Half Wave Rectifier

Following question and solution is from a textbook.


However in Wikipedia, it's given that:


Now I am confused, should we multiply V(rms) with root 2 or 2 to get V(peak)?
Kindly explain the concept. Thanks.

What do you mean by Vmax ?

from V(max), I meant V(peak). Solved as V(m) there.

To get peak voltage from sinosodial AC voltage: VpeakSine = Vrms times root 2.

If it is fully rectified, VpeakFullRect = Vpeak - Vdiode. i.e. 1.4V

If it is half rectified, VpeakHalfRect = Vpeak - Vdiode. i.e. 0.7V

For Vrms half rectified, Vrms/2, because the power transfered half of the time.

RMS can be viewed as DC equivalent of AC voltage.
e.g.: even though the Vsine peaks at 340V, the RMS is 240V because its crosses zero volt 100 times at 50Hz.

Essential difference is whether one is referring to the rms value of the ac source voltage or the rms value of the rectifier output voltage.

t_n_k said:

Essential difference is whether one is referring to the rms value of the ac source voltage or the rms value of the rectifier output voltage.

Click to expand...

So suppose, ac input gives rms voltage x and after transformation through transformer to half wave rectifier circuit it gives voltage y. Now to get peak value of this, is the answer is √2y or 2y?

pranavissmart said:

So suppose, ac input gives rms voltage x and after transformation through transformer to half wave rectifier circuit it gives voltage y. Now to get peak value of this, is the answer is √2y or 2y?

Click to expand...

If the rms value of the rectifier output voltage is known to be value 'y' then the peak value of the half wave rectified waveform would be 2y.

Assuming ideal rectifcation with zero diode forward voltage drop then one would anticipate 2y=√2x