How does the Villard circuit double voltage? I don't understand the capacitor role when the voltage sinewave goes negative.


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    \$\begingroup\$ Have you read the wiki article? \$\endgroup\$ – Dean Dec 26 '11 at 20:48
  • \$\begingroup\$ @Dean: yes, and i didn't understood it! \$\endgroup\$ – RYN Dec 26 '11 at 20:53
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    \$\begingroup\$ The Wikipedia article use[d] some rather uncommon names for these circuits, in order to pay homage to their inventors. In most textbooks this "Villard" is called just a clamp. (I've actually worked toward fixing that in Wikipedia btw). \$\endgroup\$ – Fizz Oct 8 '15 at 14:30

It doesn't actually change the peak-to-peak voltage of the AC waveform. What it does do is impose a DC offset onto that AC waveform.

So, a wave that is say +/- 12V becomes a 0-24V waveform (less a little bit for the diode voltage drop).

The capacitor is charged up when the waveform goes negative (through the diode), and releases its charge when the waveform goes positive.

Here is a link to the Falstad Circuit Simulator with a Villard circuit. You can see how the waveform stays the same but is shifted upwards.


Suppose the input signal is \$V_{\text{in}} = V_p\sin(\omega t)\$. During the negative cycle the diode conducts and the capacitor is charged to \$V_p\$ but with opposite the polarity. Then, during the positive cycle, the output would be \$V_p + V_c = 2V_p\$ because the capacitor is charged to \$V_p\$ during the negative cycle.

There you are! The voltage is doubled and DC as well for a complete cycle of input signal.

  • \$\begingroup\$ Welcome to EE Stack Exchange, Sanat. Great first answer! :) \$\endgroup\$ – Robherc KV5ROB Jan 31 '16 at 18:43

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