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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 13 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 We have 
 056.193.1874.0  PObiP VVV V 
 Then 
 (i) For 0GSV ,   06.1satVDS V 
 (ii) For 264.0
4
1
 PGS VV V, 
   792.0satVDS V 
 (iii) For 528.0
2
1
 PGS VV V, 
   528.0satVDS V 
 (iv) For 792.0
4
3
 PP VV V, 
   264.0satVDS V 
 (c) 
  











 

PO
GSbi
PD
V
VV
IsatI 3111 
 











 

PO
GSbi
V
VV
3
2
1 
  







 

93.1
874.0
3103.1 GSV
 
 











 

93.1
874.0
3
2
1 GSV
 
 (i) For 0GSV ,   258.01 satI D mA 
 (ii) For 264.0GSV V, 
   141.01 satI D mA 
 (iii) For 528.0GSV V, 
   0608.01 satI D mA 
 (iv) For 792.0GSV V, 
   0148.01 satI D mA 
_______________________________________ 
 
13.12 
 















 

2/1
1 1
PO
GSbi
Od
V
VV
Gg 
 where 
 
 
93.1
1003.133 3
1
1


PO
P
O
V
I
G 
 or 
 
3
1 1060.1 OG S 60.1 mS 
 Then 
 
 
 
 
GSV   POGSbi VVV / dg (mS) 
0 
-0.264 
-0.528 
-0.792 
-1.056 
0.453 
0.590 
0.726 
0.863 
1.0 
0.523 
0.371 
0.237 
0.114 
0 
_______________________________________ 
 
13.13 
 n-channel JFET - GaAs 
 (a) 
 
L
WaNe
G dn
O

1 
 
   
4
1619
1010
1028000106.1




 
   44 1035.01030   
 or 
 3
1 1069.2 OG S 
 (b) 
    GSbiPODS VVVsatV  
 We have 
 
s
d
PO
Nea
V


2
2
 
 
    
  14
162419
1085.81.132
1021035.0106.1




 
 or 
 69.1POV V 
 We find 
  
  
  










26
1618
108.1
102105
ln0259.0biV 
 or 
 34.1biV V 
 Then 
 35.069.134.1  PObiP VVV V 
 We then obtain 
     GSGSDS VVsatV  35.034.169.1 
 For 0GSV ,   35.0satVDS V 
 For 175.0
2
1
 PGS VV V, 
   175.0satVDS V 
 (c) 
  











 

PO
GSbi
PD
V
VV
IsatI 3111 
 











 

PO
GSbi
V
VV
3
2
1