半导体物理与器件第四版课后习题集答案解析4

Chapter 4
4.1
⎪⎪⎭
⎫ ⎝
⎛-=kT
E N N n g
c i exp 2υ ⎪⎪⎭
⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003
υ
where cO N and O N υ are the values at 300 K.
(a) Silicon
T (K) kT (eV)
i n (cm 3-) 200 400 600 01727.0 03453.0 0518.0
41068.7⨯ 121038.2⨯ 141074.9⨯
(b) Germanium (c) GaAs
T (K) i n (cm 3-) i n (cm 3-)
200 400 600
101016.2⨯ 141060.8⨯ 161082.3⨯
38.1 91028.3⨯ 121072.5⨯
_______________________________________ 4.2
Plot
_______________________________________ 4.3
(a) ⎪⎪⎭
⎫
⎝⎛-=kT E N N n g c i exp 2υ (
)(
)(
)
3
19
19
2
113001004.1108.2105⎪⎭
⎫
⎝⎛⨯⨯=⨯T
()()⎥⎦
⎤⎢⎣⎡-⨯3000259.012.1exp T
()
3
382330010912.2105.2⎪⎭
⎫
⎝⎛⨯=⨯T
()()()()⎥⎦
⎤⎢⎣⎡-⨯T 0259.030012.1exp
By trial and error, 5.367≅T K
(b)
()
252
12
2105.2105⨯=⨯=i n
(
)
()()()()⎥⎦
⎤⎢⎣⎡-⎪⎭⎫
⎝⎛⨯=T T 0259.030012.1exp 30010912.23
38
By trial and error, 5.417≅T K
_______________________________________ 4.4
At 200=T K, ()⎪⎭
⎫
⎝⎛=3002000259.0kT
017267.0=eV
At 400=T K, ()⎪⎭
⎫
⎝⎛=3004000259.0kT
034533.0=eV
()()()()
172
22102
210025.31040.11070.7200400⨯=⨯⨯=
i
i n
n
⎥
⎦
⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫
⎝⎛⎪
⎭⎫ ⎝⎛=017267.0exp 034533.0exp 3002003004003
3
g g E E
⎥⎦
⎤
⎢⎣⎡-=034533.0017267.0exp 8g g E E
()[]
9578.289139.57exp 810025.317-=⨯g E
or
()1714.38810025.3ln 9561.2817=⎪⎪⎭
⎫
⎝⎛⨯=g E or 318.1=g E eV
Now (
)
3
2
1030040010
70.7⎪⎭
⎫
⎝⎛=⨯o co N N υ
⎪⎭⎫ ⎝⎛-⨯034533.0318.1exp ()()
17
2110658.2370.210929.5-⨯=⨯o co N N υ so 371041.9⨯=o co N N υcm 6-
_______________________________________ 4.5
()()⎪⎭
⎫ ⎝⎛-=⎪⎭
⎫ ⎝⎛-⎪
⎭
⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eV For 400=T K, 034533.0=kT eV (a) For 200=T K, ()()6
10325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()4
1043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()3
1005.3034533.020.0exp -⨯=⎪⎭
⎫ ⎝⎛-=A n B n i i _______________________________________ 4.6
(a) ()⎥⎦⎤
⎢⎣⎡---∝kT E E E E f g F c F c exp
()⎥⎦
⎤
⎢⎣⎡---∝kT E E E E c c exp
()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =-
Then ⎪⎭⎫
⎝⎛-∝kT x x f g F c exp
To find the maximum value: ()⎪⎭
⎫
⎝⎛-∝-kT x x dx f g d F c exp 212/1
0exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields
2212/12
/1kT
x kT x x =
⇒= The maximum value occurs at
2
kT
E E c +=
(b)
()()⎥⎦⎤
⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ
()⎥⎦
⎤
⎢⎣⎡---∝kT E E E E υυexp
()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υ
Then ()⎪⎭
⎫ ⎝⎛-∝-kT x x f g F exp 1υ
To find the maximum value
()[]0exp 1=⎥⎦⎤
⎢
⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at
2
kT
x =
or
2
kT
E E -=υ
_______________________________________ 4.7
()
()
()()⎥⎦
⎤
⎢⎣⎡---⎥⎦
⎤
⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp exp
where
kT E E c 41+= and 2
2kT
E E c += Then
()()
()⎥⎦⎤
⎢⎣⎡--=kT E E kT kT
E n E n 2121exp 2
4
()5.3exp 22214exp 22-=⎥⎦⎤
⎢⎣⎡⎪⎭⎫ ⎝
⎛--=
or
()
()
0854.021=E n E n
_______________________________________ 4.8
Plot
_______________________________________ 4.9
Plot
_______________________________________ 4.10
⎪⎪⎭
⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E Silicon: o p m m 56.0*=, o n m m 08.1*
=
0128.0-=-midgap Fi E E eV Germanium: o p
m m 37.0*
=,
o n m m 55.0*=
0077.0-=-midgap Fi E E eV Gallium Arsenide: o p m m 48.0*=,
