半导体物理与器件第四版课后习题答案4复习进程

m* E
mo
2
o 13.6
s
0.067 13.6
2
13.1
or E 0.0053 eV
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4.17 (a) E c E F
kT ln N c no
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19
2.8 10
0.0259 ln
4.11
只供学习与交流
E E Fi
midgap
1 kT ln N
2
Nc
1
1.04 1019
kT ln 2
2.8 1019
0.4952 kT
T (K)
200 400 600
kT (eV)
0.01727 0.03453 0.0518
( E Fi E midgap )(eV)
0.0086 0.0171 0.0257
19
2.8 10 1.04 10
3
T
300
1.12 exp
0.0259 T 300
2.5 10 23 2.912 10 38
3
T
300
1.12 300 exp
0.0259 T
By trial and error, T
367.5 K
3
2.912 10 38 T exp 1.12 300
300
0.0259 T
E E Fi
midgap
0.0128 eV
*
Germanium: m p 0.37mo ,
*
m n 0.55mo
E E Fi
midgap
0.0077 eV
Gallium Arsenide: m*p 0.48mo ,
*
m n 0.067m o
E E Fi
midgap
0.0382 eV
_______________________________________
0.0259 ln
10
1.5 10
0.338 eV
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4.18 (a) E F E
N kT ln
po
19
0.0259 ln
1. 04
10
16
2 10
(b) E c E F E g E F
0.162 eV E
1.12 0.162 0.958 eV
(a) For T 200K,
ni B ni A
0.20 exp
0.017267
(b) For T 300K,
6
9.325 10
ni B ni A
0.20 exp
0.0259
(c) For T 400K,
4.43 10 4
ni B exp
0.20
ni A
0. 034533
3.05 10 3
_______________________________________
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Chapter 4
4.1
2
Eg
ni N c N exp
kT
3
T
Eg
NcO N O
exp
300
kT
(b)
2
ni
12 2
5 10
25
2.5 10
where N cO and N O are the values at 300 K.
T (K) 200 400 600
4.6 (a) gc f F
E E c exp
E EF kT
E E c exp
E Ec kT
exp
Ec EF
kT
Let E Ec x
Then g c f F
x x exp
kT
To find the maximum value:
d gc fF dx
1 x 1 / 2 exp x
2
kT
1 x1 / 2 exp x
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4.12 (a) EFi
E midgap
3
m*p
kT ln 4
mn*
3
0.70
0.0259 ln
4
1.21
10.63 meV
(b) E Fi E midgap
3
0 .75
0.0259 ln
4
0.080
43.47 meV _______________________________________
0
kT
kT
which yields
1 2x 1/ 2
x 1/ 2 kT
x kT 2
The maximum value occurs at
kT E Ec
2
(b) g 1 fF
E E exp
EF E kT
E E exp
EE kT
EF E exp
kT Let E E x
Then g 1 f F
x x exp
4.21 (a) kT
375 0.0259
300
0.032375 eV
3/2
19 375
0.28
n o 2.8 10
exp
300
0. 032375
6.86 1015 cm 3
EF E
Eg Ec EF
kT (eV) 0.01727 0.03453 0.0518
(a) Silicon ni (cm 3 )
7.68 104
12
2.38 10 9.74 1014
(c) GaAs T (K)
200 400 600
(b) Germanium
ni (cm
3
)
2.16 1010 8.60 1014
3.82 1016
E g 28.9561
17
3.025 10 ln
8
or E g 1.318 eV
38.1714
Now 7 .70 1010 2
3
N co N o 400 300
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1 .318 exp
0.034533
21
17
5.929 10 N co N o 2.370 2.658 10
400
n
2 i
200
7 .70 1010 2 1.40 10 2 2
3.025 1017
3
400
300
3
200
300
Eg exp
0.034533 Eg
exp 0.017267
Eg
Eg
8 exp
0.017267 0.034533
3.025 1017 or
8 exp E g 57 .9139 28.9578
4.14 Let g c E
Then
C1 E Ec for E Ec
no
g c E f F E dE
Ec
E Ec
C1
dE
E E c 1 exp
EF
kT
C1 E EC exp
Ec
E EF dE kT
Let
E Ec so that dE kT d kT
We can write E EF E Ec
Ec EF
4.15 We have r1 ao
For germanium,
mo
r
*
m
r
