

Problem: Explain the observed C-V characteristic.

Solution: From (1) to (2): equilibrium high-frequency C-V. At (2): ramp rate increased, driving the device into deep depletion. From (3) to (4): sweep stopped, equilibrium restored as minority carriers are generated.

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Problems — Chapter 16

Problem 16.1

Problem: For each ${\varphi }_F$, ${\varphi }_s$ parameter set, indicate the biasing condition and draw the energy band and block charge diagrams.

Solution:

(a) ${\varphi }_F=18$, ${\varphi }_s=9$ (p-type): Depletion (b) ${\varphi }_F=-12$, ${\varphi }_s=0$ (n-type): Flat band (c) ${\varphi }_F=12$, ${\varphi }_s=24$ (p-type): Inversion (d) ${\varphi }_F=-15$, ${\varphi }_s=-3$ (n-type): Accumulation (e) ${\varphi }_F=9$, ${\varphi }_s=21$ (p-type): Inversion

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Problem 16.2

Problem: Examine Fig. 16.8(c) more closely.

Solution:

(a) At the onset of inversion, the block charge diagram shows a depletion region with charge $q{N}_A{W}_T$ and a thin inversion layer.

(b) The spike-like nature near $x=0$ is the inversion layer. The value $-2$ at $x=0$ is because the inversion charge is added to the depletion charge.

(c) $W_T=\sqrt{\frac{2\times 11.8\times 8.85\times 10^{-14}}{1.6\times 10^{-19}\times 10^{15}}(2\times 0.298)}=0.882$ μm. This is consistent with Fig. 16.8(c).

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Problem 16.3

Problem: Construct a $W_T$ versus doping plot for GaAs.

Solution:

NA = logspace(14, 18);
for i = 1:length(NA)
    phiF = 0.0259*log(NA(i)/2.25e6);
    WT(i) = sqrt(2*12.85*8.85e-14/(1.6e-19*NA(i))*2*phiF);
end
loglog(NA, WT);
xlabel('NA (cm^-3)'); ylabel('WT (micrometers)');

---

Problem 16.4

Problem: MOS-C at 300 K, $x_0=0.1$ μm, $N_D=10^{15}$/cm${}^3$.

Solution:

(a) ${\varphi }_F=-0.298$ V

(b) $W_T=0.882$ μm

(c) ${\mathcal{E}}_s=1.35\times 10^4$ V/cm

(d) $V_T=-0.298+\frac{11.8\times 10^{-5}\times 1.35\times 10^4}{3.9}=-0.298+0.408=0.110$ V

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Problem 16.5

Problem: Show that the exact-solution equivalent of Eq. (16.28) is:

$$V_G=\frac{kT}{q}\left[U_s+\frac{K_s{x}_0}{K_O{L}_D}F(U_s,U_F)\right]$$

Solution: From Appendix B, the electric field at the surface is ${\mathcal{E}}_s=\frac{kT}{q{L}_D}F(U_s,U_F)$. Substituting into Eq. (16.26) yields the result.

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Problem 16.6

Problem: Derive Eq. (16.35).

Solution:

From $V_G={\varphi }_s+\frac{K_s}{K_O}{x}_0\sqrt{\frac{2q{N}_A{\varphi }_s}{K_s{\epsilon }_0}}$, solve for $\sqrt{{\varphi }_s}$:

$$\sqrt{{\varphi }_s}=\frac{-1+\sqrt{1+\frac{2K_s^2{\epsilon }_0{V}_G}{q{N}_A{K}_O^2{x}_0^2}}}{\frac{2K_s}{K_O{x}_0}\sqrt{\frac{2q{N}_A}{K_s{\epsilon }_0}}}$$

Substituting into $W=\sqrt{\frac{2K_s{\epsilon }_0}{q{N}_A}{\varphi }_s}$ gives Eq. (16.35).

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Problem 16.7

Problem: Analyze the energy band diagram for an ideal MOS-C at 300 K.

Solution:

(a) $\varphi (x)$: quadratic in the semiconductor. (b) $\mathcal{E}(x)$: constant in oxide, linear in semiconductor. (c) Yes, the semiconductor is in equilibrium. (d) $n(x)$: highest at the interface. (e) $n(0)=n_i{e}^{(E_F-E_i)/kT}=10^{10}{e}^{0.29/0.0259}=7.3\times 10^{14}$ cm${}^{-3}$ (f) $N_D=5\times 10^{15}$ cm${}^{-3}$ (g) ${\varphi }_s=0.29$ V (h) $V_G=0.56$ V (i) $\Delta {\varphi }_{ox}=0.27$ V (j) $C/C_0=0.56$

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Problem 16.8

Problem: Complete the table.

Solution:

Bias Condition	Capacitance	Band Diagram
Inversion	a	4
Depletion	b	3
Flat band	c	2
$V_T$	d	5
Accumulation	e	1

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Problem 16.9

Problem: Analyze the energy band diagram.

