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Semiconductor Devices – Solutions Manual (Physics and Technology, 3rd Ed.) | Complete Worked Solutions

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This document is the Solutions Manual for Semiconductor Devices: Physics and Technology, 3rd Edition. It provides comprehensive, step-by-step worked solutions to textbook problems, covering semiconductor fundamentals, PN junctions, diodes, transistors, MOSFETs, optoelectronic devices, and integrated circuits. Designed for students and instructors, this manual strengthens conceptual understanding, supports teaching, homework assignments, and exam preparation. Ideal for mastering device physics, semiconductor operation, and electronic applications.

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Institution
Semiconductor Devices
Course
Semiconductor Devices

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Solutions Manual to Accompany

SEMICONDUCTOR DEVICES
Physics and Technology
rd
3 Edition




S. M. SZE
Etron Chair Professor
College of Electrical and Computer Engineering
National Chaio Tung University
Hsinchu, Taiwan




M. K. LEE
Department of Electrical Engineering
National Sun Yat-sen University
Kaohsiung, Taiwan


John Wiley and Sons, Inc

New York. Chicester / Weinheim / Brisband / Singapore / Toronto



1

, Contents
Ch.0 Introduction---------------------------------------------------------------------------- 0

Ch.1 Energy Bands and Carrier Concentration in Thermal Equilibrium ------------ 1

Ch.2 Carrier Transport Phenomena ------------------------------------------------------- 9

Ch.3 p-n Junction ---------------------------------------------------------------------------18

Ch.4 Bipolar Transistor and Related Devices -------------------------------------------35

Ch.5 MOS Capacitor and MOSFET------------------------------------------------------52

Ch.6 Advanced MOSFET and Related Devices ----------------------------------------62

Ch.7 MESFET and Related Devices -----------------------------------------------------68

Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices ----------------76

Ch.9 Light Emitting Diodes and Lasers -------------------------------------------------81

Ch.10 Photodetectors and Solar Cells-----------------------------------------------------88

Ch.11 Crystal Growth and Epitaxy -------------------------------------------------------96

Ch.12 Film Formation ----------------------------------------------------------------------105

Ch.13 Lithography and Etching -----------------------------------------------------------112

Ch.14 Impurity Doping---------------------------------------------------------------------118

Ch.15 Integrated Devices-------------------------------------------------------------------126




0

, CHAPTER 1



1. (a) From Fig. 11a, the atom at the center of the cube is surround by four

equidistant nearest neighbors that lie at the corners of a tetrahedron.

Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is

1/2 [(a/2)2 + ( 2a /2)2]1/2 = 3a /4 = 2.35 Å.

(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms

each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)

for an area of a2, therefore we have

2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2

Similarly we have for (110) plane (Fig. 4a and Fig. 6)

(2 + 2 ×1/2 + 4 ×1/4) / 2a 2 = 9.6 × 1015 atoms / cm2,

and for (111) plane (Fig. 4a and Fig. 6)

⎛3⎞ 2
(3 × 1/2 + 3 × 1/6) / 1/2( 2a )( ⎜ ⎟ a ) = = 7.83 × 1014 atoms / cm2.
⎝2⎠ ⎛ 3 ⎞⎟ 2
⎜ a
⎜ 2 ⎟⎠





2. The heights at X, Y, and Z point are 3 4, 1 4, and 3 .
4

3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere.

∴ Maximum fraction of cell filled

= no. of sphere × volume of each sphere / unit cell volume

= 1 × 4π(a/2)3 / a3 = 52 %

(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the

eight corners for a total of one sphere. The fcc also contains half a sphere at

each of the six faces for a total of three spheres. The nearest neighbor

1

, distance is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ).

∴ Maximum fraction of cell filled

= (1 + 3) {4π[(a/2) / 4 ]} / a3 = 74 %.

(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a

total of three spheres, and 4 spheres inside the cell. The diagonal distance

between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is

2 2 2
1 ⎛a⎞ ⎛a⎞ ⎛a⎞ a
D= ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 3
2 ⎝2⎠ ⎝2⎠ ⎝2⎠ 4

a
The radius of the sphere is D/2 = 3
8
∴ Maximum fraction of cell filled
3
⎡ 4π ⎛a ⎞⎤ 3
= (1 + 3 + 4) ⎢ ⎜ 3 ⎟⎥ / a = π = 34 %.
⎣ 3 ⎝ 8 ⎠⎦
This is a relatively low percentage compared to other lattice structures.



4. d1 = d2 = d3 = d4 = d
d1 + d 2 + d 3 + d 4 = 0
d1 • ( d1 + d 2 + d 3 + d 4 ) = d1 • 0 = 0
2
d 1 + d1 • d 2 + d1 • d 3 + d1 • d 4 = 0

∴d2+ d2 cosθ12 + d2cosθ13 + d2cosθ14 = d2 +3 d2 cosθ= 0
−1
∴ cosθ =
3
−1
θ= cos-1 ( ) = 109.470 .
3

5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest

three integers having the same ratio are 6, 4, and 3. The plane is referred to as

(643) plane.

6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As

are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and

2

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