What will be the equivalent voltage rating of two capacitors of different voltage rating connecting in parallel ?

And how does it differ in the series connection of two capacitors with two different voltage ratings?


You cannot apply a voltage higher than the lowest rated capacitor voltage to the string. If you have 3 capacitors, one rated for 16V, one rated for 35V, and one rated for 100V, you cannot apply more than 16V to the string. Realistically you should even derate the minimum voltage. I would not apply more than 12 volts to a 16 volt capacitor.

Generally speaking, when capacitors of equal voltage ratings are placed in series the voltages are added together. That being said, if the capacitors have different voltage/capacitance ratings then there is no good way to determine the overall voltage rating. The "weak point", so to speak, could be anywhere in the string depending on the order of the capacitors. (Thanks Neil for pointing this out)

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    \$\begingroup\$ para 2, I was going to write 'only if capacities are equal', but then realised it's charges to be equal, but then realised the mid point can float around so all bets are off anyway. Modify your answer? \$\endgroup\$ – Neil_UK Jun 28 '17 at 13:55
  • \$\begingroup\$ @Neil_UK Good point! Is the update sufficient? \$\endgroup\$ – DerStrom8 Jun 28 '17 at 14:00
  • \$\begingroup\$ With the modified answer there are more questions coming out. When you say "order" is is the pcb layout that you point to @DerStrom8 ? \$\endgroup\$ – AAI Jun 28 '17 at 16:26
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    \$\begingroup\$ Not really, with two capacitors in series, the mid point voltage is in general undefined, so either capacitor could over-volt. If for some reason, they both start off at zero voltage and then charge, we can predict the mid point voltage (until leakage eventually makes that calculation moot) from equating charges, so with equal C/V ratios they would both get to peak volts at the same time. Balancing resistors, or shunt diodes and an AC drive would define mid-point voltages. \$\endgroup\$ – Neil_UK Jun 28 '17 at 17:01
  • \$\begingroup\$ @AjeyaAnand When I said "order" I meant within the string, which ones begin charging first, or which ones have residual charge, etc. It is nearly impossible to predict in the real world \$\endgroup\$ – DerStrom8 Jun 28 '17 at 17:10

When 2 capacitors are connected in parallel, the voltage rating will be the lower of the 2 values. e.g. a 10 V and a 16 V rated capacitor in parallel will have a maximum voltage rating of 10 Volts, as the voltage is the same across both capacitors, and you must not exceed the rating of either capacitors.

With capacitors in series, they will have the same charge (current/time) and as the voltage is equal to Q(charge)/Capacitance, then the higher the C, the lower the voltage. All this means is that the voltage developed across the capacitor in series will depend on its capacitance, and so it is safest to only have the lowest voltage rating across the series capacitors.

If they all have the same capacitance value, then you can use the addition of the rated voltages. I would also recommend, as real capacitors generally have poor tolerance, that you do not apply voltages at or near the rated value.

  • \$\begingroup\$ If I exceed the rated voltage of one of the capacitors; then does it make the next capacitor the dominant one in the circuit. What happens to the lower voltage rated capacitor, will it blow off ? \$\endgroup\$ – AAI Jun 28 '17 at 13:56
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    \$\begingroup\$ @AjeyaAnand It depends on the failure mode of the capacitor. If it fails dead-short, then there is a chance it would explode, cause fires, etc. If it simply burns off and opens up, then the next highest voltage would take over. There is also the chance that the capacitor would fail partially-shorted, which would load down your supply voltage and the voltage at your primary load would not be sufficient. Not worth the risk regardless - size your caps accordingly. \$\endgroup\$ – DerStrom8 Jun 28 '17 at 14:02
  • \$\begingroup\$ @DerStrom8 I read that we have to derate the capacitors so that they do not reach the maximum rated voltage. However, when it is close to 80% or more of the rated value, the capacitance itself gets degraded to a lower value. This may be the leakage or start of dielectric breakdown. I would like any pointers that can help point to the capacitance values that act on the circuit, just before capacitor breakdown. Note, this is only simulation study I would like to do. I do not want to burn caps or blast circuits. \$\endgroup\$ – AAI Jun 28 '17 at 16:23

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