HALF WAVE
RECTIFIER
CERTIFICATE
This is to certify that this project report on "HALF WAVE
RECTIFIER" is a bonafide record of work done by Arundhathy
Krishna for the requirements of AISCCE Practical examination
2013-14.
Internal External
Examiner Examiner
Principal
ACKNOWLEDGEMENTS
I would like to express my special thanks of gratitude to my teachers Ms. Divya,
Ms. Sandhya, Ms.Sivakami as well as our principal Rev. Fr. Mathew Arekulam
who gave me the golden opportunity to do this wonderful project, which also helped me in doing a lot of Research.
Secondly I would also like to thank my parents and friends who helped me a lot in finishing this project within the limited time.
Lastly, I thank almighty, my parents, brother, sisters and friends for their constant encouragement without which this project would not be possible.
CONTENTS
INTRODUCTION
AIM
APPARATUS
MECHANISM
PROCEDURE
CIRCUIT DIAGRAM
OBSERVATION
CALCULATION
RESULT
PRECAUTIONS
SOURCES OF ERROR
BIBLIOGRAPHY
INTRODUCTION
Rectification is the process of converting alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction.
The Half wave rectifier is a circuit, which converts an ac voltage to dc voltage. It has many uses, but are often found serving as components of DC power
may serve in roles other than to generate direct current for use as a source of power.
AIM
To construct a half wave rectifier using a diode and to calculate its ripple factor with and without filtering.
A diode, an electrolytic condenser 1000 µF, a condenser of 0.1 µF, a resistance 1000Ω, a step down transformer with 220V - 240V primary and 6.3V, 100mA secondary, ac and dc voltmeters or a millimeter, etc..
MECHANISM
During the positive half cycle of the input voltage the polarity of the voltage across the secondary forward biases the diode. As a result a current IL flows through the load resistor, RL. The forward biased diode offers a very low resistance and hence the voltage drop across it is very small. Thus the voltage appearing across the load is practically the same as the input voltage at every instant.
During the negative half cycle of the input voltage the polarity of the secondary voltage gets reversed. As a result, the diode is reverse biased. Practically no current flows through the circuit and almost no voltage is developed across the resistor. All input voltage appears across the diode itself.
Hence we conclude that when the input voltage is going through its positive half cycle, output voltage is almost the same as the input voltage and during the negative half cycle no voltage is available across the load. This explains the
unidirectional pulsating dc waveform obtained as output. The process of removing one half the input signal to establish a dc level is aptly called half wave
rectification.
PROCEDURE
Connections are made as shown in the circuit diagram. D is a diode. C is a 1000 µF elctrolytic condenser. C'' is a 0.1 µF condenser. RL is a 1KΩ resistor. The rectified output is obtained across RL.
To find the ripple factor (r) with filter circuit
The ac output voltage Vac is measured by an ac voltmeter or by a millimeter. The dc output voltage is also measured. The ripple factor is calculated by the equation
r = Vac/ Vdc
To find the ripple factor without filter circuit
The filter circuit is removed by disconnecting condenser C. The ripple
factor is determined as explained above.
OBSERVATIONS
(1) Rectifier with filter circuit
Vdc Vac
(2) Rectifier without filter circuit
Vdc Vac
9.8V 9V
CALCULATIONS
(1) To find ripple factor (r) of rectifier with filter circuit AC output voltage( Vac) = 3V
DC output voltage(Vdc) = 15.1V Ripple factor (r) = Vac/Vdc = 3/15.1 = 0.19
(2) To find the ripple factor (r) without filter circuit
AC output voltage( Vac) = 9V DC output voltage(Vdc) = 9.8V Ripple factor (r) = Vac/Vdc = 9/9.8 = 0.91
RESULT
The ripple factor of the half wave rectifier
(a) With filter circuit = 0.19
PRECAUTIONS
While doing the experiment do not exceed the ratings of the diode. This may lead to damage of the diode.
Connect the circuit components properly as shown in the circuit diagram.
Do not switch ON the power supply unless you have checked the circuit components a s per the cicuit diagram.
SOURCES OF ERROR
The temperature coefficient of the components will change the longer the circuit is energized.
Temperature and humidity can affect the functioning of circuit.
Human error while taking the readings should be taken into consideration.
BIBLIOGRAPHY
Reference Data for Radio Engineers
Visionics
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