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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 272.0ms V 
 Now 
      41519 10863.010106.1max  
SDQ 
 or 
   81038.1max 
SDQ C/cm 2 
 We have 
    91019 106.110106.1  
ssQ C/cm 2 
 We now find 
 
 
  
 8
14
98
10500
1085.89.3
106.11038.1 





TV 
  288.02272.0  
 or 
 07.1TV V 
_______________________________________ 
 
10.18 
 (b) 








 fp
g
mms
e
E

2
 
 where 
 20.0 m V 
 and 
   3473.0
105.1
10
ln0259.0
10
16









fp V 
 Then 
  3473.056.020.0 ms 
 or 
 107.1ms V 
 (c) For 0
ssQ 
   fpms
ox
ox
SDTN
t
QV  2max 








 
 We find 
 
   
  
2/1
1619
14
10106.1
3473.01085.87.114











dTx 
 or 
 41030.0 dTx cm 30.0 m 
 Now 
      41619 1030.010106.1max  
SDQ 
 or 
   810797.4max 
SDQ C/cm 2 
 Then 
 
 
 
 
 
 
 
  
  14
88
1085.89.3
1030010797.4




TV 
  3473.02107.1  
 or 
 00455.0TV V 0 V 
_______________________________________ 
 
10.19 
 Plot 
_______________________________________ 
 
10.20 
 Plot 
_______________________________________ 
 
10.21 
 Plot 
_______________________________________ 
 
10.22 
 Plot 
_______________________________________ 
 
10.23 
(a) For 1f Hz (low freq), 
 
  
8
14
10120
1085.89.3







ox
ox
ox
t
C 
 
710876.2  F/cm 2 
 
a
st
s
ox
ox
ox
FB
eN
V
t
C













 
  
   
  1619
14
8
14
10106.1
1085.87.110259.0
7.11
9.3
10120
1085.89.3















 710346.1 
FBC F/cm 2 
 
dT
s
ox
ox
ox
xt
C













min 
 Now 
   3473.0
105.1
10
ln0259.0
10
16









fp V 
 
   
  
2/1
1619
14
10106.1
3473.01085.87.114











dTx 
 
51000.3  cm