o n m m 067.0*= 0382.0+=-midgap Fi E E eV _______________________________________ 4.11 ()⎪⎪⎭
⎫
⎝⎛=
-c midgap Fi N N kT E E υln 21
()()kT kT 4952.0108.21004.1ln 21
1919-=⎪⎪⎭
⎫ ⎝⎛⨯⨯=
T (K) kT (eV)
(midgap Fi E E -)(eV) 200 400 600 01727.0 03453.0 0518.0 0086.0- 0171.0- 0257.0-
_______________________________________ 4.12
(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()⎪⎭
⎫
⎝⎛=21.170.0ln 0259.043
63.10-⇒meV
(b) ()⎪⎭
⎫
⎝⎛=-080.075.0ln 0259.043midgap Fi E E
47.43+⇒meV
_______________________________________ 4.13
Let ()==K E g c constant Then
()()dE E f
E g n F
E c
o c
⎰∞
=
dE kT E E K
c E F
⎰
∞
⎪⎪⎭⎫
⎝
⎛-+=exp 11
()dE kT E E K c
E F ⎰
∞
⎥⎦⎤
⎢⎣⎡--≅exp Let kT E E c
-=η so that ηd kT dE ⋅=
We can write ()()c F c F E E E E E E -+-=-
so that
()()()η-⋅⎥⎦⎤⎢⎣⎡--=⎥⎦⎤
⎢⎣⎡--e x p e x p e x p kT E E kT E E F c F The integral can then be written as
()()ηηd kT E E kT K n F c o ⎰
∞
-⎥⎦⎤
⎢⎣⎡--⋅⋅=0exp exp which becomes
()⎥⎦
⎤
⎢⎣⎡--⋅⋅=kT E E kT K n F c o exp _______________________________________ 4.14
Let ()()c c E E C E g -=1 for c E E ≥ Then
()()dE E f
E g n F
E c
o c
⎰∞
=
()
dE kT E E E E C c E F
c ⎰
∞
⎪⎪⎭⎫
⎝
⎛-+-=exp 11
()()dE kT E E E E C F E C c
⎥⎦
⎤
⎢⎣⎡
---≅⎰
∞
exp 1
Let
kT
E E c
-=
η so that ηd kT dE ⋅= We can write
()()F c c F E E E E E E -+-=-
Then
()⎥⎦⎤
⎢⎣⎡--=kT E E C n F c o exp 1
()()dE kT E E E E c E c c
⎥⎦
⎤
⎢⎣⎡
---⨯
⎰
∞
exp or
()⎥⎦⎤
⎢⎣⎡--=kT E E C n F c o exp 1 ()()()[]()ηηηd kT kT -⨯
⎰∞
exp 0
We find that
()()()11exp exp 0
+=---=-∞
∞
⎰
ηηηηηd
So
()()⎥⎦
⎤⎢⎣⎡--=kT E E kT C n F c o exp 2
1 _______________________________________ 4.15
We have ⎪⎪⎭⎫
⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then
()()()53.02955.01161=⎪⎭
⎫
⎝⎛=o a r
or
o
A r 4.151=
The ionization energy can be written as ()6.132
*⎪⎪⎭⎫
⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655
.02
=⇒=E eV
_______________________________________ 4.16
We have ⎪⎪⎭
⎫
⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r ,
o m m 067.0*= Then
()()o
A r 10453.0067.011.131=⎪⎭
⎫
⎝⎛=
The ionization energy is
()()()6.131.13067
.06.1322
*=⎪⎪⎭
⎫ ⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E or
0053.0=E eV
_______________________________________ 4.17
(a) ⎪⎪⎭
⎫
⎝⎛=-o c F c n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=1519107108.2ln 0259.0
2148.0=eV (b) ()F c g F E E E E E --=-υ
90518.02148.012.1=-=eV
(c) ()⎥⎦⎤
⎢⎣⎡--=kT E E N p F o υυexp
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.090518.0exp 1004.119 31090.6⨯=cm 3- (d) Holes
(e) ⎪⎪⎭
⎫
⎝⎛=-i o Fi F n n kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=1015105.1107ln 0259.0
338.0=eV
_______________________________________ 4.18
(a) ⎪⎪⎭
⎫
⎝⎛=-o F p N kT E E υυln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=16191021004.1ln 0259.0
162.0=eV
(b)
()υE E E E E F g F c --=- 958.0162.012.1=-=eV
(c) ()
⎪⎭
⎫
⎝⎛-⨯=0259.0958.0exp 108.219o n
31041.2⨯=cm 3-
(d) ⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=1016105.1102ln 0259.0
365.0=eV
_______________________________________ 4.19
(a) ⎪⎪⎭
⎫
⎝⎛=-o c F c n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=519102108.2ln 0259.0
8436.0=eV ()F c g F E E E E E --=-υ 8436.012.1-= 2764.0=-υE E F eV (b)
()