16 ,
*
m
0.55mo
Then
1
r 1 16
ao
0.55
29 0.53
or
o
r1 15.4 A
The ionization energy can be written as
m* E
mo
2
o 13.6 eV
s
0. 55 2 13.6
exp
300
0 .032375
4.99 103 cm 3
17
4.7 10
0.0259 ln
14
1.15 10
0.2154 eV
EF E
E g E c E F 1.42 0.2154
1.2046 eV
po 7 10 18 exp 1.2046 0.0259
4.42
10
2
cm
3
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Then no C1 exp
Ec E F kT
E Ec exp
Ec
E Ec dE kT
or
no C1 exp
Ec E F kT
kT exp
0
We find that
kT d
exp d exp
0
1
1
0
So
2
no C1 kT exp
Ec EF kT
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4.8 Plot
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4.9 Plot
_______________________________________
4.10
EFi Emidgap
3
m
* p
kT ln 4
mn*
Silicon:
m
* p
0.56 mo ,
*
mn
1.08m o
exp E EF kT
Ec EF
exp
exp
kT
The integral can then be written as
n o K kT exp
Ec EF
exp
d
kT
0
which becomes
no K kT exp
Ec EF kT
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4.13 Let gc E Then
K constant
no
gc E f F E dE
Ec
1
K
dE
Ec 1
E exp
EF
kT
K exp
Ec
E EF dE kT
Let
E Ec so that dE kT d kT
We can write
E EF Ec EF E Ec
so that
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so
N co N o
9. 41
37
10 cm
6
_______________________________________
4.5
exp ni B ni A exp
For T For T For T
1.10 kT 0.90 kT 200 K, 300 K, 400 K,
0.20 exp
kT
kT 0.017267 eV kT 0.0259 eV kT 0.034533 eV
By trial and error, T 417.5 K _______________________________________
4.4 At T 200 K, kT
200 0.0259
300
0.017267eV At T 400 K, kT
400 0.0259
300
0.034533eV
n
2 i
kT ln N c no
2.8 1019 0.0259 ln 2 10 5
EF E EF E
0.8436 eV Eg Ec EF
1.12 0.8436 0.2764 eV
19
0 .27637
(b) p o 1.04 10 exp
0. 0259
2.414 1014 cm 3
(c) p-type
_______________________________________
(c) no
19
0. 958
2.8 10 exp
0.0259
2.41 103 cm 3
(d) E Fi E F
po kT ln
ni
只供学习与交流
16
2 10
0.0259 ln
10
1.5 10
0.365eV _______________________________________
4.19 (a) E c E F
kT
To find the maximum value
d g 1 fF
d
x
x exp
0
dx
dx
kT
Same as part (a). Maximum occurs at kT
x 2
or kT
EE 2
_______________________________________
4.7
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4.20 (a) kT
375 0.0259
300
0.032375 eV
3 /2
17 375
0 .28
n o 4.7 10
exp
300
0.032375
1.15 1014 cm 3
EF E
E g E c E F 1.42 0.28 1.14 eV
18
p o 7 10 (b) E c E F
3/2
375
1.14
ni (cm
3
)
1.38 3.28 109 5.72 1012
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4.2 Plot
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4.3
2
(a) ni
Eg N c N exp
kT
11 2
5 10
19
n E1 n E2
where E1 Ec
Then n E1 n E2
E1 E c exp E2 E c exp
E1 Ec kT
E2 Ec kT
kT 4kT and E 2 E c 2
4kT exp
kT
2
E1 E2 kT
2 2 exp
1 4
2
2 2 exp 3.5
or
n E1 _________________________________
15
7 10
(b) E F E Eg E c
0.2148 eV EF
1.12 0.2148 0.90518eV
(c) p o N exp
EF E kT
19
0.90518
1.04 10 exp
0. 0259
6.90 103 cm 3
(d) Holes
(e) E F E Fi kT ln no ni
7 10 15
16
E 0.029 eV