Double voltage in full wave rectifier

Asked Nov 07 '20 at 21:09

Active Nov 07 '20 at 21:33

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I made a simple full wave rectifier using a center tapped transformer and 2 1N4007 diodes.

The DC output is around 14 volts but when I connect a 100uF capacitor with the output leads of the DC output, I get around around 21 volts DC and 44 volts AC.

Without the capacitor it works as expected but with the capacitor why is this the case?

edited Nov 07 '20 at 21:33

JRE

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asked Nov 07 '20 at 21:09

shahrOZe

With the smoothing capacitor present, and no load, you get the peak voltage on the output. Without it you measure the RMS voltage of the rectified sine. – Unimportant Nov 07 '20 at 21:22
So it means the actual peak of the pulsating dc is around 21 volts? – shahrOZe Nov 07 '20 at 21:24
Yes, peak = RMS * sqrt(2). – Unimportant Nov 07 '20 at 21:28
And what about the ac voltage why is it 44V? – shahrOZe Nov 07 '20 at 21:33
Not a clue, there shouldn't be any AC on the output. Is this a real circuit or simulator? – Unimportant Nov 07 '20 at 21:36
Its a real circuit, without the capacitor it shows 0 volts ac as expected – shahrOZe Nov 07 '20 at 21:39
Let us continue this discussion in chat. – shahrOZe Nov 07 '20 at 21:39
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Get your meter, set it to AC volts and measure the voltage across a battery. Report back with the voltage you measure or try and figure out why your meter set to AC can measure many volts of AC across a battery. – Andy aka Nov 07 '20 at 22:54
Yes its reporting, but why is it so? I am totally confused? Its reporting around 20v across a 9v battery. – shahrOZe Nov 08 '20 at 03:54
Please read https://electronics.stackexchange.com/questions/101418/measuring-ac-voltage-from-dc-battery – mhaselup Nov 08 '20 at 05:13
Ok so the multimeter measures the average which is 9V and so to get the peak value as of it were AC it multiplies it by $\pi$ then divides by $\sqrt 2$ to get te RMS value of around 20V. Am i right? – shahrOZe Nov 08 '20 at 05:20

Double voltage in full wave rectifier

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