⎪⎭
⎫
⎝⎛-⨯=0259.027637.0exp 1004.119o p
1410414.2⨯=cm 3-
(c) p-type
_______________________________________ 4.20
(a) ()032375.03003750259.0=⎪⎭
⎫
⎝⎛=kT eV
(
)
⎥⎦
⎤⎢⎣⎡-⎪⎭
⎫ ⎝⎛⨯=032375.028.0exp 30037510
7.42
/317
o n 141015.1⨯=cm 3-
()28.042.1-=--=-F c g F E E E E E υ 14.1=eV
(
)
⎥⎦
⎤⎢⎣⎡-⎪
⎭
⎫
⎝⎛⨯=032375.014.1exp 3003751072
/318
o p 31099.4⨯=cm 3-
(b) ()⎪⎪⎭
⎫
⎝⎛⨯⨯=-14171015.1107.4ln 0259.0F c E E
2154.0=eV
()2154.042.1-=--=-F c g F E E E E E υ 2046.1=eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.02046.1exp 10718o p
21042.4-⨯=cm 3-
_______________________________________ 4.21
(a) ()032375.03003750259.0=⎪⎭
⎫
⎝⎛=kT eV
(
)
⎥⎦
⎤
⎢⎣⎡-⎪
⎭
⎫
⎝⎛⨯=032375.028.0exp 300375108.22
/319
o n 151086.6⨯= cm 3-
()28.012.1-=--=-F c g F E E E E E υ 840.0=eV
(
)
⎥⎦
⎤⎢⎣⎡-⎪
⎭
⎫
⎝⎛⨯=032375.0840.0exp 3003751004.12
/319
o p 71084.7⨯=cm 3-
(b) ⎪⎪⎭
⎫
⎝⎛=-o c F c n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=151910862.6108.2ln 0259.0
2153.0=eV
9047.02153.012.1=-=-υE E F eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.0904668.0exp 1004.119o p 31004.7⨯=cm 3-
_______________________________________ 4.22 (a) p-type
(b) 28.04
12
.14===-g F E E E υeV
()⎥⎦
⎤
⎢⎣⎡--=kT E E N p F o υυexp ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.028.0exp 1004.119
141010.2⨯=cm 3- ()υE E E E E F g F c --=- 84.028.012.1=-=eV
()⎥⎦
⎤
⎢⎣⎡--=kT E E N n F c c o exp ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.084.0exp 108.219 51030.2⨯=cm 3-
_______________________________________ 4.23
(a) ⎥⎦
⎤
⎢⎣⎡-=kT E E n n Fi F
i o exp ()
⎥⎦
⎤
⎢⎣⎡⨯=0259.022.0exp 105.110
131033.7⨯=cm 3-
⎥⎦
⎤
⎢⎣⎡-=kT E E n p F Fi
i o exp ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.022.0exp 105.110
61007.3⨯=cm 3-
(b) ⎥⎦
⎤
⎢⎣⎡-=kT E E n n Fi F i o exp
()
⎥⎦
⎤
⎢⎣⎡⨯=0259.022.0exp 108.16 91080.8⨯=cm 3-
⎥⎦
⎤
⎢⎣⎡-=kT E E n p F Fi
i o exp
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.022.0exp 108.16 21068.3⨯=cm 3-
_______________________________________ 4.24
(a) ⎪⎪⎭
⎫
⎝⎛=-o F p N kT E E υυln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=151********.1ln 0259.0
1979.0=eV
(b)
()υE E E E E F g F c --=- 92212.019788.012.1=-=eV
(c) ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.092212.0exp 108.219o n
31066.9⨯=cm 3- (d) Holes
(e) ⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=1015105.1105ln 0259.0
3294.0=eV _______________________________________ 4.25
()034533.03004000259.0=⎪⎭⎫
⎝⎛=kT eV
(
)
2
/319
3004001004.1⎪
⎭
⎫
⎝⎛⨯=υN
1910601.1⨯=cm 3-
()
2
/319300400108.2⎪⎭
⎫
⎝⎛⨯=c N
19103109.4⨯=cm 3-
()()
1919210601.1103109.4⨯⨯=i n
⎥⎦
⎤⎢⎣⎡-⨯034533.012.1exp 24106702.5⨯=
1210381.2⨯=⇒i n cm 3- (a)
⎪⎪⎭
⎫ ⎝⎛=-o
F p
N kT E E υ
υln ()⎪⎪⎭
⎫
⎝⎛⨯⨯=151910510601.1ln 034533.0
2787.0=eV (b) 84127.027873.012.1=-=-F c E E eV
(c)
()
⎥⎦
⎤
⎢⎣⎡-⨯=034533.084127.0exp 103109.419o n
910134.1⨯=cm 3- (d) Holes
(e) ⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=121510381.2105ln 034533.0
2642.0=eV _______________________________________ 4.26
(a) ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.025.0exp 10718o p
141050.4⨯=cm 3-
17.125.042.1=-=-F c E E eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.017.1exp 107.417o n
21013.1-⨯=cm 3- (b) 034533.0=kT eV (
)
2
