I've recently been hired in a lab at my university to continue research on supercapacitors, specifically leakage resistance. The former student who set up the current tests was a chemist so the professor wants me to verify all of the testing procedures.

In my research about measuring capacitor leakage, manufacturers of electrometers (or other high input impedance devices) say that you must supply the capacitor with a constant voltage to measure Rleak and it will subsequently decay over time. My question is why should a constant voltage be applied and why would this make Rleak decay? Wouldn't it be constant? It seems like a better test to me would to fully charge the cap, then disconnect the power supply and measure how much current is then leaking as a function of time.

Any thoughts on this?

  • \$\begingroup\$ Just a comment - Rleak is not a real resistance, it's a model of an electrochemical process, a bit like charging a battery. So you might get different answers if you measure it with a falling voltage, compared to measuring with a constant voltage. A battery's self-discharge rate is very different to its leakage when floated. \$\endgroup\$ – tomnexus Jul 14 '15 at 4:19
  • \$\begingroup\$ If i got this right, I can't measure R_leak by using a very big resistance in series with the capacitor so R_series==R_leak - which is the case when V_cap = V_rseries \$\endgroup\$ – MdxBhmt Aug 16 '15 at 3:07

I suppose I will answer my own question. There is a constant voltage because you want to observe how the circuit reacts once the capacitor is fully charged. Theoretically there should be no current when the cap is fully charged, but you will observe with enough accuracy there will be a small current. This small current is the leakage current, and the insulation resistance can be calculated using ohm's law.


If the applied voltage changes, then current will flow into or out of the capacitor. Since q=CV, then changing V while C stays constant means q (the charge) has to change. In English, this means that if the voltage changes, then the voltage on the capacitor is different than that of the supply, and current will flow into or out of the cap in order to equalize the difference.

If your current measuring setup has any significant resistance, then it will take time for the voltages to come into equilibrium. Especially since the capacitance (of a supercap) is so large. I think that's what the decaying Rleak is about. Unless the cap voltage precisely matches the power supply voltage at the start of the measurement, there will be a decaying current which settles out to just that due to Rleak.

Charging the cap and watching the voltage decay by itself has two problems. One, it could take weeks to measure it. And two, you have to measure the voltage on the cap without your voltmeter drawing off any current.


I will try to explain briefly. Theoretically capacitors should not allow DC current, but in practice they are not perfect (actually some of them are). When you apply a DC voltage to a supercapacitor in transient response phase there will a resistance appear. So you will need to make a transient analysis of the component.

First, you will have an RC circuit in that transient phase. And you need to find R in this circuit (because you know the C value). Measure the time that the capacitor fully charges and apply the formula \$V_c=V_s \cdot e^{-t/RC}\$. Formula may slightly change I don't remember clearly, but you got the point.

  • \$\begingroup\$ I feel like this is not totally answering my question above. It is not so much as to how to measure the tau value using an oscope, I know I could do that with a single trigger mode (although accuracy is a concern), my question is more along the lines of using a high input impedance instrument (like an electrometer), why is the test done with a constant voltage rather then the power being removed? \$\endgroup\$ – Mtk59 Jul 14 '15 at 16:03

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