ELE3805 Power Electronics: Rectifiers, Modulation USQ Assignment

This assignment solution provides detailed answers to a power electronics quiz, focusing on rectifiers and modulation techniques. It includes waveform analysis for single-phase fully controlled and half-controlled rectifiers, considering continuous load current and ideal component conditions. The solution also addresses thyristor commutation, calculating notch depth and illustrating waveform overlapping. Furthermore, it covers bipolar sinusoidal pulse width modulation (SPWM), determining output voltage, current, and THD using full-bridge inverters. The assignment utilizes given parameters such as voltage, inductance, and frequency to calculate relevant electrical characteristics. Desklib offers a wide range of solved assignments and study resources for students.

ASSIGNMENT
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QUESTION ONE
Solution:
The waveforms will be as shown in figure below:
Given that there is large inductance the load current will be continuous. Voltage VT4 will be
maximum during the positive cycle of the source voltage and zero in the negative cycle.
Given also that the elements are ideal there is no voltage drop. Therefore:
V T 4=V m sin α =351 sin 10=60.95 v

QUESTION TWO
Solution:
The waveforms will be as shown in figure below:

Given that there is large inductance the load current will be continuous. Voltage VD2 will be
maximum during the positive cycle of the source voltage and zero in the negative cycle.
Given also that the elements are ideal there is no voltage drop. Therefore:
V D 2=V m sin α=352 sin 22=131.86 v
QUESTION THREE
Solution
When thyristor commute the two phase are shorted through secondary transformer reactance
and as a result they cause a notch on the alternating voltage. The notch has both maximum
depth and width which depends on the reactance, delay angle and average of rectified current.

Notch depth =√ 2.Vf-f x sin α
Where Vf-f- phase to phase voltage
=√ 2x339x√3 x sin110
=158.44
Notch depth= 2 πX fX 2 LXI
√2 . V f −f x sin α
Where L- transformer secondary reactance
I- rectified current
f = ω /2 π =100 /2 π
=15.92Hz
Notch depth= 2 πX 15.92 X 2 X 43 X 3.13
√2 x 339 x √3 x sin 44
=26925.62/158.44
= 169.94
Waveform
For instance, during the commutation time, two switches are conducting at the same time
creating short circuit and reduction in the rectified voltage and in its average resulting to
overlapping of waveforms as shown below,

QUESTION FOUR
V AB=V m sin ( α + μ ) =339sin (2.81+43)=243.07 v
QUESTION FIVE

Solution
In Bipolar sinusoidal pulse with modulation mode, if the instantaneous value of signal is
larger than the triangular carrier the output is positive and when less than the carrier it is
negative and we use full bridge inverter to determine it.
Vo =
4 V D
π √2 but VDC= MavD
=0.88X200
=176V
= 4 X 176
π √2
= 704
4.443
=158.46 v
i0= V o
Z = V O
√R2 +¿ ¿ ¿

i0= V O
√R2+¿ ¿ ¿
ω=2πf
=2x πx49
= 307.88rad/sec
Z=√ R2 +ωL2
=√ 202 +(307.88 X 0.0148)2
=√ 400+20.76
=20.51Ω
i0= 158.45
20.51
=7.73A
QUESTION SIX
Solution:

V o =V AD−V AB =∑
1,3,5
∞ 4 Vdc
nπ sin nωt
V orms= 4 Vdc
π √2 =0.9 Vdc=0.9∗200=180 v
QUESTION SEVEN
Solution
In Bipolar sinusoidal pulse with modulation mode, if the instantaneous value of signal is
larger than the triangular carrier the output is positive and when less than the carrier it is
negative and we use full bridge inverter to determine it.
Vo =
4 V D
π √2
VDC= MA xVD
=0.88X200
=176v
= 4 X 176
π √2

= 704
4.443
=158.45 v
i0= V o
Z = V O
√R2 +¿ ¿ ¿
i0= V O
√ R2+¿ ¿ ¿
ω=2πf
=2x πx33
= 207.35rad/sec
Z=√ R2 +ωL2
=√92 +( 207.35 X 0.068)2
=√81+198.80 .
=16.71Ω
i0= 158.45
16.71
=9.482A
Current THD can be obtained by replacing the harmonic voltage with harmonic current:
THDi= √ ∑
N =2
∞
¿ ¿ ¿ ¿ ¿
¿ √ 9.482× 9.482
22.22
¿ 0.4267 A
QUESTION EIGHT

Fundamental peak line volatge ¿ V L=ma
V D
2 √3
¿ 0.88∗ √3∗200
2 =152.42V V Lrms= V L
√2 =152.42
√2 =107.48
Impedance
Z= √ R2+ ( ωL ) 2
¿ √ 202 + ( 2∗π∗49∗0.0148 ) 2
¿ 20.5 Ω
Line current:
ioL= V ph
Z =107.48
20.5 =5.24 A
QUESTION NINE