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Investigatory Project on Full Wave Rectifier

Investigatory Project on Centre—Tap Full Wave Rectifier


Aim: To construct a low voltage full wave rectifier using two identical Junction diodes and to calculate the ripple factor with and without filtering.


Materials and apparatus


(i) A step-down transformer with 220 to 240 V primary and 6-0-6 or 9-0-9 centre-tap secondary

(ii) Two identical diodes (eg: 1N 4001)

(iii) A 1000 or 5000 µF electrolytic condenser, a 0.1µF condenser

(iv) A 1 kΩ resistor, ac and dc voltmeters or electronic multimeter etc.


Working


During the half cycle of the ac input when the diode D1 is forward biased, a current i conducts in the circuit in the direction AD1T1T2OA (See Figure below). At this instance the diode D2 is reverse biased. So no electric current flows. During the next half cycle the diode D2 is forward biased and the diode D1 is reverse biased. Hence in diode D2 current flows in the direction BD2T1T2OB and the diode D1 does not carries current. In successive half cycles of the ac input the above procedure are repeated. In both the half cycles of the ac, it is obvious that the electric current flows through RL only and it would be in one direction; That is, from T1 to T2.


The dc output is produced across the load resistor, RL. Although the voltage across the resistor RL is unidirectional, it is pulsating. It can be made smooth using filter circuits.


The theoretical maximum efficiency of the rectifier is 81.2%.


Note: Since the dc output of a rectifier is pulsating, it contains a steady dc component and an ac component. The ac component is called ripples. To eliminate the ripples, filter circuit is used. The frequency of the ripples is same as the frequency of ac input for a half wave rectifier. It is twice the frequency of the input for a full wave rectifier.


Procedure


Connections are made as shown in the figure. D1 and D2 are two identical diodes. The P-regions of the diodes are connected to the secondary terminals A and B of the transformer.


The n-regions of the diodes are connected together. A wire DE is connected to it. The centre-tap O of the secondary of the transformer is connected to a wire FG. The electrolytic capacitor C in the filter circuit is of 1000 or 500 µF. Its positive terminal is connected to the wire DE and negative to the wire FG. The output load RL (1000Ω) is connected to the wires DE and FE, parallel to the condenser. C is a condenser of 0.1 µF.

 

(a) To find the ripple factor (r) with the filter circuit

 

Specification of the diodes (from data book)

Type number of the diode : ........

Maximum forward current rating = ...... mA

Peak Inverse Voltage (PIV) rating = ...... V


Circuit Diagram

The ac output voltage Vac is measured across the point T3 and T4 by the ac voltmeter V1 (or the multimeter). The de output voltage Vdc is measured across the points T1 and T2. The ripple factor of the rectifier with the filter is calculated by the equation,

r = ac output voltage/dc output voltage = Vac/Vdc

ac output voltage, Vac = ... V

dc output voltage, Vdc = ... V

Ripple factor, r = Vac/Vdc = ...

 

(b) To find the ripple factor (r) without the filter circuit


The above experiment is repeated without the filter circuit. That is by disconnecting the condenser C.


ac output voltage, Vac = ...V

dc output voltage, Vdc = ...V

Ripple factor, r = ...

 

Result


Ripple factor of the rectifier

(a) with the filter circuit = ...

(b) without the filter circuit = ...


Note: Ripple factor (r) may be calculated for different loads RL. A graph can be drawn with RL and r.

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