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Bridge Full Wave Rectifier

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3 views4 pages

Bridge Full Wave Rectifier

Uploaded by

yashk2107
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Expt. No.

: 6 -Date:
. Bridge Full Wave Rectifier

Objective:
i. To implement and verify the working of a Bridge Full Wave Rectifier
ii. Toobtain the values of ripple factor (r) and rectification efficiency (n) from the
experimental setup and verify with theoretical values.
Equipment:

Instruments: Step Down Transformer (12-0-12), Oscilloscope, Digital Multimeter


(DMM)
Components:Diodes: Silicon (D1N4007), Resistor: 1 KA

Theory :

Most electronic eguipments have atransformer at the input. The transformer serves two
purposes. First, it allows us to step the voltage up or down. This way we can get the
desired level of dc voltage. Second, it provides isolation from the power line and reduces
the risk of electrical shock. The primary of the transformer is connected to the power
mains. An ac voltage is induced across the secondary of the step down transformer. It has
alternate positive and negative half-cycles.

A more widely used full-wave rectifier circuit the bridge rectifier shown in figure 6.1.
It requires 4 diodes instead of two, but avoids the need for a center tap transformer.

During the positive half-cycles of the secondary voltage, diodes D2 and D4 are conducting
and diode D1 and D3 are non-conducting. Therefore, the current flows though the
secondary winding, diode D2, loadresistor R, and D4 as shown in figure 6.2 (a).During
negative half-cycles of thesecondary voltage, diodes D1 and D3 are conducting and diode
D2 and D4 are non-conducting. The current therefore flows through the secondary
winding,diode D3, load resistor RL and diode D1 as shown in figure 6.2 (b).
In both cases, the current passes through the load resistor in the same direction.
Therefore, a fluctuating, unidirectional voltage is developed across the load. The load
voltage waveform isshown in fîgure 6.3.

The voltage waveform in figure 6.3 is exactly the same as that in figure 5.3. In both the
rectifier circuits, the load voltage is the same. However there is one difference. In the
bridge rectifier, Vm is the maximum voltage across the secondary winding. But in center
tap rectifier, Vm represents the maximum voltage across half the secondary winding.
Circuit Diagram:

D) D2
Power
mains
200 V
50 Hz
D4
D3
Fig. 6.1The Bridge Full Wave Rectifier Circuit

iL DI D2 D2
IL D)

R
D4
D3 D4
D3

(a) Positive Half Cycle (b) Negative Half Cycle


Fig. 6.2 Half Cycles of Bridge Full Wave Rectifier
Circuit
Procedure, Observations and Calculations:
1. Measure the accurate value of the load resistor using DMM.
RL = KO
2. Connect the transformer primary to the power mains,
connect the
secondary terminals to Channel 1 of the oscilloscope using probes, transformer
power mains and observe this transformer secondary signal on the turn ON the
3. This is the input to the Bridge Full oscilloscope.
Wave Rectifier circuit. Note that, at any point in
the operation of this rectifier, full winding of the
input and rectified. Plot this input waveform withtransformer secondary is used as
4. Measure the peak to peak voltage of the correct scale on the graph page.
Using this, calculate the value of Vm.
transformer secondary using oscilloscope.
Vpp = div x V/div =
Vm(osc) = Vpp/2 =
5. Measure the fYequency of the transformer secondary using
T= divx msec/div = msec.
oscilloscope.
f= 1/T = Hz.
6. Observe the transformer secondary voltage on the oscilloscope first in ac mode,
then in ac + dc mode. Comment on the observation.

E&TC Dept,Goa College of Engineering 2


7. Using the ac voltage. (V) setting on the DMM, measure the voltage across the
transformer secondary between one of the terminals and the center tap. This is
the RMS representation of the ac voltage through this part of the the secondary.
Calculate Vm(DMM) using this value.
V
Vms(sec) =
Vm(DMM) =V2 x Vyms (sec) =
Circuit Diagram of
8. Connect the Bridge FullWave Rectifier circuit as shown in the
Fig. 6.1.
terminals on Channel1 of the
9. Observe the transformer secondary signal between
oscilloscope. Plot this
oscilloscope and across the load resistor on Channel 2 of the
signal, on the
output waveform with correct scale corresponding to the input
graph page.
first in ac mode, then in ac+
10. Observe the load resistor voltage on the oscilloscope
dc mode. Comment on the observation.

Measuring the ripple factor (r) using DMM


voltage across the load
11. Using the acvoltage (V) setting on the DMM, measure thc
the
resistor. This is the rms representation of the pure ac component of voltage in
load resistor.
Vrms(R) =
across the load
12. Using the dc voltage (V) setting on the DMM, measure the voltage
resistor.
resistor. This is the pure dccomponent of voltage in the load
Vpc(R)=
13. Calculate the ripple factor using values from DMM:
TDMM Vyms(RL)
Vpc(RL)

Measuring the ripple factor () using Oscilloscope Reading of Vm


values of Iros
14. Using the value Vmiosc) from step 4 and RL from step 1, calculate the
and ldc. Using these values, calculate the ripple factor .
Im=Vm(osc)/Ri = mA.
Ims = Im /W2= mA
ldc = 2*1m/T= mA

-1=
Tosc
3
E&TC Dept, Goa College of Engineering
E&TC circuits.
Therefore, improvement
We
the transformer In
rectifier 2. weConclusion:
1. A Waveforms
Typical
this Bridge-FWR
Assee As
load
observe 18. 17. 15.
Dept,Goa wwe we 16.
way, that values.
Verify Calculate Pac = RL
=1dcRL=Calculate
=1rmsRLCalculate Pacfrom
step1.
as rectifies
observe observe
Bridge from
compared that, at there
College secondary
over Fig.
any is that step Measuring
Bridge a existsefficient
fromn from 5.3 the the the
rectifier CT-FWR point high the
of to Rectification 1. ac dc
Bridge
Engineering both, FWR percentage
is in the more the D4D2 ripple power power the
circuit
is and used the value values Rectification
dc Full Nosc
produces
the CT-FWR operation factor input delivered
HWR. as in Wave
of because:
of =
Efficiency
most input the D3DI
efficiency,
of ripple Pac Pas to
the output values
and Rectifier 2
the
commonly almost Efficiency
to
HWR. and of input 100 Bridge
the factor, (n) the
than D4D2 and
double both % load
Bridge signal nosc Output using
the = FWR (n)
ac.TpMM
used half resistor using
and efficiency values
the FWR Volatge circuit
cycles =
rectifier amount gives % Oscillloscope
rectifier, from using
% using
are aThis Waveform matches
pulsating step ldc
circuit of rectified. Tosc
and Irms
full indicates from
DC 15 Reading
from
Voltage winding the and step
in dc.
electronic This theoretical 16: step
that 14 of
4
across 14 and Vn
is of
the and
anthe R.

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