# Unbalanced super capacitors in series

#### pgib8

Jul 26, 2015
107
Something new to me is that when you put 2 (or more) supercaps in series that one of them could reach a higher voltage than the other and possibly exceed its rated voltage.
It appears there are 2 mechanisms by which this happens, one is variance in capacitance and the other is variance in leakage current.
I've been searching the internets for hours and all I ever see is they talk a little bit about it and jump right into how to balance the supercapacitors.

What I would like to know is the long-term effect. For example does the leakage continue to increase voltage to one or does it reach an equilibrium.
Let's say I had a battery with 4V exact and would stay like that forever and I connect 2 super caps in series rated at 2.7V each. Let's say I let this experiment go for a whole year. Would the imbalance continue to build up until one of the capacitors fails? Or will the imbalance reach some type of equilibrium where it settles and with an extra 1.4V margin across both capacitors maybe it would ultimately settle still within the rated voltage.

I'll also report back if I find out more, and I'll be doing some experimentation as well.

#### Alec_t

Jul 7, 2015
3,330
It's most unlikely that the two caps would have exactly the same capacitance, so they won't share the 4V equally even when brand new. It's possible too that they will age differently, increasing the imbalance. As you can't predict their behaviour you have to design around it, assuming a worst case scenario.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,508
Yes, if one had higher leakage than the other, the charge will be slowly shifted from one to the other.

There are multiple methods of balancing supet capacitors in series. I'm sure you will find several as you look.

They can be as simple as resistors across the capacitors, or zener diodes in parallel.

One interesting option involves switching a capacitor alternately across the capacitors

There are also active methods

#### pgib8

Jul 26, 2015
107
I'm not looking for advice on how to balance them, so far I'm only reading the same stuff I read online.
How far will the leakage take the imbalance, what are the long term effects (no balancing circuit applied), will they reach equilibrium?
Bonus: Describe the difference in the way variation in capacitance versus variation in leakage current contribute to the imbalance.

#### pgib8

Jul 26, 2015
107
I don't think it's correct.
Vcap1=Vsupply x (Ccap1/(Ccap1 + Ccap2)
Assuming Vsupply=5.4V
Vcap1=5.4 x (12/(12+9)) = 3.08V

In this formula the capacitor with the higher capacitance is getting the higher voltage, however in the same page it says this:
a cell with a lower capacitance will charge to a higher voltage in a series string

Perhaps I'm not getting it, I contacted them about it.

#### Alec_t

Jul 7, 2015
3,330
How far will the leakage take the imbalance
It can result in one cap having virtually all the 4V across it, unless both caps have identical leakage.

#### pgib8

Jul 26, 2015
107
It can result in one cap having virtually all the 4V across it, unless both caps have identical leakage.
Currently my theory is that the leakage will increase with voltage so that they will never reach a 0V to 4V imbalance; but if I'm wrong then it would just be a matter of time before one cap will be at 4V and the other one at 0V

#### Alec_t

Jul 7, 2015
3,330
Currently my theory is that the leakage will increase with voltage
That seems reasonable (I haven't researched the topic) and can be modelled (at least approximately) as a parasitic resistor in parallel with a perfect cap. So if each cap, as modelled, has a parallel resistor then the voltage at the junction of the two caps in series will tend towards the voltage at a simple resistive voltage divider.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,073
In this formula the capacitor with the higher capacitance is getting the higher voltage, however in the same page it says this:
a cell with a lower capacitance will charge to a higher voltage in a series string
The equation is wrong, the text is correct. From C = Q/V = (I × t) / V follows V = (I × t) / C. As the product (I × t) is the same for each capacitor in series, the variation is in C only and therefore a smaller C results in a higher voltage.

Currently my theory is that the leakage will increase with voltage so that they will never reach a 0V to 4V imbalance; but if I'm wrong then it would just be a matter of time before one cap will be at 4V and the other one at 0V
During charging, leakage current is subtracted from the charge current, effectively bypassing the capacitor. Worst case scenario: leakage current = charge current -> no charge stored in the capacitor, see @Alec_t ' s post #6.
Oh, by the way for those inclined to comment on "charge stored in a capacitor": please don't. I do know that it is technically not correct as charge is only redistributed within the capacitor, but discussing this detail will not help in furthering this thread - not again, please.

When not used or during discharge:
The leakage current will discharge each capacitor at its own rate. This leakage doesn't flow out of one capacitor and into the other capacitor, charging this other capacitor. What happens is that both capacitors discharge internally with time at different rates depending on true capacitance, charge voltage and, of course, leakage current. Compare this to @Alec_t 's model of the internal resistor in parallel.

#### pgib8

Jul 26, 2015
107
Thanks Herald. What you are saying about the leakage is also the way I understand it. The leakage is like a resistor across that capacitor. This will place a greater share of the voltage on the other one. Like you said in the worst case scenario with 100% leakage is essentially bypassing one of the capacitors.

Since I started my thread I started a little test, this is not a good test yet because I'm using 5V super caps and I'm using a bench power supply. I will be starting a real test soon where I will use single cells (2.7V) and a battery. However my rough test is already showing some results. First one of the capacitors was slowly increasing in voltage while the other one was decreasing. I didn't take notes, maybe 0.05V per day. I had both of them very close to their rating. Then about 2 days ago, the one that was increasing in voltage started to decrease. My guess is that one of its cells got overcharged and took some damage, increasing its leakage. If what I'm thinking is right, then now the other one will increase until it takes damage and that may keep going back and forth until both capacitors are toast. Like I said, I'm going to do a real test (waiting for parts), take proper notes and report back (it will be a long test though).

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