I attached a picture of the Villard circuit bellow. I can't seem to understand why the capacitor stays charged-up. Since I = CdV/dt, a capacitor finishes being charged at the crests of the voltage sine wave. A capacitor alone would then begin uncharging (see the second picture). The thing that is confusing me is that for the Villard, after the peak voltage, the voltage is obviously still positive (since the voltage and current are 90 degrees out of phase). Wouldn't this mean that the diode is forward-biased? Isn't it supposed to act like a plain conductor when it is forward-biased? If so then why doesn't the capacitor just discharge through it? BTW with the voltage and current 90 degrees out of phase, the current can flow in both directions when the voltage is above zero, right? (One direction while voltage is rising, the other while it is falling).

Perhaps the explanation is that a diode only lets current flow in one direction, if it is forward biased too.

Obviously this goes toward understanding the fundamentals. Feel free to throw what you like at me.

Villard Circuit Villard Circuit

AC Voltage source and capacitor AC Voltage source and capacitor

  • \$\begingroup\$ Yup it's all about the diode, (well cap blocks the DC current). Current has to flow in a loop, the diode allows charge to be pulled out in one direction, but not pushed back in the other.. so you get a DC voltage. \$\endgroup\$ – George Herold Aug 5 '15 at 20:22

The capacitor stays charged because there is no current path for it to discharge (the diode only allows the current to pass one way). Here's what happens when you plug this in to 320Vpk (~230V RMS) AC power supply:

Initial condition, the power supply instantaneous voltage is zero, the capacitor is discharged: initial condition

1st 90 degrees of the power supply phase - the voltage rises to 320V. The diode is reverse biased, so the capacitor cannot charge, therefore the voltage at the output is also 320V.


2nd 90 degrees of the power supply phase - the voltage drops down to 0V. The capacitor voltage stays at 0V, so the output drops to 0V.


3rd 90 degrees of the power supply phase - the voltage goes negative and reaches -320V. Now the diode is forward biased and the capacitor can charge (current goes CCW), which means that the output stays at 0V. The capacitor charges to 320V


4th 90 degrees - the voltage does down to zero again, but the capacitor cannot discharge (for the capacitor to discharge, the current has to flow CW, however, the diode prevents it), so the voltage across it stays. The output voltage rises to 320V


5th 90 degrees (first 90 degrees of the second cycle), the voltage rises to 320V, but since the capacitor is charged to 320V too, the output rises to 640V.


6th 90 degrees - the voltage drops to zero, but the output only drops to 320V, since the cap is still charged.


7th 90 degrees - the voltage goes to -320V, the output to zero, the capacior is charged, so no current flows.


From here the 4th - 7th parts loop.

Essentially, the capacitor charges and provides a voltage offset to the power supply, look at this graph, the red line is input, the green line is output, vertical scale for both is 500V/div.


  • \$\begingroup\$ I believe @Pentium100's analysis is correct. One small point is that the output is actually shifted down by the forward bias voltage of the diode (~0.7V). This would be hardly noticeable for an input of 320V, but for say 5V it would be more apparent. This makes me think this circuit could actually be useful for something, there are times when one wants a smallish negative voltage, perhaps to get a single-supply op amp to get all the way to zero. Something like this with perhaps two diodes in series could give you a 1.2-1.4V source without having to drop all the voltage of a full negative rail. \$\endgroup\$ – scanny Aug 5 '15 at 21:41
  • \$\begingroup\$ Thanks, but I got stuck on the first 90 degrees in your explanation. I think for it to work there has to be a load so that current can flow and charge up the capacitor. On the better-understanding side, I notice that the Villard is actually not really a voltage doubler, rather it moves the bottom of the output wave up to zero (so it's a rectifier, I guess). \$\endgroup\$ – Elliot Aug 7 '15 at 14:16
  • \$\begingroup\$ The capacitor charges through the diode in the 3rd step. \$\endgroup\$ – Pentium100 Aug 7 '15 at 15:09
  • \$\begingroup\$ OK was just looking at a simulation with a resistor load on it and got mixed up, sry... \$\endgroup\$ – Elliot Aug 7 '15 at 18:55
  • \$\begingroup\$ At the moment I think that the output will be zero for the first half cycle (before the capacitor starts charging up). Explantion: with the diode not conducting due to being reverse-biased, and hence no current flowing and no charging of the capacitor, the output looks like being isolated from the voltage source to me. I did a simulation: tinyurl.com/o2t7sad Seems necessary to press reset for some reason or else the result is more like what you have said with output voltage from the start.... \$\endgroup\$ – Elliot Aug 8 '15 at 21:58

yeh wat he said. maybe i can add a few things though to make it a little more simple. its an ac or pulse circuit. when the current flows from - to + on the input voltage, the circuit is completed on the cap side and the energy is stored into the capacitor. When the current flows from + to - the energy isn't released from the capacitor because it has no load to drain to. it becomes a voltage booster/doubler when the circuit its complete. the current will fill the capacitor flowing up through the diode. when the current switches directions, both voltages from the source and voltage from the capacitor are released into the load. im a nooby to this but maybe it helps! :) enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.