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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 By trial and error, 402.0mV V 
 








t
m
SLm
V
V
III exp 
   





 
0259.0
402.0
exp10210180 93 
 11069.1  A 169 mA 
    9.67402.0169  mmm VIP mW 
(c)  379.2
169.0
402.0
m
m
L
I
V
R 
(d)     568.3379.25.1LR 
 Now 
 








t
SL
L V
V
II
R
V
I exp 
   





 
0259.0
exp10210180
568.3
93 VV
 
 By trial and error, 444.0V V 
 Then 1244.0
568.3
444.0

LR
V
I A 
    2.55444.04.124  IVP mW 
_______________________________________ 
 
14.16 
(a)   






 



10
3
10
10100
1ln0259.0ocV 
 5367.0 V 
(b) 
S
L
t
m
t
m
I
I
V
V
V
V
















 1exp1 
 
10
3
10
10100
1


 
 
910 
 By trial and error, 461.0mV V 
 Then 
   





 
0259.0
461.0
exp1010100 103
mI 
 
210463.9  A 63.94 mA 
    62.43461.063.94  mmm VIP mW 
(c) 227.21
461.0
10
 nn cells 
(d) Now    14.10461.022 V V 
 IVP  
   5128.014.102.5  II A 
 Then 642.5
09463.0
5128.0
 nn 
 
(e) Then    5678.009463.06 I A 
So  86.17
5678.0
14.10
I
V
RL 
_______________________________________ 
 
14.17 
 Let 0x correspond to the edge of the space 
 charge region in the p-type material. Then in 
 the p-region 
 
 
0
2
2

n
p
L
p
n
n
G
dx
nd
D


 
 or 
 
 
n
L
n
pp
D
G
L
n
dx
nd

22
2 
 
 where 
    xxG OL   exp 
 Then we have 
 
 
 x
DL
n
dx
nd
n
O
n
pp




 exp
22
2
 
 The general solution is of the form 
   






 







 

nn
p
L
x
B
L
x
Axn expexp 
  x
Ln
nO 





 exp
122
 
 As x , 0pn so that 0B . Then 
    x
LL
x
Axn
n
nO
n
p 


 









 
 exp
1
exp
22
 
 We also have  
1
00
22 


n
nO
p
L
An


 , 
 which yields 
 
122 


n
nO
L
A


 
 We then obtain 
    















 


 x
L
x
L
xn
nn
nO
p 


 expexp
122
 
 where O is the incident flux at 0x . 
_______________________________________ 
 
14.18 
 For 90% absorption, we have 
 
 
  10.0exp 


x
x
O
 
 Then