Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 14
Problem Solutions
Chapter 14
Problem Solutions
14.3
g =
14.1
(a) =
1.24
E
m
I ( x ) = 10 W / cm
we obtain
2
1.24
g = 1.44 x10 cm s
.
15 x10 cm
b
13
n-type GaAs,
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 13
Problem Solutions
Chapter 13
Problem Solutions
(iii) For VGS = 1 V , VDS = 5 V
a h = 0.045 m
13.1
Sketch
which implies no undepleted region.
13.2
Sketch
13.4
p-ch
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 15
Exercise Solutions
Chapter 15
F 2 I I (max) = 1.33 A
H 3K
At the maximum power point,
F 2I R
15 = 30
H 3K
Exercise Solutions
I C ( max ) = 2
E15.1
(a) Collector
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 12
Exercise Solutions
Chapter 12
Exercise Solutions
E12.1
I D1
I D2
Now
14
FG V IJ
H V K = expFG V
=
F V IJ H
expG
HV K
F I IJ
V = V lnG
HI K
GS 1
exp
VGS 2
GS 1
t
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 10
Problem Solutions
Chapter 10
Problem Solutions
i E = 17.64 mA
10.1
Sketch
10.5
(a) =
10.2
Sketch
=
10.3
IS =
(a)
b
=
1.6 x10
19
g b gb g
=
14
14
exp
4
A
F 0.5 I
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 8
Problem Solutions
Chapter 8
Problem Solutions
8.4
The cross-sectional area is
8.1
In the forward bias
eV
I f I S exp
kT
Then
F I
H K
F eV I
exp
I
H kT K = expL e a
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 9
Problem Solutions
Chapter 9
Problem Solutions
9.2
(a) BO = m = 51 4.01
.
or
BO = 1.09 V
9.1
(a) We have
FG N IJ
HN K
F 2.8x10 IJ = 0.206 eV
= (0.0259 ) lnG
H 10 K
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 7
Problem Solutions
Chapter 7
Problem Solutions
(c)
Vbi = Vt ln
FG N N IJ
H n K
a
d
2
i
where Vt = 0.0259 V and ni = 1.5 x10 cm
We find
(a)
10
For N d = 10 cm
15
(i
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 11
Exercise Solutions
Chapter 11
Exercise Solutions
E11.5
fp = 0.376 V
E11.1
ms = (0.555 0.376) ms = +0.179 V
F 3x10 IJ = 0.376 V
(a) = (0.0259 ) lnG
H 15x10 K
.
R
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 6
Exercise Solutions
Chapter 6
Exercise Solutions
E6.1
n(t ) = 5 x10
FG t IJ
H K
F t IJ
expG
H 1 s K
14
no
n(t ) = 10
15
(a) t = 0 ; n = 10 cm
15
LM1 expFG t IJ OP
N
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 4
Exercise Solutions
Chapter 4
Exercise Solutions
Now
E4.1
ni = 2.8 x10
b
2
no = 2.8 x10 exp
19
F 0.22 I
H 0.0259 K
15
ni = 5.65 x10
2
ni = 2.38 x10 cm
12
E F E v =
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 3
Exercise Solutions
Chapter 3
Exercise Solutions
gT =
E3.1
1 = 10
sin a
h
so
2
(5.305) h
E2 =
2 ma
2
b
b
2 9.11x10
31
25
34
gb5x10
g
g
10
b g
4 2 m p
*
2
gT =
h
2
h
Semiconductor Physics and Devices: Basic Principles, 3rd edition
Solutions Manual
Chapter 5
Exercise Solutions
Chapter 5
Exercise Solutions
a
e n N d N a
(b)
b
= 1.6 x10
E5.1
no = 10 10 = 9 x10 cm
15
so
2
po =
Now
ni
no
14
14
.
b15x10 g
=
10
9 x10
b
=
=