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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 7 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 Then 
 
a
d
n
p
N
N
x
x
 25.0 
 Now 
 









2
ln
i
da
tbi
n
NN
VV 
 so 
  
  









26
2
108.1
25.0
ln0259.0137.1 aN
 
 We can then write 
 
 





0259.02
137.1
exp
25.0
108.1 6
aN 
 which yields 
 161023.1 aN cm 3 
 and 
 151007.3 dN cm 3 
_______________________________________ 
 
7.23 
  
  
  










26
1516
108.1
105102
ln0259.0biV 
 162.1 V 
 
Rbi VV
C


1
 
 So 
 
  1
2
2
1
Rbi
Rbi
R
R
VV
VV
VC
VC





 
 
5.0162.1
162.1
50.1 2


 RV
 
  
662.1
162.1
50.1 22 RV
 
 which yields 58.22 RV V 
_______________________________________ 
 
7.24 
(a)  
  
  










210
1615
105.1
104102
ln0259.0biV 
 6889.0 V 
 
  
2/1
2 








daRbi
das
NNVV
NNe
ACAC 
 
 
 
 
 
      
 







RV6889.02
1085.87.11106.1
105
1419
4 
 
  
 
2/1
1615
1615
104102
104102





 
 
RV
C




6889.0
102806.6 12
 
 (i) For 0RV , 
 567.7C pF 
 (ii) For 5RV V, 
 633.2C pF 
(b)  
  
  










26
1615
108.1
104102
ln0259.0biV 
 157.1 V 
 
  
2/1
2 








daRbi
das
NNVV
NNe
ACAC 
      
 







RV157.12
1085.81.13106.1
105
1419
4 
 
  
 
2/1
1615
1615
104102
104102





 
 
RV
C




157.1
106457.6 12
 
 (i) For 0RV , 
 178.6C pF 
(ii) For 5RV V, 
 678.2C pF 
_______________________________________ 
 
7.25 
  
  
  










210
1517
105.1
105102
ln0259.0biV 
 7543.0 V 
(a) 
  
2/1
2 








daRbi
das
NNVV
NNe
ACAC 
      
 







107543.02
1085.87.11106.1
108
1419
4 
 
  
 
2/1
1517
1517
105102
105102





 
 
1210904.4 C F 
 
 2
2
1
2
1
fC
L
LC
f

