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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 (b) 
 (i) For 0s , 
   14100 p cm 3 
 (ii) For 2000s cm/s, 
   1410833.00 p cm 3 
 (iii) For s , 
   00 p 
_______________________________________ 
 
6.42 
   710525  nOnn DL  
 
4104.35  cm 
(a) At 0x , 
    15721 10105102  
nOg  cm 3 
 or 
   15100  nOgn  cm 3 
 For 0x 
 
   
00
22
2
2
2

nnO
n
L
n
dx
ndn
dx
nd
D



 
 The solution is of the form 
 






 







 

nn L
x
B
L
x
An expexp 
 At 0x , 
   BAnn  0 
 At Wx  , 
 






 







 

nn L
W
B
L
W
An expexp0 
 Solving these two equations, we find 
 
   
 n
n
LW
LWn
A
2exp1
2exp0




 
 and 
 
 
 nLW
n
B
2exp1
0



 
 Substituting into the general solution, we find 
 
 















 







 

nn L
W
L
W
n
n
expexp
0
 
 
   













 





 

nn L
xW
L
xW
expexp 
 which can be written as 
 
 
 
 
 
 
 
 











 

n
n
L
W
L
xW
n
n
sinh
sinh0
 
 where 
   15100 n cm 3 and 4.35nL m 
(b) If nO , we have 
 
 
0
2
2

dx
nd 
 
 so the solution is of the form 
 DCxn  
 Applying the boundary conditions, we find 
   






W
x
nn 10 
_______________________________________ 
 
6.43 
 For pO , we have 
 
 
0
2
2

dx
pd 
 
 So the solution is of the form 
 BAxp  
 At Wx  
 
 
 
WxWx
p ps
dx
pd
D

 

 
 or 
  BAWsADp  
 which yields 
  sWD
s
A
B p 

 
 At 0x , the flux of excess holes is 
 
 
AD
dx
pd
D p
x
p 
0
1910

 
 so that 
 18
19
10
10
10


A cm 4 
 and 
   





 W
s
sW
s
B
10
1010
10 18
18
 
 The solution is now 
 






s
xWp
10
1018