Soluções de Problemas em Física de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
14.13 
 








S
L
tOC
J
J
VV 1ln 
   






 


SJ
31030
1ln0259.0 
 where 
 









p
p
dn
n
a
iS
D
N
D
N
enJ

112 
 which becomes 
   2619 108.1106.1  
SJ 
 












 8198 105
7
10
1
105
2251
aN
 
 or 
   







  15
4
7 10183.1
10708.6
10184.5
a
S
N
J 
 Then 
aN (cm 3 ) SJ (A/cm 2 ) OCV (V) 
1510 
1610 
1710 
1810 
1710477.3  
1810478.3  
1910484.3  
2010539.3  
891.0 
950.0 
01.1 
07.1 
_______________________________________ 
 
14.14 
 (a) 
    33 105021025   AJI LL A 
 We have 
 









p
p
dn
n
a
iS
D
N
D
N
enJ

112 
 or 
   21019 105.1106.1  
SJ 
 












 719616 105
6
10
1
105
18
103
1
 
 which becomes 
 1210289.2 SJ A/cm 2 
 or 
 1210579.4 SI A 
 We have 
 
















 1exp
t
SL
V
V
III 
 or 
 
 
  
















  1exp10579.41050 123
tV
V
I 
 We see that when 0I , 599.0 OCVV V. 
 We find 
 
V (V) I (mA) 
0 
0.1 
0.2 
0.3 
0.4 
0.45 
0.50 
0.55 
0.57 
0.59 
50 
50 
50 
50 
49.98 
49.84 
48.89 
42.36 
33.46 
14.19 
 
 (b) The voltage at the maximum power point 
 is found from 
 
S
L
t
m
t
m
I
I
V
V
V
V














 1exp1 
 
12
3
1058.4
1050
1




 
 
1010092.1  
 By trial and error, 
 520.0mV V 
 At this point, we find 
 6.47mI mA 
 so the maximum power is 
   520.06.47 mmm VIP 
 or 
 8.24mP mW 
 (c) We have 
 
3106.47
520.0


m
m
I
V
I
V
RIRV 
 or 
  9.10R 
_______________________________________ 
 
14.15 
(a)   












9
3
102
10180
1ln0259.0ocV 
 474.0 V 
(b) 
S
L
t
m
t
m
I
I
V
V
V
V
















 1exp1 
 
9
3
102
10180
1





7109