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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
8.47 
 (a) If 2.0
F
R
I
I
 
 Then we have 
 
2.01
1
1
1






F
RRF
F
pO
s
I
III
It
erf

 
 or 
 833.0
pO
sterf

 
 We find 
 956.0978.0 
pO
s
pO
s tt

 
 (b) If 0.1
F
R
I
I
, then 
 50.0
11
1



pO
sterf

 
 which yields 
 228.0
pO
st

 
_______________________________________ 
 
8.48 
(a) erf
RF
F
p
s
II
It



 
 erf 3.0 = erf  5477.0 
  erf   56332.055.0  
 Then 
F
R
I
I


1
1
56332.0 
 775.01
56332.0
1

F
R
I
I
 
(b) erf   























 

F
R
p
p
p I
I
t
t
t
1.01
exp
0
2
0
2
0
2




 
   775.01.01 
 0775.1 
 By trial and error, 80.0
0
2 
p
t

 
_______________________________________ 
 
 
8.49 
 18jC pF at 0RV 
 2.4jC pF at 10RV V 
 We have 
 
710  pOnO  s , 2FI mA 
 and 
 1
10
10

R
V
I R
R mA 
 So 
   













 
1
2
1ln101ln 7
R
F
pOs
I
I
t  
 or 
 7101.1 st s 
 Also 
 1.11
2
2.418


avgC pF 
 The time constant is 
   124 101.1110  avgS RC 
 
71011.1  s 
 Now, the turn-off time is 
   71011.11.1  Ssoff tt  
 or 
 
71021.2 offt s 
_______________________________________ 
 
8.50 
  
 
 
136.1
105.1
105
ln0259.0
210
219











biV V 
 We find 
 
 
2/1
2















 

da
daabis
NN
NN
e
VV
W 
 
   








19
14
106.1
40.0136.11085.87.112
 
 
 
2/1
219
1919
105
105105














 
 which yields 
 
71017.6 W cm
o
A7.61 
_______________________________________ 
 
8.51 
 Sketch 
_______________________________________