Solution:

(a) Yes. (b) ${\varphi }_F=-0.30$ V (c) ${\varphi }_s=0.30$ V (d) $V_G=0.30+\frac{11.8\times 10^{-5}\times 1.35\times 10^4}{3.9}=0.71$ V (e) $x_0=0.1$ μm (f) Block charge diagram: positive gate charge, negative depletion charge. (g) C-V characteristic: high at accumulation, decreasing through depletion, low at inversion. (h) $C=\frac{C_0}{\sqrt{1+\frac{V_s}{V_{\delta }}}}$ (iii)

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Problem 16.10

Problem: An ideal MOS capacitor is constructed on intrinsic silicon.

Solution:

(a) Flat band: $E_F$ at midgap. (b) Positive bias: accumulation of electrons. Negative bias: accumulation of holes. (c) C-V: high at both accumulation and inversion (low frequency), low at depletion.

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Problem 16.11

Problem: SOS capacitor.

Solution:

(a) Energy band diagrams for various biases. (b) Block charge diagrams. (c) C-V characteristic: similar to MOS-C but with different accumulation/inversion behavior.

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Problem 16.12

Problem: (a) For which curve is an equilibrium inversion layer present when $V_G>V_T$?

Solution: Curve a (low-frequency).

(b) The MOS-C exhibiting curve b has (thicker oxide) and (higher doping).

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Problem 16.13

Problem: The C-V characteristic exhibited by an MOS-C.

Solution:

(a) n-type (accumulation at positive bias, inversion at negative bias). (b) Energy band diagram at point 2: inversion. (c) Block charge diagram at point 1: accumulation. (d) $x_0=\frac{K_O{\epsilon }_{0}A}{C_0}=\frac{3.9\times 8.85\times 10^{-14}\times 3\times 10^{-3}}{100\times 10^{-12}}=1.04\times 10^{-5}$ cm = 0.104 μm (e) $W_T=0.882$ μm, $N_D=10^{15}$ cm${}^{-3}$

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Problem 16.14

Problem: An ideal MOS-C at 300 K.

Solution:

(a) High-frequency C-V: high at accumulation, decreasing through depletion, low at inversion. (b) $C_{\max }=C_0=\frac{K_O{\epsilon }_{0}A}{x_0}=345$ pF (c) $C_{\min }=\frac{C_0{C}_{s\min }}{C_0+C_{s\min }}=69$ pF (d) ${\varphi }_s=2{\varphi }_F=0.712$ V (e) $V_T=0.712+\frac{11.8\times 10^{-5}\times 1.35\times 10^4}{3.9}=1.12$ V (f) Energy band diagram at ${\varphi }_s=1.5{\varphi }_F$. (g) Block charge diagram at ${\varphi }_s=2.5{\varphi }_F$. (h) Deep depletion C-V.

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Problem 16.15

Problem: Analyze the block charge diagram.

Solution:

(a) n-type (positive charge in semiconductor). (b) Inversion (thin inversion layer at interface). (c) Energy band diagram: strong band bending. (d) High-frequency a.c. signal: inversion layer charge fixed, depletion width fluctuates. (e) C-V: high at accumulation, decreasing through depletion, low at inversion. (f) Deep depletion: wider depletion region, no inversion layer.

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Problem 16.16

Problem: Write a MATLAB program for total-deep-depletion C-V.

Solution: Use the exact-charge relationships from Appendix C.

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Problem 16.17

Problem: Determine $x_0$ and $N_A$ from the C-V data.

Solution:

(a) $x_0=\frac{K_O{\epsilon }_{0}A}{C_0}=\frac{3.9\times 8.85\times 10^{-14}\times 4.75\times 10^{-3}}{82\times 10^{-12}}=0.020$ μm

(b) $C/C_0=0.25$, $W_{eff}=\frac{K_s{\epsilon }_{0}A}{C_{\min }}=2.0\times 10^{-4}$ cm

$W_{eff}/L_D=2.0\times 10^{-4}/2.91\times 10^{-3}=68.7$

(c) Solving iteratively: $U_F=14.5$, $N_D=1.0\times 10^{16}$ cm${}^{-3}$

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Chapter 17: MOSFETs — The Essentials

17.1 Qualitative Theory of Operation

When $V_G>V_T$, an inversion layer forms a conducting channel between source and drain.

As $V_D$ increases, the channel narrows near the drain. Pinch-off occurs when the inversion layer vanishes at the drain.

