PCE3601/201/0/2016
Tutorial Letter 201/0/2016
Power Electronics III (Theory)
PCE3601
Year module
Department
Engineering
of
Electrical
and
Mining
This tutorial letter contains important information
about your module.
BARCODE
MEMO ASSIGNMENT 2 MEMO
Question 1
A power control module of which the schematic is shown below in Figure 1 must be built.
The only transformer available with a 10 VA rating steps 220 V down to 12 V at 50 Hz.
The secondary of the transformer is represented by V1 in the circuit.
D1
D2
12k
R1
D3
1k
R2
X1
R3
1K
R4
48
R5
X2
10k
R6
X3
10k
C1
R7
V1
D4
C2
D5
150
R8
12k
R9
50%
10k
VR1
500m
R11
Q2
Q1
1K
R10
Figure 1
1.1 Identify and write down the labels of the components that form the synchronized
ramp generator.
(2)
R2, R9, R3, C2, Q2, X1
1.2 Calculate the maximum voltage (VDD of OPAMPs) over capacitor C1 taking a diode
volt drop to be 0.7 V.
(2)
VDD  2Vs  3VD  2(12)  3(0,7)  14,9 V
1.3 What is the practically measured amplitude of the zero-crossing-pulses if each of
these pulses is 50 us wide?
(1)
Vz  VDD  2  14,9  2  12,9 V
1.4 Calculate the pedestal voltage present at the non-inverting input of OPAMP X1. (2)
V 
R9
12k
VDD 
14,9  13,75 V
R9  R2
12k  1k
1.5 Determine the time that Q2 is OFF in each half cycle.
tOFF 
(2)
1
1
 tz 
 0,05  103  9,95 ms
2f
2  50
1.6 Suggest suitable values for R3 and C2. Show your reasoning and all design
calculations to support the values you propose for these two components.
(4)
C2 must not be too big nor too small, thus select 100 nF. Q  C2V  I R3 tOFF where
I R3 
VDD  V
R3
 100  109  13,75 
14,9  13,75
9,95  103
R3
 R3  8322 
1.7 Sketch a functional block diagram of the power control module in Figure 1.
2
(6)
PCE3601/201/0/2016
PSU
From 12 V
50 Hz source
Bridge
rectifier
Zero
Crossing
Detector
SynchronisedR
amp Generator
PWM
Power
Stage
[19]
Question 2
Four topologies of DC to DC regulators are considered to regulate a load voltage. They
are the BUCK, BOOST, BUCK-BOOST and Ҫuk regulators.
2.1 Sketch the basic topology of the BUCK regulator.
(1)
2.2 Sketch the basic topology of the BOOST regulator.
(1)
2.3 Sketch the basic topology of the BUCK-BOOST regulator.
(1)
2.4 Sketch the basic topology of the ҪUK regulator.
(1)
3
Figure 2
2.5
If the inductor current is as shown in Figure 2, determine the
2.5.1
average inductor current in a BUCK regulator,
I ave
Area


period
2.5.2
I ave
1
4  40  106   40  106  1
160  106  20  106 180  106
2


 4,5 A
40  106
40  106
40  106
average switch current in a BOOST regulator and
(2)
1
4  20  106   20  106  1
Area
80  106  10  106 90  106
2




 2, 25 A
period
40  106
40  106
40  106
2.5.3
I ave
(2)
average diode current in a BUCK-BOOST regulator.
Area


