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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 11 
By D. A. Neamen Problem Solutions 
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  16875.0
115.080.0
80.0







 
 19708.0 mA 
 (ii)  510077.2 L 
  6.03758.043758.0  
 510293.2  cm 2293.0 m 
 DD I
LL
L
I 






 
  16875.0
2293.080.0
80.0







 
 23655.0 mA 
(b) 
 
1
3
1
24
1019708.023655.0






















DS
D
o
V
I
r
 41007.5  or 7.50 k 
(c)   6.14.00.2  TGSDS VVsatV V 
(i)  510077.2 L 
  6.13758.023758.0  
 
610819.2  cm 02819.0 m 
 DD I
LL
L
I 






 
  16875.0
02819.080.0
80.0







 
 17491.0 mA 
 (ii)  510077.2 L 
  6.03758.043758.0  
 
510425.1  cm 1425.0 m 
 DD I
LL
L
I 






 
  16875.0
1425.080.0
80.0







 
 20532.0 mA 
 
1











DS
D
o
V
I
r 
 
 
1
3
24
1017491.020532.0










 
 410577.6 or 77.65 k 
_______________________________________ 
 
 
 
 
11.7 
 (a) 
 (i)  2
2
TGS
n
D VV
L
Wk
I 

 
   2
35.08.010
2
075.0






 
 07594.0 mA 94.75 A 
 (ii)  DSDD VII  1 
      5.102.019375.75  
 22.78 A 
 (iii) 
  94.7502.0
11

D
o
I
r

 
 658.0 M 658 k 
 (b) 
 (i)   2
35.025.110
2
075.0






DI 
 30375.0 mA 
 (ii)      5.102.0130375.0 
DI 
 3129.0 mA 
 (iii) 
  
165
30375.002.0
1
or k 
_______________________________________ 
 
11.8 
 Plot 
_______________________________________ 
 
11.9 
 (a) Assume   1satVDS V. Then 
 
 
L
satVDS
sat  
 We find 
 L (  m) 
sat (V/cm) 
 3 
 1 
 5.0 
 25.0 
 13.0 
31033.3  
4101 
4102 
4104 
41069.7  
 (b) 
 Assume 500n cm 2 /V-s, we have 
 satn  
 Then 
 For 3L m, 
61067.1  cm/s 
 For 1L m, 
6105 cm/s 
 For 5.0L m, 
710 cm/s 
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