What I need help designing is a circuit that will be capacitor voltage divider that will be fed by a high Ac input voltage, but be able to discharge each capacitors through diodes at a set lower voltage for dc output. My main goal, which I suspect this voltage divider/charge pump set will do, is to charge the capacitors in series at a higher voltage, and discharge each capacitor through diodes in parallel. Assuming this functions as I suspect it might, I will want to be able to sum up all the parallel capacitor discharges(after each capacitors respective diode). Will this function as I think it might? Thank you for the advice and I apologize for the shadows in the crude wiring diagram.

I added a 2nd diagram, after thinking about this design, does this diagram make more sense? enter image description here

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    \$\begingroup\$ This site includes a schematic drawing tool. Please post schematics using this tool or post higher quality images for schematics. \$\endgroup\$ Mar 11, 2021 at 5:20
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    \$\begingroup\$ CircuitLab is very limited and not so great to use. I have mostly given up on it. The OP's drawing is fine. \$\endgroup\$ Mar 11, 2021 at 18:14
  • \$\begingroup\$ I assume your battery cell symbol is a capacitor! \$\endgroup\$ Mar 11, 2021 at 21:04

2 Answers 2


I kind of get what you're trying to do. The flaw in your design is that the back-to-back diodes creates a DC path between each capacitor junction, which defeats the voltage divider function you were trying to make with the caps in series.

So I played around with this and came up with something that kind of works, as below:

enter image description here

Simulate it here: (Falstad sim)

What I did differently is to block the DC path to the diodes with caps, allowing each series-connected cap to float. It's driven with 1000V AC (+-1000V peak), and as designed each cap only sees 1/5 of that, which I believe was your intention. Note the cap values are selected to balance their voltage drop and to equalize the currents through the diodes.

This delivers about 15mA at ~16VDC, dropped from 1000V AC. The Zener diodes are necessary to prevent the output from floating up to a very high value (700V or so) without a load.

To modify the input voltage while maintaining the current, scale the caps up. So for 230VAC rms (+-325V peak), scale them up by a factor of about 3.

For what it's worth it's very inefficient: the current flowing in the HV side is about the same as that in the LV side. It's not behaving like a charge pump at all; there's no impedance conversion going on. This is where a some kind of a flying-cap charge transfer would help.

Note that this is only a theoretical exercise. I wouldn’t actually ever use the circuit in a working system - much too inefficient, and very dangerous.

BONUS: A circuit that charges in series, discharges in parallel (and it's efficient, too!):

enter image description here

Simulate it here: (Falstad sim)

This circuit uses FETs to steer the cap charge. It works in two phases:

  • Charge: caps are charged in series. At the end of the charge cycle, each cap has about 70V max on it (out of 325V)
  • Discharge: Caps are dumped into the load in parallel. Not all of the charge is dumped, only a little. This is the key to the efficiency.

The control clocks are 30KHz, as a two-phase non-overlap clock to switch the FETs. I'll leave the actual design of this as an exercise for the student; I just want to illustrate the switched-cap principle here. It's important that they not overlap to avoid shoot-through.

One more subtle thing: N-FETs on the discharge path work as followers, with a gate control pulse height to roughly set the output voltage. The FETs turn off when Vout is (pulse height - Vgs), or with no load, 10V-1.5V = 8.5V. A real system would probably use duty cycle. You can play around with this with the sliders.

This is making about 9W @ 14V power.

  • \$\begingroup\$ this is very interesting. If used instead with 180Vpk input, considering the isolation provided by the capacitors, do you still consider it dangerous? Even with the resistor grounded, as in your circuit? \$\endgroup\$
    – devnull
    Mar 11, 2021 at 21:54
  • \$\begingroup\$ Yes. In general, capacitors can't be relied upon to provide a galvanic isolation from the line, at least with values large enough to give meaningful power transfer. By definition they form a leakage path to line. \$\endgroup\$ Mar 11, 2021 at 21:57
  • \$\begingroup\$ Thats a great simulation, thank you. A question with your design, would it function if there was simply one forward biased diode coming off between each capacitor junction, producing a 1/2 wave rectifier? Then again paralleling the all the cathode ends of the diode for a total output at the same voltage but massed current? \$\endgroup\$ Mar 12, 2021 at 3:40
  • \$\begingroup\$ I don’t think so, not without some active control. You’re welcome to try it by editing the sim yourself. \$\endgroup\$ Mar 12, 2021 at 3:55
  • \$\begingroup\$ Bonus content - check it out. \$\endgroup\$ Mar 12, 2021 at 5:53

That's not going to work, consider when the input is a large neagtive voltage, there's a direct path through two diodes to the output.

One way to fix that would be to replace the diodes with thyristors and trigger them one pair at a time. This makes the circuit much more complicated.

  • \$\begingroup\$ That is an interesting idea with the Thyristor, even though it would be more involved. Do you see anyway to do this with diodes assuming it will not function is drawn? How would this work with thyristors? One thyristor per capacitor? Would I be able to mass up all the capacitors in parallel? \$\endgroup\$ Mar 11, 2021 at 6:05
  • \$\begingroup\$ no way with diodes because diodes will conduct when you don't want them to, and thus most of the capacitors will be bypassed. \$\endgroup\$
    – Jasen
    Mar 11, 2021 at 8:44

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