Soluções de Problemas de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 7 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
     44
2 10210726.70  d 
 or 
 45.152  V 
 By symmetry, the potential difference across 
 the p-region space-charge region is also 
 3.863 V. The total reverse-bias voltage is 
 then 
   2.2345.15863.32 RV V 
_______________________________________ 
 
7.35 
(a) 
B
crits
B
eN
V
2
2
 
 or 
   
  40106.12
1041085.87.11
2 19
25142







B
crits
B
eV
N 
 Then 1610294.1  aB NN cm 3 
(b) 
   
  20106.12
1041085.87.11
19
2514




BN 
 Or 161059.2  aB NN cm 3 
_______________________________________ 
 
7.36 
 
   
  80106.12
1041085.87.11
2 19
25142







B
crits
a
eV
N 
 
151047.6  cm 3 
_______________________________________ 
 
7.37 
(a) For 1610dN cm 3 , from Figure 7.15, 
 75BV V 
(b) For 1510dN cm 3 , 
 450BV V 
_______________________________________ 
 
7.38 
(a) From Equation (7.36), 
 
 
2/1
max
2



















da
da
s
Rbi
NN
NNVVe
 
 Set critmax and BR VV  
  
  
  










210
1616
105.1
102102
ln0259.0biV 
 7305.0 V 
 
 
 Then 
 
  
  








14
19
5
1085.87.11
106.12
104 Rbi VV
 
 
  
2/1
1616
1616
102102
102102












 
 77.51 Bbi VV V 
 So 04.51BV V 
(b) 
  
  
  










210
1515
105.1
105105
ln0259.0biV 
 6587.0 V 
 Then 
 
  
  








14
19
5
1085.87.11
106.12
104 Rbi VV
 
 
  
2/1
1515
1515
105105
105105












 
 1.207 Rbi VV 
 So 206RV V 
_______________________________________ 
 
7.39 
 For a silicon np  junction with 
 15105dN cm 3 and 100BV V, then, 
 neglecting biV we have 
 
2/1
2





 

d
Bs
n
eN
V
x 
 
   
  
2/1
1519
14
105106.1
1001085.87.112











 
 or 
   41009.5min nx cm 09.5 m 
_______________________________________ 
 
7.40 
 We find 
  
  
 
933.0
105.1
1010
ln0259.0
210
1818










biV V 
 Now 
 
s
nd xeN

max 
 so