Problem in Voltage Doubler

[dangan1993]

Can someone please to help me explain why the DC output is limited by the input power or the DC output get saturated as the input power is high enough? Thank you.

 

[BradtheRad]

Is your question about a plain voltage doubler made from diodes and capacitors? With no load there is a maximum voltage you get at the output. The incoming waveform charges the capacitors to its maximum peak amplitude. There is nothing which discharges them. (Is this what you mean when you say 'saturated'?)

As you add load the capacitors (all of them or some of them) drop to a lesser voltage. It is because the load discharges them somewhat, yet they are not replenished sufficiently during the brief time the incoming waveform exceeds their charge level.

The Farad value of the capacitors has a lot to do with it. Usually by increasing their value, the voltage multiplier can sustain a heavier load.

 

[dangan1993]

Is your question about a plain voltage doubler made from diodes and capacitors? With no load there is a maximum voltage you get at the output. The incoming waveform charges the capacitors to its maximum peak amplitude. There is nothing which discharges them. (Is this what you mean when you say 'saturated'?)

As you add load the capacitors (all of them or some of them) drop to a lesser voltage. It is because the load discharges them somewhat, yet they are not replenished sufficiently during the brief time the incoming waveform exceeds their charge level.

The Farad value of the capacitors has a lot to do with it. Usually by increasing their value, the voltage multiplier can sustain a heavier load.

Yes, I concern about a plain voltage doubler loaded with a resistor. In theory, DC output voltage = 2Vpeak. there is no constraint with the peak amplitude of input signal, but I read some papers, Dc voltage is saturated as input power is high. Is it because the non-linear property of diode?

 

[BradtheRad]

Is it because the non-linear property of diode?

Not that, but because diodes don't conduct when volt levels at each side are the same. During each cycle there is an idle period when the capacitors drop in voltage.

Then a peak waveform comes through. It causes a brief burst of current through the diodes, charging the capacitors. (A simulator demonstrates this behavior.)

Often the burst of current lasts a small fraction of a cycle. However a heavy load draws so much current that the capacitors cannot maintain the same as the no-load peak, no matter how long the diodes conduct.