Soluções de Problemas de Física de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 13 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 so 
     1619 1074500106.156.88  L 
   44 105103.0   
 or 
 41067.0 L cm 67.0 m 
_______________________________________ 
 
13.35 
 Considering the capacitance charging time, 
 we have 
 
G
m
T
C
g
f
2
 
 where 
 
a
WL
C s
G

 
 
    
4
4414
103.0
105.11051085.81.13




 
 or 
 15109.2 GC F 
 We must use mg  , so we obtain 
 
  
 15
3
109.22
80.01082.2






Tf
1110238.1  Hz 
 We can also write 
 
T
C
C
T
f
f


 2
1
2
1
 
 so 
 
 
12
11
10285.1
10238.12
1 



 C s 
 The channel transit time is 
 11
7
4
105.1
10
105.1 



tt s 
 The total time constant is 
 
1211 10285.1105.1   
 
1110629.1  s 
 Taking into account the channel transit time 
 and the capacitance charging time, we find 
 
 1110629.12
1
2
1



Tf 
 or 
 91077.9 Tf Hz 77.9 GHz 
_______________________________________ 
 
 
 
 
 
 
 
13.36 
(a) For constant mobility 
 
2
2
2 L
aNe
f
s
dn
T




 
 
    
   2414
241619
102.11085.81.132
1030.01047500106.1






 
 111012.4 Tf Hz 412 GHz 
 
(b) For saturation velocity model 
 
 4
7
102.12
10
2 



L
f sat
T 
 101033.1 Tf Hz 3.13 GHz 
_______________________________________ 
 
13.37 
 
2
2
2 L
aNe
f
s
dn
T




 
 
    
   214
241619
1085.87.112
1040.01021000106.1
L





 
 
2
975.786
L
fT  
(a) 
 24103
975.786

Tf 
 
91074.8  Hz 74.8 GHz 
(b) 
 24105.1
975.786

Tf 
 
101050.3  Hz 0.35 GHz 
_______________________________________ 
 
13.38 
 
2
2
2 L
aNe
f
s
ap
T




 
 or 
 
2/1
2
2 









Ts
ap
f
aNe
L


 
 
    
  
2/1
14
241619
1085.87.112
1040.0102420106.1













Tf
 
 
Tf
L
18.18
