180499717 Half Wave Rectifier

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    H LF W VE

    RECTIFIER

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    CERTIFICATE

    This is to certify that this project report on "HALF WAVE

    RECTIFIER" is a bonafide record of work done by Arundhathy

    Krishnafor the requirements of AISCCE Practical examination 2013-

    14.

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    Internal External

    Examiner Examiner

    Principal

    ACKNOWLEDGEMENTS

    I would like to express my special thanks of gratitude to my teachers Ms. Divya,

    Ms. Sandhya, Ms.Sivakamias 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 infinishing 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.

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    CONTENTS

    INTRODUCTION

    AIM

    APPARATUS

    MECHANISM

    PROCEDURE

    CIRCUIT DIAGRAM

    OBSERVATION

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    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

    supplies and high-voltage direct current power transmission systems. Rectification

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    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.

    APPARATUS

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    A diode, an electrolytic condenser 1000 F, a condenser of 0.1 F, a resistance

    1000, 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 ILflows 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.

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    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.

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    PROCEDURE

    Connections are made as shown in the circuit diagram. D is a diode. C is a 1000 F

    elctrolytic condenser. C'' is a 0.1 F condenser. RLis a 1K 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

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    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.

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    OBSERVATIONS

    (1) Rectifier with filter circuit

    Vdc Vac

    15.1 3V

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    (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

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    (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

    (b) Without filter circuit = 0.91

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    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.

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    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.

    Error can occur in the resolution of equipments used.

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    BIBLIOGRAPHY

    Reference Data for Radio Engineers

    Visionics

    ebooks.com

    chestofbooks.com

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