Supercapacitors in parallel

Thread Starter

ttf

Joined Mar 11, 2015
5
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.
 

wayneh

Joined Sep 9, 2010
17,496
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.
 

nsaspook

Joined Aug 27, 2009
13,079
The first thing to do is to repeat the experiment after double checking all measurement devices, values and procedures independently.
 

DickCappels

Joined Aug 21, 2008
10,153
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)
 

Thread Starter

ttf

Joined Mar 11, 2015
5
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.

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

Could be one cap esr is higher than the other.
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.
 

DickCappels

Joined Aug 21, 2008
10,153
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.
 

MrChips

Joined Oct 2, 2009
30,706
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.
Your observation is correct. Do the math!
 

Roderick Young

Joined Feb 22, 2015
408
@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.
 

WBahn

Joined Mar 31, 2012
29,976
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.
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.
 

MrChips

Joined Oct 2, 2009
30,706
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.
 

WBahn

Joined Mar 31, 2012
29,976
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?
 

AnalogKid

Joined Aug 1, 2013
10,986
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
 

wayneh

Joined Sep 9, 2010
17,496
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.
 

MrChips

Joined Oct 2, 2009
30,706
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
 

WBahn

Joined Mar 31, 2012
29,976
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
Huh?

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