Supercapacitor balancing?

Discussion in 'General Electronics Chat' started by snowdrifter, Aug 23, 2013.

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Aug 13, 2013
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I have 14, 3000f 2.7f capacitors. Wired in parallel pairs then wired in series together.

I'm looking for something simple yet effective in terms of a balancing circuit. I'm using 50 ohm bleeding resistors right now, but those drain the cap down over a couple days and because it's wired in parallel with a battery, drains the battery.

The resistors work well, but I'm trying to get to the point where I don't have to tend to the setup and either wire the battery/cap to a charger or unhook the capacitor then recharge it later if I won't be there using it for a few days.

I've looked into Zener diodes to only bleed the capacitors when their voltage gets too high, but it seems like they don't allow enough current though to do this effectively. At least not the ones I was using. 8, 2.5v zeners in parallel would only allow a bleed current of 35ma with a capacitor pair charged to 2.65v

Ideally: I'd like to limit the maximum voltage of any individual capacitor to 2.4v, mainly for my own OCD. But also so I have some overhead since the capacitors will be charged with a 400 amp charge, and discharged with an approx. 1000 amp load.

If that's not practical, then something that will allow me to discharge the bank from 16.8v down to 11.5v safely without risking "reversal" of any of the individual cells.

Here is a picture of said bank. Excuse the odd configuration, it was the only way I could get it to fit within my space requirements lol

Any ideas? I've done some searching for balancing circuitry before, but given that I'm really new to this whole circuit diagram thing, I just end up glazing over when I see the diagram.

Last edited: Aug 23, 2013

Aug 13, 2013
43
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Just as an experiment I upped the amount of diodes to 14 in parallel and got a 42ma draw with a single cell at 2.6v

Apr 5, 2008
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4. Rbeckett Member

Sep 3, 2010
205
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I think he meant Obsessive Compulsive disorder referring to being extra cautious and insuring he doesn't get bit by a bunch of charged up caps. Definitely an attention getter if you get across a fully charged supercap... Don't ask me how I know, it was not pretty

Wheelchair Bob

5. wayneh Expert

Sep 9, 2010
12,361
3,220
Define "balancing". I don't really understand the problem. Is this an issue during charging only, or while the capacitors sit charged but idle?

If you are charging at 400A and potentially need to shunt 400A around a capacitor to prevent any additional charging, that's going to be quite a challenge.

6. #12 Expert

Nov 30, 2010
16,655
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Yes. My first impression is, "You gotta be nuts". That's going to need a thousand watts worth of regulators.
Don't forget to read my signature line.

Aug 13, 2013
43
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Here's what I was thinking: Since what I'm charging is limited to 16.8v, that **should** give me a bit of leadway right? So it's not like I need to be able to dissipate all 400 amps of current as much as I can take advantage of current over time. With the maximum voltage at 2.4v, perhaps that .3v leadway before I reach the max capacitor voltage would allow the use of something smaller to dissipate say.. A couple amps of current over time to bring the capacitor voltage down.

Or is this a poor assumption/idea? I'm not sure

@wayneh it's just making sure all the capacitors read roughly around the same voltage so that none get overcharged, or "reversed" during a heavy discharge. 50 ohm resistors keep them all within +/- .03v of each other when I've measured, but I run into the issue of unintended leakage current

8. #12 Expert

Nov 30, 2010
16,655
7,293
If you put a 2.4V limiter on each capacitor, and one of the capacitors has a bit less capacitance than the others, the limiter will have to pass that 400 amps around the lesser capacitor. Therefore, all bypass regulators must be able to pass 400 amps WITHOUT HAVING A SIGNIFICANT VOLTAGE DROP CAUSED BY THE CURRENT. Where are you going to find a semiconductor device with less than 1 milliohm of inherent resistance?

9. BLUESHIFT New Member

Aug 23, 2013
24
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Hey guys -
I just posted something about this on my site yesterday. Snowdrifter, it sounds like you are on a similar page, but why wait until you're at 2.4 V?? You can start balancing from OV, Q=CV holds through the whole voltage range.

http://blueshiftpdx.com/blog/2013/a-new-method-for-capacitor-balancing

Basically, I think the best way to do it is to dynamically balance the capacitors as they charge - this would limit the worst-case balancing current for capacitor #n to I*(C(n)-C(avg))/(C(avg)) which is to say 20% of I(total) if they're in spec, which I'm sure they are.

I'm a little unclear as to whether a transistor, comparator -> transistor, or op-amp -> transistor would be the best system, but it seems clear that balancing over the full charge is the way to go.

10. #12 Expert

Nov 30, 2010
16,655
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I don't think this will handle a 400 amp charge rate.

11. wayneh Expert

Sep 9, 2010
12,361
3,220
I don't think that circuit will work at all. As drawn, T2 would always be on. Neither transistor would have anywhere near enough base current to support 100A in the collector-emitter circuit. Since it is only responding to an imbalance, perhaps the balancing current can be far less. I think some modeling is called for here; it would be nice to know the maximum balancing current that is needed.

