Power Electronics
School of Electrical Engineering & Telecommunications
University of New South Wales
ELEC4614 - Power Electronics
Solution of Tutorial 1 - Basic Concepts in Power Electronics
Question 1.
(a)
Vd VT 300 I T Rds
200 A This neglects the on-state voltage drop of the
R
1.5 switch. This is only acceptable when the on-state voltage is small compared to the DC supply voltage Vd (= 300V in this case).
t
During turn-on time, vT (t ) Vd 1 ; again neglecting the on-state voltage.
t ri
IT
iT (t ) I d (
t t ri
)
t ri
1
Won vT (t )iT (t )dt Vd I d t ri
6
0
t
During turn-off time, vT (t ) Vd
t fi
t iT (t ) I d 1
t fi
t fi
1
Woff vT (t )iT (t )dt Vd I d t fi
6
0
Wtotal Won Woff
1
1
Vd I d ( tri t fi ) 300 200 100 200 10 9 3 mJ
6
6
(b)
Ptotal Wtotal f s 300W
(c)
After turn-on, on-state voltage drop Von I d Rds 200 0.02 4 V
Solution of Tutorial 1
1
April, 2015
ELEC4614
Power Electronics
ton
Won Von I d dt Von I d ton 4 200 5 6 4 mJ
0
Pon Won f s 4 10 3 100,000 400W
(d)
Ploss( total ) Ptotal Pon 300 400 700W
Question 2
(a) Assume that the load current at turn off is Vd/R = 200A.
vds Vd L
did
I
300 /1.5
Rid Vd L 0 300 0.010
10,000,300V 10 MV dt t ri
200 109
(b) No practical semiconductor switch can withstand 10MV voltage across it’s ds terminals.
The device will most likely be destroyed. This overvoltage is due to the inductance in the circuit. Even for a purely resistive load, there is always some inductance in the connecting wires (the so-called parasitic inductance). So, some form of voltage clamping is always necessary.
Solution of Tutorial 1
2
April, 2015
ELEC4614
Power Electronics
Question 3.
Circuit A
Vds ,max Vd I o R f 24 20 50 1024 V id Io
Highly
Inductive
Load
if
Rf = 50
Vd = 24V
vds iT Vd + Vf = 1024 V vds I 0 + If
Vd