/318
300400107⎪
⎭
⎫ ⎝⎛⨯=υN
1910078.1⨯=cm 3- (
)
2
/317
30040010
7.4⎪⎭
⎫
⎝⎛⨯=c N
1710236.7⨯=cm 3- ⎪⎪⎭⎫
⎝⎛=-o F p N kT E E υυln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=14191050.410078.1ln 034533.0
3482.0=eV
072.13482.042.1=-=-F c E E eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=034533.007177.1exp 10236.717o n 41040.2⨯=cm 3-
_____________________________________ 4.27
(a) ()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.025.0exp 1004.119o p
141068.6⨯=cm 3-
870.025.012.1=-=-F c E E eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=0259.0870.0exp 108.219o n
41023.7⨯=o n cm 3- (b)
034533.0=kT eV
(
)
2
/319
3004001004.1⎪
⎭
⎫
⎝⎛⨯=υN
1910601.1⨯=cm 3- ()
2
/319
300400108.2⎪⎭
⎫ ⎝⎛⨯=c N
19
10311.4⨯=cm 3
-
⎪⎪⎭⎫
⎝⎛=-o F p N kT E E υυln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=14191068.610601.1ln 034533.0
3482.0=eV
7718.03482.012.1=-=-F c E E eV
()
⎥⎦
⎤
⎢⎣⎡-⨯=034533.077175.0exp 10311.419o n
91049.8⨯=cm 3-
_______________________________________ 4.28
(a) ()F c o F N n ηπ
2/12
=
For 2kT E E c F +=,
5.02
==-=kT
kT kT E E c F F η Then ()0.12/1≅F F η
()()0.1108.22
19
⨯=
π
o n
191016.3⨯=cm 3-
(b) ()F c o F N n ηπ
2/12
=
()()0.1107.42
17
⨯=
π
171030.5⨯=cm 3-
_______________________________________ 4.29
()F o F N p ηπ
υ'=
2/12
()()F
F ηπ
'⨯=
⨯2
/119
191004.12
105
So ()26.42/1='F
F η We find kT
E E F
F
-=≅'υη0.3
()()0777.00259.00.3==-F E E υeV
_______________________________________ 4.30
(a) 44==-=kT
kT
kT E E c F F η
Then ()0.62/1≅F F η ()F c o F N n ηπ
2/12
=
()()0.6108.22
19
⨯=
π
201090.1⨯=cm 3-
(b) ()
()0.6107.42
17⨯=π
o n
181018.3⨯=cm 3-
_______________________________________ 4.31
For the electron concentration ()()()E f E g E n F c =
The Boltzmann approximation applies, so ()()
c n
E E h
m E n -=
3
2
/3*
24π
()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or
()()
()⎥⎦
⎤⎢⎣⎡--=
kT E E h m E n F c n
exp 243
2
/3*
π
()⎥⎦
⎤
⎢⎣⎡---⨯kT E E kT E E kT
c c exp Define
kT
E E x c
-= Then
()()()x x K x n E n -=→exp To find maximum ()()x n E n →, set
()()x x K dx x dn -⎢⎣⎡==-exp 2
1
02/1 +()()⎥⎦
⎤--x x exp 12/1
or
()⎥⎦
⎤
⎢⎣⎡--=-x x Kx 21exp 02/1
which yields
kT E E kT E E x c c 2
1
21+=⇒-==
For the hole concentration ()()()[]E f E g E p F -=1υ
Using the Boltzmann approximation ()()
E E h m E p p
-=
υπ3
2
/3*24
()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or
()(
)
()⎥⎦
⎤⎢⎣⎡--=
kT E E h m
E p
F p
υπexp 243
2
/3*
()⎥⎦
⎤
⎢⎣⎡---⨯kT E E kT E E kT
υυexp Define
kT
E
E x -='υ
Then
()()x x K x p '-''='exp To find maximum value of ()()x p E p '→,
set
()0='
'x d x dp Using the results from above,
we find the maximum at
kT E E 2
1
-=υ
_______________________________________
4.32 (a) Silicon: We have
()⎥⎦⎤
⎢⎣⎡--=kT E E N n F c c o exp We can write
()()F d d c F c E E E E E E -+-=- For
045.0=-d c E E eV and
kT E E F d 3=-eV we can write
()
⎥⎦
⎤
⎢⎣⎡--⨯=30259.0045.0exp 108.219o n
()
()737.4exp 108.219-⨯= or
171045.2⨯=o n cm 3- We also have
()⎥⎦⎤
⎢⎣⎡--=kT E E N p F o υυexp Again, we can write
()()υυE E E E E E a a F F -+-=- For
kT E E a F 3=- and
045.0=-υE E a eV Then
()
⎥⎦
⎤⎢⎣⎡
--⨯=0259.0045.03exp 1004.119o p ()
()737.4exp 1004.119-⨯= or
161012.9⨯=o p cm 3- (b) GaAs: assume 0058.0=-d c E E eV Then
