POWER ELECTRONICS
BPB23204
L01-T02
LAB 1
(LINE COMMUTATED SINGLE PHASE UNCONTROLLED
RECTIFIERS)
LECTURER: SIR MOHAMED RAZIF BIN NOORDIN
DATE SUBMISSION: 11TH SEPTEMBER 2020
NAME
MOHAMAD HARITH BIN ZAIFULLIZAN
ID NUMBER
51212119250
MOHD ADHAM BIN AHMAD DALI
51211119293
NURUL NADIRAH BINTI ADNAN
51211119255
CHE NUSAIBAH BINTI CHE ROHIM
51212119262
1
1. TABLE OF CONTENT
1. TABLE OF CONTENT
2
2. INTRODUCTION
3
3. RESULT
4. ANALYSIS OF RESULT
4-41
42-43
5. CONCLUSION
44
6. REFERENCES
45
2
2.
INTRODUCTION
Rectifier circuits are widely used in power electronic circuit to convert ac to
dc. The word rectification is used not because it produces dc, but rather
because the current flow in one direction which only the average output signals
(voltage or current) has a dc component. When rectifier circuits are used solely,
their outputs consist of dc along with high-ripple ac components. To significantly
reduce or eliminate the output ripple, additional filtering circuitry is added to the
output. In the majority of applications, diode rectifier circuit are placed at the
front end of the power electronic 60Hz systems, interfaced with the sine-wave
voltage produce by the electric utility. In dc-to-dc application, at the rectified side
or the dc side, a large capacitor is added to reduce the rectified voltage ripple.
This dc voltage maintained across the output capacitor is known as raw dc or
uncontrolled dc.
There are two types of single-phase diode rectifier which are single-phase
half-wave rectifiers and single-phase full-wave rectifiers. Single-phase halfwave rectifier is an electronic circuit which converts only one-half of the AC
cycle for the conversion process. On the other hand, single-phase full-wave
rectifies converts entire cycle of AC into Pulsating DC.
In this lab experiment, a simulation on line commutated single-phase
uncontrolled rectifiers had been identified. In part A, the circuit simulations of
the operational of a single phase half wave without inductor is investigated at
task 1 while with inductor in task 2. In part B also, single phase full wave is
carried out to analyse the simulated output signals when each of the circuits is
connected to a resistive load. The effect on the rectification circuit output signals
when adding an inductor to the resistive load in each rectifier also is identified.
At the end of the experiments, the simulation result were verify with the
calculation.
3
3. RESULT

