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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 15 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
(c)  CEQCT VIP 





 max,
2
1
 
 CEQV






2
2
30 
 30 CEQV V 
  30
1
30
CQ
CEQ
L
I
V
R 
   603022max,  CEQCE VV V 
(d) Same as part (b) 
_______________________________________ 
 
15.13 
(a) CQCEQT IVP  CQ
CC I
V







2
 
 667.1610  CQCQ II A 
  60.3
667.1
6
CQ
CEQ
L
I
V
R 
(b)   333.3667.122max,  CQC II A 
_______________________________________ 
 
15.14 
 If 25CCV V, then 
   25.0
100
25
max 
L
CC
C
R
V
I A ratedCI , 
 The power 
  LCCCCCEC RIVIVIP  
 Now, to find the maximum power point 
   10022520 CLCCC
C
IRIV
dI
dP
 
 which yields 
 125.0CI A 
 So 
        100125.025125.0max P 
 or 
   56.1max P W TP 
 So maximum CCV is 25CCV V 
_______________________________________ 
 
15.15 
 Now 
D
DS
on
I
V
R  
 Power dissipated in the transistor 
 
on
DS
DSD
R
V
VIP
2
 
 
 
 We have 
 
100
200 DS
D
V
I

 
 so we can write 
 
on
DS
DS
DS
R
V
V
V
P
2
100
200





 
 
 For  25T C,  2onR . 
 Then 
 
2100
200 2
DS
DS
DS V
V
V





 
 
 which yields 
 92.3DSV V 
 The power is 
   69.792.3
100
92.3200





 
P W 
 We then have 
T (C) onR ( ) DSV (V) P (W) 
25 
50 
75 
100 
2.0 
2.33 
2.67 
3.0 
3.92 
4.56 
5.19 
5.83 
7.69 
8.91 
10.1 
11.3 
_______________________________________ 
 
15.16 
 (a) We have, for three devices in parallel, 
   551.15
2.228.1
 V
VVV
 
 or 
 311.3V V 
 Then, 
R
V
I  , so that 
 839.11 I A 
 656.12 I A 
 505.13 I A 
 Now, IVP  , so 
 09.61 P W 
 48.52 P W 
 98.43 P W 
 (b) Now 
 882.35
2.2
1
6.3
1
8.1
1






 VV V 
 Then 
 157.21 I A, 37.81 P W 
 078.12 I A, 19.42 P W 
 765.13 I A, 85.63 P W 
_______________________________________