Soluções de Problemas em Física de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 7 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 which yields 
 18.7RV V 
(c)  
  
  










210
1717
105.1
104104
ln0259.0biV 
 or 
 886.0biV V 
 We have 
     
  








14
19
25
1085.87.11
106.12
103 Rbi VV
 
 
  






1717
1717
104104
104104
 
 so that 
   456.1 Rbi VV V 
 which yields 
 570.0RV V 
_______________________________________ 
 
7.21 
 (a) 
 
 
 
 
 
2/1
2/1
2
2















 















 

dBa
dBaRbiBs
dAa
dAaRbiAs
NN
NN
e
VV
NN
NN
e
VV
BW
AW
 
 or 
 
 
 
 
 
 
 
2/1























dA
dB
dBa
dAa
RbiB
RbiA
N
N
NN
NN
VV
VV
BW
AW
 
 We find 
  
  
 
7543.0
105.1
1010
ln0259.0
210
1518










biAV V 
 
  
  
 
8139.0
105.1
1010
ln0259.0
210
1618










biBV V 
 We find 
 
 
  




















1618
1518
1010
1010
8139.5
7543.5
BW
AW
 
 
2/1
15
16
10
10












 
 or 
 
 
 
13.3
BW
AW
 
 
 
 
 (b) 
 
 
 
 
 
 
 
 
  RbiB
RbiA
RbiB
RbiA
VV
VV
AW
BW
BW
VV
AW
VV
B
A








2
2
 
 












8139.5
7543.5
13.3
1
 
 or 
 
 
 
316.0


B
A
 
 (c) 
 
 
 
  
  
2/1
2/1
2
2



















dBaRbiB
dBas
dAaRbiA
dAas
j
j
NNVV
NN
NNVV
NN
BC
AC
 
 
2/1





































dAa
dBa
RbiA
RbiB
dB
dA
NN
NN
VV
VV
N
N
 
 
2/1
1518
1618
16
15
1010
1010
7543.5
8139.5
10
10
































 
 or 
 
 
 
319.0


BC
AC
j
j
 
_______________________________________ 
 
7.22 
 (a) We have 
 
 
 
  
  
2/1
2/1
2
2
10
0



















daRbi
das
dabi
das
j
j
NNVV
NN
NNV
NN
C
C
 
 or 
 
 
 
2/1
13.3
10
0







 



bi
Rbi
j
j
V
VV
C
C
 
 For 10RV V, we find 
   1013.3
2
 bibi VV 
 or 
 137.1biV V 
 (b) 
  npp xxWx  2.02.0