(PART A –TASK 1)
 Circuit construction in Multisim
Figure 3.1: Circuit construction in Multisim
 Output and Input voltage waveforms
Figure 3.2: Waveform in oscilloscope
4
Figure 3.3: Output and Input Voltage waveforms
 Peak and RMS values of input voltage waveforms
 Circuit construction in Multisim for input
Figure 3.4: Circuit construction in Multisim
5
 Simulation on peak input voltage
Figure 3.5: Peak input voltage waveform
 Result from simulation: Peak input voltage = 2 x 50V
Vpeak(input) =100V
 Result from oscilloscope: Peak input voltage = 98.995V
 Calculation on peak input voltage
Vp = √
Vp = √
Vp = 1.414 x 70
Vp = 98.98 V
6
 Simulation on RMS values of input voltage
Figure 3.6: RMS values of input voltage
 Result from simulation: RMS values of input voltage = 70.144V
 Calculation on RMS values of input voltage
Vs =
= 70 V
7
 Peak, average and RMS values of output voltage waveforms
 Circuit construction in Multisim for load
Figure 3.7: Circuit construction in Multisim for output
 Simulation on peak output voltage
Figure 3.8: Peak output voltage waveform
 Result from simulation: Peak output voltage = 2 x 50V
Vpeak(output) = 100V
 Result from oscilloscope: Peak output voltage = 98.171V
8
 Calculation on peak output voltage
Vs=Vrms, Vp= Vm, Vrms=70V
Vm
= √2
Vrms
Vm = 1.414 x Vrms
Vm = 1.414 x 70V
Vm = 98.98V
 Simulation on average output voltage
Figure 3.9: Average output voltage waveform
 Result from simulation: Average output voltage = 39.743V
9
 Calculation on average output voltage
∫ 𝑖𝑛𝜔𝑡 𝑑𝜔𝑡
Vo(avg) =
2𝜋
Vo(avg) =
(-cos 𝜔𝑡) ;limitation
2𝜋
from 0 to 𝜋
Vo(avg) = 0.318 x
Vo(avg) = 0.318 x 98.98V
Vo(avg) = 31.48V
 Simulation on RMS values of output voltage
.
Figure 4.0: RMS values of peak output voltage
 Result from simulation: RMS output voltage = 49.415V
10
 Calculation on RMS values of output voltage
Vo(rms) = √ 1 ∫ (
2
𝑆𝑖𝑛𝜔𝑡)²𝑑𝜔t
Vo(rms) = 0.5 x Vm
Vo(rms) = 0.5 x 98.98V
Vo(rms) = 49.49V
 Circuit construction in Multisim at input to detect current
Figure 4.1: Circuit construction in Multisim at input
11
 Output and Input current waveforms
Figure 4.2: Peak input current waveform
 Peak and RMS values of input current waveforms
 Circuit construction in Multisim to find input current
Figure 4.3: Circuit construction in Multisim at input
12
 Simulation on peak input current
Figure 4.4: Peak input current waveform
 Result from simulation: Ipeak(input) = 654.321A
 Calculation on peak input current
Im =
R
Im =
Im =
Im =
13
 Simulation on RMS values of input current
Figure 4.5: RMS values of input current
 Result from simulation: RMS values of input current = 332.626A
 Calculation on RMS values of input current
Irms =
2
Irms =
Irms =
Irms =
14
 Peak, average and RMS values of output current waveforms
 Circuit construction in Multisim at output
Figure 4.6: Circuit construction in Multisim at output
 Simulation on peak output current
Figure 4.7: Peak output current waveform
 Result from simulation: Peak output curent = 654.473V
15
 Calculation on peak output current
Im =
R
Im =
Im =
Im =
 Simulation on average output current
Figure 4.8: Average output current waveform
 Result from simulation: Average output current = 273.736A
16
 Calculation on average output current
Io(avg) =
𝜋
Io(avg) =
𝜋
Io(avg) =
Io(avg) =
 Simulation on RMS values of peak output current
Figure 4.9: RMS values of peak output current
 Result from simulation: Average output current = 329.126A
17
 Calculation on RMS values of peak output current
Io(rms) =
2
Io(rms) =
.
2
Io(rms) =
Io(rms) =
18