period
2.5.4
(2)
1
4  20  106   20  106  1
80  106  10  106 90  106
2


 2, 25 A
40  106
40  106
40  106
peak-to-peak current in inductor L1 of a ҪUK regulator.
(2)
Making the assumption that the ripple current is the most logical interpretation of the question
I ripple  I peak  I min  5  4  1 A
2.6
Referring to Figure 2, determine the voltage over a 1,5 mH inductor between 0 us
and 20 μs on the graph, if the circuit is a
2.6.1
BUCK regulator,
vL
4
di
54
 1,5  103
 75 V
dt
20 106  0
(2)
PCE3601/201/0/2016
2.6.2
vL
BUCK-BOOST regulator and
(2)
di
54
 1,5  103
 75 V
dt
20 106  0
2.6.4
vL
(2)
di
54
 1,5  103
 75 V
dt
20 106  0
2.6.3
vL
BOOST regulator,
ҪUK regulator.
(2)
di
54
 1,5  103
 75 V
dt
20 106  0
2.7
Referring to Figure 2, calculate the voltage over the same inductor between 20 μs
and 40 μs if the circuit is a
2.7.1
BUCK regulator,
vL
di
45
 1,5  103
 75 V
6
dt
40  10  20  106
2.7.2
vL
(2)
BUCK-BOOST regulator and
(2)
di
45
 1,5  103
 75 V
6
dt
40  10  20  106
2.7.4
vL
BOOST regulator,
di
45
 1,5  103
 75 V
6
dt
40  10  20  106
2.7.3
vL
(2)
ҪUK regulator.
(2)
di
45
 1,5  103
 75 V
6
dt
40  10  20  106
2.8
Use the information in the previous two questions to determine the load voltage if the
circuit is a
2.8.1
BUCK regulator,
Vo  vswitchopen   L
2.8.2
(2)
di
45
 1,5  103
   75  75 V
6
dt
40  10  20  106
BOOST regulator,
(4)
Vo  vswitchclosed  vswitchopen  75  75  150 V
2.8.3
BUCK-BOOST regulator and
Vo  vswitchopen  L
2.8.4
(4)
di
45
 1,5  103
 75 V
6
dt
40  10  20  106
ҪUK regulator.
(2)
Assuming both inductors have the same current wave shape
Vo  vswitchopen  L
di
45
 1,5  103
 75 V
dt
40  106  20  106
5
2.9 Calculate the frequency of operation of the circuits.
f 
(2)
1
1

 25 kHZ
T 40  106
[42]
Question 3
3.1 Explain why power electronic circuits are sources of RFI.
(3)
Power electronics makes use of static switches (no moving parts). This enables fast
switching of the devices from conduction state to blocking state. Further, the devices
are capable of conducting high currents and blocking high voltages. This causes rapid
change in the current and or voltage levels over components in the circuit. These rapid
changes are called transients. When ever a wave form does not conform to a pure sine
or cosine wave it contains harmonics. Harmonics are multiples of the switching
frequency in the case of power electronics.
3.2 Define EMC.
(2)
Electromagnetic compatibility (EMC) is defined as the ability of equipment or a system
to function satisfactorily in its electromagnetic environment (EME) without introducing
intolerable electromagnetic disturbance to any equipment in that environment (even
itself).
3.3 How can EMC be achieved in power electronic circuits?
(3)
Electromagnetic compatibility can be achieved by the following actions:

Reduce the level of EMI emissions from the source.

Reduce the effectiveness of the path of the EMI.

