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Solution of Tutorial 1 Basic Concepts in Power Electronics

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Solution of Tutorial 1 Basic Concepts in Power Electronics
ELEC4614

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  109

(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

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