(PART A –TASK 2)
 Circuit construction in Multisim
Figure 1.1: Circuit construction in Multisim
 Output and input voltage waveforms
Figure 1.2: Output and Input Voltage waveforms
19
 Peak and RMS values of input voltage waveforms
 Circuit construction in Multisim for input
 Simulation on peak input voltage
Figure 1.3: Peak input voltage waveform
 Result from oscilloscope: Peak input voltage = 98.437V
20
 Calculation on peak input voltage
Vp =Vm
Vp = √
Vp = √
Vp = 1.414 x 70
Vp = 98.98 V
Vm = 98.98 V
 Simulation on RMS values of input voltage
Figure 1.4: RMS values of input voltage
 Result from simulation: RMS values of input voltage = 70.172V
 Calculation on RMS values of input voltage
Vs =
= 70 V
21
 Peak, average and RMS values of output voltage waveforms
 Circuit construction in Multisim for load
Figure 1.5: Circuit construction in Multisim for output
 Simulation on peak output voltage
Figure 1.6: Peak output voltage waveform
 Result from oscilloscope: Peak output voltage = 97.742V
22
 Calculation on peak output voltage
Vs=Vrms, Vp= Vm, Vrms=70V
Vm
= √2
Vrms
Vm = 1.414 x Vrms
Vm = 1.414 x 70V
Vm = 98.98V
 Simulation on average output voltage
Figure 1.7: Average output voltage waveform
 Result from simulation: Average output voltage = 32.239V
23
 Calculation on average output voltage
(
Vo,(avg) =
(
=
= 26.69˚
)(
)
=
(
)
)
(
(
))
Vo,(avg) = 29.83 V
𝜋
= 206.69
 Simulation on RMS output voltage
Figure 1.8: RMS output voltage
 Result from simulation: RMS output voltage = 49.969V
24
 Calculation on RMS output voltage
Vo(rms) = √ 1 ∫ (
2
𝑆𝑖𝑛𝜔𝑡)²𝑑𝜔𝑡
Vo(rms) = 0.5 x Vm
Vo(rms) = 0.5 x 98.98V
Vo(rms) = 49.49V
 Circuit construction in Multisim at input to detect current
Figure 1.9: Circuit construction in Multisim at input
25
 Peak and RMS values of input current waveforms
 Simulation on peak input current
Figure 2.0: Peak input current waveform
 Result from simulation: Peak input current = 587.312A
 Calculation on peak input current
XL = 𝜋
= 2𝜋( )(
= 24𝜋 Ω
Im=
)
)
Z = √(
(
= 150.79 Ω
𝜋)
=
= 0.66 A x 1000
= 660 A
26
 Simulation on RMS values of input current
Figure 2.1: RMS values of input current
 Result from simulation: RMS values of input current = 329.845A
 Calculation on RMS values of input current
Irms =
2
Irms =
Irms =
=
27
 Peak, average and RMS values of output current waveforms
 Circuit construction in Multisim at output
Figure 2.2: Circuit construction in Multisim at output
 Simulation on peak output current
Figure 2.3: Peak output current waveform
 Result from simulation: Peak output curent = 585.730A
28
 Calculation on peak output current
XL = 𝜋
= 2𝜋( )(
= 24𝜋 Ω
Im=
=
)
)
Z = √(
(
= 167.88 Ω
𝜋)
= 0.59 A x 1000
= 590 A
 Simulation on average output current
Figure 2.4: Average output current waveform
 Result from simulation: Average output current = 267.995A
29
 Calculation on average output current
(
Io(avg)=
=
(
)
(1 -
)
)
= 0.2A x 1000
= 200A
 Simulation on RMS values of peak output current
Figure 2.5: RMS values of peak output current
 Result from simulation: Average output current = 329.845A
30
 Calculation on RMS values of peak output current
Io(rms) =
2
Io(rms) =
2
.
Io(rms) =
=
31

(PART B –TASK 1)
Input
Voltage
Current
Peak
98.5V (peak)
0.645A (peak)
RMS
70V
(rms)
0.457A (rms)
Output
Voltage
Current
Peak
98.4V (peak)
0.583A (peak)
RMS
49V
(rms)
0.457A (rms)
Average
30.8 V
0.411 A
32
Input Waveform
Output Waveform
33

(PART B –TASK 2)
Input
Voltage
Current
Peak
97.5V (peak)
0.575A (peak)
RMS
70V
(rms)
0.433A (rms)
Output
Voltage
Current
Peak
97.7V (peak)
0.362A (peak)
RMS
49.0V (rms)
0.433A (rms)
Average
31.0 V
0.412 A
34
Input Waveform
Output Waveform
35
 Calculation (Part B) Task 1
I. The value for the input

The peak input voltage
Vp= √
Vp=Vm
Vp= √
Vp= 1.414 x 70
Vp= 98.98 V
Vm= 98.98 V

RMS input voltage
Vs= VRMS
VRMS = 70 V

Peak input current
Im =
=
= 0.66 A
Im=0.66 A

RMS input current
Io(RMS) =
√
=
√
= 0.467 A
36
II.