Reduce the susceptibility of the receptor.
3.4 Explain which precautions must be taken to limit RFI in power electronic circuits.
Include basic circuits for RFI suppression.
(6)
Radio frequency interference must be prevented from entering into a circuit and must be
eliminated if it enters the circuit. This is done by screening sensitive circuit sections and
filtering of RFI frequencies. Harmonics between lines are removed by placing T-, - or
L-type filters between the supply and the switching element. Further screening of the
circuit and supplying a good earth to the screen prevents radiation of the RFI.
L
Switch
N
LOA
D
Scree
n
Earth
Figure 5.2: Removing harmonics between the lines (symmetrical voltages)
A symmetrical voltage (between a line and the earth), is attenuated by introducing a
coupled inductor in series between the supply and the circuit. Screening will eliminate
any radiation.
6
PCE3601/201/0/2016
Coupled inductor
L
LOA
D
220 V
50 Hz
Switch
N
Earth
Figure 5.3: Removing harmonics between line and earth (asymmetrical)
LOA
D
L
2 F
Switch
N
0,1 F
2 F
Earth
Figure 5.4: Minimal harmonic reduction scheme
Limiting conductive interference is done by using filters and snubbers (transient
suppression). Limiting radiative interference is done by using screening, earthing,
correct printed circuit board (PCB) techniques and layout as well as other means of
preventing the equipment becoming a transmitter of high-frequency emissions.
3.5 Up until 2008 the SABS was the guardian of EMC standards in the SADC region.
Which entity is now the guardian according to a law passed in 2008?
(1)
ICASA
[14]
Question 4
Consider a DC load of 100 V. Ignore overlap and other losses.
4.1
Determine the harmonic content (up to 24th) of a single-phase bridge rectifier. (6)
4.2
Determine the harmonic content (up to 24th) of a three pulse zig-zag rectifier.
(6)
4.3
Determine the harmonic content (up to 24th) of a three-phase bridge rectifier.
(6)
4.4
Determine the harmonic content (up to 24th) of a twelve-pulse rectifier.
(6)
VDC  100 V ,thus
 200 
an   2 2    cos m 
 m p  1
m  1, 2,3,...
7
[24]
8
PCE3601/201/0/2016
ASSIGNMENT 3
Question 1
1.1
Sketch a diagram to illustrate the concept of dynamic braking of a DC motor.
(2)
1.2
Explain the use of dynamic braking of a DC motor.
(4)
To apply dynamic braking, the armature circuit is isolated from the supply and then a
resistor of suitable rating is connected to the armature. The motor now acts as a
generator and the kinetic energy of the motor is now dissipated in the resistor. The
work done by the motor slows it down rapidly. As the motor speed drops, the braking
power also drops. Dynamic braking is also called resistive braking. When the motor
stops all the energy has been dissipated and the motor will not reverse.
1.3
Sketch a diagram to illustrate the concept of plugging to rapidly slow down a DC
motor.
(2)
1.4
Explain the application of plugging on a DC motor.
(4)
The supply terminals of the motor are swopped while the motor is still running. This
causes the back-emf and the supply voltage to add, resulting in a high current and
thus high counter torque which slows down the motor rapidly. If this process is not
controlled by means of a PWM control scheme, serious damage can be caused to the
motor and/or the load. A speed sensor must be used to sense zero speed to prevent
reversal of the motor direction.
1.5
Sketch a diagram to illustrate the concept of regenerative braking of a DC motor.
(2)
9
1.6
Explain the application of regenerative braking of a DC motor.
(4)
Regenerative braking implies that the motor is acting as a generator and its kinetic
energy is not wasted in a resistor as in dynamic braking, but it is returned to the
source. E must be greater than V for regenerative braking to take place. The control
of the motor is used to track the back-emf of the motor in such a way that the back-emf
is larger than the voltage supplied to the motor to allow power to be transferred from
the motor to the supply. The motor is acting as a generator and the motor drive is
operating in the inverting mode. This is done by means of phase angle control with
   2 . As soon as the motor speed falls too low to provide a voltage E greater than
V, dynamic braking ceases and another form of braking must be employed. It is a very
energy efficient method of braking but much more sophisticated control circuitry is
needed than with the other types of braking.
[18]
Question 2
Sketch a block diagram of a variable speed drive suitable for controlling the speed of an
squirrel cage motor.
[9]
10
PCE3601/201/0/2016
Question 3
A circuit of a functional block in the DC Drive Trainer and one cycle of the voltage applied to
the transistor is shown in Figure 3.
Vzero-cross (V)
20
18
+ Vcc = 20 V
1k
10k
16
Vzero-cross
iin
14
12
if
10
_
C = 0,1 μF
+
8
eout
6
4
9k
2
0
1
2
3
Figure 3
4
5
6
7
8
9
10
t (ms)
3.1 Calculate eout just before application of a zero-crossing pulse.
eout  20 
(6)
9k
(20  18)(9 103 )

 18  18  0 V
9k + 1k
0,1 106  10k
3.2 Sketch one cycle of eout .
(4)
[10]
Question 4
Consider a DC load of 100 V. Ignore overlap and other losses.
4.1
Determine the ripple factor of a single-phase bridge rectifier.
(2)
4.2
Determine the ripple factor of a three pulse zig-zag rectifier.
(2)
4.3
Determine the ripple factor of a three-phase bridge rectifier.
(2)
4.4
Determine the ripple factor of a twelve-pulse rectifier.
(2)
The ripple factor in the load voltage is defined as the ratio of the total rms value of all
the alternating components to the mean value. Thus
11
ripple factor 
2
V1rms
 V22rms    Vn2rms
Vmean
2
2
Vrms
 Vmean

