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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 and 
 21
2
2
CxCx
D
G
p
p
o 

 
 For LxL 3 , 0g so we have 
 
 
0
2
2

dx
pd 
 so that 
 
3C
dx
pd


 and 
 43 CxCp  
 For LxL 3 , 0g so that 
 
 
0
2
2

dx
pd 
 so that 
 
5C
dx
pd


 and 
 65 CxCp  
 The boundary conditions are: 
 (1) 0p at Lx 3 
 (2) 0p at Lx 3 
(3) p continuous at Lx  
(4) p continuous at Lx  
(5) 
 
dx
pd 
 continuous at Lx  
(6) 
 
dx
pd 
 continuous at Lx  
 Applying the boundary conditions, we find 
  225
2
xL
D
G
p
p
o 

 for LxL  
  xL
D
LG
p
p
o 

 3 for LxL 3 
  xL
D
LG
p
p
o 

 3 for LxL 3 
_______________________________________ 
 
6.27 
 20
4.0
8
0 
L
V
V/cm 
 
  6
00 103220
25.0




t
d
p 
 6.390 cm 2 /V-s 
 
   
0
22
0
16t
t
D
p
p



 
 
     
 6
262
103216
1035.9206.390




 
 42.10pD cm 2 /s 
 
 
 
 
 
 We find 
 02668.0
6.390
42.10

p
pD

V 
 This value is very close to 0.0259 for 
 300T K. 
_______________________________________ 
 
6.28 
 (a) 
 Assume that     






 


Dt
x
Dttxf
4
exp4,
2
2/1
 
 is the solution to the differential equation 
 
t
f
x
f
D












2
2
 
 To prove: we can write 
   






 





 


 
Dt
x
Dt
x
Dt
x
f
4
exp
4
2
4
2
2/1
 
 and 
   






 








 


 
Dt
x
Dt
x
Dt
x
f
4
exp
4
2
4
22
2/1
2
2
 
 











 





 

Dt
x
Dt 4
exp
4
2 2
 
 Also 
   






 





 







 


 
Dt
x
tD
x
Dt
t
f
4
exp
1
4
4
2
2
2
2/1
 
   






 





 
 
Dt
x
tD
4
exp
2
1
4
2
2/32/1
 
 Substituting the expressions for 
2
2
x
f


 and 
 
t
f


into the differential equation, we find 
 0 = 0. 
 Q.E.D. 
 (b) 
 Consider 
 dx
Dt
x










 
4
exp
2
 
 Let 
2xu  , then dxxdu  2 or 
 
u
du
x
du
dx
22
 
 Let 
Dt
a
4
1
 
 Now