17.2 Quantitative $I_D$-$V_D$ Relationships

17.2.1 Preliminary Considerations

Threshold voltage:

$$V_T=2{\varphi }_F+\frac{K_s{x}_0}{K_O}\sqrt{\frac{4q{N}_A{\varphi }_F}{K_s{\epsilon }_0}}$$

Effective mobility:

$${\bar{\mu }}_n=\frac{{\mu }_0}{1+\theta (V_G-V_T)}$$

17.2.2 Square-Law Theory

$$I_D=\frac{Z{\bar{\mu }}_n{C}_0}{L}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right]$$

for $0\le V_D\le V_{Dsat}$.

$$V_{Dsat}=V_G-V_T$$ $$I_{Dsat}=\frac{Z{\bar{\mu }}_n{C}_0}{2L}(V_G-V_T{)}^2$$

17.2.3 Bulk-Charge Theory

$$I_D=\frac{Z{\bar{\mu }}_n{C}_0}{L}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}-\frac{2}{3}\frac{\sqrt{2q{N}_A{K}_s{\epsilon }_0}}{C_0}{V}_D^{3/2}\right]$$

17.2.4 Charge-Sheet and Exact-Charge Theories

These theories remove the limitations of the square-law and bulk-charge theories, including subthreshold current and self-saturation.

17.3 a.c. Response

17.3.1 Small-Signal Equivalent Circuits

$$i_d=g_d{v}_d+g_m{v}_g$$ $$g_d=\frac{\partial I_D}{\partial V_D}{|}_{V_G},g_m=\frac{\partial I_D}{\partial V_G}{|}_{V_D}$$

17.3.2 Cutoff Frequency

$$f_{\max }=\frac{g_m}{2\pi C_G}$$

17.3.3 Small-Signal Characteristics

$g_d$ versus $V_G$ is used to determine $V_T$. $C_G$ versus $V_G$ gives a low-frequency type characteristic.

---

Exercise 17.1

Problem: Indicate how the figures must be modified for a p-channel device.

Solution: Reverse all voltage polarities and current directions.

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Exercise 17.2

Problem: Construct a “universal” plot of $I_D/(Z{\bar{\mu }}_n{C}_0/L)$ versus $V_D$.

Solution:

for VGT = 4:-1:1
    VD = linspace(0, VGT);
    ID = VGT.*VD - VD.*VD./2;
    IDsat = VGT*VGT/2;
    VD = [VD, 9]; ID = [ID, IDsat];
    plot(VD, ID); hold on;
end
xlabel('VD (volts)'); ylabel('ID/(ZunCo/L)');

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Exercise 17.3

Problem: Determine $I_D$ when gate and drain are tied together.

Solution:

(a) $V_G=V_D=1$ V. $V_T=1.42$ V. Since $V_G<V_T$, $I_D=0$.

(b) $V_G=V_D=3$ V. $I_D=0.382$ mA (bulk-charge theory).

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Problems — Chapter 17

Problem 17.1 — Quick Quiz

(a) Why are the current-carrying contacts referred to as source and drain?

Solution: Carriers enter through the source and leave through the drain.

(b) What is the channel?

Solution: The inversion layer connecting source and drain.

(c) What is the saturation region of operation?

Solution: The region where $V_D>V_{Dsat}$ and $I_D$ is approximately constant.

(d) What is the relationship between $V_T$ in the MOS-C discussion and the threshold voltage in the MOSFET discussion?

Solution: They are the same — the voltage at which inversion occurs.

(e) Why is the mobility in the surface channel different from the bulk?

Solution: Additional surface scattering reduces the mobility.

(f) Why is the formulation referred to as the square-law theory?

Solution: Because $I_{Dsat}\propto (V_G-V_T{)}^2$.

(g) Why is the bulk-charge theory so named?

Solution: Because it accounts for the variation in the bulk depletion charge along the channel.

(h) What variables are displayed in subthreshold transfer characteristics?

Solution: $\log (I_D)$ versus $V_G$.

(i) What is the mathematical definition of drain conductance and transconductance?

Solution: $g_d=\partial I_D/\partial V_D{|}_{V_G}$, $g_m=\partial I_D/\partial V_G{|}_{V_D}$

(j) Why is the MOSFET $C$-$V_G$ curve typically a low-frequency characteristic?

Solution: Because the source and drain supply the minority carriers needed for the inversion layer to follow the a.c. signal.

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Problem 17.2 — Sample Calculations

Problem: An ideal n-channel MOSFET at 300 K with $Z=50$ μm, $L=5$ μm, $x_0=0.05$ μm, $N_A=10^{15}$/cm${}^3$, ${\bar{\mu }}_n=800$ cm${}^2$/V-sec.

Solution:

(a) $V_T=2{\varphi }_F+\frac{K_s{x}_0}{K_O}\sqrt{\frac{4q{N}_A{\varphi }_F}{K_s{\epsilon }_0}}=0.596+0.234=0.83$ V

(b) $I_{Dsat}=\frac{Z{\bar{\mu }}_n{C}_0}{2L}(V_G-V_T{)}^2=3.31\times 10^{-4}$ A (square-law)

(c) $I_{Dsat}=2.92\times 10^{-4}$ A (bulk-charge)

(d) $g_d=\frac{Z{\bar{\mu }}_n{C}_0}{L}(V_G-V_T)=2.21\times 10^{-4}$ S

(e) $g_m=\frac{Z{\bar{\mu }}_n{C}_0}{L}(V_G-V_T)=2.21\times 10^{-4}$ S

(f) Similar for bulk-charge.

