Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 1 Problem Solutions
Chapter 1
Problem Solutions
1.1 (a) fcc: 8 corner atoms 1/8 = 1 atom 6 face atoms = 3 atoms Total of 4 atoms per unit cell (b) bcc: 8 c
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 12 Problem Solutions
Chapter 12
Problem Solutions
12.1 (a) 2 eN a
15
L 2(11.7)b8.85x10 g OP =M N b1.6x10 gb10 g Q
14 19 16
1/ 2
LM V OP N (2.1)V Q For V
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 10 Problem Solutions
Chapter 10
Problem Solutions
10.1 Sketch 10.2 Sketch 10.3 (a) IS = eDn ABE n BO xB
19
i E = 17.64 mA 10.5 (a) = iC iB = 510 6 85 86
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 9 Problem Solutions
Chapter 9
Problem Solutions
9.1 (a) We have 9.2 (a) BO = m  = 51  4.01 . or BO = 1.09 V (b)
e n = eVt ln
FG N IJ HN K F 2.8x10 IJ
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 8 Problem Solutions
Chapter 8
Problem Solutions
8.4 The crosssectional area is J We have A= I = 10 x10 20
D t 3
F I H K F eV I exp I H kT K = expL e aV  V
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 7 Problem Solutions
Chapter 7
Problem Solutions
7.1 Vbi = Vt ln (c) N d = 10 cm , N a = 10 cm Then Si: Vbi = 0.814 V , Ge: Vbi = 0.432 V ,
17 17 10 3 3 3
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 6 Problem Solutions
Chapter 6
Problem Solutions
6.1 ntype semiconductor, lowinjection so that R = or R = 5 x10 cm s
19 3 1
G= (c)
2.25 x10 20 x10
4

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 5 Problem Solutions
Chapter 5
Problem Solutions
5.1 (a) nO = 10 cm
16 3
N d = 4.63 x10 cm
13
3
(d) e p pO
and pO = ni
2
0.01 = 1.6 x10
6 2
nO
=
b
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 4 Problem Solutions
Chapter 4
Problem Solutions
4.1 ni
2
F  E IJ = N N expG H kT K
g C V
FG 5.83x10 IJ = F 300I expL E F 1  1 I O H 182 x10 K H 200K MN H
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 3 Problem Solutions
Chapter 3
Problem Solutions
3.1 If a o were to increase, the bandgap energy would decrease and the material would begin to behave less lik