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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 13 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
(b)   0713.0
103
107.4
ln0259.0
16
17










n V 
 0713.087.0  nBnbiV  
 7987.0 V 
 
pObiT VVV  
 or  1103.07987.0  TbipO VVV 
 909.0 V 
 Then 
 
2/1
2





 

d
pOs
eN
V
a 
 
   
  
2/1
1619
14
103106.1
909.01085.81.132











 
 
510095.2  cm 2095.0 m 
_______________________________________ 
 
13.20 
 POnBnPObiT VVVV   
 We want 5.0TV V, so 
 POn V 85.05.0 
 Now 
   






 

d
n
N
17107.4
ln0259.0 
 and 
 
s
d
PO
Nea
V


2
2
 
 
  
  14
2419
1085.81.132
1025.0106.1




 dN
 
 or 
   dPO NV 171031.4  
 Then 
   






 

dN
17107.4
ln0259.085.05.0 
   dN171031.4  
 By trial and error 
 151045.5 dN cm 3 
_______________________________________ 
 
13.21 
 n-channel MESFET - silicon 
 (a) For a gold contact, 82.0Bn V. 
 We find 
   206.0
10
108.2
ln0259.0
16
19







 
n V 
 
 and 
 614.0206.082.0  nBnbiV  V 
 With 0DSV and 35.0GSV V, we find 
 
410075.0  ha 
 
 
2/1
2





 

d
GSbis
eN
VV
a 
 so that 
 410075.0 a 
 
   
  
2/1
1619
14
10106.1
35.0614.01085.87.112











 
 or 
 
41026.0 a cm 26.0 m 
 Now 
 
s
d
PObiT
Nea
VVV


2
614.0
2
 
 or 
 
    
  14
162419
1085.87.112
101026.0106.1
614.0




TV 
 We obtain 
 092.0TV V 
 (b) 
    GSbiPODS VVVsatV  
    GSbiTbi VVVV  
 or 
   092.035.0  TGSDS VVsatV 
 which yields 
   258.0satVDS V 
_______________________________________ 
 
13.22 
(a)   










16
17
102
107.4
ln0259.0n 
 0818.0 V 
 (i) 0818.090.0  nBnbiV  
 818.0 V 
 (ii) 
s
d
pO
Nea
V


2
2
 
 
    
  14
162419
1085.81.132
1021065.0106.1




 
 83.5 V 
 (iii) 83.5818.0  pObiT VVV 
 012.5 V