(g) $f_{\max }=\frac{g_m}{2\pi C_G}=1.1$ GHz

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Problem 17.3

Problem: Sketch diagrams for a p-channel MOSFET at threshold.

Solution: Similar to n-channel but with opposite polarities.

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Problem 17.4

Problem: Construct a plot of $V_T$ versus $N_A$ for various $x_0$.

Solution:

NA = logspace(14, 18);
for x0 = [0.01 0.02 0.05 0.1]*1e-4
    phiF = 0.0259*log(NA/1e10);
    VT = 2*phiF + 11.8*x0/3.9.*sqrt(4*1.6e-19*NA.*phiF/(11.8*8.85e-14));
    semilogx(NA, VT); hold on;
end

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Problem 17.5

Problem: Temperature dependence of $V_T$.

Solution: $V_T$ decreases with increasing temperature because ${\varphi }_F$ decreases.

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Problem 17.6

Problem: Derive $V_{Dsat}$ from the condition that $I_D$ is maximum.

Solution: Setting $\partial I_D/\partial V_D=0$ gives $V_{Dsat}=V_G-V_T$.

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Problem 17.7

Problem: Modify the equations for a p-channel MOSFET.

Solution: Reverse all voltage polarities and current directions.

---

Problem 17.8

Problem: Incorporate the ${\bar{\mu }}_n$-$V_G$ dependence.

Solution: Replace ${\bar{\mu }}_n$ with ${\mu }_0/[1+\theta (V_G-V_T)]$.

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Problem 17.9

Problem: Sketch $I_D$ versus $V_D$ with a battery between gate and drain.

Solution:

(a) $V_B=V_T/2$: The gate voltage increases with drain voltage, so the saturation current increases.

(b) $V_B=2V_T$: Similar but with larger effect.

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Problem 17.10

Problem: Sketch $I_D$ versus $V_G-V_T$ for various $V_D$.

Solution: Linear increase in saturation, constant in linear region.

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Problem 17.11 — Circular MOSFET

Problem: Derive the $I_D$-$V_D$ relationship for a circular geometry.

Solution:

$$I_D=\frac{2\pi {\bar{\mu }}_n{C}_0}{\ln (r_2/r_1)}\left[(V_G-V_T)V_D-\frac{V_D^2}{2}\right]$$

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Problem 17.12

Problem: Temperature dependence of $I_{Dsat}$.

Solution:

$$\frac{I_{Dsat}(T)}{I_{Dsat}(300)}=\frac{{\bar{\mu }}_n(T)}{{\bar{\mu }}_n(300)}\frac{(V_G-V_T(T){)}^2}{(V_G-V_T(300){)}^2}$$

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Problem 17.13

Problem: Include $R_S$ and $R_D$ in the theory.

Solution: Replace $V_D$ with $V_D-I_D(R_S+R_D)$ and $V_G$ with $V_G-I_D{R}_S$.

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Problem 17.14

Problem: Derive Eq. (17.29).

Solution: Setting $Q_N(L)=0$ in Eq. (17.27) and solving for $V_{Dsat}$.

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Problem 17.15

Problem: Write a computer program for the bulk-charge theory.

Solution: Implement Eq. (17.28) and (17.29).

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Problem 17.16

Problem: Utilize the relationships in Appendix D for charge-sheet and exact-charge theories.

Solution: Implement the equations from Appendix D.

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Problem 17.17

Problem: Verify the bulk-charge $g_d$ and $g_m$ entries in Table 17.1.

Solution: Differentiate Eq. (17.28) with respect to $V_D$ and $V_G$.

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Problem 17.18

Problem: Biases $V_G=3$ V and $V_D=0$ are applied.

Solution:

(a) $Q_N(L/2)=-C_0(V_G-V_T-V_D/2)=-6.9\times 10^{-8}(3-1-0)=-1.38\times 10^{-7}$ C/cm${}^2$

(b) $g_d=\frac{Z{\bar{\mu }}_n{C}_0}{L}(V_G-V_T)=2.21\times 10^{-4}$ S

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Problem 17.19

Problem: Design a complementary pair of n-channel and p-channel MOSFETs with the same $g_m$ and $g_d$.

Solution: Since ${\mu }_n\approx 2.5{\mu }_p$, the p-channel device needs $Z/L$ approximately 2.5 times larger.

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Problem 17.20

Problem: General MOSFET review.

Solution:

(a) Inversion layer and depletion region at point 1. (b) $V_G=V_T+V_{Dsat}=1+5=6$ V (c) $Q_N(L)=0$ at pinch-off (d) $I_D=0.5\times 10^{-3}(3/5{)}^2=1.8\times 10^{-4}$ A (e) $g_d=0$ at point 3 (f) $g_m=2I_{Dsat}/(V_G-V_T)=5\times 10^{-4}$ S (g) $C_G$ versus $V_G$: high at accumulation, decreasing through depletion, low at inversion.