()
⎥⎦
⎤
⎢
⎣⎡--⨯=30259.00058.0exp 107.417o n ()
()224.3exp 107.417-⨯= or
161087.1⨯=o n cm 3-
Assume 0345.0=-υE E a eV Then
()
⎥⎦
⎤
⎢
⎣⎡--⨯=30259.00345.0exp 10718o p ()
()332.4exp 10718-⨯=
or
161020.9⨯=o p cm 3-
_______________________________________ 4.33
Plot
_______________________________________ 4.34
(a)
151510310154⨯=-⨯=o p cm 3- ()
4152
10105.710
3105.1⨯=⨯⨯=o n cm 3-
(b)
16103⨯==d o N n cm 3-
()
3162
10105.710
3105.1⨯=⨯⨯=o p cm 3-
(c)
10105.1⨯===i o o n p n cm 3-
(d) ()(
)
3
19
19
23003751004.1108.2⎪⎭
⎫
⎝⎛⨯⨯=i
n
()()()()⎥⎦
⎤⎢⎣⎡-⨯3750259.030012.1exp
1110334.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-
()815
2
111034.110
410334.7⨯=⨯⨯=
o
n cm 3-
(e) (
)(
)
3
19
19
23004501004.1108.2⎪⎭
⎫
⎝⎛⨯⨯=i
n
()()()()⎥⎦
⎤⎢⎣⎡-⨯4500259.030012.1exp
1310722.1⨯=⇒i n cm 3-
()
2
132
14
1410722.1210210
⨯+⎪⎪⎭
⎫ ⎝⎛+=
o n
1410029.1⨯=cm 3-
()
1214
2
131088.210
029.110722.1⨯=⨯⨯=
o p cm 3-
_______________________________________ 4.35
(a)
151510104-⨯=-=d a o N N p
15103⨯=cm 3-
()
315
2
6
21008.1103108.1-⨯=⨯⨯=
=o i o p n n cm 3-
(b)
16103⨯==d o N n cm 3-
()416
2
61008.1103108.1-⨯=⨯⨯=
o p cm 3-
(c)
6108.1⨯===i o o n p n cm 3-
(d) ()(
)
3
18
17
2300375100.7107.4⎪⎭
⎫
⎝⎛⨯⨯=i
n
()()()()⎥⎦
⎤⎢⎣⎡-⨯3750259.030042.1exp
810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-
()215
2
81044.110
410580.7⨯=⨯⨯=
o
n cm 3-
(e) (
)(
)
3
18
17
2300450100.7107.4⎪⎭
⎫
⎝⎛⨯⨯=i
n
()()()()⎥⎦
⎤⎢⎣⎡-⨯4500259.030042.1exp
1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-
()714
2
101048.110
10853.3⨯=⨯=
o
p cm 3-
_______________________________________ 4.36 (a) Ge: 13104.2⨯=i n cm 3-
(i)22
22i d
d o n N N n +⎪⎪⎭
⎫ ⎝⎛+=
()
2
132
1515
104.221022102⨯+⎪⎪⎭
⎫ ⎝⎛⨯+⨯=
or
15102⨯=≅d o N n cm 3-
()
15
2
13
2102104.2⨯⨯==o i o n n p
111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- ()
15
2
13
2103104.2⨯⨯==o i o p n n
111092.1⨯=cm 3-
(b) GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm
()315
2
61062.110
2108.1-⨯=⨯⨯=
o
p cm 3-
(ii)15103⨯=-≅d a o N N p cm 3-
()315
2
61008.1103108.1-⨯=⨯⨯=
o
n cm 3-
(c) The result implies that there is only one minority carrier in a volume of 310cm 3. _______________________________________ 4.37
(a) For the donor level
⎪⎪⎭⎫ ⎝⎛-+=kT E E N n F d d d exp 21
11
⎪⎭
⎫ ⎝⎛+=
0259.020.0exp 2111
or
41085.8-⨯=d
d N n
(b) We have
()⎪
⎪⎭
⎫ ⎝⎛-+=kT E E E f F F exp 11
Now
()()F c c F E E E E E E -+-=- or
245.0+=-kT E E F
Then
()⎪
⎭
⎫ ⎝⎛
++=
0259.0245.01exp 11
E f F
or
()51087.2-⨯=E f F
_______________________________________ 4.38 (a) ⇒>d a N N p-type (b) Silicon:
1313101105.2⨯-⨯=-=d a o N N p or
13105.1⨯=o p cm 3- Then
()
713
2
10
2105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:
22
22i d
a d a o n N N N N p +⎪⎪⎭
⎫ ⎝⎛-+-=
()
2
132
1313104.22105.12105.1⨯+⎪⎪⎭
⎫
⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯=
or
131026.3⨯=o p cm 3- Then
()
13132
13
21076.110
264.3104.2⨯=⨯⨯==o i o p n n cm 3-
Gallium Arsenide:
13105.1⨯=-=d a o N N p cm 3- and
()
216.010
5.1108.1132
6
2=⨯⨯==o i o p n n cm 3- _______________________________________ 4.39 (a) ⇒>a d N N n-type
(b)
1515102.1102⨯-⨯=-≅a d o N N n
14108⨯=cm 3-
()
514
2
10
21081.2108105.1⨯=⨯⨯=
=o i o n n p cm 3-
(c)
()d a a
o N N N p -+'≅ 151515
102102.1104⨯-⨯+'=⨯a
N 15108.4⨯='⇒a
