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I have a big 500 F 2.7 V capacitor, and a module of six 120 F 2.7 V capacitors. If I calculated the capacitance correctly, the module has 20F total capacity.

If I calculate the energy stored in them by E = 0.5 * C * V^2, I get:

  • For the module: E = 0.5 * 20 * 16.2^2 = 2624 J
  • For the big cap: E = 0.5 * 500 * 2.7^2 = 1822 J

So far, so good.

I bought a battery tester device and charged the capacitors to 2.7 V and 15 V, respectively (my DC source only goes to 15 V), and I've connected them like a battery to the tester device. I set 0.5 V as the lower cutoff voltage. The results I got are:

  • The module yielded about 58 mAh.
  • The big cap yielded about 220 mAh.

These results are in the ballbark of what an "Ah equivalence" equation, Ah = Farads * DeltaVolts / 3600 gives, but both the measurements and this equation are in conflict with the capacitor energy equation, which shows the energy stored in the module should be greater than in the big cap. The battery tester puts the current through a (constant) resistor, so it should be fairly simple as far as calculations go.

So my question is: how to reconcile the two?

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  • \$\begingroup\$ As the voltage and current fall exponentially (unlike a battery) how did you actually calculate the two equivalent capacities? \$\endgroup\$ – JIm Dearden Dec 16 '16 at 11:41
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    \$\begingroup\$ The module has about 1/4 the capacity (Ah) at 6* the voltage, or 1.5* the energy storage. So, where's the conflict? \$\endgroup\$ – Brian Drummond Dec 16 '16 at 12:13
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    \$\begingroup\$ You can't reconcile capacity with energy. \$\endgroup\$ – Dmitry Grigoryev Dec 16 '16 at 12:18
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    \$\begingroup\$ Important lesson: Your comment "But it doesn't matter, even if it's completely incorrect,..." is completely incorrect, and it matters that it is :-). The reason for the swapping of order is that your calculations have excluded a relevant numerical factor (Voltage) which is substantially different in the two examples and which skewed the result by a factor of 15/2.7 = 5.56. | I'm writing this not so much to note that you missed something out as to note that in life it's easy to use a "proxy" for some measure which works well in some cases and fails totally in others. ... \$\endgroup\$ – Russell McMahon Dec 16 '16 at 13:17
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    \$\begingroup\$ ... Being aware of where it is and isn't OK to do this is crucial. In this case the trap was to use mAh as a measure of energy capacity. In eg 12V lead acid batteries you get a reasonably (not perfect) linear relationship between Ah and Wh of discharge - with a factor of about 12 (no surprise) between them. With eg LiIon cells where Vat 100% chg = 4.2V and V at ~= 1% charge ~= 3V the relationship between mAh and Wh is less linear. With capacitors with - in one of your examples a 30:1 ratio between Vmax and Vmin the relationship between mAh and Wh is extremely non linear. \$\endgroup\$ – Russell McMahon Dec 16 '16 at 13:21
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Your calculations are consistent, though your measurements of capacity look a bit low.

The 20F, 16.2v capacitor stores 20*16.2 = 324 Coulombs, 324 Ampere seconds, or about 0.09 ampere hours. As you correctly calculate, 2624 Joules.

The 500F, 2.7v capacitor stores 500*2.7 = 1350 Coulombs, 1350 ampere seconds, 0.375 ampere hours. As you say, 1822 Joules.

You can also get the capacitor energy by multiplying the charge stored by the average voltage. So for the 20F cap, 324C * 8.1v = 2624 Joules, and the 500F cap, 1350C * 1.35v = 1822J. This is actually the same equation as \$0.5CV^2\$, but it's easier to see how energy depends on both voltage and charge stored.

As you can see, energy is related to ampere hours through voltage. They are different voltage caps, so you would expect the higher voltage cap to have a higher energy to charge stored ratio.

As an interesting thought experiment, let's connect the series capacitors in parallel. Now it should be easier to compare what's happening. You have a 720F cap and a 500F cap, both rated at 2.7v. You would hope they store the same energy as before with the series connection.

\$0.5CV^2\$ for each gives 0.5*2.7*2.7*720 = 2624J, and 0.5*2.7*2.7*500 = 1824J, as before. Yes, the energy sum comes out exactly as before.

However, now the charge in the 720F is 2.7v * 720F = 1944C. The energy is the same, but as the voltage is lower, the charge is higher.

When capacitors are in series, the same charge passes through each. The total charge in the whole series string is the same as for one capacitor. When capacitors are in parallel, the charges add, just like current does.

The same thing confuses people with batteries, especially Lithiums sold as series blocks!

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  • \$\begingroup\$ It's a bit counter-intuitive to me. On the one side I have a big 500 F cap rated at 2.7 V, and on the other hand, 6 smaller 120 F caps also rated at 2.7 V; if those were connected in parallel, I'd get 6x120 F = 720 F of capacity. It seems to me (conservation of energy and all that), that the same caps connected in series should also absorb more energy (a factor of 720 / 500, to be somewhat exact), than the single 500 F cap. Right? Or am I thinking about it the wrong way around? \$\endgroup\$ – Ivan Voras Dec 16 '16 at 11:53
  • \$\begingroup\$ @IvanVoras added that to my answer. \$\endgroup\$ – Neil_UK Dec 16 '16 at 11:59
  • \$\begingroup\$ I've seen your edit, and that's kind of what my original question was about: when connected in parallel, the energy is the same. So, since the battery tester puts the current through a simple resistive load, and measures voltage and current through it, why should it get a so low result for the 6 caps? If I got a 720 F 2.7 V cap, the battery tester would almost certainly show me a larger mAh result, with a factor of 720/500. \$\endgroup\$ – Ivan Voras Dec 16 '16 at 12:02
  • \$\begingroup\$ Actually, I see your latest edit. I think it answers this question. \$\endgroup\$ – Ivan Voras Dec 16 '16 at 12:04
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    \$\begingroup\$ It depends what current your load draws at those voltages. Let's pretend that SMPS converters are 100% efficient. If your load has one of those on the input, and draws constant power regardless of the input voltage, then it doesn't really matter, as you are using all the energy of your capacitors. If your load has a linear regulator on its input, and draws constant current, so draws 3x the power at 15v input to 5v input, and wastes the voltage in excess to 5v, then it's much better to feed it the minimum voltage it will work at, the minimum power input, and use an external SMPS. \$\endgroup\$ – Neil_UK Dec 16 '16 at 13:58
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Your confusion comes from trying to compare two basically different variables.

mAh is and indication of a reservoir's energy capacity only either at a constant voltage or, if the voltage falls during discharge, at some mean value.

Using simplistic calculations of mAh x max voltage or mAh x mean voltage the energy in Joules is higher for the module than for the individual capacitor, as you'd expect.

This was going to be a longer answer but I see that Neil has posted similar to what I was going to so I'll quit while I'm behind :-).

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    \$\begingroup\$ lol. +1 for knowing when to quit. \$\endgroup\$ – JIm Dearden Dec 16 '16 at 11:53

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