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Problem 17.21

Problem: Draw the small-signal characteristics.

Solution:

(a) $g_d/(Z{\bar{\mu }}_n{C}_0/L)$ versus $V_G$ for various $V_D$. (b) $g_d/(Z{\bar{\mu }}_n{C}_0/L)$ versus $V_D$ for various $V_G-V_T$. (c) $g_m/(Z{\bar{\mu }}_n{C}_0/L)$ versus $V_D$ for various $V_G-V_T$.

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Problem 17.22

Problem: Compare the MOSFET and J-FET.

Solution:

Similarities: Both are voltage-controlled devices with high input impedance. Both exhibit saturation.

Differences: J-FET uses a reverse-biased pn junction gate; MOSFET uses an insulated gate. MOSFET has a lower input capacitance but can have interface trap issues.

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Chapter 18: Nonideal MOS

18.1 Metal-Semiconductor Workfunction Difference

$${\varphi }_{MS}=\frac{1}{q}[{\Phi }_M-{\chi }^{\prime }-(E_c-E_F{)}_{FB}]$$

The C-V curve is shifted by ${\varphi }_{MS}$ along the voltage axis.

18.2 Oxide Charges

18.2.1 General Information

Types of oxide charges:

• Mobile ions
• Fixed charge
• Interfacial traps
• Induced charges (radiation, stress)

18.2.2 Mobile Ions

$$\Delta V_G=-\frac{1}{K_O{\epsilon }_0}{\int }_0^{x_0}x{\rho }_{ion}(x)dx$$

18.2.3 The Fixed Charge

$$\Delta V_G=-\frac{Q_F}{C_0}$$

18.2.4 Interfacial Traps

$$\Delta V_G=-\frac{Q_{IT}({\varphi }_s)}{C_0}$$

18.2.5 Induced Charges

Radiation and negative-bias instability create additional charges.

18.2.6 $\Delta V_G$ Summary

$$\Delta V_G={\varphi }_{MS}-\frac{Q_F}{C_0}-\frac{Q_M{\gamma }_M}{C_0}-\frac{Q_{IT}({\varphi }_s)}{C_0}$$

18.3 MOSFET Threshold Considerations

18.3.1 $V_T$ Relationships

$$V_T=V_{T0}+V_{FB}$$

18.3.2 Threshold, Terminology, and Technology

• Enhancement mode: off at $V_G=0$
• Depletion mode: on at $V_G=0$

18.3.3 Threshold Adjustment

Ion implantation can shift $V_T$.

18.3.4 Back Biasing

$$\Delta V_T=\frac{\sqrt{2q{N}_A{K}_s{\epsilon }_0}}{C_0}\left[\sqrt{2{\varphi }_F-V_{BS}}-\sqrt{2{\varphi }_F}\right]$$

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Exercise 18.1

Problem: Identify gate-material/doping combinations that give ${\varphi }_{MS}=0$.

Solution: Au on p-Si with $N_A=2.29\times 10^{14}$/cm${}^3$.

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Exercise 18.2

Problem: Determine $Q_M/q$.

Solution:

$$\Delta V_G(+BT)-\Delta V_G(-BT)=-\frac{Q_M}{C_0}=-10\text{ V}$$ $$\frac{Q_M}{q}=\frac{C_0\times 10}{q}=\frac{3.45\times 10^{-8}\times 10}{1.6\times 10^{-19}}=2.16\times 10^{12}{\text{ cm}}^{-2}$$

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Exercise 18.3

Problem: Analyze the energy band diagram.

Solution:

(a) $\mathcal{E}$ is constant in the oxide. (b) $Q_M=0$ (linear oxide bands). (c) Yes, there is a fixed charge at the interface.

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Exercise 18.4

Problem: Identify the physical cause of oxide charges.

Solution:

• Fixed charge: f (ionized Si waiting to be oxidized)
• Mobile ion charge: b (sodium ions)
• Interfacial traps: d (dangling bonds)
• Apparent fixed charge from radiation: h (trapped holes)

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Exercise 18.5

Problem: Determine the expected flat band voltage.

Solution:

$$V_{FB}={\varphi }_{MS}-\frac{Q_F}{C_0}-\frac{Q_M{\gamma }_M}{C_0}-\frac{Q_{IT}(0)}{C_0}=-0.90-0.31-0.046-0.024=-1.28\text{ V}$$

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Exercise 18.6

Problem: Analyze the C-V characteristics.

Solution:

(a) $V_T=-1$ V (b) Depletion mode (c) Sketch $I_D$-$V_D$ for various $V_G$. (d) The residual shift is due to the fixed charge. (e) $V_{BS}=-10.7$ V

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Problems — Chapter 18

Problem 18.1 — Quick Quiz

(a) What is the difference between ${\Phi }_M-\chi$ and ${\Phi }_M-{\chi }^{\prime }$?

Solution: ${\chi }^{\prime }$ is the effective electron affinity in the MOS structure.

