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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
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(a) 0SGV , 0DI 
 6.0SGV V,   25.0satVSD V 
     2
35.06.0961.0 satI D 
 060.0 mA 
 2.1SGV V,   85.0satVSD V 
     2
35.02.1961.0 satI D 
 694.0 mA 
 8.1SGV V,   45.1satVSD V 
     2
35.08.1961.0 satI D 
 02.2 mA 
 4.2SGV V,   05.2satVSD V 
     2
35.04.2961.0 satI D 
 04.4 mA 
(c) 0DI for 35.0SGV V 
 6.0SGV V 
        2
1.01.035.06.02961.0 DI 
 0384.0 mA 
 2.1SGV V 
        2
1.01.035.02.12961.0 DI 
 154.0 mA 
 8.1SGV V 
        2
1.01.035.08.12961.0 DI 
 269.0 mA 
 4.2SGV V 
        2
1.01.035.04.22961.0 DI 
 384.0 mA 
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10.39 
(a) From Problem 10.37, 111.1nK mA/V 2 
 For 8.0GSV V, 0DI 
 0GSV ,   8.0satVDS V 
     2
8.00111.1 satI D 
 711.0 mA 
 8.0GSV V,   6.1satVDS V 
     2
8.08.0111.1 satI D 
 84.2 mA 
 6.1GSV V,   4.2satVDS V 
     2
8.06.1111.1 satI D 
 40.6 mA 
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10.40 
 Sketch 
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10.41 
 Sketch 
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10.42 
 We have 
   TDSTGSDS VVVVsatV  
 so that 
   TDSDS VsatVV  
 Since  satVV DSDS  , the transistor is always 
 biased in the saturation region. Then 
  2
TGSnD VVKI  
 where, from Problem 10.37, 
 111.1nK mA/V 2 and 45.0TV V 
 Then 
 GSDS VV  DI (mA) 
 0 
 1 
 2 
 3 
 4 
 5 
 0 
 0.336 
 2.67 
 7.22 
 14.0 
 23.0 
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10.43 
 From Problem 10.38, 961.0pK mA/V 2 
    22 SDSDTSGpD VVVVKI  
  TSGp
VSD
D
d VVK
V
I
g
SD





2
0
 
 For 35.0SGV V, 0dg 
 For 35.0SGV V, 
   35.0961.02  SGd Vg 
 For 4.2SGV V, 
   35.04.2961.02 dg 
 94.3 mA/V 
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