N cm 3-
()415
2
1010625.510
4105.1⨯=⨯⨯=
o
n cm 3-
_______________________________________ 4.40
()
155
2
10
210125.110
2105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type
_______________________________________ 4.41
()()
3
18192300250100.61004.1⎪⎭
⎫
⎝⎛⨯⨯=i n
()()⎥⎦
⎤⎢⎣⎡-⨯3002500259.066.0exp
24
108936.1⨯=
1210376.1⨯=⇒i n cm 3- 22
22414i o o i o i o n n n n p n n =⇒==
i o n n 2
1
=⇒
So 111088.6⨯=o n cm 3-,
Then 121075.2⨯=o p cm 3-
2
2
22i a
a o n N N p +⎪⎪⎭
⎫ ⎝⎛+= 2
12
210752.2⎪⎪⎭
⎫ ⎝⎛-⨯a N
242
108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N
(
)
2
12
242
10
752.2105735.7⎪⎪⎭
⎫ ⎝⎛+⨯-⨯a
a N N
242
108936.12⨯+⎪⎪⎭
⎫ ⎝⎛=a
N so that 1210064.2⨯=a N cm 3-
_______________________________________ 4.42
Plot
_______________________________________ 4.43
Plot
_______________________________________ 4.44
Plot
_______________________________________ 4.45
2
2
22i a
d a d o n N N N N n +⎪⎪⎭
⎫ ⎝
⎛-+-= 2
102.1102101.114
1414
⨯-⨯=⨯
22
14142102.1102i n +⎪⎪⎭
⎫
⎝⎛⨯-⨯+
()()
22
132
13
14
10410410
1.1i n +⨯=⨯-⨯
22727106.1109.4i n +⨯=⨯ so 131074.5⨯=i n cm 3-
1314
27
210310
1.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________ 4.46 (a) ⇒>d a N N p-type
Majority carriers are holes
1616105.1103⨯-⨯=-=d a o N N p
16105.1⨯=cm 3-
Minority carriers are electrons
()
416
2
10
2105.110
5.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be added
d a a
o N N N p -+'=
161616105.1103105⨯-⨯+'=⨯a
N So 16105.3⨯='a
N cm 3-
()316
2
10105.4105105.1⨯=⨯⨯=
o
n cm 3-
_______________________________________ 4.47 (a) ⇒<<i o n p n-type (b) o
i o o i o p n n n n p 2
2=⇒=
o
n ()164
2
1010125.1102105.1⨯=⨯⨯=
cm 3-
⇒electrons are majority carriers
4102⨯=o p cm 3-
⇒holes are minority carriers (c) a d o N N n -= 151610710125.1⨯-=⨯d N so 1610825.1⨯=d N cm 3-
_______________________________________ 4.48
⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln For Germanium T (K) kT (eV)
i n (cm 3-) 200
400 600
01727.0 03453.0 0518.0
101016.2⨯ 141060.8⨯ 161082.3⨯
22
22i a a o n N N p +⎪⎪⎭
⎫
⎝⎛+=
and 1510=a N cm 3- T (K) o
p (cm 3-)
()F Fi E E -(eV)
200
400 600
15100.1⨯
151049.1⨯ 161087.3⨯
1855.0 01898.0 000674.0
_______________________________________ 4.49
(a) ⎪⎪⎭
⎫
⎝⎛=-d c F c N N kT E E ln
()⎪⎪⎭⎫
⎝⎛⨯=d N 19108.2ln 0259.0 For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV
(b) ⎪⎪⎭⎫
⎝⎛=-i d Fi F n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯=10105.1ln 0259.0d N For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________ 4.50 (a)
22
22i d d o n N N n +⎪⎪⎭
⎫
⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()
2
1515105.01005.1⨯-⨯
()
22
15
105.0i n +⨯=
so 2821025.5⨯=i n
Now
()()
3
191923001004.1108.2⎪⎭
⎫ ⎝⎛⨯⨯=T n i
()()⎥⎦
⎤⎢⎣⎡-⨯3000259.012.1exp T
(
)
3
38
28
30010912.21025.5⎪⎭
⎫ ⎝⎛⨯=⨯T
⎥⎦
⎤⎢⎣⎡-⨯T 973.12972exp By trial and error, 5.536=T K (b) At 300=T K,
⎪⎪⎭
⎫
⎝⎛=-o c F c n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯=-151910108.2ln 0259.0F c E E
2652.0=eV At 5.536=T K,
()046318.03005.5360259.0=⎪⎭
⎫
⎝⎛=kT eV
()
2
/3193005.536108.2⎪
⎭
⎫
⎝⎛⨯=c N
1910696.6⨯=cm 3-
⎪⎪⎭
⎫
⎝⎛=-o c F c n N kT E E ln
()⎪⎪⎭
⎫
⎝⎛⨯⨯=-15191005.110696.6ln 046318.0F c E E
5124.0=eV then ()2472.0=-∆F c E E eV (c) Closer to the intrinsic energy level.
