Soluções de Problemas de Semicondutores

Prévia do material em texto

Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 Then 
    1174 101010   ss AJI mA 
 Case 4: 
















0259.0
72.0
exp
20.1
exp
t
a
s
V
V
I
I 
 1210014.1 sI mA 
 
8
12
102
10014.1





s
s
J
I
A 
 51007.5  cm 2 
_______________________________________ 
 
8.11 
(a) 
p
nop
n
pon
n
pon
pn
n
L
peD
L
neD
L
neD
JJ
J



 
 
d
i
po
p
a
i
no
n
a
i
no
n
N
nD
N
nD
N
nD
22
2





 
 










d
a
pon
nop
N
N
D
D


1
1
90.0 
 1
90.0
1









d
a
pon
nop
N
N
D
D


 
 





 1
90.0
1
nop
pon
d
a
D
D
N
N


 
 
  
  
 1111.0
10510
1025
7
7



 
 07857.0
d
a
N
N
or 73.12
a
d
N
N
 
(b) From part (a), 
 





 1
20.0
1
nop
pon
d
a
D
D
N
N


 
 
  
  
 4
10510
1025
7
7



 
 828.2
d
a
N
N
or 354.0
a
d
N
N
 
_______________________________________ 
 
 
8.12 
 The cross-sectional area is 
 4
3
105
20
1010 




J
I
A cm 2 
 We have 
 














0259.0
65.0
exp20exp S
t
D
S J
V
V
JJ 
 which yields 
 1010522.2 SJ A/cm 2 
 We can write 
 









pO
p
dnO
n
a
iS
D
N
D
N
enJ

112 
 We want 
 10.0
11
1



pO
p
dnO
n
a
nO
n
a
D
N
D
N
D
N


 
 or 
 
77
7
105
101
105
251
105
251







da
a
NN
N
 
 =
 
10.0
10472.410071.7
10071.7
33
3



d
a
N
N
 
 which yields 
 23.14
d
a
N
N
 
 Now 
   2101910 105.1106.110522.2  
SJ 
 
  











 77 105
101
105
25
23.14
1
dd NN
 
 We find 
 141009.7 dN cm 3 
 and 
 161001.1 aN cm 3 
_______________________________________ 
 
8.13 
 Plot 
_______________________________________