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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
    dxxWeI
Ox
pOpnL  
0
exp  
   pOpnWe   
  
Ox
x
0
exp
1






 

 
 which becomes 
     OpOpnL xWeI   exp1 
 Now 
     504501200106.11050 194  
LI 
        44716 10105exp110210   
 or 
 131.0LI A 
_______________________________________ 
 
14.22 
  
  
  










210
1516
105.1
10210
ln0259.0biV 
 6530.0 V 
   7
0 10525  nnn DL  
 
310536.3  cm 
   7
0 1010  ppp DL  
 
310 cm 
 
 
2/1
2















 

da
daRbis
NN
NN
e
VV
W 
 
   
 







19
14
106.1
5653.01085.87.112
 
 
  
2/1
1516
1516
10210
10210












 
 Then 
 
410095.2 W cm 
(a) AeWGI LL 1 
     321419 101010095.2106.1   
 
510352.3  A 52.33 A 
(b) In n-region, 
    14721
0 101010  
pLGp  cm 3 
 In p-region, 
   721
0 10510  nLGn  
 
14105 cm 3 
 
 
 
 
(c)  pnLL LLWAeGI  
    32119 1010106.1  
   4100.1036.35095.2  
 410593.7 LI A 7593.0 mA 
_______________________________________ 
 
14.23 
 In the n-region under steady state and for 
 0 , we have 
 
 
0
2
2

 p
n
L
n
p
p
G
xd
pd
D


 
 or 
 
 
p
L
p
nn
D
G
L
p
xd
pd

 22
2 
 
 where ppp DL  and where x  is 
 positive in the negative x direction. The 
 homogenerous solution is found from 
 
 
0
22
2


p
nhnh
L
p
xd
pd 
 
 The general solution is found to be 
  







 








 

pp
nh
L
x
B
L
x
Axp expexp 
 The particular solution is found from 
 
p
L
p
np
D
G
L
p

2

 
 which yields 
 pL
p
pL
np G
D
LG
p  
2
 
 The total solution is the sum of the 
 homogeneous and particular solutions, so we 
 have 
   pL
pp
n G
L
x
B
L
x
Axp  







 








 
 expexp 
 One boundary condition is that np remains 
 finite as x which means that 0B . 
 Then at 0x ,     nOnn ppp  000  , so 
 that   nOn pp 0 . 
 We find that 
  pLnO GpA  
 The solution is then written as 
    







 

p
nOpLpLn
L
x
pGGxp exp