Soluções de Problemas de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
6.40 
 We have that 
 
 
   ppCnnC
nnpNCC
R
pn
itpn



2
 
 
 
   inOipO
i
npnn
nnp




2
 
 If nnn o  and npp o  , then 
 
  
   ionOiopO
ioo
nnpnnn
nnpnn
R




 2
 
 
   
   ionOiopO
ioooo
nnpnnn
nnpnnpn




 22
 
 If inn  , we can neglect  2
n : also 
 2
ioo npn  
 Then 
 
 
   ionOiopO
oo
npnn
pnn
R





 
(a) For n-type; Oo pn  , io nn  
Then 
 710
1 
pOn
R

s 1 
(b) For intrinsic, ioo npn  
Then 
 
   inOipO
i
nn
n
n
R
22
2
 
 
 or 
 




 77 10510
11
nOpOn
R

 
 
61067.1 
n
R

s 1 
 
(c) For p-type; oo np  , io np  
Then 
 
6
7
102
105
11 




nOn
R

s 1 
_______________________________________ 
 
6.41 
(a) From Equation (6.56) 
 
 
0
2
2

pO
p
p
g
dx
pd
D


 
 Solution is of the form 
 







 








 

pp
pO
L
x
B
L
x
Agp expexp 
 
 At x , 
pOgp   so that 0B , 
 Then 
 







 

p
pO
L
x
Agp exp 
 We have 
 
 
 
00 

xx
p ps
dx
pd
D 

 
 We can write 
 
 
px L
A
dx
pd 

0

 and   Agp pO
x



0
 
 Then 
  Ags
L
AD
pO
p
p


 
 Solving for A , we find 
 
s
L
D
gs
A
p
p
pO




 
 The excess concentration is then 
 
  














 



ppp
pO
L
x
sLD
s
gp exp1 
 where 
    37 101010   pOpp DL  cm 
 Now 
   721 1010 p 
 
  














 




pL
x
s
s
exp
1010
1
3
 
 
 or 
 















 



pL
x
s
s
p exp
10
110
4
14 
 (i) For 0s , 
 1410p cm 3 
 (ii) For 2000s cm/s, 
 















 

pL
x
p exp167.011014 
 (iii) For s , 
 















 

pL
x
p exp11014