Kevin Weddle

I think at 1Khz, the capacitor could not completely charge in one cycle. No, that's not it. You have to make the resistance higher than the impedance of the capacitor.

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Something I noticed about voltage multipliers. I've never seen two equally charged capacitors reverse bias a diode. But as long as the capacitors are completely charged the diode is reverse biased.

I know there is a simpler 15V to 45V voltage multiplier than the one I posted. It might be adequate. I won't bother posting that one.

Wow this Windows 10 is great.

Thanks guys. I love to hear you.

They always conduct. No matter what. These two circuits are different. Resistors don't clamp a voltage. The voltages halve, halve, and halve again until Vp. ac.pdf

The other equation is Xc - Xl. So a series LRC is only inductive or capacitive.

In the quation v = L di/dt. The voltage goes up and the current goes up. In the equation i = C dv/dt. The voltage goes up and the current goes up.

The only addition I'll add to my post is the current of the capacitor must at least be C dv/dt.

Capacitor's pull current through a resistance. Then you have voltage momentarily.

I went ahead and built this circuit today. The input resistor voltage changed very little. The output transistor voltages changed quite a bit.

The schematic shown has a gain of 10. I changed the gain to 10,000 in this circuit. I don't know how unstable that would make it. I guess I'll have to match the opamp and output transistors.

It is versatile.