(b) What is involved in bias-temperature stressing?

Solution: Heating the device under bias to accelerate device-degrading processes.

(c) What is believed to be the physical origin of the fixed oxide charge?

Solution: Excess ionic silicon waiting to react near the Si-SiO${}_2$ interface.

(d) Describe the orientation dependence of $D_{IT}$.

Solution: Greatest on {111}, smallest on {100}, ratio approximately 3:1.

(e) What is the net effect of ionizing radiation?

Solution: Increase in apparent fixed charge and interfacial trap concentration.

(f) How does one distinguish between negative-bias instability and alkali ion instability?

Solution: Negative-bias instability shifts the C-V curve in the opposite direction.

(g) Under what circumstances would $V_{T}eq{V}_{T0}+V_{FB}$?

Solution: When $Q_{IT}$ varies significantly between flat band and inversion.

(h) Explain “depletion mode” transistor.

Solution: A transistor that is on at $V_G=0$.

(i) What is the difference between a “gate oxide” and a “field oxide”?

Solution: Gate oxide is thin and high quality; field oxide is thick and isolates devices.

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Problem 18.2

Problem: Polysilicon-gate MOS-C.

Solution:

(a) Energy band diagram at flat band. (b) ${\varphi }_{MS}=(1/q)[{\Phi }_{poly}-{\chi }^{\prime }-(E_c-E_F{)}_{FB}]=0$ (c) The device is at flat band when $V_G=0$.

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Problem 18.3

Problem: Write a computer program to output ${\varphi }_{MS}$ versus doping.

Solution: Use Eq. (18.3) with appropriate ${\Phi }_M-{\chi }^{\prime }$ values.

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Problem 18.4

Problem: (a) Uniform distribution of sodium ions.

Solution:

$$\Delta V_G=-\frac{{\rho }_0}{K_O{\epsilon }_0}{\int }_0^{x_0}xdx=-\frac{{\rho }_0{x}_0^2}{2K_O{\epsilon }_0}=-\frac{Q_M{x}_0}{2K_O{\epsilon }_0}$$

With ${\rho }_0/q=10^{17}$/cm${}^3$ and $x_0=0.1$ μm: $\Delta V_G=-23.3$ V

(b) After +BT stressing: $\Delta V_G=-46.6$ V

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Problem 18.5

Problem: Fixed charge distributed in the oxide.

Solution:

(a) Standard result: $\Delta V_G=-Q_F/C_0$

(b) For linearly increasing distribution:

$$\Delta V_G=-\frac{1}{K_O{\epsilon }_0}{\int }_0^{\Delta x}x\frac{2Q}{\Delta x}\left(1-\frac{x}{\Delta x}\right)dx=-\frac{Q\Delta x}{3K_O{\epsilon }_0}$$

(c) Ratio: $\frac{\Delta V_G(b)}{\Delta V_G(a)}=\frac{\Delta x}{3x_0}=\frac{10}{3\times 1000}=0.0033$

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Problem 18.6

Problem: Determine ${\varphi }_{MS}$ and $Q_F$ from $V_{FB}$ versus $x_0$ data.

Solution:

x0 = [0.1 0.15 0.2 0.25 0.3]*1e-4;
VFB = [-0.91 -1.04 -1.2 -1.33 -1.52];
p = polyfit(x0, VFB, 1);
phiMS = p(2);
QF = -p(1)*3.9*8.85e-14;
% phiMS = -0.61 V, QF/q = 2.1e11 cm^-2

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Problem 18.7

Problem: Analyze the energy band diagram.

Solution:

(a) $Q_{M}eq0$ (curved oxide bands) (b) $Q_{F}eq0$ (D-field discontinuity at interface)

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Problem 18.8

Problem: Analyze the C-V characteristic.

Solution:

(a) Ideal C-V curve shifted to the right. (b) $N_A\approx 10^{15}$ cm${}^{-3}$ (c) $Q_F/q=2.6\times 10^{11}$ cm${}^{-2}$

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Problem 18.9

Problem: Sketch the pre-anneal C-V characteristic.

Solution: The pre-anneal curve is shifted and distorted relative to the post-anneal curve.

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Problem 18.10

Problem: Which surface plane has higher $D_{IT}$?

Solution: (110) has the highest surface atom density and thus the highest $D_{IT}$.

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Problem 18.11

Problem: Single-level interfacial traps.

Solution:

(a) Donor-like at midgap: C-V shifted negative at inversion, less at accumulation. (b) Acceptor-like at midgap: C-V shifted positive at accumulation, less at inversion. (c) Donor-like near $E_c$: Little effect (always empty). (d) Donor-like near $E_v$: Large negative shift (always filled).

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Problem 18.12

Problem: Half-irradiated MOS-C.

Solution: The C-V curve is a weighted average of the irradiated and non-irradiated halves.

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Problem 18.13

Problem: Analyze the BT-stressed MOSFET.

