Soluções de Problemas de Dispositivos Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 10 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
10.26 
 (a) We have  x
Q
t
SS  


 
 Then 
 V
C
x x
t
dx
FB
ox ox
oxt
 
 z1
0

 
  
F
HG
I
KJ
F
H
I
K

z1
C
t
t
Q
t
dx
ox
ox
oxox
ox
SS
t t
t
 b g
 
  

   
F
H
I
K
1
C
Q
t
t t t
Q
C
ox
SS
ox ox
SS
ox

a f 
 or 
 V Q
t
FB SS
ox
ox
  

F
HG
I
KJ 
 
   

 
 
16 10 8 10 200 10
3 9 8 85 10
19 10 8
14
.
. .
b gb gb g
b g 
 or 
 V
FB
 0 0742. V 
 
 (b) 
 We have 
 x
Q
t
SS
ox
  


 



16 10 8 10
200 10
19 10
8
.b gb g
 
   

6 4 10
3
. 
O
 
 Now 
 V
C
x x
t
dx
C t
xdx
FB
ox ox
ox
O
ox ox
oxt t
   
 z z1
0 0
 
 
 or 
 V
t
FB
O ox
ox
 


2
2
 
 
  

 
 
6 4 10 200 10
2 3 9 8 85 10
3 8 2
14
.
. .
b gb g
b g 
 or 
 VFB  0 0371. V 
 (c) 
  x
x
t
O
ox
 
F
HG
I
KJ 
 We find 
 
 
 
 
 
 
1
2
2 16 10 8 10
200 10
19 10
8
t Q
ox O SS O
    
 



.b gb g
 
 or 
O
 

1 28 10
2
. 
 Now 
 V
C t
x
x
t
dx
FB
ox ox
O
ox
tox
   
F
HG
I
KJz1 1
0
 
    z1
2
2
0
C t
x dx
ox
O
ox
oxt

af 
 which becomes 
 V
t
t
x t
FB
ox
ox
O
ox
ox
O ox
ox
t
 

   
F
HG
I
KJ
1
3 3
2
3
0
2
 
af 
 Then 
 V
FB

  

 
 
1 28 10 200 10
3 3 9 8 85 10
2 8 2
14
.
. .
b gb g
b g 
 or 0494.0 FBV V 
_______________________________________ 
 
10.27 
 Sketch 
_______________________________________ 
 
10.28 
 Sketch 
_______________________________________ 
 
10.29 
 (b) 
 









2
ln
i
da
tbiFB
n
NN
VVV 
  
  
  









210
1616
105.1
1010
ln0259.0 
 or 
 695.0FBV V 
 (c) Apply 3GV V, 3oxV V 
 
 For 3GV V, 
 
sdx
d


 
 
 n-side: deN 
 1C
xeNeN
dx
d
s
d
s
d 




