Supppose i have the following circuit. And I complelty discharge all the capacitors by having a load on them for a long time, so there will be 0V overall. Let's assume the one in the middle has a slightly different capacity. My question is will C5 have a negative Voltage because it has less capacity ?


simulate this circuit – Schematic created using CircuitLab

I'm asking this because in my setup i have 3 Supercaps (which have polarity) connected in series and I'm wondering if I can discharge them all the way without giving one of the caps a reverse polarity.

Some thoughts:

This seems like a very basic question, but i've never asked myself this until now. So from a logical point of view it must get negative, because let's say they all start at 3V when discharging . Then the voltage starts to drop , then one of the caps must reach 0V first because it hass less capacity and then the current should still continue to flow because the overall potential on the load is > 0V .

But my first instinct was that it shouldn't get below 0 , although I have no argument for that it's just a feeling. So that's why I'm a bit skeptical what will happen.

  • \$\begingroup\$ Voltage distribution across capacitors in series, where caps are not of equal value, is different. Start with that modifier and see what you get. \$\endgroup\$ – Kripacharya Aug 2 '19 at 19:52
  • \$\begingroup\$ Why do you believe that the voltage wouldn't "average out" across the caps, such that they all reach 0 at the same time? Caps in series have a much lower capacitance value, (around 330nF in your case). If you want them to act together, you'd have to connect them in parallel. \$\endgroup\$ – Ron Beyer Aug 2 '19 at 19:54
  • \$\begingroup\$ @RonBeyer No idea why , it was just my initial feeling , but I just ran an experiment and the smaller cap got a a negative voltage. Thx guys \$\endgroup\$ – KoKlA Aug 2 '19 at 20:02
  • \$\begingroup\$ @Kripacharya Sry I don't understand what you mean with : "Start with that modifier and see what you get." But i just tried it out with some smaller caps as mentioned in the comment before and it got a negative Voltage . So I'm happy now, thx \$\endgroup\$ – KoKlA Aug 2 '19 at 20:04

It depends on how they were charged to begin with.

If you charge the capacitors to the same voltage individually, then connect them in series and discharge them, then when the total voltage reaches zero, you'll find that the largest capacitor has a small positive bias on it and the smaller capacitors all have small negative biases.

On the other hand, if you connect the discharged capacitors in series and then charge them, you'll find that the largest capacitor ends up with the smallest relative voltage, and all of the smaller capacitors have larger voltages across them. When you discharge them, they all reach zero volts at the same time.

  • \$\begingroup\$ Thank you , very good response, i upvoted but it will only count once my account reaches enough reputation. Just one additional thing I'd like to point out , even if you charge them all together (2nd Scenario) , they might still end up at different voltages due to different self discharge over time. Also this assumes that all the individual caps have the same ratio of charge-effiiciency/discharge-efficiency lets call this er though I have no idea if there can be deviations of er in practice. But i guess electrochemical reactions can have weird side effects. \$\endgroup\$ – KoKlA Aug 2 '19 at 20:22

If the capacitors are ideal, then no, current through them will always be equal and they all get charged with the same charge Q=It=CU but the voltage will be different as the capacitances are different. And by discharging them, they also hit zero voltage simultaneously.

But you are asking about real capacitors, they have leakage current through them that varies. In general, electrolytic capacitor banks that have capacitors in series sometimes have equalizing resistors over the caps to prevent excess imbalance of charge. Assume two identical supercaps, of which one has leakage current of 6uA and the other 4uA are put in series and charged to 10V and kept there. Initially they both charge to 5V, but due to difference in the leakage currents, there will be voltage imbalance, and thus when rapidly discharging, there will be negative voltage over the one with less voltage.

Supercapacitor application notes do suggest using active or passive balancing when capacitors are connected in series.


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