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I have a built a simple voltage doubler. the circuit is as below:

schematic

simulate this circuit – Schematic created using CircuitLab

This represents only one stage, and my circuit has 4 stages. The capacitors are 1nF 1kV ceramic capacitors and the diodes are BY228GP-E3/54 (2.5A Iavg 1500V silicon diode with max 1.6V forward voltage drop) the circuit is shown below:

enter image description here

So this circuit is supposed to be connected to mains (230V) to generate about 2.6 KV open circuit output voltage. Before doing so, i tried connecting a 24Vrms 50Hz sine wave to the input to make sure the circuit is working fine, and I simulated this in Ltspice to have sth to compare my results with.

The circuit acts completely different! The first capacitor in my circuit and the following diode (C2 and D1 in the schematic) form a voltage clamp. the voltage across C2 in the simulation was as follows (as one would easily guess):

enter image description here

the voltage measured across the first cap on the proto board is shown below:

enter image description here

This cap seems to consume all of the energy of the injected signal somehow! I am a bit lost, and could use some help, thank you in advance.

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    \$\begingroup\$ I do not see the wire which connects the anode of D2 to C2. Without it the voltage increase does not reach D2. I might miss something, but even the simulation does not meet your expectation, doesn't it? \$\endgroup\$ Jul 14, 2020 at 14:32
  • \$\begingroup\$ I can't tell from your pictures. Is the cathode from D1 connected to the anode of D2 (for all stages)? \$\endgroup\$ Jul 14, 2020 at 14:32
  • \$\begingroup\$ Are you sure you measured the voltage across the capacitor? Isn't it the ground reference voltage of the other side of the first capacitor? \$\endgroup\$ Jul 14, 2020 at 14:37

2 Answers 2

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The 1 nF capacitor has an impedance of 3.18 MΩ at 50 Hz and you are connecting this to your oscilloscope probe that probably has an input resistance of between 1 MΩ and 10 MΩ. Do you see the problem here?

The BY228GP diode has a specified leakage of typically 5 uA. That's barely below the current flow into the capacitor at 50 Hz from 24 volts so, do you see this problem too?

I tried connecting a 100KHz square wave @ 50% duty cycle and 5V peak voltage. so at this frequency, the impedance of the caps would be 1.6K, conciedarably lower than the measuring instrument (1-10 MOhm), but I am still not getting my output correctly

Now your problem is the forward volt-drop of the diode. If we consider it to be circa 1 volt, then, because there are two diode drops per stage, the 5 volt source is eroded by 2 volts. This is equivalent to a 3 volt driving source using ideal diodes. Does that tally with what you found?

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As Andy aka says, your capacitors are simply too small.

The output voltage won't meet your expectations, anyway, even if you had capacitors that were large enough.

This equation gives you the theoretical output voltage of an n-stage Cockcroft-Walton half wave voltage multiplier:

$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{2 \pi fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$

Given:

  • \$E_{pk}\$ the peak voltage of the AC input (not the peak to peak voltage, but the simple peak voltage.)
  • \$I_{load}\$ - the load current.
  • \$f\$ - the frequency of the AC input
  • \$C\$ - the capacitance of your individual capacitors (in farads.)
  • \$n\$ - the number of stages in your multiplier.
  • \$V_{f}\$ - the forward voltage of your diodes.

The theoretical unloaded output voltage of a four stage multiplier would be about 2900 V, given a peak voltage at the input of 365 V peak (230VAC.)

Any load will cause the output voltage to drop, and will increase the voltage ripple on the output.

You need to figure out if the extra 300 volts will be a problem. You also need to figure out how much current you will be drawing from your high voltage output - that will determine the size of the capacitors you will need.


I borrowed the above equation and description from my blog. There's more there about various types of voltage multipliers.

I did many of my experiments using 100nF capacitors at 50Hz. I found that the output begins to "sag" quite badly with just 3 stages. The voltage when using just the "load" of my oscilloscope (1MOhm) was noticeably lower than when measuring the output voltage with a voltmeter (10MOhm.)

To get usable voltage multiplier at 50Hz, you'll probably want to go with 1µF capacitors. The alternative is to use a higher frequency.


I recommend you do not connect that thing straight to mains. With any capacitor value large enough to be useful, touching any part of the circuit would amount to touching the mains directly.

Use an isolation transformer at the very least. If that's not possible, enclose any exposed leads in a box so you can't possibly touch them.

I did my experiments at low voltage using a 230VAC to 9VAC transformer. You may want to do any tuning or testing at low voltage, then switch to high voltage when you have the multiplier itself working properly.

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  • \$\begingroup\$ I tried connecting a 100KHz square wave @ 50% duty cycle and 5V peak voltage. so at this frequency, the impedance of the caps would be 1.6K, conciedarably lower than the measuring instrument (1-10 MOhm), but I am still not getting my output correctly, could it be entirely because of the leakage out of the diodes then as "Andy aka" said. I am planning on using a auto transformer to get different voltages and feed them into a full wave rectifier and chop that off with a mosfet at 100KHz. \$\endgroup\$
    – NeuroEng
    Jul 15, 2020 at 7:48
  • \$\begingroup\$ The leakage was a problem at 50Hz because the diode leakage was a large portion of the current available through the capacitors. The available current through the capacitors at the higher frequency should be high enough that the diode leakage isn't your major problem. \$\endgroup\$
    – JRE
    Jul 15, 2020 at 9:47
  • \$\begingroup\$ Given a 4 stage multiplier, and an impedance of 1.6k for each capacitor, the total impedance of your multiplier will be around 480k. That is significant compared to the load of your measuring instrument. \$\endgroup\$
    – JRE
    Jul 15, 2020 at 9:50

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