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I got two 2.7V 100F supercapacitors. I have wired them in series and applied 5V.

One capacitor I have tried have been about 2V. I put it on my desk without any connection.

A couple of hours later, when I measured the voltage of capacitor, it was about 0.5V.

So, energy formula of a capacitor says that it is proportional with 2nd power of capacitor voltage. In my case, voltage dropped 4 times and rate of energy loss is about 16 times.

I cannot figure out where this energy has gone.

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  • \$\begingroup\$ Equivalent parallel resistance is what you want to look at. \$\endgroup\$
    – Matt Young
    Jun 24, 2014 at 2:04
  • \$\begingroup\$ Properly stated, the voltage has dropped to 1/4 of its initial value, therefore the stored energy has dropped to 1/16 of its initial value. This is the total energy loss, not the rate of energy loss. \$\endgroup\$
    – user28910
    Jun 25, 2014 at 18:20

2 Answers 2

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What you are noticing is known as leakage current or self discharge. It is to be expected. A supercap is not like a battery that uses a chemical reaction to supply electrical energy, it discharges like a capacitor does.

Here is an interesting plot from Tecate group enter image description here

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What you have is not really a supercapacitor in the traditional sense. A supercap will have a very low self discharge in order to replace a battery. You just have a very high capacitance capacitor that has a fairly average self discharge rate because it is designed to provide smoothing and high currents, rather than long term battery backup.

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  • \$\begingroup\$ Can I find a supercapacitor in order to replace a battery? Does it provide high current too? \$\endgroup\$ Jun 26, 2014 at 22:30

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