# Supercapacitor connection

1. Jan 22, 2010

### JLauro

I am familiar with changes that occur with capacitance and ESR when you connect supercapacitors in series or in parallel.

My question is along those lines, but the answer is elusive to me...

From schematics I have seen about supercapacitors imbedded in circuitry, the supercaps always have both electrodes(+,-) connected inside the circuit. Depending on how the electric circuitry is engineered, the supercaps can be in series or in parallel.

What happens if each supercapacitor were connected independently directly to a power source with one electrode, and then to a circuit with a load with the other, and this circuit provided the exit for the current to close the total circuit? Think of power going into each supercap directly from a powersource into one electrode, and out of each supercap by connecting its other electrode to a circuit with some kind of load. Is this a parallel circuit? Is it something different? What happens to voltage across the supercaps, and the ESR?

The reason I am baffled is because in connections both in parallel and series, the current going into the supercaps is always coming from the circuit. There is some kind of branching of the current pathway, but the current going into the supercap is the current moving through the initial wire coming from the source. In my scenario, each supercap has an independent connection to the source outside the general circuit, right?

2. Jan 22, 2010

### Bob S

Never put caps, especially supercaps, in series, unless you fully understand what you are doing, and the two caps have the same Farad rating, and are the same mfgr, design, and have adequate voltage rating.
Bob S

3. Jan 22, 2010

### JLauro

Yes, I am aware that variations in characteristics of the individual supercaps can cause differences in voltage, potentially damaging the electrolytes. But I am asking something else. Thanks.

4. Jan 22, 2010

### Bob S

My above comment was just a reminder, for safety reasons.

If You connect an uncharged supercapacitor C1 in series with a power supply (turned on) and with a resistor R1 (ESR plus whatever) in series with a supercap C2 which is in parallel with a load R2, This is a capacitor-series circuit. C1 will eventually charge to the power supply voltage, and C2 will eventually be uncharged (after transient).

In the short term, both C1 and C2 will partially charge and/or discharge, depending on time constants.

Is this what you were asking?

Bob S

5. Jan 22, 2010

### es1

JLauro, do you know how to predict what would happen if you used an ideal capacitor instead of a real super capacitor in your setup?

6. Jan 22, 2010

Hello JLauro,

Could you please post a diagram of your proposed circuit? This would help people to assist you, as it is not particularly easy to understand a circuit idea described only in words.

If you have not drawn it up before, you may well find that doing so will help your own understanding of it, quite apart from what other people may advise you. Try it!

7. Jan 23, 2010

### sophiecentaur

It isn't clear why you would want to do this. Supercapacitors are not chep so why would you want to connect them int series? You will end up with a resulting capacity less than the smaller one.
Capacitor or super capacitor, the formula is the same for the total C value.
However, the actual voltage on the mid point can't really be determined unless you know the actual details of leakage current, internal resistance and inductance.

8. Jan 23, 2010

### cjameshuff

One reason is to get a higher voltage capacitor. The energy stored is E = 0.5*C*V^2, so it is the same whether you're doubling the voltage and halving the capacitance (putting them in series) or doubling the capacitance and keeping voltage the same (parallel). Circuits powered by supercapacitors typically stop operating at a certain minimum voltage, leaving the capacitors still partially charged. Putting the capacitors in series discharges each capacitor to half that voltage, getting more of the stored energy out and extending run time (given a circuit that doesn't simply draw more power at a higher voltage). For constant power draw, the higher voltage also means lower current and less stress on the capacitors.

JLauro's actual question seems to be a basic misunderstanding about capacitors, rather than anything about supercapacitors. I'm not sure what the exact question is though...there's nothing special about the current the capacitor is charged with that would change how it later discharges, it can charge from one part of the circuit and later discharge through another. Current is just motion of charge, and capacitors accumulate differences in charge. Power doesn't go in one terminal and come out the other...a voltage develops across the capacitor that opposes the current flowing through the capacitor as charge builds up on one plate and is depleted from the other.

9. Jan 23, 2010

### mheslep

For supercaps especially, I'm not sure that's a good idea, at least not without some precautions. The voltage across each capacitor of the series chain is determine the ESR's of each cap acting as voltage divider. ESR's can vary widely even between otherwise identical devices, and also due to temperature. Thus it is still possible get a large majority of the circuit voltage across one capacitor, causing it to fail catastrophically, and in turn causing the circuit voltage to fall across the remaining cap at least momentarily so that it might fail as well.

10. Jan 23, 2010

### cjameshuff

This is simple enough to solve. Charge the capacitors through a resistor that drops enough voltage at the initial high currents to limit that applied to the capacitors. As the capacitors charge, current through the capacitors and resistor and the voltage across that resistor will drop, with the capacitors eventually reaching full charge. For large banks of capacitors that store significant amounts of energy, a switching regulator and more complex control circuitry could be used as a more efficient option.

Leakage is a bigger problem than ESR, allowing some capacitors to drift up to higher voltages over time and charge/discharge cycles, possibly eventually exceeding their ratings. Balancing the voltage with voltage dividers and using diodes to protect against reverse charging of high leakage capacitors can be of use in preventing this, but it's not an ideal solution...especially for supercaps, where the forward voltage drop of a protection diode can be a good chunk of the maximum voltage of the capacitor it's protecting. Depending on the application and capacitors, it may be enough to just charge through a resistor divider. There are certainly issues to consider, but using series strings of capacitors to achieve higher voltages is doable and it is done.

11. Jan 27, 2010

### sophiecentaur

The fact that there is still energy left in the capacitor which you can't get at is also a feature of may batteries (many lead acid batteries don't like being taken below half charge capacity).

