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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 
 Now TTOT VVV  
 114.05.0  TOV 
 386.0 TOV V 
_______________________________________ 
 
10.52 
(a) 
  
8
14
10200
1085.89.3




oxC 
 710726.1  F/cm 2 
 
ox
ds
C
Ne

2
 
 
    
7
151419
10726.1
1051085.87.11106.12




 
 2358.0 V 2/1 
(b)   3294.0
105.1
105
ln0259.0
10
15










fn V 
  fnBSfnT VV  22  
     BSV 3294.022358.022.0 
  3294.02 
 39.2 BSV V 
_______________________________________ 
 
10.53 
 (a) 
n poly-to-p-type 0.1 ms V 
   288.0
105.1
10
ln0259.0
10
15









fp V 
 also 
 
2/1
4





 

a
fps
dT
eN
x

 
 
   
  
2/1
1519
14
10106.1
288.01085.87.114











 
 or 
 410863.0 dTx cm 
 Now 
      41519 10863.010106.1max  
SDQ 
 or 
   81038.1max 
SDQ C/cm 2 
 Also 
 
  
8
14
10400
1085.89.3







ox
ox
ox
t
C 
 or 
 
 
 81063.8 oxC F/cm 2 
 We find 
    91019 108105106.1  
ssQ C/cm 2 
 Then 
 
 
fpms
ox
ssSD
T
C
QQ
V  2
max


 
  288.020.1
1063.8
1081038.1
8
98













 
 or 
 357.0TV V 
(b) For NMOS, apply SBV and TV shifts in a 
positive direction, so for 0TV , we want 
357.0 TV V. 
So 
  fpSBfp
ox
as
T V
C
Ne
V  22
2


 
 or 
    
8
151419
1063.8
101085.87.11106.12
357.0





     288.02288.02  SBV 
 or 
  576.0576.0211.0357.0  SBV 
 which yields 
 43.5SBV V 
_______________________________________ 
 
10.54 
 Plot 
_______________________________________ 
 
10.55 
 (a) 
  TGS
oxn
m VV
L
CW
g 

 
  TGS
ox
oxn VV
tL
W




 
 
    
 65.05
10475
1085.89.340010
8
14






 
 or 
 26.1mg mS 
 Now 
 
smm
m
sm
m
m
rgg
g
rg
g
g






1
1
8.0
1
 
 which yields

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