Half
wave rectifier circuit diagram is shown below.
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.
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.
Lets calculate various quantities for hwr. One
quantity will be used to calculate next quantity.
1. Average output current ( IDC) :-
2. RMS output current Irms or (IL)rms :-
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
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.
= 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
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
7. Efficiency (η) :-
It is a measure of the ability of rectitifer to convert i/p ac power into dc power.
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.
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.
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.
