1 (98)

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
(b) We have 
 
 
dx
pd
eDJ n
pp

 
 
 











 44
14
10828.2
exp
10828.2
10808.3 xeD p
 
 At 4103 x cm, 
 
   





 





828.2
3
exp
10828.2
10808.38106.1
3
4
1419
pJ 
 or 
   5966.03 pJ A/cm 2 
 (c) We have 
 








t
a
n
pon
no
V
V
L
neD
J exp 
 We can determine that 
 
3105.4 pon cm 3 and 72.10nL m 
 Then 
 
   
4
319
1072.10
105.423106.1




noJ 
 






0259.0
610.0
exp 
 or 
 2615.0noJ A/cm 2 
 We can also find 
 724.1poJ A/cm 2 
 Then at 3x m, 
    33 pponon JJJJ  
 5966.0724.12615.0  
 or 
   39.13 nJ A/cm 2 
_______________________________________ 
 
8.18 
 (a) Problem 8.7 
 








t
a
pop
V
V
nn exp 
 or 
 


















ai
a
t
po
p
ta
Nn
N
V
n
n
VV
2
1.0
lnln 
 
 









2
21.0
ln
i
a
t
n
N
V 
  
  
  










213
215
104.2
1041.0
ln0259.0 
 205.0 V 
 
 
(b) Problem 8.8 
 








t
a
non
V
V
pp exp 
 or 






no
n
ta
p
p
VV ln
 









di
d
t
Nn
N
V
2
1.0
ln 
 
 









2
21.0
ln
i
d
t
n
N
V 
  
  
  










210
215
105.1
1081.0
ln0259.0 
 623.0 V 
_______________________________________ 
 
8.19 
 The excess electron concentration is given by 
 popp nnn  
 






 


















nt
a
po
L
x
V
V
n exp1exp 
 The total number of excess electrons is 
 dxnAN pp 


0
 
 We may note that 
 n
n
n
n
L
L
x
Ldx
L
x







 







  


0
0
expexp 
 Then 
 
















 1exp
t
a
ponp
V
V
nALN 
 We find that 
 25nD cm 2 /s and 0.50nL m 
 Also 
 
  4
15
2102
1081.2
108
105.1




a
i
po
N
n
n cm 3 
 Then 
    443 108125.2100.5010  
pN 
 
















 1exp
t
a
V
V
 
 or 
  
















 1exp1406.0
t
a
p
V
V
N 
 Then, we find the total number of excess 
 electrons in the p-region to be: