Power Electronics Systems 3
Tutorial 2 Solutions
Devices
1. Voltage across the MOSFET:
60
0.25
15
Voltage across the motor:
100
15
85
60
85
5.1
0.25
900
Power in the motor:
Power loss in the MOSFET:
60
2.
a)
The mean power loss due to switching is a product of total energy loss due switching by
the switching frequency.
Energy loss due to switching ON:
300 200
1 10
0.01
_
6
6
Energy loss due to switching OFF:
300 200
2 10
0.02
_
6
6
Total energy loss due switching:
0.01 0.02 0.03
_
_
_
Therefore, the mean power loss due to switching is:
0.03 10 10
300
_
_
b) Conduction Loss:
_
200
2.1
0.5
210
c) Total Loss:
_
300
_
210
510
10
1
3. The average current:
1
sin 2
sin 2
1
2
2
cos 2
cos 2
cos 2
2
500
cos 2
2
3.895
cos 0
500
100
The RMS current:
1
sin 2
1/4
1
sin 2
1
1
1
2
cos 2 2
2
1
cos 2 2
1
2
cos 2 2
1
1
2
1
1
2
4
500
100
2
500
sin 4
4
sin 4
10
1
4
500
sin 4
500
100
10
20.08
The peak current:
500
4. The described situation is shown in Fig. 1.
a) To calculate the RMS current
calculate the peak current .
‐ In general, for a sinusoidal function:
1
sin
2
‐ In this case:
1
sin
2
2
2
2
⟹
we need to
cos
cos
cos
1
30
0
2
2
2
188.4
30
Fig.1
‐
In general, RMS is given by:
1
2
‐
RMS current in this case:
2/4
1
2
2
1
1
2
2
1
4
cos 2
cos 2
4
1
sin 2
2
4
1
sin 2
2
0
4
2
2
2
4
1
sin
2
0
1
8
8
66.6
b) For the RMS value given above, the level current (DC
current) the thyristor could be expected to carry for one‐
third of a cycle, as shown in Fig. 2, and can be calculated
as the following:
1
2
2
2
Fig. 2
3
2
2
3
√3
⟹
√3
√3 66.6 115.4
c) The mean power loss of the thyristor is given by:
1
Or, in terms of angles, the mean power can be written as:
1
2
1
2
1
2
2
2
3/4
‐
Where
1
0.007, and the voltage‐
current characteristic is shown in Fig. 3.
For the sinusoidal waveform given in (a):
188.4 sin
⟹
1
2
2
2
188.4 sin
0.007
2
188.4
2
188.4 sin
sin
0.007 188.4
2
188.4
2
sin
cos
0.007 188.4
2
188.4
2
‐
Fig. 3
0.007 188.4
4
2
1
sin 2
4
2
61
For the level waveform given in (b):
115.4
⟹
2
2
2
2
2
2
2
2
2
115.4
2
69.5
2
3
0.007 115.4
2
3
4/4