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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 11 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
11.10 
 
ox
oxn
oxnn
t
Ck



 
 
   
8
14
10110
1085.89.3425




 
 410334.1  A/V 2 1334.0 mA/V 2 
  2
2
TGS
n
D VV
L
Wk
I 










 
 
  2
45.08.0
2.1
20
2
1334.0












 
 1362.0 mA 
 
















 D
DSDS
D
o
I
LL
L
VV
I
r
1
 
   1



 LL
V
LI
DS
D 
    











DS
D
V
L
LLLI
2
1 
 
  DS
D
V
L
LL
LI





2
 
 Now 
   satVV
eN
L DSfpDSfp
a
s 

 
2
 
   2/1
2
12 





DSfp
a
s
DS
V
eNV
L
 
 We find 
 
  
  1619
14
103106.1
1085.87.1122






a
s
eN
 
 
510077.2  cm/V 2/1 
   3758.0
105.1
103
ln0259.0
10
16










fp V 
   DSTGSDSDSDS VVVVsatVV  
 35.2245.08.0  V 
 Now 
 
  6
5
10290.6
35.23758.02
10077.2 







DSV
L
cm/V 
   35.23758.010077.2 5  L 
 35.03758.0  
 
510660.1  cm 166.0 m 
 
 
 
 
 
 (a) 
 
  DS
D
o V
L
LL
LI
r 




2
1
 
 
  
  
 6
24
43
10290.6
10166.02.1
102.1101362.0 





 
 610615.9  
 51004.1  or 104 k 
 (b) 
 
  
  
 6
24
43
10290.6
10166.08.0
108.0101362.01 






or
 
 
510705.1  
 65.5810865.5 4  or k 
_______________________________________ 
 
11.11 
 (a) 
    2
2
TGS
oxn
D VV
L
CW
satI 

 
    28 1109.6500
2
10






 
GSV 
 or 
    2
1173.0  GSD VsatI (mA) 
 and 
    1173.0  GSD VsatI (mA) 2/1 
 (b) 
 Let 
3/1











C
eff
Oeff  
 Where 1000O cm 2 /V-s and 
 4105.2 C V/cm 
 Let 
ox
GS
eff
t
V
 
 We find 
 
  
8
14
109.6
1085.89.3









ox
ox
ox
ox
ox
ox
C
t
t
C 
 or 
 
o
ox At 500 
 Then