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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
8.43 
 (a) p-region: 
 
 ANe
L
A
L
A
L
R
app
p
p


 
 so 
 
    21619 1010480106.1
2.0

pR 
 or 
  26pR 
 n-region: 
 
 ANe
L
A
L
A
L
R
dnn
n
n


 
 so 
 
    21519 10101350106.1
10.0

nR 
 or 
  3.46nR 
 The total resistance is 
 3.4626 np RRR 
 or 
  3.72R 
 (b) 
  3.721.0 IIRV  
 which yields 
 38.1I mA 
_______________________________________ 
 
8.44 
 
 
 
 
 pA
pL
nA
nL
R
pn

 
 
     
5
2
5
2
102
101.0
102
102.0







 
 or 
 150R 
 We can write 
 








S
D
tD
I
I
VRIV ln 
 (a) (i) For 1DI mA, 
      











10
3
3
10
10
ln0259.015010V 
 or 567.0V V 
 (ii) For 10DI mA, 
      











10
2
2
10
10
ln0259.015010V 
 or 98.1V V 
 
 
 (b) Set 0R 
 (i) For 1DI mA, 
   










10
3
10
10
ln0259.0V 
 or 417.0V V 
 
 (ii) For 10DI mA, 
   










10
2
10
10
ln0259.0V 
 or 477.0V V 
_______________________________________ 
 
8.45 
(a) 
32
0259.0

d
t
D
D
t
d
r
V
I
I
V
r 
 or 41009375.8 DI A 
 








s
D
ta
I
I
VV ln 
   












12
4
105
1009375.8
ln0259.0 
 4896.0aV V 
(b) 4103167.4
60
0259.0 
d
t
D
r
V
I A 
   












12
4
105
103167.4
ln0259.0aV 
 4733.0 V 
_______________________________________ 
 
8.46 
 (a) 
















t
a
t
S
a
D
d V
V
V
I
dV
dI
r
exp
11
 
 or 
 







0259.0
020.0
exp
0259.0
101 13
dr
 
 which yields 
  11102.1dr 
 (b) 
 




 


0259.0
02.0
exp
0259.0
101 13
dr
 
 which yields 
  11106.5dr 
_______________________________________