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Semiconductor Physics and Devices: Basic Principles, 3
rd
edition
Chapter 14
Solutions Manual
Problem Solutions
215
Chapter 14
Problem Solutions
14.1
(a)
λµ
=
124
.
E
m
Then
Ge:
Ee
V
g
=⇒
066
.
=
188
.
m
Si:
V
g
112
.
=
111
.
m
GaAs:
V
g
142
=
0873
.
m
(b)
E
=
.
λ
For
570
nm
V
=
2.18
For
700
nm
V
=
177
.
14.2
(a) GaAs
he
V
ν
=
2
⇒=
062
.
m
so
α
≈
−
15 10
41
.
xc
m
Then
Ix
I
xx
x
O
()
=−
−
exp
exp
.
.
035 10
44
bg
b
g
or
I
O
=
059
.
so the percent absorbed is (10.59), or
41%
(b) Silicon
Again
V
=
2
.
m
So
≈
−
0
31
m
Then
I
x
O
−
exp
exp
.
4 10
0 35 10
34
b
g
or
I
O
=
087
.
so the percent absorbed is (10.87), or
13%
14.3
′ =
g
h
For
V
m
νλ
µ
=
=
13
095
.
.
.
For silicon,
≈
−
310
21
m
,
Then for
W cm
() =
−
10
22
/
we obtain
′
−
−
g
x
x
310 10
16 10
19
b
g
..
′ =
−−
gx
c
m
s
144 10
19
3
1
.
The excess concentration is
δτ
ng
x
=
′
−
10
19
6
.
b
g
δ
nx
c
m
=
−
13
3
.
14.4
ntype GaAs,
τ
=
−
10
7
s
(a)
We want
δδ
np
c
m
g
g
==
=
′
=
′
10
10
15
3
7
or
′
−
gc
m
s
10
10
10
15
7
22
3
1
We have
V
m
=
=
19
065
.
.
.
so that
≈
−
13 10
.
m
Then
′
=
′
g
h
gh
=
−
10
22
19
4
b
gb
g
.
x
x
or
IW
c
m
I
O
00
2
3
4
2
./
(b)
I
O
==−
020
4
.e
x
p
.
We obtain
xm
=
.
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View Full Document Semiconductor Physics and Devices: Basic Principles, 3
rd
edition
Chapter 14
Solutions Manual
Problem Solutions
216
14.5
GaAs
(a)
For
he
V
m
νλ
µ
=⇒
=
165
075
..
So
α
≈
−
07 10
41
.
xc
m
For 75% obsorbed,
Ix
I
x
O
()
==−
025
.e
x
p
Then
xx
x
=
F
H
I
K
F
H
I
K
ln
ln
.
1
1
1
4
or
xm
=
198
.
(b)
For 75% transmitted,
I
O
4
x
p.
bg
we obtain
=
041
.
14.6
GaAs
For
c
m
==
−
11
0
4
, we have 50% absorbed
or 50% transmitted, then
I
x
O
050
x
p
We can write
=⋅
=
⋅
F
H
I
K
F
H
I
K
F
H
I
K
−
05
1
10
2
4
x
ln
.
ln
or
=
−
069 10
.
m
This value corresponds to
λµ
.,
.
mE
e
V
14.7
The ambipolar transport equation for minority
carrier holes in steady state is
D
dp
dx
G
p
p
n
L
n
p
2
2
0
δδ
τ
af
+−
=
or
dx
p
L
G
D
nn
p
L
p
2
22
−=
−
where
LD
pp
p
2
=
The photon flux in the semiconductor is
ΦΦ
O
( )
=−
exp
and the generation rate is
Gx
x
LO
−
αα
exp
so we have
dx
p
x
p
O
p
2
−
−
Φ
exp
The general solution is of the form
δ
pA
x
L
B
x
L
n
=
−
+
+
F
H
G
I
K
J
F
H
G
I
K
J
exp
exp
−
−
−
ατ
Φ
Op
p
L
x
1
exp
At
xp
n
→∞
=
,
0
So that
B
=
0 , then
x
LL
x
n
p
p
=
−
−
−
−
F
H
G
I
K
J
exp
exp
Φ
1
At
x
=
0 , we have
D
dx
sp
p
n
n
a
f
=
00
so we can write
L
n
x
p
=
−
0
1
Φ
and
dx
A
n
p
p
x
=
+
−
0
2
1
Φ
Then we have
−+
−
−
AD
L
D
L
sA
s
L
p
p
Op p
p
p
2
Solving for
A
, we find
A
L
sD
sDL
p
p
=
−
⋅
+
+
L
N
M
O
Q
P
Φ
1
b
g
The solution can now be written as
p
L
x
L
n
p
p
p
=
−
⋅
+
+
⋅
−
R
S
T
F
H
G
I
K
J
Φ
1
exp
−−
exp
x
r
Semiconductor Physics and Devices: Basic Principles, 3
rd
edition
Chapter 14
Solutions Manual
Problem Solutions
217
14.8
We have
D
dn
dx
G
n
n
p
L
p
n
2
2
0
δδ
τ
bg
+−
=
or
dx
n
L
G
D
pp
n
L
n
2
22
−=
−
where
LD
nn
n
2
=
The general solution can be written in the form
δτ
nA
x
L
B
x
L
G
p
Ln
=+
+
F
H
G
I
K
J
F
H
G
I
K
J
cosh
sinh
For
s
=∞
at
x
=
0 means that
δ
n
p
00
() =
,
Then
0
⇒=
−
AG
A G
ττ
At
xW
=
,
==
D
dx
sn
n
p
op
Now
nW
G
W
L
B
W
L
G
pL
n
()
F
H
G
I
K
J
F
H
G
I
K
J
=−
+
+
cosh
sinh
and
dx
G
L
W
L
B
L
W
L
p
n
n
=
+
F
H
G
I
K
J
F
H
G
I
K
J
sinh
cosh
so we can write
GD
L
W
L
BD
L
W
L
Ln n
n
sinh
cosh
F
H
G
I
K
J
F
H
G
I
K
J
−
F
H
G
I
K
J
sG
W
L
oL
n
n
cosh
++
F
H
G
I
K
J
B
W
L
G
n
sinh
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This note was uploaded on 11/06/2008 for the course ECE 103 taught by Professor Wang during the Fall '08 term at UCSD.
 Fall '08
 WANG

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