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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 Then 
 
  
 58
14
min
1000.3
7.11
9.3
10120
1085.89.3










C 
 
810083.3  F/cm 2 
 C  (inv) 710876.2  oxC F/cm 2 
(b) 1f MHz (high freq), 
 710876.2 oxC F/cm 2 (unchanged) 
 710346.1 
FBC F/cm 2 (unchanged) 
 8
min 10083.3 C F/cm 2 (unchanged) 
 C  (inv) 8
min 10083.3 C F/cm 2 
(c) 10.1 msFBV  V 
 
 
fpFB
ox
SD
TN V
C
Q
V 2
max


 
 Now 
   dTaSD xeNQ  max 
    51619 1000.310106.1   
 
81080.4  C/cm 2 
  3473.0210.1
10876.2
1080.4
7
8






TNV 
 2385.0TNV V 
_______________________________________ 
 
10.24 
(a) 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













 
  
   
  1419
14
8
14
105106.1
1085.87.110259.0
7.11
9.3
10120
1085.89.3















 810726.4 
FBC F/cm 2 
 
dT
s
ox
ox
ox
xt
C













min 
 Now 
   2697.0
105.1
105
ln0259.0
10
14










fn V 
 
 
 
   
  
2/1
1419
14
105106.1
2697.01085.87.114











dTx 
 410182.1  cm 
 Then 
 
  
 48
14
min
10182.1
7.11
9.3
10120
1085.89.3










C 
 910504.8  F/cm 2 
 C  (inv) 710876.2  oxC F/cm 2 
(b) 1f MHz (high freq), 
 710876.2 oxC F/cm 2 (unchanged) 
 810726.4 
FBC F/cm 2 (unchanged) 
 9
min 10504.8 C F/cm 2 (unchanged) 
 C  (inv) 9
min 10504.8 C F/cm 2 
(c) 95.0 msFBV  V 
 
 
fnFB
ox
SD
TP V
C
Q
V 2
max


 
 Now 
   dTdSD xeNQ  max 
    41419 10182.1105106.1   
 
910456.9  C/cm 2 
 Then 
  2697.0295.0
10876.2
10456.9
7
9






TPV 
 378.0TPV V 
_______________________________________ 
 
10.25 
 The amount of fixed oxide charge at x is 
   xx  C/cm 2 
 By lever action, the effect of this oxide charge 
 on the flatband voltage is 
   xx
t
x
C
V
oxox
FB 







 
1
 
 If we add the effect at each point, we must 
 integrate so that 
 
 
dx
t
xx
C
V
oxt
oxox
FB 
0
1 
 
_______________________________________