Bearing in mind all the hassle of control circuitry to get 'all the juice' out of a capacitor, why not just use more capacitors in parallel or a higher maximum voltage and accept that you cant get at all the stored energy?
Low tech has a lot to recommend it.

12. Jan 27, 2010

### cjameshuff

An odd argument. Batteries are almost always wired in series to get a higher voltage...it's so common that even single cells are nearly universally referred to as "batteries", which more technically means a collection of cells in series. It's a bit easier for batteries, their charge characteristics make them self leveling for the most part because individual cells don't continue to increase in voltage when overcharged.

Farad-range "high" voltage supercapacitors are low tech? A string of resistors isn't?

Given the expense of supercapacitors, it sometimes makes a lot of sense to try and get as much out of them as practical. You don't have to accept that you can't get the energy out because it's simply not true, you can get at more of the stored energy with little effort, while adding more supercapacitors adds expense, mass, and bulk.

A resistor charge leveling string and a constant-power load is in fact a low tech, less efficient way of doing this. You can instead use a boost converter to get the voltage you need, as is commonly done with lithium batteries...there would be little reason to put the capacitors in series then, and if you put them in parallel you have no losses from protection diodes or leveling resistors. You would need to add inductors for the switching regulators, however...you need to raise the voltage, so you couldn't simply use PWM and a filter capacitor, or just a linear regulator to get the voltage you want.

13. Jan 27, 2010

### sophiecentaur

I have never really been an advocate of super capacitors.
Batteries are 'stacked up' in series because they are constant voltage devices. Capacitors are not - are they? There are only a few applications where it's even worth giving them a second glance afaik.
If you are suggesting that the energy storage limit of super C is its max voltage then you have a point but shunt resistors will drain energy constantly. otoh, you can't afford to take them out or the voltages will soon become unbalanced and could cause damage.
I don't think the actual problem has been stated fully. You won't get more energy from a capacitor than its voltage limit will allow. The difference in extractable stored energy that you can get from fancy connection methods seems to me to be not a lot. If someone could actually do the sums and show that the saving is significant then this argument might progress fruitfully. Otherwise. . . .

14. Jan 27, 2010

### sophiecentaur

I did think of this significant fact. The Energy stored in a capacitor is CV2/2 so, being proportional to the square of the voltage, it becomes rapidly less and less significant as the volts drop. If the voltage drops to half value you have used 3/4 of the stored energy. If it drops to 1/4 then you have used 15/16 of the stored energy.
I don't know the maximum operating voltage of a super capacitor but, if it's over 100V and fully charged and you need a 12V supply, there's only 1.4% left in there when it's down to 12V

15. Jan 27, 2010

### sophiecentaur

P.S. Having looked at some specs, it seems we're in the ballpark of lower voltages than 100V and 12V but the square law argument still applies.

16. Jan 27, 2010

### cjameshuff

Batteries are not constant voltage. Internal resistance increases as the battery discharges, so the output voltage at a given current draw drops. They do have different discharge curves from capacitors, and some chemistries give a very flat curve that only drops off at the very end. However, stacking for higher voltage has nothing to do with them being constant voltage devices or not, only with the need for a voltage higher than the cells provide.

As for supercapacitors, there are several applications where they have major advantages. Those requiring particularly long lifetimes without battery replacement, or with particularly large numbers of charge/discharge cycles, for example. I advocate using the right device for the job.

You can take them out, using them only for charging. Leakage will cause the voltages to become unbalanced, but won't cause any capacitor to exceed its rated voltage. It may cause a particularly leaky capacitor to experience a reverse voltage as the string discharges, but that's what protection diodes are for...if the amount of leakage is low enough and the capacitors are left with enough charge, they can be omitted.

Yes, I pointed this out in my first post in this thread. A capacitor at 1/2 its fully charged voltage contains only 25% of the energy of one at full charge. At 1/4 max voltage, 6.25%. At 3/4, 56% of the energy remains in the capacitor.

If you have a couple 5.5 V supercapacitors and a circuit that requires 4.125 V to operate, in parallel you will only be able to use 44% of the stored energy. In series, given a constant current draw independent of voltage, you will discharge each to a bit over 2 V, you will be able to use 86% of the stored energy, and will get nearly double the run time.

100 V? Typical voltages are 2.5-5.5 V.

17. Jan 27, 2010

### mheslep

I'd suggest that any application requiring very high specific power, high cycle life, or wide temperature range warrants a look at ultracaps.

18. Jan 27, 2010

### mheslep

Jumping in here ...
Yes, but there goes the simple, ('low tech')solution. I've designed in voltage dividers in parallel with high voltage series capacitor chains for, e.g., radar gear, but never for longer term energy storage solutions.

19. Jan 27, 2010

### mheslep

All the high voltage ultracaps are integrated packs of typical voltage cells, as you say.

20. Jan 27, 2010

### sophiecentaur

Most of your comments are quite correct. Having looked at specs, I can see that the exponential voltage characteristic of supercaps is a serious drawback because of their very low Vmax. (I can't agree that chemical cells are not, inherently, constant voltage. You choose your cell, normally, to suit the application in which it can be treated as one. e.g. car starter battery is near enough 12v to do it's job and a watch battery maintains its voltage for five years)
The few volts you can get from a super C means that V drop in switching devices is going to be an embarrassment. We can only hope for an electrolyte to come along that will allow some higher voltages to be used which will make the (useful) energy densities more like those which are quoted- "better" than most chemical cells, it's claimed.