# Half Wave Rectifier

Discussion in 'Homework Help' started by pranavissmart, Feb 23, 2014.

1. ### pranavissmart Thread Starter New Member

Feb 23, 2014
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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.

Last edited: Feb 23, 2014
2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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What do you mean by Vmax ?

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3. ### pranavissmart Thread Starter New Member

Feb 23, 2014
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from V(max), I meant V(peak). Solved as V(m) there.

4. ### AfdhalAtiffTan Active Member

Nov 20, 2010
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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.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

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6. ### pranavissmart Thread Starter New Member

Feb 23, 2014
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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?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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

Last edited: Feb 24, 2014
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