The value for the output
Peak output voltage
Vs=Vrms
Vrms=70 V
V p = Vm
Vp= √
or
Vm= 1.414 x Vrms
= 98.98V
= 1.414 x 70
= 98.98 V

Average output voltage
Vo=Vavg=
= 0.636 x Vm
= 0.636 x 98.98V
= 62.95V

Rms output voltage
Vo(rms)= √
=
√
(
)
=
∫
(
𝜔𝑡)𝑑𝜔𝑡 =
√
√
= 70 V
37

Peak output current
Im =
=
= 0.66 A
Im=0.66 A

Average output current
Io(avg) =
=
(
)
= 0.42A
Io(avg)= 0.42 A
 RMS output current
Io(RMS) =
√
=
√
= 0.467 A
Io(RMS) = 0.467A
38
 Calculation (Part B) Task 2
I. The value for the input

The peak input voltage
Vp= √
Vp= √
Vp= 1.414 x 70
Vp= 98.98 V

RMS input voltage
Vs= VRMS
VRMS = 70 V

Peak input current
XL
=
Z = √(
𝜋
= 2𝜋(
)(
)
)
(
𝜋)
= 167.883
= 24𝜋
Im=
=
= 0.59 A
Im=0.59 A
39

RMS input current
Io(RMS) =
√
=
√
= 0.417 A
Io(RMS) = 0.417A
II.

The value for the output
Peak output voltage
Vs=Vrms
Vrms=70 V
V p = Vm
Vp= √
or
Vm= 1.414 x Vrms
= 98.98V
= 1.414 x 70
= 98.98 V

Average output voltage
Vo(avg) = ∫
(
𝜋
𝑖𝑛𝜔𝑡 𝑑𝜔𝑡 =
=
(
(
𝜋
)
)
Vo(avg) = 63.01 V
40

Rms output voltage
Vo(rms)= √
Vo(rms)=
(
)
=
√
(
∫
𝜔𝑡)𝑑𝜔𝑡 =
√
√
Vo(rms)= 70.0 V

Peak output current
Z = √(
XL = 𝜋
)(
= 2𝜋(
)
)
(
𝜋)
= 150.79
= 24𝜋
Im=
=
= 0.66 A
Im=0.66 A

