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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
10.4 
 p-type silicon 
 (a) Aluminum gate 
 
















 fp
g
mms
e
E

2
 
 We have 
 









i
a
tfp
n
N
V ln 
   334.0
105.1
106
ln0259.0
10
15










 V 
 Then 
   334.056.025.320.3 ms 
 or 
 944.0ms V 
 (b) 
n polysilicon gate 
 








 fp
g
ms
e
E

2
 334.056.0  
 or 
 894.0ms V 
 (c) p polysilicon gate 
  334.056.0
2









 fp
g
ms
e
E
 
 or 
 226.0ms V 
_______________________________________ 
 
10.5 
   3832.0
105.1
104
ln0259.0
10
16










fp V 
 








 fp
g
mms
e
E

2
 
  3832.056.025.320.3  
 9932.0ms V 
_______________________________________ 
 
10.6 
(a) 17102dN cm 3 
(b) Not possible - ms is always positive. 
(c) 15102dN cm 3 
_______________________________________ 
 
 
 
 
 
 
10.7 
 From Problem 10.5, 9932.0ms V 
 
ox
ss
msFB
C
Q
V

  
(a) 
  
8
14
10200
1085.89.3







ox
ox
ox
t
C 
 710726.1  F/cm 2 
 
  
7
1910
10726.1
106.1105
9932.0




FBV 
 040.1 V 
(b) 
  
8
14
1080
1085.89.3




oxC 
 
710314.4  F/cm 2 
 
  
7
1910
10314.4
106.1105
9932.0




FBV 
 012.1 V 
_______________________________________ 
 
10.8 
(a) 42.0ms V 
 42.0 msFBV  V 
(b) 
   7
8
14
10726.1
10200
1085.89.3 





oxC F/cm 2 
 (i)
  
7
1910
10726.1
106.1104







ox
ss
FB
C
Q
V 
 0371.0 V 
 (ii)
  
7
1911
10726.1
106.110




 FBV 
 0927.0 V 
(c) 42.0 msFBV  V 
 
   7
8
14
10876.2
10120
1085.89.3 





oxC F/cm 2 
 (i)
  
7
1910
10876.2
106.1104




 FBV 
 0223.0 V 
 (ii)
  
7
1911
10876.2
106.110




 FBV 
 0556.0 V 
_______________________________________