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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 
 











 1
8.0
1
26.1
1
1
8.0
11
m
s
g
r 
 or 
 198.0sr k 
 (b) For 3GSV V, 683.0mg mS 
 Then 
 
  
602.0
198.0683.01
683.0


mg mS 
 or 
 88.0
683.0
602.0


m
m
g
g
 
 which is a 12% reduction. 
_______________________________________ 
 
10.56 
 (a) The ideal cutoff frequency for no overlap 
 capacitance is, 
 
 
222 L
VV
C
g
f TGSn
gs
m
T




 
 
  
 241022
75.04400




 
 or 
 17.5Tf GHz 
 (b) Now 
 
 MgsT
m
T
CC
g
f


2
 
 where 
  LmgdTM RgCC  1 
 We find 
   44 10201075.0   oxgdT CC 
 
  
8
14
10500
1085.89.3




 
   44 10201075.0   
 or 
 
1410035.1 gdTC F 
 Also 
  TGS
oxn
m VV
L
CW
g 

 
 
    
  84
144
10500102
1085.89.34001020




 
  75.04 
 or 
 
3108974.0 mg S 
 
 
 
 
 Then 
  1410035.1 MC 
    33 1010108974.01   
 or 
 1310032.1 MC F 
 Now 
   WLCC oxgsT
41075.0  
 
  
8
14
10500
1085.89.3




 
   444 10201075.0102   
 or 
 
1410797.3 gsTC F 
 We now find 
 
 MgsT
m
T
CC
g
f


2
 
 
 1314
3
10032.110797.32
108974.0






 
 or 
 01.1Tf GHz 
_______________________________________ 
 
10.57 
 (a) For the ideal case 
 
 4
6
1022
104
2 




L
f ds
T 
 or 
 18.3Tf GHz 
(b) With overlap capacitance (using the 
values from Problem 10.56), 
 
 MgdT
m
T
CC
g
f


2
 
 We find 
 dsoxm WCg  
 
    
8
6144
10500
1041085.89.31020




 
 or 
 
3105522.0 mg S 
 We have 
  LmgdTM RgCC  1 
  1410035.1  
    33 1010105522.01   
 or 
 
1410750.6 MC F