Vmean
where
Vrms is the rms value of the total wave.
Vmcost
Vm
0,866Vm
0,5Vm
pt
0


Vrms 
1 T 2
1  2
v
(
t
)
dt

Vm cos2 t dpt


0
0
T

 
Vm2

Vm2
 
2


0


0
1
V2
(1  cos 2t )dpt  m
2
2
Vm2
(1  cos 2t )dpt 
2


0
(1  cos 2t )dpt

p sin 2t 

 pt 

2
0


p sin 2 

V
1 p
2
p
 

sin
 
  Vm
2 
2
2 4
p



2
m
Vmean  VDC 
pVm

sin

p

 Vm  sin 
p
p

2-pulse:
 Vrms  Vm
1
p
2
1 2
2 Vm

sin
 Vm

sin

2 4
p
2 4
2
2
p

  2V
2
Vmean  Vm  sin   Vm  sin   m
p
2 


2
2
Vrms
 Vmean
ripple factor 
Vmean
12
Vm2 4Vm2
 2
 Vm2 4Vm2
2



 2
2Vm
2Vm 2


PCE3601/201/0/2016


Vm 1
2Vm
2

4
2

Vm 1
2Vm
2

4
 0,483
2
3-pulse:
Vrms  Vm
1
p
2
1 3
2

sin
 Vm

sin
 0,841Vm
2 4
p
2 4
3
p


3
Vmean  Vm  sin   Vm  sin   0,827Vm
p
3


2
2
Vrms
 Vmean
ripple factor 

Vmean
 0,841
2
Vm2   0,827  Vm2
2
0,827Vm
 0,185
6-pulse:
Vrms  Vm
1
p
2
1 6
2

sin
 Vm

sin
 0,9557Vm
2 4
p
2 4
6
p


6
Vmean  Vm  sin   Vm  sin   0,955Vm
p
6


2
2
Vrms
 Vmean
ripple factor 

Vmean
 0,9557 
2
Vm2   0,955 Vm2
2
0,955Vm
 0,038
12-pulse:
Vrms  Vm
1
p
2
1 12
2

sin
 Vm

sin
 0,9887Vm
2 4
p
2 4
12
p

 
 12
Vmean  Vm  sin   Vm  sin   0,9886Vm
p
12 


2
2
Vrms
 Vmean
ripple factor 

Vmean
 0,9887 
2
Vm2   0,9886  Vm2
2
0,9886Vm
 0,014
[8]
Question 5
5.1
What is the cause of overlap in power electronic converters?
(1)
The source inductance of the supply lines causes a delay in the transfer of current
from an outgoing device to an incoming device. This causes two devices to be on
simultaneously which in three phase systems boils down to a line to line short of
limited duration.
5.2
Sketch a circuit consisting of a Thévenin equivalent of a three phase source supplying
a three-pulse controlled rectifier.
(3)
13
5.3
Sketch voltage and current waveforms illustrating the effect of overlap in a three-phase
half-wave controlled rectifier.
(7)
5.4
The short circuit impedance of a transformer feeding a three pulse rectifier is 0,645 Ω
per phase. The load current is 40 A feeding into a highly inductive load. The
transformer secondary supplies 220 V 50 Hz to the rectifier. The delay angle is 15˚.
Determine the angle of overlap in degrees under these circumstances.
(5)
Note:
IL 
IL 
3Vm
cos   cos     
2 L 
3Vm
cos   cos     
2 L 
 40 
3(220 2) 



cos

cos




2(0,645)  12
 12

 2(0,645)  40 


 cos      cos 
12
3(220 2)
 12

14
PCE3601/201/0/2016


 cos      0,966  0,09575
 12

   cos1 (0,87025) 

