Soluções de Problemas de Dispositivos Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 (b) For 80.0BOx m 
 For 5CBV V, 
 6125.01875.080.0 Bx m 
 Then 
 2.214
106125.0
10312.1
4
2






J A/cm 2 
 For 10CBV V, 
 5431.02569.080.0 Bx m 
 and 
 6.241
105431.0
10312.1
4
2






J A/cm 2 
 Now 
 
510
2.2146.241








CBCE V
J
V
J
 
 48.5 A/cm 2 /V 
 We can write 
  ACE
CE
VV
V
J
J 


 
 or 
   AV 7.548.52.214 
 which yields 
 4.33AV V 
 (c) For 60.0BOx m 
 For 5CBV V, 
 4125.01875.060.0 Bx m 
 Then 
 1.318
104125.0
10312.1
4
2






J A/cm 2 
 For 10CBV V, 
 3431.02569.060.0 Bx m 
 and 
 4.382
103431.0
10312.1
4
2






J A/cm 2 
 Now 
 
510
1.3184.382








CBCE V
J
V
J
 
 86.12 A/cm 2 /V 
 We can write 
  ACE
CE
VV
V
J
J 


 
 or 
   AV 7.586.121.318 
 which yields 
 0.19AV V 
_______________________________________ 
 
12.39 
 (a) 
 
 
 
2/1
12


















CBB
CBCbis
dB
NNN
N
e
VV
x 
 
   









19
14
106.1
1085.87.112 BCbi VV
 
 
 
2/1
161516
15
1010
1
10
10











 
     2/110101766.1 BCbi VV   
 Now 
 









2
ln
i
CB
tbi
n
NN
VV 
  
  
  









210
1615
105.1
1010
ln0259.0 
 6350.0 V 
 For 1BCV V, 1387.0dBx m 
 For 5BCV V, 2575.0dBx m 
 Then 1188.01387.02575.0  dBx m 
 
(b) 








t
EB
B
BEBB
C
V
V
x
ApeD
I exp0 
 We find 
 
 
16
2102
0
10
105.1 

B
i
B
N
n
p 
 
41025.2  cm 3 
 Then 
 
    
B
C
x
I
4419 101025.210106.1  
 
 






0259.0
625.0
exp 
 
Bx
7100874.1 
 A 
 For 1BCV V, 
  4
7
101387.070.0
100874.1




CI 
 
310937.1  A 937.1 mA 
 For 5BCV V, 
  4
7
102575.070.0
100874.1




CI 
 
310456.2  A 456.2 mA 
 Then 
 519.0937.1456.2  CI mA