Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 5 Exercise Solutions
Chapter 5
Exercise Solutions
E5.1 no = 10 10 = 9 x10 cm
15 14 14 3
(b)
= 1.6 x10
b
e n N d N a
19
a
g(1000)b3x10 g
16 1
f
= 4.8 ( cm) =
1
so
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 6 Exercise Solutions
Chapter 6
Exercise Solutions
E6.1
n(t ) = 5 x10
14
n(t ) = n(0) exp
so
n(t ) = 10
15
FG t IJ H K F t IJ expG H 1 s K
no 15 3 14 3 13 3
LM1 expFG
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 7
Exercise Solutions
Chapter 7
Exercise Solutions
6
x n = 4.11x10 cm
R 2(11.7)b8.85x10 g(0.718) FG 5x10 IJ
=S
T b1.6x10 g H 5x10 K
F 1 IU
H 5x10 + 5x10 K V
W
14
E7.1
Semiconductor Physics and Devices: Basic Principles
Solutions Manual
Chapter 8
Exercise Solutions
Chapter 8
Exercise Solutions
E8.1
2
pno =
Then
ni
Nd
.
b15x10 g
=
10
5 x10
af
which yields
Va = Va ( max) = 1.067 V
2
= 4.5 x10 cm
3
16
3
n po
af
.
b15x10 g
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 9
Exercise Solutions
Chapter 9
Exercise Solutions
max =
E9.1
(a) Bo = 4.5 4.01 Bo = 0.49 V
(d)
.
Rb1.6x10 g(131)b8.85x10 gb3x10 g U
C = S
V
2(0.919 + 5)
T
W
19
1/ 2
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Exercise Solutions
Chapter 2
Exercise Solutions
E2.1 (a) or E = 1.99 x10 Also E= (b) E= or E = 1.99 x10 Also E= E2.2 (a) or p = 3.68 x10 Then p
2 26 16 19
=
34 10
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 3 Exercise Solutions
Chapter 3
Exercise Solutions
E3.1 1 = 10 sin a + cos a gT = 4 2(1.08) 9.11x10
34
a By trial and error, a = 5.305 rad Now
2 mE 2 so E2 = or E 2 =