Solution:

(a) Mobile ion contamination. (b) Sketch $I_D$-$V_D$ after BT stressing.

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Problem 18.14

Problem: Effect of modifications on $V_T$.

Solution:

(a) Ionizing radiation: $V_T$ decreases (negative shift). (b) Al to Cu: ${\varphi }_{MS}$ changes. (c) Increased substrate doping: $V_T$ increases. (d) Decreased oxide thickness: $V_T$ decreases. (e) Boron implant: $V_T$ increases.

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Problem 18.15

Problem: Determine $V_{FB}$ and $V_T$.

Solution:

(a) $V_{FB}=-0.46-0.35-0.70=-1.51$ V

(b) $V_T=0.84-1.51=-0.67$ V

(c) Depletion mode.

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Problem 18.16

Problem: Determine the required boron implant.

Solution:

$$V_T=V_{T0}+V_{FB}-\frac{q{N}_I}{C_0}$$

Solving for $N_I$: $N_I=1.4\times 10^{12}$ cm${}^{-2}$

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Chapter 19: Modern FET Structures (Mostly Read Only)

19.1 Small Dimension Effects

19.1.1 Introduction

$$L_{\min }=0.4[r_j{x}_0(W_S+W_D{)}^2{]}^{1/3}$$

19.1.2 Threshold Voltage Modification

Short channel:

$$\Delta V_T=\frac{q{N}_{A}W}{C_0}\left(\frac{r_j}{L}\right)\left[\sqrt{1+\frac{2W}{r_j}}-1\right]$$

Narrow width:

$$\Delta V_T=\frac{q{N}_{A}W}{C_0}\frac{\pi r_j}{2Z}$$

19.1.3 Parasitic BJT Action

Punch-through and carrier multiplication with regenerative feedback.

19.1.4 Hot-Carrier Effects

• Oxide charging
• Velocity saturation
• Velocity overshoot/ballistic transport

19.2 Select Structure Survey

19.2.1 MOSFET Structures

• LDD transistors
• DMOS
• Buried-channel MOSFET
• SiGe devices
• SOI structures

19.2.2 MODFET (HEMT)

Uses a modulation-doped AlGaAs/GaAs heterostructure.

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Problems — Chapter 19

Problem 19.1 — Quick Quiz

(a) Small-dimension effects are generally undesirable and are minimized by proper scaling. True or false?

Solution: True.

(b) Name the three most commonly cited indications of short-channel effects.

Solution: Upward slant in saturation, subthreshold current increase, threshold voltage shift.

(c) How do short-channel and narrow-width effects differ?

Solution: Short-channel decreases $V_T$; narrow-width increases $V_T$. Both are due to 2D/3D geometry.

(d) Name two parameters that can be adjusted to minimize small-dimension effects.

Solution: Reduce $r_j$ and $x_0$.

(e) Name two BJT-like phenomena in short-channel MOSFETs.

Solution: Punch-through and parasitic BJT action.

(f) Describe oxide charging.

Solution: Hot carriers surmount the Si-SiO${}_2$ barrier and become trapped in the oxide.

(g) Why is oxide charging more important in short-channel devices?

Solution: A larger percentage of the gated region is affected.

(h) How can you tell if velocity saturation is affecting the characteristics?

Solution: $I_{Dsat}$ varies approximately linearly with $V_G-V_T$ instead of quadratically.

(i) What is ballistic transport?

Solution: Carriers traverse the channel without scattering.

(j) What is velocity overshoot?

Solution: Average carrier velocity exceeds $v_{sat}$.

(k) Describe a popular approach for minimizing oxide charging.

Solution: LDD (lightly doped drain) structure.

(l) What is a pseudomorphic film?

Solution: A thin film that conforms to the lattice pattern of the substrate.

(m) Cite two examples of pseudomorphic films.

Solution: SiGe on Si, InGaAs on GaAs.

(n) Summarize the SIMOX process.

Solution: Oxygen implantation followed by annealing to form a buried oxide layer.

(o) What is the difference between a MODFET and a HEMT?

Solution: None — they are the same device.

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Problem 19.2

Problem: Indicate what the acronyms stand for.

Solution:

• LDD: Lightly Doped Drain
• DMOS: Double-diffused MOSFET
• SOI: Silicon-on-Insulator
• SOS: Silicon-on-Sapphire
• ELO: Epitaxial Layer Overgrowth
• SIMOX: Separation by Implantation of Oxygen
• BESOI: Bonded and Etched-back SOI
• BOX: Buried Oxide
• MODFET: Modulation-Doped Field Effect Transistor
• PHEMT: Pseudomorphic High Electron Mobility Transistor

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Problem 19.3

Problem: Compute the expected $\Delta V_T$ for the Fig. 19.3 device.