Average output current
Io(avg)=
Io(avg)=
=
(
)
(
)
Io(avg)= 0.42 A

RMS output current
Io(RMS) =
√
=
√
= 0.467 A
Io(RMS) = 0.467A
41
4. ANALYSIS OF RESULT
In this experiment, operational of a single-phase half wave and full wave
uncontrolled rectification circuit are investigated by doing a circuit simulation.
The simulation output is compared with the calculation of simulations result in
each of the rectifier circuits.
Based on the simulation result for part A (task 1), it can be analyse that the
simulation output waveform is change from the AC signal to DC output. In this
task 1, the half wave rectifier is being used to rectify only half cycle of the
waveform. In the simulation, AC input is being used as a supply where it has
positive and negative cycles. The half wave rectifier works only on positive half
cycles while it eliminated the negative cycle. This is because the diode is works
as a switch where it can only allow current flow in one direction. When the
anode voltage is greater than the cathode voltage, diode will turn to the forward
bias where it will conduct current with a small voltage drop across it. Otherwise,
if the cathode voltage (negative supply) is more than anode voltage,
automatically the diode will turn to the reverse bias and it will become the open
switch condition or blocking state while the forward bias it becomes the
conducting state or close switch condition. Hence, the simulation result is
formed as half wave dc output waveform. Otherwise, the output voltage is in
phase with output current. Apart from that, the result of average, the peak and
RMS values in the waveform is almost similar with the theoretical calculation.
In part A task 2, the simulation result that been recorded is a bit different
with the result in Task 1. This is because the inductor is added on the circuit as
series with resistor. This means the loads current flows not only during the
simulations result in each of the rectifier circuits. The diode is kept in the on
state by the inductor’s voltage, which offsets the negative voltage of Vs(t). The
waveform comes out with some lagging. That situation is occurred when the
voltage will not stop until the inductor current become zero. Therefore, the
current will lag the voltage until its current that been stored in the inductor reach
zero. Hence, the output voltage waveform will have an extinction angle, ( ) and
the peak of Idc would be a bit flattened compare to the peak of Vdc. As a result,
42
the average, the peak and RMS value is almost the same with theoretical
calculation. It can be concluded that objective for this experiment was
successful when it wants to confirm the effect on the rectification circuit output
signal when adding an inductor to the resistive load.
In part B task 1, it focuses on the single phase uncontrolled full wave
diode rectification circuit. A full wave rectifier on the other hand utilizes both the
negative and positive portions of input waveform while the half wave
rectification is one half of the source waveform is not utilized. This means, there
is no power from the source will be converted into DC during the negative
cycles. In the simulation, the full wave rectifiers can obtain output voltage during
the positive and negative half cycles. Therefore, it delivers improved efficiency
than the half wave rectifiers. It produces an output voltage that is purely DC. For
the full wave rectifiers, the average direct current output voltage is higher than
that of half wave, the output of the full wave rectifier has much less ripple than
that of the half wave rectifier producing a smoother output waveform. For the
simulation result are almost the same as the theoretical calculation.
Lastly, the simulation of the part B task 2 as same as part A task 2 where it
has some added inductor to the circuit as series to the resistor. The inductor
can give some effect to the output voltage waveform where the current will lag
the output voltage, Vdc in certain time but it cannot be reached to zero because
the full wave rectification can be continuously having an input supply compare
to the single phase. Therefore, it cannot happened to produce an extinction
angle where it was lagging between positive and negative output waveform.
This is because the diode is kept in the on state by the inductor’s voltage.
Otherwise, it will produce the dc output continuously. And for the simulation
result, the peak, average and RMS value is almost the same but it needs to
measure in the different screen to get the maximum value of current. This occur
when the peak current is a not reach to the maximum amplitude compare to the
output voltage.
43
5.
CONCLUSION
In conclusion, it is concluded that half wave and full wave rectifier circuits
can be built. It can measure and record their output voltages and curves
systematically, such as for single phase uncontrolled half wave rectifier is the
simplest and probably the most widely used rectifier circuit albeit at relatively small
power levels. The output voltage and current of this rectifier are strongly influenced
by the type of the load. In this section, operation of the rectifier with resistives,
inductives and capacitives loads are discussed. Single phase uncontrolled half wave
rectifiers suffer from poor output voltage or input current ripple factor. In addition, the
input current contains a dc component which may cause problem in the power
supply system. The output dc voltage is also relatively less.
Meanwhile, for single phase uncontrolled full wave rectifier have two types of
full wave uncontrolled rectifiers commonly in use. If a split power supply is available,
only two diode will be required to produce a full wave rectifier. These are called split
secondary rectifiers and are commonly used as the input stage of a linear dc voltage
regulator. However, if no split supply is available the bridge configuration of the full
wave rectifier is used. This is the more commonly used full wave uncontrolled
rectifier configuration.
Based on the result for simulation, it can be conclude that the experiment
was successful. The operational of a single phase half wave and full wave rectifier
are analysed. At the end of the experiments, the simulation result were verify with
the calculation. All of the task result is almost similar with the calculation. It means
the modelling rectifier circuit are function as the real time rectifier circuit.
44
6. REFERENCES
1) Jojo (2017, December), Half wave rectifiers. Retrieved by,
https://www.circuitstoday.com/half-wave-rectifiers.
2) Victoria Cherksova (2018, April), What is a single phase full wave rectifier,
Retrieved by, https://www.student-circuit.com/learning/year2/powerelectronics-year2/what-is-a-single-phase-full-wave-rectifier/
3) Dr.Prof.Mohammed Tawfeeq, LECTURE .4 : Single-phase full-wave
uncontrolled rectifiers, Rectified by,
http://www.philadelphia.edu.jo/academics/mlazim/uploads/PE%20Lecture%20
No.4.pdf
4) Working of diode rectifiers uncontrolled rectifiers, Rectified by,
https://www.electricalclassroom.com/working-of-diode-rectifiers-uncontrolledrectifiers/
45