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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 9 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 or 
 
 
 
310856.4
300
400

ST
ST
J
J
 
 Then 
    834 10410856.4107  I 
 






03453.0
4447.0
exp 
 or 
 3.53I mA 
_______________________________________ 
 
9.24 
 Plot 
_______________________________________ 
 
9.25 
(a) 


1.0
10
10
3
4
A
R
R c 
(b) 


1
10
10
4
4
A
R
R c 
(c) 


10
10
10
5
4
A
R
R c 
_______________________________________ 
 
9.26 
(a) 




5
10
105
5
5
A
R
R c 
 (i)    551  IRV mV 
 (ii)    5.051.0  IRV mV 
(b) 50
10
105
6
5





R  
 (i)    50501  IRV mV 
 (ii)    5501.0  IRV mV 
_______________________________________ 
 
9.27 
 
2
exp
TA
V
V
R
t
Bn
t
c 










 
 or 










t
c
tBn
V
TAR
V
2
ln 
(a)  
   







 


0259.0
300120105
ln0259.0
25
Bn 
 258.0 V 
 
 
 
(b)  
   







 


0259.0
300120105
ln0259.0
26
Bn 
 198.0 V 
_______________________________________ 
 
9.28 
 (b) We need 20.00.42.4   mn V 
 And 
 









d
c
tn
N
N
V ln 
 or 
   






 

dN
19108.2
ln0259.020.0 
 which yields 
 161024.1 dN cm 3 
 (c) 
 Barrier height = 0.20 V 
_______________________________________ 
 
9.29 
 We have that 
  xx
eN
n
s
d 


 
 Then 
 2
2
2
C
x
xx
eN
dx n
s
d 









  
 Let 0 at 00 2  Cx , so 
 










2
2x
xx
eN
n
s
d 
 At nxx  , biV , so 
 
2
2
n
s
d
bi
xeN
V 

 
 or 
 
d
bis
n
eN
V
x


2
 
 Also 
 nBObiV   
 where 
 








d
c
tn
N
N
V ln 
 Now for 
 35.0
2
70.0
2
 BO
 V 
 we have