_______________________________________ 4.51
⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eV
At 200=T K,
(
)(
)
3
19
19
23002001004.1108.2⎪⎭
⎫
⎝⎛⨯⨯=i
n
⎥⎦
⎤⎢⎣⎡-⨯017267.012.1exp
410638.7⨯=⇒i n cm 3- At 400=T K,
(
)(
)
3
19
19
2
3004001004.1108.2⎪⎭
⎫
⎝⎛⨯⨯=i
n
⎥⎦
⎤⎢⎣⎡-⨯034533.012.1exp 1210381.2⨯=⇒i n cm 3- At 600=T K,
(
)(
)
3
19
19
23006001004.1108.2⎪⎭
⎫
⎝⎛⨯⨯=i
n
⎥⎦
⎤⎢⎣⎡-⨯0518.012.1exp
1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,
22
22i a a o n N N p +⎪⎪⎭⎫
⎝⎛+=
()
2
142
1515
10740.921032103⨯+⎪⎪⎭
⎫ ⎝⎛⨯+⨯=
1510288.3⨯=cm 3-
Then, 200=T K, 4212.0=-F Fi E E eV 400=T K, 2465.0=-F Fi E E eV
600=T K, 0630.0=-F Fi E E eV
_______________________________________ 4.52
(a)
()⎪⎪⎭⎫
⎝⎛⨯=⎪⎪⎭
⎫ ⎝⎛=-6108.1ln 0259.0ln a i a F Fi N n N kT E E
For 1410=a N cm 3-,
4619.0=-F Fi E E eV
1510=a N cm 3-,
5215.0=-F Fi E E eV
16
10=a N cm 3-,
5811.0=-F Fi E E eV
1710=a N cm 3-,
6408.0=-F Fi E E eV (b)
()⎪
⎪⎭
⎫
⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-a a F N N N kT E E 18100.7ln 0259.0ln υυ For 1410=a N cm 3-,
2889.0=-υE E F eV
1510=a N cm 3-,
2293.0=-υE E F eV
16
10=a N cm 3-,
1697.0=-υE E F eV
1710=a N cm 3-,
1100.0=-υE E F eV
_______________________________________ 4.53
(a) ⎪⎪⎭
⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()()10ln 0259.043
= or
0447.0+=-midgap Fi E E eV
(b) Impurity atoms to be added so 45.0=-F midgap E E eV (i) p-type, so add acceptor atoms (ii)
4947.045.00447.0=+=-F Fi E E eV Then
⎪⎪⎭
⎫
⎝⎛-=kT E E n p F Fi i o exp
()
⎪⎭
⎫
⎝⎛=0259.04947.0exp 105
or
131097.1⨯==a o N p cm 3-
_______________________________________
4.54
()⎥⎦⎤
⎢⎣⎡--=-=kT E E N N N n F c c a d o exp so
()
⎪⎭
⎫
⎝⎛-⨯+⨯=0259.0215.0exp 108.21051915d N
15151095.6105⨯+⨯=
or
16102.1⨯=d N cm 3-
_______________________________________ 4.55 (a) Silicon
(i)⎪⎪⎭
⎫
⎝⎛=-d c F c N N kT E E ln
()2188.0106108.2ln 0259.01519=⎪⎪⎭
⎫
⎝⎛⨯⨯=eV
(ii)1929.00259.02188.0=-=-F c E E eV
()⎥⎦⎤
⎢⎣⎡--=kT E E N N F c c d exp
()
⎥⎦⎤
⎢⎣⎡-⨯=0259.01929.0exp 108.219
1610631.1⨯=d N cm 3-15106⨯+'=d
N 1610031.1⨯='⇒d
N cm 3- Additional
donor atoms (b) GaAs
(i)()⎪⎪⎭
⎫
⎝⎛⨯=-151710107.4ln 0259.0F c E E
15936.0=eV
(ii)13346.00259.015936.0=-=-F c E E eV
()
⎥⎦⎤
⎢⎣⎡-⨯=0259.013346.0exp 107.417d N
1510718.2⨯=cm 3-1510+'=d
N 1510718.1⨯='⇒d
N cm 3- Additional
donor atoms _______________________________________
4.56 (a) ⎪⎪⎭⎫ ⎝⎛=-a F Fi N N kT E E υln
()⎪⎪⎭⎫
⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV
(b) ⎪⎪⎭⎫
⎝⎛=-d c Fi F N N kT E E ln ()1876.0102108.2ln 0259.01619=⎪
⎪⎭
