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I am building a hobby project - a sort of supercapacitor powerbank, where I basically connected twelve 500F 2.7V supercapacitors in series. Despite these capacitors being from same manufacturer and same batch, I instantly hit the disbalancing issue, that was mentioned, for example, here, and figured out that for this kind of stuff a balancing circuit is needed.

The simplest solution is just using "balancing resistors", for example like this:

I would like to understand how this works. I found many places that explain that this is needed, but nowhere that explains how such circuit actually works.

  • How are these resistors actually balancing anything?
  • What is the current path?

The way I understand it, they are basically connecting + and - of each capacitor, which probably increases its self-discharge rate, with higher voltage capacitor having higher self discharge. Is that it? Is this how it "balances" them? By discharging more charged capacitors faster than others? Is there some other path of current that actually does balance them (eg. moves charge from higher charged to less charged)? Can someone explain this balancing in simple way?

Also if balancing of power sources that are connected in series is so important, why is it not implemented inside of batteries that consist of multiple cells? For example, a 12V battery contains six individual cells connected in series, but there is no balancing circuit for them, yet they remain balanced and don't explode or catch fire over time. Why is this different from connecting two same batteries in series? I found that batteries suffer from same problem as capacitors and should be balanced when connected in series.

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    \$\begingroup\$ Resistors with more voltage across them have more current through them, so capacitors with more voltage discharge faster, which makes the array more balanced. This is a super cheap and junky way to balance things - don't expect it to work well. \$\endgroup\$ Commented Mar 28, 2023 at 11:00

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What you might be missing is the fact that the leakage resistance of one super-cap might be 1 MΩ whereas for another, it might be 9 MΩ. So, if you connected these two in series across a 5 volt supply, the voltage on each would start off equal (balanced by their respective capacitances being equal) then drift away from each other as determined by their individual leakage resistances.

In the example I gave above, you would end up with 0.5 volts across the capacitor with 1 MΩ leakage resistance and, 4.5 volts across the capacitor with 9 MΩ leakage resistance.

So, the added external resistors are present to equalize the leakage resistances.

Also if balancing of power sources that are connected in series is so important, why is it not implemented inside of batteries that consist of multiple cells?

It is on cells that are extremely vulnerable to over-charge scenarios i.e. lithium cells. For lead acid cells, it's not so important or critical.


Mini calculation based on external balancing resistors of value 100 kΩ

If you placed 100 kΩ resistors in parallel with the super-cap with 1 MΩ leakage resistance it's net leakage would be 91 kΩ. For the super-cap with 10 MΩ leakage, the net value would become 99 kΩ and, across a 5 volt supply you would get: -

  • 2.395 volts on one super-cap
  • 2.605 volts on the other super-cap

This is a much more balanced scenario. Of course, an active balancing circuit would make both super-caps have very nearly the same voltage across them but, that's quite an expense to go to and, best left for the guys trying to balance hundreds of series super-caps (there are a few applications).

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  • \$\begingroup\$ I still don't understand few things, correct me if I am wrong - so let's say I put these capacitors in series, 2 of them for simplicity, both fully discharged with 0V on them. Then I connect 5V power supply. They get gradually charged to ~2.5V each of them because capacity is roughly same. If I kept them plugged to that 5V supply, over long time the voltage would change as you said to 0.5V on one and 4.5V on the other? Or when would that happen? Would this happen even if used in this "power bank" scenario when they are not connected to power supply until discharge? \$\endgroup\$
    – Petr
    Commented Mar 28, 2023 at 14:16
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    \$\begingroup\$ Actually they will instantly charge to 2.5 volts each based on the capacitance being the same. Then over a few seconds they will gradually settle to a different value based on their internal discharging resistance being unequal. So, they pretty much instantly charge and take a big pulse of current from the applied 5 volts supply in doing so. And, yes, based on my numbers, they will gradually migrate from equal voltage at the application of the 5 volts to 0.5 volts on one and 4.5 volts on the other. It happens over the equivalent RC time constant. \$\endgroup\$
    – Andy aka
    Commented Mar 28, 2023 at 14:48
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    \$\begingroup\$ (RC time constant: 5 farads * 1 megaohm = about 2 months... but yours are 500 farads so 100 times longer... if the leakage is 1 megaohm) \$\endgroup\$ Commented Mar 28, 2023 at 15:51
  • \$\begingroup\$ @user253751 please use the @somebody to indicate who your comment is for. You might also suggest what point you are trying to make. \$\endgroup\$
    – Andy aka
    Commented Mar 28, 2023 at 16:02
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    \$\begingroup\$ I am not saying that these balancing resistors aren't needed, but I am just not able to replicate the described effect, which I think is simply due to the fact it would take many months for it to be visible at all. Actually given the datasheet numbers, the internal resistance should be around 3.5MOhm, if that RC constant calculation by @user253751 is correct, that would be some 50 years until this effect is visible. \$\endgroup\$
    – Petr
    Commented Mar 28, 2023 at 19:58

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