# Solution of Tutorial 1 Basic Concepts in Power Electronics

Topics: 2007, April, 1965 Pages: 18 (2441 words) Published: April 22, 2015
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

if
Rf = 50
Vd = 24V

vds
iT

Vd + Vf = 1024 V
vds
I 0 + If

Vd

I0

iT

Von

t

t fi

t doff
Vd+IoxRf
vds

iT

Io + 24/50

tri

Switch-off transient

Switch-on transient

(b) Turn-off switching loss
At turn-off, the initial current through the switch is If = Vd/Rf + Io. This current falls to zero in time tfi. The switch voltage is Vds = Vf + Vd where Vf = - IfRf is the voltage across the clamping resistor Rf. After tfi, the load current circulates in Rf and falls exponentially with time. Because of high frequency switching, the off period is short compared to the time constant of the load and the clamping resistor, so the exponential fall of voltage across the switch after tfi may be neglected.

With time origin starting from tdoff,

Solution of Tutorial 1

3

April, 2015

ELEC4614

Power Electronics

iT  I o  I f 

vds 

Vd  I o R f
t fi

 Io  I f  t
t fi

t

ps , off  it  vds
Ws , off  

t fi

0

iT  vds dt  

t fi

0

1024
1 
 t  20.48  1  t  dt
 t fi 
t fi

 t2
1024
t3 

 20.48   
 2 3t fi 
t fi

t fi

0

= 0.699 mJ
(b) Turn-on switching loss
At turn-on, the initial current through Rf is assumed to be Io. Prior to turn-on, the load was circulating through Rf and was decaying exponentially. If the switching frequency is high, it can be assumed that the decay in current during toff is negligible. Switch current iT rises from zero to Io + Vd/Rf in time tri. Note that tfv is smaller than tri, so tri determines voltage rise also.

iT 

I o  Vd / R f
t ri

t

1 
vds  Vd  I o R f  1  t 
 t ri 

ps , on

 I o  Vd / R f
 Vd  I o RF  
t ri

Ws , on  

tri

0


t2 
  t  
  t ri 

 I o  Vd / R f
t ri

Vd  I o RF  

1024  20.48  t 2 t 3 


 
t ri
 2 3t...