Help me to understand the simplest voltage multiplers

Question

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

AC Voltage source and capacitor

Answer 1

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:

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.

Answer 2

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! :)