1 (134)

Prévia do material em texto

Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 
10.44 
(a) 
GS
D
m
V
I
g


 
     22 DSDSTGSn
GS
VVVVK
V



 
  DSn VK 2 
   05.0225.1 nK 
 5.12 nK mA/V 2 
(b)        2
05.005.03.08.025.12 DI 
 594.0 mA 
(c)   2
3.08.05.12 DI 
 125.3 mA 
_______________________________________ 
 
10.45 
 We find that 2.0TV V 
 Now 
    TGS
oxn
D VV
L
CW
satI 
2

 
 where 
 
  
8
14
10425
1085.89.3







ox
ox
ox
t
C 
 or 
 81012.8 oxC F/cm 2 
 We are given 10LW . From the graph, for 
 3GSV V, we have 
   033.0satI D , 
 then 
  2.03
2
033.0 
L
CW oxn 
 or 
 
310139.0
2

L
CW oxn
 
 or 
     38 10139.01012.810
2
1  n 
 which yields 
 342n cm 2 /V-s 
_______________________________________ 
 
10.46 
 (a) 
   TGSDS VVsatV  
 or 
 8.48.04  GSGS VV V 
 
 
 (b) 
      satVKVVKsatI DSnTGSnD
22
 
 so 
  24 4102 nK  
 which yields 
 5.12nK A/V 2 
 (c) 
   2.18.02  TGSDS VVsatV V 
 so  satVV DSDS  
     25 8.021025.1  satI D 
 or 
   18satI D A 
 (d) 
  satVV DSDS  
   22 DSDSTGSnD VVVVKI  
        25 118.0321025.1   
 or 
 5.42DI A 
_______________________________________ 
 
10.47 
(a) 
  
8
14
10180
1085.89.3




oxC 
 
7109175.1  F/cm 2 
 (i)   7109175.1450  oxnn Ck  
 
510629.8  A/V 2 
 or 29.86nk A/V 2 
 (ii)    2
2
TGS
n
D VV
L
Wk
satI 










 
 
  2
4.02
2
08629.0
8.0 












L
W
 
 24.7
L
W
 
(b) (i)   7109175.1210  oxpp Ck  
 
510027.4  A/V 2 
 or 27.40pk A/V 2 
 (ii)    2
2
TSG
p
D VV
L
Wk
satI 












 
 
  2
4.02
2
04027.0
8.0 












L
W
 
 5.15
L
W