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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 7 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 Then 
  
  









19
1617
10780.2
104102
ln024778.0biV 
 82494.0 V 
 We find 
 
   
 
%100
300
300287


bi
bibi
V
VV
 
 %08.2%100
80813.0
80813.082494.0


 
 %2 
_______________________________________ 
 
7.11 
 









2
ln
i
da
tbi
n
NN
VV 
  
  







 







2
1516 102104
ln
300
0259.0550.0
in
T
 
 Using the procedure from Problem 7.10, we 
 can write, for 300T K, 
  2102 105.1 in 
    




 

0259.0
12.1
exp1004.1108.2 1919K 
 659.4 K 
 At 300T K, 
  
  
  










210
1516
105.1
102104
ln0259.0biV 
 68886.0 V 
 For 550.0biV V, 300T K 
 At 380T K, 032807.0kT eV 
 Also 
    
3
19192
300
380
1004.1108.2659.4 





in 
 




 

032807.0
12.1
exp 
 
2410112.4  
 Then 
  
  









24
1516
10112.4
102104
ln032807.0biV 
 5506.0 V 550.0 V 
_______________________________________ 
 
 
 
 
 
 
7.12 
(b) For 1610dN cm 3 , 
 








i
d
FiF
n
N
kTEE ln 
   









10
16
105.1
10
ln0259.0 
 or 
 3473.0 FiF EE eV 
 For 1510dN cm 3 
   









10
15
105.1
10
ln0259.0FiF EE 
 or 
 2877.0 FiF EE eV 
 Then 
 28768.034732.0 biV 
 or 
 0596.0biV V 
_______________________________________ 
 
7.13 
(a) 









2
ln
i
da
tbi
n
NN
VV 
  
  
  









210
1612
105.1
1010
ln0259.0 
 or 
 456.0biV V 
 (b) 
 
   








19
14
106.1
456.01085.87.112
nx 
 
2/1
161216
12
1010
1
10
10


















 
 or 
 71043.2 nx cm 
 (c) 
 
   








19
14
106.1
456.01085.87.112
px 
 
2/1
161212
16
1010
1
10
10


















 
 or 
 
31043.2 px cm