ECE320 Home Work # 8 – SOLUTIONS
PROBLEM 1
Explain in your own words what Power Electronics is. Give one or two power electronics
product examples that you can think of and explain why they are power electronics and
what’s inside to your knowledge.
SOLUTION:
Power electronics is power conversion circuits and their control, from one available form of
power/energy source to a desired from of power.


Examples are:
Computer power supply; switch-mode power supply (inside they have ac-dc rectifier and dcdc converter)
Motor drive controller: Inverter inside to produce variable ac (both in frequency and in
voltage amplitude) to control motor speed and torque
Vdc
PROBLEM 2
The figure below shows an inverter circuit
using four switches to converter a dc source
into ac. Draw a control sequence for each of
the four switches by using “0 (for off)” and
“1 for on”, respectively, to complete the
waveform figure.
S1
S3
0
Vac
Vdc
0
Vdc
S1
Load vac
vdc
S2
S2
S4
S3
S4
SOLUTION:
Two solutions.
PROBLEM 3
S
In the given circuit, assume the switch is
used to dim the light and turned on and off
in a sequence as shown. a). Derive the
power loss equation expressed in dc
voltage Vdc, on-current IL
= Vdc / RL, on-drop voltage Von, switching
frequency fsw, turn-on time ton, and turn-off
time toff (assuming that the current
changes linearly during the switching and
ton and toff are very small compared with
the switchingcycle) . b). calculate power
loss using the derived equation given that
a 50% duty cycle, Vdc= 100 V, on-current
IL = 10 A, on-drop voltage Von= 1 V,
switching frequency fsw= 1 kHz, turn-on
time ton= 1 s, and turn-off time toff = 1 s.
c) Calculate the efficiency for the power
conversion described in b).
Vdc
is
vs
IL
Vdc
ton
Conduction power loss:
TON
TSW
When the switch is in off state, current is zero, thus no off-state loss.
Turn –ON power loss PT-ON :
Assume VON  Vdc
PT ON
1

TSW
=
tON

0
1
(vs is ) dt 
TSW
VON I L tON
……. (1)
2
TSW
Von
toff
1/fsw
Derive the power loss equation
PON  (VON I L )
Light
Bulb
RL
vS
SOLUTION:
a)
iS
tON

0
VON
t
tON
I L dt
Turn-off power loss PT-OFF :
1
PT-off =
Tsw
toff
 (v
s
. is ) dt
0
1
=
Tsw
=
toff
 (v
dc
)(
0
t
toff
) I L dt
Vdc I L toff
2
TSW
 Total power loss Ploss  PON  PT ON  PT off
=
VON I L tON
V I t
TON
 dc L off

2
TSW
2 TSW
TSW
VON I L
calculate power loss using the derived equation
b)
Vdc  100V
I L  10 A ,
TSW 
1
f SW 
1 ms ,
TON  50% TSW  .5n ,
tON  1  s , and
toff  1  s
Ploss  PON  PT ON  PT off = VON I L
=
=
c)
(1)(10)(
VON I L tON
V I t
TON
 dc L off

2
TSW
2 TSW
TSW
6
0.5
(100)(10)  1106 
 (1)(10)   110 
)  



3 
3 
1
2
 2   110 
 110 
5  (5 103 )  (0.5)  5.505 W
Calculate the efficiency for the power conversion
 
P P
POUT
V I (50%)  PLoss
 IN  Loss  dc L
PIN
PIN
Vdc I L (50%)
=
(100)(10)(0.5)  5.505
(100)(10)(0.5)
= 98.89%