Half
wave rectifier circuit diagram is shown below.
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It consists of a single phase transformer with turns
ratio 22√2 :1 (let). Let transformer
primary be supplied with
220√2 sin wt , then, secondary will have instantaneous
voltage 10 sin wt .
Half-Wave Rectifier Working
Secondary voltage waveform is shown below.
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From 0 to π :-
Diode is forward biased , therefore, conducts and
whole Vs appears across RL . IL is given by Vs/RL
.
From π to 2π :-
Diode is reverse biased, therefore, it acts like an
open switch and hence IL=0 .
We get waveform for IL for a complete cycle
as shown below.
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Lets calculate various quantities for hwr. One
quantity will be used to calculate next quantity.
1. Average output current ( IDC) :-
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2. RMS output current Irms or (IL)rms :-
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3. RMS ac output
current :-
As IL is a periodic waveform, therefore, fourier series expansion
can be written for it which will have a dc component and various harmonics i.e
ac components.
Therefore, from fourier series,
IL = IDC + IAC
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4. Ripple
factor :-
The output current IL is not pure dc. It is
a unidirectional periodically fluctuating current. It is for this reason that
filters are used to block the fluctuating component and let pass through them the
dc component.
A quantity called ripple factor(R.F), gives a measure
of the amount such fluctuating component present in the waveform.
Ripple
factor of a waveform is defined
as the ratio of, rms value of ac components present in the waveform, to the,
average value of the waveform.
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= 1.21
Qualitatively, R.F indicates the deviation of the waveform from
pure dc. Ripple factor is 0 for pure dc.
We found R.F of hwr equal to 1.21 , indicating it has large
amount of harmonics in the rectifier output waveform and is much deviated from
pure dc. So, we would need a large filter (costly and bulky) ,hence it is one
of the major drawback of half wave
rectifier.
5. Input power :-
It is the average of the instantaneous input (i/p) power, over a
cycle.
Instantaneous input power,
Pi = instantaneous i/p voltage × instantaneous i/p
current
= Vs IL
= (Rf+RL) IL × IL
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where, Rf = resistance offered by the diode in forward
biased condition.
6. output power Po :-
Average of the instantaneous power delivered to the load.
Po = IDC2 RL
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7. Efficiency (η) :-
It is a measure of the ability of rectitifer to convert i/p ac power into dc power.
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If we take ideal diode i.e
Rf=0 , we get maximum efficiency of Half wave rectifier,
equal to 40.6% .
8. Peak Inverse Voltage (PIV) :-
During rectifier operation, diode is subjected to a maximum
negative voltage, called Peak Inverse
Voltage.
For hwr PIV=Vm
9. Form Factor
(F.F) :-
Form factor of any waveform is defined as the ratio of, rms
value of waveform, to the, average value of the waveform.
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Qualitatively, F.F indicates the smoothness of waveform. As F.F
decreases and approaches 1 , smoothness of waveform improves towards pure dc.
For hwr we got F.F=1.57 ,
indicating that waveform is much deviated from pure dc.
10. Transformer
utilisation factor (TUF) :-
Transformer utilisation factor is the ratio of dc o/p power to the ac rating of transformer winding.
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If we assume transformer and diode to be ideal, then, Rf
and Rs =0 , we get maximum
TUF = 28.6 %
A high value of TUF is desired.
Lets take an example to understand why we want high TUF.
Lets say we want dc o/p power = 57.2 Watt.
Case (1) :- If we have a TUF of 28.6% then we would need 200KVA
transformer to get 57.2 Watt of power.
Case (2) :- If we have a
TUF of 57.2% then we would need only 100KVA transformer to get 57.2 Watt of
power.
So, with high TUF we need low rating transformer to get same
output dc power and since a transformer cost is directly proportional to its
rating , therefore with high TUF we would need to invest less money on the
transformer hence its makes the system cheaper. So, system with high TUF is preferred.