Electronic – Order capacitors by leakage rate in Cockcroft-Walton voltage multiplier

multipliervoltage

Looking at a system like the Cockcroft-Walton multiplier posted here: https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/voltage-multipliers/

To use 400V 100uF electrolytic capacitors running at 110VAC (the first capacitor cannot be electrolytic: it has to run in both polarities), do you think there is any advantage of measuring the leakage in each capacitor, and ordering them in the chain?

The reason to do this would in theory be to reduce the build-up of voltage within the capacitor chain: Supposing I have N capacitors, and I order them from lowest leakage to highest, would that give me a more stable combination than mixing them up in a more alternating order (hi, lo, hi, lo, etc)?

I do understand that there should be balancing resistors in parallel. Just wondering if there is any value in ordering them, since there is a huge variation in the capacitors I have looked at.

Best Answer

Cockcroft-Walton Amplifier Cockcroft-Walton Amplifier - Wikipedia

For a stable output, you at least want to have low leakage in the bottom capacitors. If any of these capacitors has leakage, then it will cause \$V_o\$ to decrease continuously as if an extra load would be present. When \$V_i\$ transitions from low to high, the output will be bumped up again. The output ripple is affected immediately.

The top capacitors do not influence \$V_o\$ directly so their effect on \$V_o\$ is more limited. The leakage will only slightly increase the output ripple which is caused when the bottom capacitors are used to charge the top capacitors. The lost charges cause a slight excess drop of the output voltage.

C1 is probably the least important, as it is not charged by a capacitor on the bottom but by the input directly. Of course, leakage is still to be avoided. Leakage in any capacitor will decrease the maximum output voltage and efficiency.

I suggest using simulations to find something that works for you. In my opinion, this type of multiplier is easy to understand if you don't think too hard about it. Once you try to go a bit more in-depth it quickly gets messy (to give some indication of what I mean: If \$V_i\$ transitions from high to low, then C1 is not only charged via \$C1 \rightarrow V_i \rightarrow D1 \rightarrow C1\$, but also via \$C1 \rightarrow V_i \rightarrow C2 \rightarrow D3 \rightarrow C3 \rightarrow C1\$).