Supercapacitors in parallel

Discussion in 'General Electronics Chat' started by ttf, Jul 3, 2015.

1. ttf Thread Starter New Member

Mar 11, 2015
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0
Hi, I did an experiment to compare voltage drop for two cases. First case, a supercapacitor (1F rated at 3.6V) is charged up to 3.6v and connected to a load. Second case, two supercapacitors (1F rated at 3.6V each) connected in parallel, are charged up to 3.6v and connected to the same load. In the first case, I observed a drop of 0.37v after 24 hours. In the second case, I observed a drop of only 0.05v after 24 hours. I am a novice at this and find this quite puzzling. What are the possible explanations for the superior performance for the parallel configuration? Thank you.

2. wayneh Expert

Sep 9, 2010
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I'm willing to bet that 1) the load was not really the same, or 2) the first capacitor has a much lower capacity and/or higher leakage rate than the second one.

Test #2 on it's own.

3. nsaspook AAC Fanatic!

Aug 27, 2009
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The first thing to do is to repeat the experiment after double checking all measurement devices, values and procedures independently.

4. DickCappels Moderator

Aug 21, 2008
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Or there was a temperature difference between the two tests.

Having two capacitor in parallel not loosing as much charge as a single capacitor brings to mind something similar. Cleaning out our desks for summer vacation at the end of the 5th grade, Cliff showed me two 9 volt batteries he had in his desk. Then he plugged them together and said "Feel this...they are getting warm because they are charging each other." (True story)

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5. R!f@@ AAC Fanatic!

Apr 2, 2009
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Could be one cap esr is higher than the other.

6. ttf Thread Starter New Member

Mar 11, 2015
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Thanks guys for the comments so far. I will definitely repeat the experiments again to check through. In the meantime, I have a couple of questions below.

Are you suggesting that the leakage currents from the supercapacitors could be charging each other? Thanks.

In case 1, the supercapacitor I used has an esr of 40 ohms. In case 2, I used 2 of the same supercapacitors in parallel; hence effective esr is 20 ohms. This will account for the better performance? Thanks.

7. DickCappels Moderator

Aug 21, 2008
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I did not mean the recounting of my vaguely related boyhood experience to suggest that the capacitors are charging each other. Although Cliff was sure the batteries were charging each other, it was clear that they were actually getting hot while discharging one-another.

The result of your experiment showed that two 1F capacitors in parallel retained their charge better than a single 1F capacitor. The result is counter-intuitive, so it is a good idea to look for flaws in the experiment or test for the result in another way.

8. MrChips Moderator

Oct 2, 2009
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Your observation is correct. Do the math!

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9. Roderick Young Member

Feb 22, 2015
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@MrChips makes a good suggestion.

The energy stored in a capacitor is
$
E = 0.5 CV^2
$

So the initial energy in the 1 F capacitor is
$0.5 * 1 * 3.6^2 = 6.48 joules$
The energy left after discharge is
$0.5 * 1 * (3.6 - 0.37)^2 = 5.22 joules$
So the net energy used up is
$6.48 - 5.22 = 1.26 joules$
Applying similar reasoning to the 2 F case, the energy used up would be
$0.5 * 2 * 3.6^2 - 0.5 * 2 * (3.6 - 0.05)^2 = 0.36 joules$
There does seem to be quite a discrepancy.

10. WBahn Moderator

Mar 31, 2012
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What do you think the drop should have been (in the second case)?

If I scribbled the math correctly, I would expect the voltage to have a drop of about 0.09V in the second test - which is to say that the drop should be the square root of the drop seen in the first test. Part of that is due to the charge coming from twice the capacitance and part is coming from the time constant being twice as large.

11. MrChips Moderator

Oct 2, 2009
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There is an exponential in there. e^(-t/RC)

Time constant equals R x C. Hence twice C will result in twice the time constant.

Here is a simple comparison test: do the second case for 48 hours instead of 24 hours. The voltage drop should be the same.

12. WBahn Moderator

Mar 31, 2012
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There are some puzzling, almost vexing, questions that one can ponder, particularly if they are phrased so as to start you off down the wrong path.

For instance, assume for the moment that the load was a constant current source. Then doubling the capacitance would lead to a voltage drop that was half what it would have been at any moment in time before doubling it. Now consider that, because time constant (for a resistive load) has doubled, the voltage doesn't drop as much and therefore, at any moment in time, the current that is flowing is more with the doubled capacitance than with the non-doubled. Therefore, shouldn't the voltage drop be at least one half of the original voltage drop? If not, why not? What is the fallacy in this reasoning?

13. AnalogKid Distinguished Member

Aug 1, 2013
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Another possible variable is the load. If it is an active device rather than a simple resistor, it might have a non-linear relationship with its applied voltage. For example, DC/DC converter with a regulated output and a constant load appears as a negative resistance to its source.

ak

14. wayneh Expert

Sep 9, 2010
12,388
3,244
I don't see this as vexing at all. C = Q/V = dQ/dV or by rearranging ∆V = ∆Q/C. Doubling C should halve ∆V.

In this particular case ∆V is smallish and we assume C is constant. IF the load was truly the same in both tests, then ∆Q is roughly the same in both tests; the load draws nearly the same current-times-time in both tests. Admittedly ∆Q should be a bit higher in the second case where voltage on the load averages a bit higher. But this error should be close to the other experimental errors.

15. MrChips Moderator

Oct 2, 2009
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I stand to be corrected. I must have made an error in my math.

In case #1 the voltage drop is about 10%.
In case #2 the voltage drop is about 5%, i.e. half of that in case #1

16. WBahn Moderator

Mar 31, 2012
18,087
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Huh?

Case #1: 0.37V from 3.6V => 10.3%
Case #2: 0.05V from 3.6V => 1.4%