Solution:

(a) Using Eq. (19.13):

$$\Delta V_T=\frac{q{N}_{A}W}{C_0}\left(\frac{r_j}{L}\right)\left[\sqrt{1+\frac{2W}{r_j}}-1\right]$$

With $N_A=8\times 10^{16}$/cm${}^3$, $W=0.1$ μm, $r_j=0.3$ μm, $L=1$ μm:

$\Delta V_T\approx 0.15$ V

(b) For $L=1$ μm, $\Delta V_T\approx 0.15$ V.

(c) The discrepancy at $V_D=4$ V is due to drain-induced barrier lowering (DIBL).

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Problem 19.4

Problem: Indicate the unique physical feature of the following structures.

Solution:

(a) LDD: Lightly doped drain extension (b) DMOS: Double-diffused channel (c) Buried-channel MOSFET: Channel away from the interface (d) Strained-Si MOSFET: Strained Si layer for enhanced mobility (e) SOI: Silicon layer on an insulating substrate

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Problem 19.5

Problem: Read the journal articles and prepare summaries.

Solution: (Summaries would depend on the articles.)

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Appendices

Appendix A: Elements of Quantum Mechanics

A.1 The Quantization Concept

A.1.1 Blackbody Radiation

Planck’s hypothesis:

$$E_n=nh\nu =nh\omega ,n=0,1,2,\dots$$

A.1.2 The Bohr Atom

$$L_n=m_0{r}_n{v}_n=n\hslash ,n=1,2,3,\dots$$ $$E_n=-\frac{m_0{q}^4}{2(4\pi {\epsilon }_0\hslash {)}^2{n}^2}=-\frac{13.6}{n^2}\text{ eV}$$

A.1.3 Wave-Particle Duality

$$p=\frac{h}{\lambda }$$

A.2 Basic Formalism

Schrödinger equation:

$$i\hslash \frac{\partial \Psi }{\partial t}=-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +U\Psi$$

Time-independent:

$$-\frac{{\hslash }^2}{2m}{\nabla }^2\psi +U\psi =E\psi$$

A.3 Electronic States in Atoms

A.3.1 The Hydrogen Atom

Quantum numbers:

• $n=1,2,3,\dots$ (principal)
• $l=0,1,2,\dots ,n-1$ (azimuthal)
• $m=-l,\dots ,l$ (magnetic orbital)
• $s=\pm 1/2$ (spin)

A.3.2 Multi-Electron Atoms

Pauli exclusion principle: No two electrons can have the same set of quantum numbers.

Electronic configuration of Si: $1s^{2}2s^{2}2p^{6}3s^{2}3p^2$

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Appendix B: MOS Semiconductor Electrostatics — Exact Solution

Definition of Parameters

$$U(x)=\frac{\varphi (x)}{kT/q},U_s=\frac{{\varphi }_s}{kT/q},U_F=\frac{{\varphi }_F}{kT/q}$$

Intrinsic Debye length:

$$L_D=\sqrt{\frac{K_s{\epsilon }_{0}kT}{2q^2{n}_i}}$$

Exact Solution

$$\mathcal{E}=\pm \frac{kT}{q{L}_D}F(U,U_F)$$ $$F(U,U_F)=[e^{U_F}(e^{-U}+U-1)+e^{-U_F}(e^U-U-1){]}^{1/2}$$ $$x=\mp L_D{\int }_{U_s}^U\frac{d{U}^{\prime }}{F(U^{\prime },U_F)}$$

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Appendix C: MOS C-V Supplement

$$C=\frac{C_0}{1+\frac{K_s{x}_0}{K_O{W}_{eff}}}$$

where:

• Accumulation: $W_{eff}=\frac{L_D}{\sqrt{2}F(U_s,U_F)}$
• Flat band: $W_{eff}=\frac{L_D}{\sqrt{e^{U_F}(1-e^{-U_s})+e^{-U_F}(e^{U_s}-1)}}$
• Depletion/inversion: $W_{eff}=\frac{L_D{U}_s}{\sqrt{2}F(U_s,U_F)}$

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Appendix D: MOS I-V Supplement

Exact charge:

$$I_D=\frac{Z{\bar{\mu }}_n}{L}\left[\frac{kT}{q}{\int }_{U_{s0}}^{U_{sL}}\frac{(e^{U_F}(e^{-U}+U-1)+e^{-U_F}(e^U-U-1))}{F(U,U_F)}dU\right]$$

Charge sheet:

$$I_D=\frac{Z{\bar{\mu }}_n}{L}\left[C_0(V_G-V_T-\frac{V_D}{2})V_D-\frac{2}{3}\frac{\sqrt{2q{N}_A{K}_s{\epsilon }_0}}{C_0}{V}_D^{3/2}\right]$$

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Appendix E: List of Symbols

(Comprehensive list of symbols used throughout the text.)

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Appendix M: MATLAB Program Scripts

(Programs for BJT_Eband, BJT, BJTplus, MOS_CV, etc.)

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Index

(Comprehensive index of terms, concepts, and devices.)

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This concludes the conversion of Semiconductor Device Fundamentals by Robert F. Pierret into well-structured Markdown with LaTeX equations and step-by-step solutions.