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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 13 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
(b)    GSbipODS VVVsatV  
 (i)     0.1818.083.5 satVDS 
 01.4 V 
 (ii)     0.2818.083.5 satVDS 
 01.3 V 
 (iii)     0.3818.083.5 satVDS 
 01.2 V 
_______________________________________ 
 
13.23 
(a) 
aL
W
k sn
n
2



 
 
    
  44
414
105.11025.02
10121085.81.136500




 
 
310206.1  A/V 2 206.1 mA/V 2 
(b)    2
1 TGSnD VVksatI  
 (i)     2
1 15.025.0206.1 satI D 
 01206.0 mA 06.12 A 
 (ii)     2
1 15.045.0206.1 satI D 
 1085.0 mA 
(c)   TGSDS VVsatV  
 (i)   10.015.025.0 satVDS V 
 (ii)   30.015.045.0 satVDS V 
_______________________________________ 
 
13.24 
(a)   2
TGSn
GSGS
D
ms VVk
VV
I
g 





 
  TGSn VVk  2 
  15.045.0225.1  nk 
 083.2 nk mA/V 2 
 
aL
W
k sn
n
2



 
 
   
  44
14
3
105.11025.02
1085.81.136500
10083.2






W
 
 
310073.2 W cm 73.20 m 
(b)    2
1 TGSnD VVksatI  
 (i)     2
1 15.025.0083.2 satI D 
 02083.0 mA 83.20 A 
 (ii)     2
1 15.045.0083.2 satI D 
 1875.0 mA 
_______________________________________ 
 
 
 
13.25 
 Plot 
_______________________________________ 
 
13.26 
 Plot 
_______________________________________ 
 
13.27 
  
  
  










210
1618
105.1
10310
ln0259.0biV 
 8424.0 V 
 
s
d
pO
Nea
V


2
2
 
 
    
  14
162419
1085.87.112
1031050.0106.1




 
 795.5 V 
(a)    GSbipODS VVVsatV  
 953.48424.0795.5  V 
 
  
2/1
2





 

d
DSDSs
eN
satVV
L 
 
   
  
2/1
1619
14
103106.1
953.4101085.87.112











 
 
510666.4 L cm 
 Now 
 
 
90.0
21
1 



L
L
L
L
 
 
   10.02
10666.4
10.02
5



L
L 
 
410333.2 L cm 333.2 m 
(b)    GSbipODS VVVsatV  
  38424.0795.5  
 953.1 V 
 
   
  
2/1
1619
14
103106.1
953.1101085.87.112











L 
 
510892.5  cm 
 Then 
 
   10.02
10892.5
10.02
5



L
L 
 
410946.2  cm 946.2 m 
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