ELEX_4420_-_Lab_04_-_Single-Phase_FW_Rectifiers
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Lab 4: Single-Phase Full-Wave
Rectifiers
Marcel Moreno
BCIT ELEX 4420 – Power Electronics and Renewable Energy Applications
Lab 4: Single-Phase Full-Wave
Rectifiers
Objective
To study the behavior of various single-phase rectifier circuits and illustrate some key principles.
Equipment
Three-Phase Rectifier (External Rectifier) (used as single-phase rectifier in this lab)
Capacitors: DC-Link Capacitor (
850
μF
¿
and EMI Filter Bank capacitors
Three-Phase Load Bank (used as DC load)
Power Analyzer: For all the parts we use the power analyzer for measurements.
DMM (x2): One used as current sensor, one as a voltage sensor.
Pre-Lab (2 marks):
A Simulink file is provided (FW_rectifier_LCfilter.slx).
You need to change the transformer winding ratio so that the transformer’s secondary
voltage is 20 V rms.
Add your initials on the input lines of the scope.
RLC values need to be changed according to the following table and for each case, run the
simulation and fill out the table. You will need to remove the L and/or C from the circuit for
some cases.
For each case, capture the scope results. Scale the axes limits as shown below and copy the
plots into your pre-lab report. V
dc
V
rms
THD
v
I
dc
I
rms
THD
i
R = 83 Ω No Filter
16.65
18.77
52.03
0.201
0.226
52.03
R = 62.5 Ω
No Filter
16.59
18.72
52.24
0.265
0.299
52.24
R = 62.5 Ω
C = 40 μ
F
17.84
19.13
38.76
0.285
0379
87.26
R = 62.5 Ω
C = 120 μ
F
20.65
21.09
20.68
0.330
0.560
137.0
R = 62.5 Ω
C = 40 μ
F
L = 1.5 mH
17.98
19.34
39.58
0.288
0.406
99.61
R = 62.5 Ω
18.19
19.75
42.28
0.291
0.424
106
P. Taheri
BCIT ELEX 4420 – Power Electronics and Renewable Energy Applications
C = 40 μ
F
L = 4.5 mH
R = 83 Ω
No Filter
P. Taheri
BCIT ELEX 4420 – Power Electronics and Renewable Energy Applications
R = 62.5 Ω
No Filter
R = 62.5 Ω
C = 40 μ
F
R = 62.5 Ω
C = 120 μ
F
P. Taheri
BCIT ELEX 4420 – Power Electronics and Renewable Energy Applications
P. Taheri
BCIT ELEX 4420 – Power Electronics and Renewable Energy Applications
R = 62.5 Ω
C = 40 μ
F
L = 1.5 mH
R = 62.5 Ω
C = 40 μ
F
P. Taheri
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Related Questions
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Shown in he figure:
1. Derive a fomu la for mean load Voltage.
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Full Wave Rectifier Circuit
The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier.
Advantages of Full Wave Rectifier
The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%.
The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21.
The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers.
Question:
What is the main disadvantage of a conventional full-wave…
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Deternine:
A} the average voltage and average current in the load.
B) The form factor and ripple factor.
C) Peak inverse voltage of the thyristor.
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Vm.
Vmean-
i. Draw the rectifier circuit.
un
%3D
ii. Draw the input and the output voltages waveforms.
iii. Determine the mean and rms values of the output voltage. s Ec
iv. If the load is (R=10 S), then determine the mean and rms values of the current.
2019
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Facts
Full Wave Rectifier Circuit
The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre
tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier.
The second method uses a standard transformer with four diodes arranged as a bridge. This is known
as a bridge rectifier.
Advantages of Full Wave Rectifier
• The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The
efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is
81.2%.
• The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple
factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21.
• The output voltage and the output power obtained in full wave rectifiers are higher than that
obtained using half wave rectifiers.
Question:
What is the main disadvantage of a conventional full-wave…
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Secondary
D D D,
R
AD. Z D. D,
Select one:
O a. 99.83%, 100.08%, 4%
O b. 99.83%, 55.08%, 4%
O C. 99.83%, 100.08%, 16%
O d. 87.83%, 100.08%, 4%
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3
Given the bridge-type full-wave rectifier circuit
shown
220V
60Hz
220V:Es
Es
D1 D2
D3 D4
The load resistor is 220 and the transformer
average current is 300mA.
Determine:
(a) Average value, RMS value, and frequency
of load voltage VRL.
(b) Average current and PIV of each diode.
RL
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C) none
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Vms = 280V
Vin
RL= 12ks2
Vrms = 280V
Figure 1
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C4
R1
Vout
For the precision rectifier
circuit shown in Figure C4
what is the correct operation
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D1
R2
LM324 VEE
applied to the input signal
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Vin
D2
10k
OUT
V3
U1A
Figure C4
A. When VIn is negative the circuit operates as unity voltage follower
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circuit conducts but only to one diode drop,
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(c) 120.98V
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Related Questions
- 92) For three - phase half-wave Controlleed rectifier Shown in he figure: 1. Derive a fomu la for mean load Voltage. 2. praw the wave form of vo, i,,i, iy and voltage across thyristor 1 (r.) assuming highly inductive load and d = 120arrow_forwardA single-phase half wave uncontrolled rectifier circuit is operating with a purely inductive load and a source voltage of 100 sin (277t) V. An ideal PMMC ammeter is connected in series with the inductive load reads 20 A. Calculate the distortion factor of the rectifier. Answer Choices: a. 0.866 b. 0.707 c. 0.577 d. 0.816arrow_forward92) For three- phase half-wave Controlled rectifier Shown in the figure: I- Derive. a formula for mean load Voltage. 2- Draw the wave forms of vo, Li,i, o and voltage across thyristor1 (r.) assuming highly inductive loud and d = 120 Voarrow_forward
- Full Wave Rectifier Circuit The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier. Advantages of Full Wave Rectifier The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%. The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21. The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers. Question: What is the main disadvantage of a conventional full-wave…arrow_forwarda) Draw the 3-phase half-wave uncontrolled rectifier circuit for resistive load.b) Briefly explain the purpose of use.c) Draw the sinusoidal input voltages and the output voltage on the load.arrow_forwardIn the circuit in the figure, an AC voltmeter will be made with full wave rectifier circuit structure. R = 50ohm and diodesResistance values in the direction of transmission are Rd = 100ohm. Inside of the DC ammeter to be used as indicatorthe resistance is very, very small. The AC mark to be measured is Vs = 100 Sinwt Volts.a- Draw the shape of the current passing through the DC Ammeter and calculate its maximum value.b- Find the average value of the current passing through the DC Ammeter.c- Analyze the voltage seen at the ends of diode D1 for both alternans.d- Find the Rms value of the voltage seen at the ends of the diode D1.arrow_forward
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- Draw the circuit diagram of a half-wave rectifier for producing a nearly steady dc voltage from an ac source. Draw two different full-wave circuits.arrow_forwardDesign a single-phase full wave bridge rectifier with input voltage of 100v and inductive load of 10ohm and 40mH a. Show the waveform relation between the I/P and O/P b. Calculate the average O/P voltage c. Calculate the rms currentarrow_forwardA three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%arrow_forward
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ISBN:9781337399128
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Publisher:Cengage Learning