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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
(a) For s , 
  xp  418 102010 cm 3 
 Then 
 
 
dx
pd
eDJ pp

 
    1819 1010106.1   
 or 
 6.1pJ A/cm 2 
(b) For 
3102s cm/s, 
  xp  418 107010 cm 3 
 Also 
 6.1pJ A/cm 2 
_______________________________________ 
 
6.44 
 For 0 xW 
 
 
0
2
2
 on G
dx
nd
D

 
 so that 
 
 
1Cx
D
G
dx
nd
n
o 



 
 and 
 21
2
2
CxCx
D
G
n
n
o 

 
 For Wx 0 , 
 
 
0
2
2

dx
nd 
 
 so that 
 43 CxCn  
 The boundary conditions are 
 (1) 0s at Wx  so that 
 
 
0
 Wxdx
nd 
 
 (2) s at Wx  so that 
   0Wn 
 (3) n continuous at 0x 
 (4) 
 
dx
nd 
 continuous at 0x 
 Applying the boundary conditions, we find 
 
n
o
D
WG
CC

 31 and 
n
o
D
WG
CC
2
42

 
 
 
 
 
 
 
 Then for 0 xW 
  22 22
2
WWxx
D
G
n
n
o 

 
 and for Wx 0 
  xW
D
WG
n
n
o 

 
_______________________________________ 
 
6.45 
 Plot 
_______________________________________ 
 
6.48 
(a) GaAs: 
 



6
6
10
102
2
I
V
R 
 
 A
L
R

 and   pe pn   
    13821
0 10510510  
pgp  cm 3 
 For 1610dN cm 3 , from Figure 5.3, 
 7000n cm 2 /V-s, 310p cm 2 /V-s 
    1319 1053107000106.1   
 05848.0 ( -cm) 1 
 Let 20W m 
 Then   44 1041020  WdA 
 
81080  cm 2 
 So 
  8
6
108005848.0
10


L
R 
 Which yields 
21068.4 L cm 
(b) Silicon: 
  610R , 13105p cm 3 
 For 1610dN cm 3 , from Figure 5.3, 
 1300n cm 2 /V-s, 410p cm 2 /V-s 
    1319 1054101300106.1   
 01368.0 ( -cm) 1 
 Let 20W m 
 Then   44 1041020  WdA 
 
81080  cm 2 
 So 
  8
6
108001368.0
10


L
R 
 Which yields 
21009.1 L cm 
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