# 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 D_{1} is forward biased, a
current i conducts in the circuit in the direction AD_{1}T_{1}T_{2}OA (See Figure below).
At this instance the diode D_{2 }is reverse biased. So no electric
current flows. During the next half cycle the diode D_{2} is forward
biased and the diode D_{1} is reverse biased. Hence in diode D_{2}
current flows in the direction BD_{2}T_{1}T_{2}OB and the
diode D_{1} 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 R_{L} only and it
would be in one direction; That is, from T_{1} to T_{2}.

The dc
output is produced across the load resistor, R_{L}. Although the
voltage across the resistor R_{L} 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. D_{1} and D_{2} 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 R_{L} (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 V_{ac} is measured across the point T_{3 }and T_{4}
by the ac voltmeter V_{1} (or the multimeter). The de output voltage V_{dc
}is measured across the points T_{1} and T_{2}. The ripple
factor of the rectifier with the filter is calculated by the equation,

r = ac
output voltage/dc output voltage = V_{ac}/V_{dc}

ac output
voltage, V_{ac} = ... V

dc output
voltage, V_{dc} = ... V

Ripple factor,
r = V_{ac}/V_{dc} = ...

**(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, V_{ac} = ...V

dc output
voltage, V_{dc} = ...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 R_{L}. A graph
can be drawn with R_{L} and r.

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