You might be able to design a something based on that dynamic balancing concept though, using comparators and big MOSFETs as switches. No, wait, I think you probably do need a proportional response, so maybe some PWM approach. Definitely a big project.

12. wayneh Expert

Sep 9, 2010
12,361
3,220
Instead of balancing, don't you "just" need a way to stop charging when any individual cell reaches the target voltage?

13. ian field Distinguished Member

Oct 27, 2012
4,447
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I've seen a collection of photos of supercaps like that used to start and run a car.

You'd need awfully big zeners to clamp each cell pair within their rated voltage - you could try a series chain of diodes per pair (sum of Vf per chain), you'd probably want to use fairly large diodes to handle the anticipated charge current, but at 0.7V per diode you can probably get nearer the target voltage than an off the shelf zener.

If you do replace your car battery with that stack - a PV panel on the roof isn't such a bad idea!

14. BLUESHIFT New Member

Aug 23, 2013
24
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I don't see any reason it wouldn't. Not yet at least...This is the wrong format for doing math but here goes:

Q=CV

Error voltage of capacitor(n) = Q/C(n) - Q/C(avg)
= Q*(C(n) - C(avg))/(C(n)*(C(avg))
approximately = Q*(difference from average capacitance) / C(avg)

approximate maximum error voltage V(e) of cap(n) at charge=Q
= .33*Q / (C(avg)) given: C(avg) = +20% and C(n)= -10%
= .33 V(avg)

dV(e)/dt = .33 dV(avg)/dt

dV(avg)/dt = 1/C(avg) dQ/dt = I/C

Cap bank:
= 3000F 2 parallel, 7 series (maximum voltage 2.7 V)
= 6000F, 7 series
= 1/7*6000F = 857 F, 1 series (maximum voltage 18.9 V)

V(total) = Q/C(total)
dV(total)/dt = I/C(total) = 400A/857F = .46V/sec

but this is distributed over the 7 series cell, so the average cell charges at .46/7 = .07V/sec = dV(avg)/dt

From above, the maximum change in error voltage/second in any one cell is:
dV(e)/dt = .33*dV(avg)/dt = .021 V/S

I(balance) = Current to reduce the voltage by .021 V / sec in one 6000F 'cell'

Q = CV
dQ/dt = C dV(e)/dt
I(balance) = 6000F * dV(e)/dt = 6000F * .021 V/sec = 130A

(I guess I could have just said the maximum is .33 * I(total) but that was more informative for me...)

So you need 7 x transistors that can pass up to (worst case) 130A.

One problem in this setup - if you are draining each cell into the next one down, and adjacent cells are over voltage, that would stack all that current into the lowest over-voltage transistor, so ones farther down potentially could see double, triple, etc. current depending on where the under-spec caps happen to be.

Maybe the best solution would be to drain directly to ground? In that case the top-cell transistor (and possibly series power resistor) would need to dissipate up to 18V * 130A = 2400W. That is a lot, but there are transistors that get close to that rating, you could use two in parallel.

You'd need to amplify the error voltage above G(th) for each cell but that should be doable...

Can I ask what you're using all that power for?

__________
Sam Beck
Blueshift: Battery-Free Portable Electronics
www.blueshiftPDX.com

15. BLUESHIFT New Member

Aug 23, 2013
24
4

Um, right - thing I drew up is totally wrong, that's what I get for getting excited and drawing pictures.

But what I meant I think could work - you're right, it would need comparators (or op-amps?) as drivers for the MOSFETs, whereas I put in only the mosfets. Will try to make sense of that, thanks.

-sam

16. BLUESHIFT New Member

Aug 23, 2013
24
4

Aug 13, 2013
43
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Big amplifier. Audio competition stuff.

Reason I picked 2.4v is suppose I were to limit each cell to 2.7v, would it not be possible, in theory, to have the bank charged to 16.8v, 6 capacitors charged to 2.7v, 1 charged to .6v. then if the bank were to drop below 12.6v, it would cause the final capacitor to reverse charge, damaging it I've been able to do that when playing around with smaller capacitors, but haven't had the balls to try it with these

Aug 13, 2013
43
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As a FYI: I got some new hardware for the capacitors so there isn't quite as much space between them to squeeze things

And I was thinking: What if we take advantage of current over time? Instead of trying to pass all 400 amps of current through a capacitor, what if we were to pass say... 10 or 20 amps through them over the course of a couple seconds? Allow a capacitor to be charged up a bit beyond the 2.4v limit to 2.5-2.6v, then redistribute the charge over the course of a couple seconds.

Seems like it would make it a bit more compact and make thermal management a bit easier? Or is this not a good idea because it wouldn't be fast enough?

Pics of said hardware

Aug 13, 2013
43
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I REALLY like this. If I'm understanding it right it will bleed the capacitor with the highest voltage into the one with the lowest voltage, correct?

How hard would this be to expand it out into a circuit of 7 series'd caps?