⎫ ⎝⎛⨯⨯=eV (c) For part (a); 16102⨯=o p cm 3-
()
162
10
2102105.1⨯⨯==o i o p n n
410125.1⨯=cm 3-
For part (b): 16102⨯=o n cm 3
-
()
16210
2102105.1⨯⨯==o i o n n p
410125.1⨯=cm 3-
_______________________________________
4.57
⎥⎦
⎤⎢⎣⎡-=kT E E n n Fi F i o exp ()
⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.16
15100.3⨯=cm 3-
Add additional acceptor impurities
a d o N N n -= a N -⨯=⨯1515107103
15104⨯=⇒a N cm 3-
_______________________________________ 4.58
(a) ⎪⎪⎭⎫
⎝⎛=-i o F Fi n p kT E E ln ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫
⎝⎛⨯⨯=eV (b) ⎪⎪⎭
⎫
⎝⎛=-i o Fi F n n kT E E ln ()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫
⎝⎛⨯⨯=eV
(c) Fi F E E =
(d) ⎪⎪⎭⎫
⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=111510334.7104ln 3003750259.0 2786.0=eV
(e) ⎪⎪⎭⎫ ⎝⎛=-i o Fi F n n kT E E ln
()⎪
⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=131410722.110029.1ln 3004500259.0 06945.0=eV _______________________________________ 4.59 (a) ⎪⎪⎭⎫ ⎝⎛=-o F p N kT E E υυln
()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV
(b) ()⎪⎪⎭
⎫ ⎝⎛⨯⨯=--4181008.1100.7ln 0259.0υE E F 360.1=eV
(c) ()⎪⎪⎭
⎫
⎝⎛⨯⨯=-618108.1100.7ln 0259.0υE E F
7508.0=eV
(d) ()⎪⎭
⎫
⎝⎛=-3003750259.0υE E F
()
()⎥⎥⎦
⎤⎢⎢⎣⎡⨯⨯⨯15
2/318104300375100.7ln 2526.0=eV
(e) ()⎪⎭
⎫
⎝⎛=-3004500259.0υE E F
()
()⎥⎥⎦
⎤⎢⎢⎣⎡⨯⨯⨯7
2/3181048.1300450100.7ln 068.1=eV
_______________________________________ 4.60
n-type
⎪⎪⎭
⎫
⎝⎛=-i o Fi F n n kT E E ln
()3504.0105.110125.1ln 0259.01016=⎪⎪⎭
⎫
⎝⎛⨯⨯=eV ______________________________________ 4.61
2
2
22i a
a o n N N p +⎪⎪⎭
⎫ ⎝
⎛+= 2
1051008.515
15
⨯=⨯
2
2
152105i n +⎪⎪⎭
⎫ ⎝⎛⨯+
()
2
1515105.21008.5⨯-⨯
()
22
15
105.2i n +⨯=
230301025.6106564.6i n +⨯=⨯
29210064.4⨯=⇒i n
⎥⎦
⎤
⎢⎣⎡-=kT E N N n g c i exp 2υ
()030217.03003500259.0=⎪⎭
⎫
⎝⎛=kT eV
(
)
19
2
19
10633.1300350102.1⨯=⎪⎭
⎫ ⎝⎛⨯=c N cm 3-
(
)
192
19
1045.230035010
8.1⨯=⎪⎭
⎫ ⎝⎛⨯=υN cm 3- Now
()()
1919291045.210633.110064.4⨯⨯=⨯
⎥⎦⎤⎢⎣⎡-⨯030217.0exp g E
So
()()()
⎥⎦
⎤
⎢⎣⎡⨯⨯⨯=29
191910064.41045.210633.1ln 030217.0g E 6257.0=⇒g E eV
_______________________________________ 4.62 (a) Replace Ga atoms ⇒Silicon acts as a
donor
()()
1415105.310705.0⨯=⨯=d N cm 3-
Replace As atoms ⇒Silicon acts as an
acceptor
()()
15151065.610795.0⨯=⨯=a N cm 3-
(b) ⇒>d a N N p-type
(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-
()
415
2
6
21014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3- (d)
⎪⎪⎭
⎫
⎝⎛=-i o F Fi n p kT E E ln
()5692.0108.1103.6ln 0259.0615=⎪⎪⎭
⎫
⎝⎛⨯⨯=eV
_______________________________________。