Part of Lecture series on EE321N, Power Electronics-I delivered by me during Fifth Semester of B.Tech. Electrical Engg., 2012 Z H College of Engg. & Technology, Aligarh Muslim University, Aligarh Please comment and feel free to ask anything related. Thanks!

1. Solution 1
The basic circuit equation is
Subsituting the given values
Peak Power loss = Vg Ig = 5
Solving above quadratic eqn for Ig and taking the
+ve value we get Ig = 0.7 A. Using this value in (1)
Rs = 7 Ω
s s g gE R I V 
 12 1.5 8 8 1.5 ...(1)s g g s gR I I R I     
  2
5 1.5 8 8 1.5g g g gI I I I    
12-Oct-12 1EE-321N, Lec-10

2. Solution 1 Contd...
Average Power Dissipation
Pgav = δ·Pgm = 0.2 × 5 = 1 W
12-Oct-12 2EE-321N, Lec-10

3. Solution 2
Given Ig = 500 mA = 0.5 A, Vg/Ig = 16 V/A
So, Vg = 16 × 0.5 = 8 V
Again using the basic circuit equation
Rs is obtained as
s s g gE R I V 
15 8
14
0.5
s g
s
g
E V
R
I
 
   
12-Oct-12 3EE-321N, Lec-10

4. Solution 2 ... contd
Power dissipation Pg = Vg Ig = 8 × 0.5 = 4 W
Also Pgav = δ·Pgm = Pgm = f ·TON ·Pgm
Thus,
ONT
T
6
0.3
18.75 kHz 19 kHz
4 10 4
f 
  
 
12-Oct-12 4EE-321N, Lec-10