12
 0,2533 rad  0,2533(57,3)  14,5
[16]
Question 6
A DC link voltage of 220 V supplies a single pulse-width-modulated single-phase inverter.
6.1
Derive an equation for the r.m.s. voltage over a load in terms of the pulse width δ.
(Note: No circuit or wave form = no marks)
(7)
6.2
Plot the harmonic profile up to the 5th for a pulse width δ = 72˚.
vLoad (t ) 


n 1,3,5...
an cos nt 
V1 
4(220) 1(72)
sin
 165 V
1
2
V3 
4(220)
3(72)
sin
 89 V
3
2
V5 
4(220)
5(72)
sin
0 V
5
2
(8)
4E
n
sin cos nt
2
n 1,3,5 n


15
6.3
Determine the r.m.s. voltage of the fundamental component.
V1rms 
Note:
V1max
2

165
2
(2)
 117 V
vLoad (t ) 
4 E n
sin cos nt
n

2
n 1,3,5


[17]
Question 7
7.1
16
Determine the power factor of a single-phase bridge rectifier using the load voltage
wave forms as reference. Sketch the wave forms. Assume level load current.
(7)
PCE3601/201/0/2016
Determine the power factor of a single phase bridge rectifier assuming level
load current.
Consider the circuit of a single phase bridge rectifier in figure 1. The load is very
heavy inductive and thus the assumption of level load current can be made. The wave
forms applicable to the output side and input side of the bridge rectifier is shown in
figure 2 and figure 3 respectively.
The power factor can be determined as follows:
Considering the output waveforms:
I o  I mean  I rms and Vmean 
2Vm


2 2V

where V is the rms value .
Considering the input waveforms:
2
I 2    I mean 
2 I mean
V
 mean

 I mean and Vrms  m  V
2
2
2
2
I rms
By definition
V I
PF  mean mean 
Vrms  I rms
2 2V

 Imean
V  I mean

2 2

 0.9
The current is not displaced with respect to the voltage, thus 1  0 .
PF 
I1rms
I rms
cos 1
Where
I1rms
I rms
= the input distortion factor
The input distortion factor is thus
I1rms
I rms
7.2

PF
PF
PF


 0.9
cos 1 cos 0
1
Determine the power factor of a three-phase bridge rectifier using the load voltage
wave forms as reference. Sketch the wave forms. Assume level load current.
(7)
17
Considering the output waveforms:
I L  I mean  I rms and Vmean  VDC 
3 3Vm


3 3 2V

where V is the rms value .
Considering the input waveforms:
2
I 2    I mean 
2 I mean
2
 mean


I mean
3
3
3
2
I rms
and Vrms 
Vm
V
2
By definition
V I
PF  mean mean 
3  Vrms  I rms
3 3 2V

3 V 
 Imean
2
I mean
3

3 3 2


3
3
  0,955
3 2 
The current is not displaced with respect to the voltage, thus 1  0 .
PF 
I1rms
I rms
cos 1
Where
I1rms
I rms
= the input distortion factor
The input distortion factor is thus
I1rms
I rms

PF
PF
PF


 0,955
cos 1 cos 0
1
[14]
18
PCE3601/201/0/2016
Question 8
A three-phase rectifier’s load voltage is shown below in Figure 3. Use a Fourier expansion to
determine harmonic content and also to plot the harmonic profile. The ripple frequency is
150 Hz. [8]
/100/
VLOAD
V
311
0
Figure 3
t
VLoad3 pulse  0.827Vm  0.2067Vm cos 3t  0.04726Vm cos 6t  0.0207Vm cos 9t 
0.011567Vm cos12t  0.007384Vm cos15t  0.00512Vm cos18t 
0.003759Vm cos 21t  ....
19
 VLoad3 pulse  0.827(311)  0.2067(311) cos 3t  0.04726(311) cos 6t  0.0207(311) cos 9t 
0.011567(311) cos12t  0.007384(311) cos15t  0.00512(311) cos18t 
 VLoad3 pulse
20
0.003759(311) cos 21t  ....
 257  64cos3t  14,7 cos 6t  6, 4cos9t  3,6cos12t  2,3cos15t  1,6cos18t 
1,17 cos 21t  ....