# Half Wave Rectifier With and Without Filters

Objective:

To study the operation of Half- Wave Rectifier with and without filter and to find its:

1. Percentage Regulation
2. Ripple Factor
3. Efficiency

Components:

S.No. Name Quantity
1 Bread board 1 (One) No.
2 Diodes (1N4007) 1 (One) No.
3 Resistor (1K) 1 (One) No.
4 Capacitor 100F 1 (One) No.

Equipment:

S.No. Name Quantity
1 Transformer with Center Tapped Secondary ( 9 - 0 - 9 ) V 1 (One) No.
2 Digital Multimeter/ Digital Voltmeter (0-20V) 1 (One) No.
3 Cathode Ray Oscilloscope (CRO) (0-20MHz) 1 (One) No.
4 Connecting wires (Single Strand)

Operation:

The conversion of AC into DC is called Rectification. Electronic Devices can convert AC power into DC power with high efficiency.

Consider the given circuit. Assume the diode to be ideal i.e., Vf = 0, Rr = Infinite, Rf = 0. During the positive half cycle, the diode is forward biased and it conducts and hence a current flows through the load resistor. During the negative half cycle, the diode is reverse biased and it is equivalent to an open circuit, hence the current through the load resistance is zero. Thus the diode conducts only for one half cycle and results in a half wave rectified output.

Percentage of Regulation:

It is a measure of the variation of DC output voltage for variations in the load.

Percentage of regulation $= \frac{V_{NL} - V_{FL}}{V_{FL}} * 100$%

VNL = DC voltage across load resistance, when minimum current flows through it.

VFL = DC voltage across load resistance, when maximum current flows through.

For an ideal half-wave rectifier, the percentage regulation is 0 percent.

For a practical half-wave rectifier.

$V_{NL} = \frac{V_{m} }{\Pi }$

$V_{FL} = \frac{V_{m} }{\Pi } - I_{dc}(R + R_{f})$

Converting Idc into its corresponding Im value and substituting in the percentage of regulation formula we get.

Percentage of regulation $= \frac{R_{f} + R}{R_{L}} * 100$%     (where R is the winding resistance)

Since Rf + R is small as compared to RL. The percentage of regulation is very small for half-wave rectifier.

Circuit Diagram:

Half Wave Rectifier (without filter):

Half Wave Rectifier (with filter):

Note: Third Terminal at the secondary of the transformer is not connected.

Procedure:

1. Connect the circuit as shown in the circuit diagram.
2. Connect the primary side of the transformer to AC mains and the secondary side to rectifier input.
3. Using a CRO, measure the maximum voltage Vm of the AC input voltage (at the anode) of the rectifier and AC voltage (at the cathode) at the output of the rectifier.
4. Using a DC voltmeter, measure the DC voltage at the load resistance.
5. Observe the Waveforms at the secondary windings of transformer and across load resistance for a load of 1K.
6. Calculate the ripple factor (), percentage of regulation and efficiency () with the below given formulae.

Observations:

1. Peak Voltage, Vm =               (From CRO for HWR with and without filter)
2. DC Voltage, VDC(full load) =               (From Voltmeter/ Multimeter for HWR with and without filter)
3. No Load DC Voltage, VDC(No load) =               (From Voltmeter/ Multimeter for HWR with and without filter)
4. Ripple Voltage, Vr =               (From CRO for HWR with filter)

Calculations:

Without Filter:

Vrms = Vm / 2

Ripple factor (Theoretical)

Ripple Factor(practical)   where

With Filter:

Ripple Factor (Theoretical) $r = \frac{1}{2\sqrt{3}fCR}$

Where f = 50Hz, R = 1K, C = 1000F

Ripple Factor(practical)

Percentage Regulation $= \frac{V_{NL} - V_{FL}}{V_{FL}} * 100$%

VNL = DC voltage at the load without connecting the load (Minimum current).

Efficiency

PAC = V2rms / RL

PDC = Vdc / RL

Expected Waveforms:

Result:

The operation of Half Wave rectifier is studied and the following are calculated.

 Type of Rectifier Ripple factor Efficiency % Regulation Theoretical Practical HWR without filter HWR with filter

Outcomes: Students are able to

1. analyze the operation of Half Wave rectifier with and without filter.
2. calculate its performance parameters-ripple factor, percentage regulation, efficiency with and without filter.

Viva Questions:

1. What is a Rectifier?

Ans: A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction. The process is known as rectification.

2. What is a Ripple Factor ()?

Ans: Ripple factor can be defined as the variation of the amplitude of DC (Direct current) due to improper filtering of AC power supply. it can be measured by RF = vrms / vdc

3. What is Efficiency ()?

Ans: Rectifier efficiency is the ratio of the DC output power to the AC input power.

4. What is PIV?

Ans: The peak inverse voltage is either the specified maximum voltage that a diode rectifier can block, or, alternatively, the maximum that a rectifier needs to block in a given application.

5. What are the applications of rectifier?

Ans: The primary application of rectifiers is to derive DC power from an AC supply. Virtually all electronic devices require DC, so rectifiers are used inside the power supplies of virtually all electronic equipment. Rectifiers are also used for detection of amplitude modulated radio signals. ectifiers are used to supply polarised voltage for welding.

6. What is dot convention?

Ans: Dot is convention is a method to find the voltage polarity of mutual inductance of two components those are 1) If current enters from dotted terminal of one coil then the e.m.f produced in the other coil is positive at the dotted terminal of the second coil. 2) If current leaves from dotted terminal of one coil then the e.m.f produced in the other coil is negative at the dotted terminal of the second coil.

7. What is the principle of transformer operation?

Ans: Transformer consists of two coils primary coil and secondary coil which are electrically not conneted but there is

8. Why step down transformer is used in HWR?

Ans:In step down transformer the primary coil winding turns are greater than secondary coil turns,which reduces the secondary coil voltage where this voltage is connected to diode in HWR circuit,  because the voltage needed for the diode is very small. Applying a large AC voltage without using transformer will permanently destroy the diode. So we use step-down transformer in half wave rectifier.

9. What is the output of HWR? Is it unidirectional or constant?

Ans: The output of HWR is . It is unidirectional output.

10. Which is preferable- High regulation or low regulation?

Ans:Voltage regulation means how much voltage devication from no load to full load.if VFL and VNL are equal voltage regulation is zero,no voltage drop across the load, this preferable means low regulation is p[referable.

11. What are the different types of filters used for the rectifiers?

Ans:the filter circuts used for filters are capacitive filter,Inductive filter,LC filter,CLC filter.

12. What is the % of load regulation ?

Ans:Percentage load Regulation $= \frac{V_{NL} - V_{FL}}{V_{FL}} * 100$%

13. Define line and load regulation.

Ans:Line regulation is the ability  to maintain specified output voltage over changes in the input line voltage.

Load regulation is the ability to maintain specified output voltage given changes in the load.

14. What is load regulation and line regulation in power supplies?

Ans: Line regulation is the ability of the power supply to maintain its specified output voltage over changes in the input line voltage.

Load regulation is the ability of the power supply to maintain its specified output voltage given changes in the load.

15. What is TUF?

Ans:How much is the utilization of transformer in the rectifier circuit is called Transformer utilization factor(TUF).Which can be obtained by DC power delivered to load to AC power rated of transformer. TUF =

• Created
Jun 26, 2013
• Updated
Oct 29, 2019
• Views
205,362
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