I want to make capacitor with large capacity which can hold both positive and negative charges. Is it possible to combine elecrolytic capacitors somehow to achieve this?

  • \$\begingroup\$ Yes, of course it is possible, somehow. But then what? There's not enough info how would you like to use such a cap, what would it achieve and how would you like to charge and discharge the capacitance? \$\endgroup\$
    – Justme
    Commented Nov 1, 2021 at 8:35
  • \$\begingroup\$ Two electrolitics each with a protection diode would get you most of the way there, though the reverse voltage through the diodes would significantly reduce the life of the caps. Why are you trying to do this? What's wrong with unpolarised caps? Or it might be easier to rectify the input voltage, and then carry on with a polarised cap. Depends on why you are doing this and what the rest of the system is. \$\endgroup\$
    – Puffafish
    Commented Nov 1, 2021 at 8:36
  • \$\begingroup\$ Yes, by putting them in anti-series. See here: newbedev.com/… \$\endgroup\$
    – winny
    Commented Nov 1, 2021 at 8:40

1 Answer 1


Yes. There are two ways to do this. The rough way, and the nice way.


simulate this circuit – Schematic created using CircuitLab

The rough way is to omit the positive bias and R1. Diodes D1 and D2 will conduct when required to rectify the input signal and 'pump up' the middle node to keep the capacitors from becoming more than one diode drop reverse biassed, which is usually OK for aluminium electrolytics. While the initial charging is happening, the overall capacitor will appear to have a value of C1 or C2, with a diode drop in series. Once the middle node is charged, the capacitor will behave as C1 in series with C2, capacity = C1*C2/(C1+C2).

The nice way is to use a positive bias to pre-charge the middle node to more than the expected signal swing. This will keep both capacitors correctly biassed throughout. The diodes are still required for safety.

Usually, you'll make C1 equal to C2, though there's no requirement to.

  • \$\begingroup\$ How to make C1 equal to C2 if tolerance +80%/-20%? Why just not a bridge? \$\endgroup\$
    – user263983
    Commented Nov 1, 2021 at 12:31
  • \$\begingroup\$ How does the bias work if the capacitors are discharged? You'd have to charge the capacitors through the bias before the bias becomes effective, not? \$\endgroup\$
    – RJR
    Commented Nov 1, 2021 at 12:39
  • \$\begingroup\$ @RJR which bit of 'precharge' don't you understand? \$\endgroup\$
    – Neil_UK
    Commented Nov 1, 2021 at 13:56

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