# Calculating capacitor leakage current

I'm trying to calculate the approximate leakage current for some capacitors I have. They are 5F 2.7v supercapacitors. I charged them up to 2.69v using a precision bench supply through a 68 ohm resister, then I disconnected them and left them disconnected. I found that their voltage dropped from 2.69 to 2.40 volts in 3.5 days. That's a 10% drop in 3.5 days, extrapolating to a 63% drop (1TC), it would be 22 days (1.9E6 sec).

I'm trying to calculate leakage current by using the Time Constant formula to calculate the resistance of the RC circuit, assuming the leakage current is roughly equivalent to the current through a resistor shorting out the capacitor.

The capacitor lost 10% of its voltage in 3.5 days, I extrapolate the time constant to 22 days to lose 63% of its voltage (it may not be linear, but I'm only need an approximation). 22 days is 1.9 x 10^6 sec.

Given T=RC, then R=T/C

With T=1,900,000 sec, C=5F, that's 1.9E6/5 gives R of 380,000 ohms.

Given E=IR, then I=E/R, that's 2.7v / 380k gives a leakage current of 7 uA.

Is this reasonable? Is my method correct?

I looked at this existing question/answer but I don't see how to apply it here.

• Which supercapacitors, make and model? Do they have a datasheet, and does the datasheet contain a self discharge rate or leakage current? Oct 4, 2021 at 18:32
• I got them on Amazon, no data sheet provided. Oct 4, 2021 at 18:35
• No datasheet? No purchase! =-D Oct 4, 2021 at 18:39
• Even if I had a data sheet, it would be useful to be able to verify that the particular piece meets its specs Oct 4, 2021 at 18:55
• A datasheet guarantees the part to be within spec (provided genuine parts are obtained.) Without a datasheet (or with a vague datasheet), one part can vary wildly compared to another, even from the same vendor. Not an issue for experimentation and learning, but keep this in mind if you ever decide to design something which must be in spec. Oct 4, 2021 at 20:41

## 1 Answer

The capacitor lost 10% of its voltage in 3.5 days

Discharge from (2.69V to 2.4) volts in 3.5 days (302,400 seconds).

$$\I=C\times{{\Delta V}\over{\Delta t}} = 5 \times {{0.29}\over {302400}} = 4.8 uA\$$

The assumption here is that self-discharge current is constant, so that the voltage rate-of-discharge is a linear slope. These super-capacitors may not self-discharge this way. Your time-to-discharge-10% is a reasonable experimental test. I would hope that a voltmeter hasn't been attached all during this test-time, and has only briefly made a measurement after 3.5 days. This measurement should be done at maximum expected ambient temperature.

The "time-constant" model, where leakage is approximated by a parallel resistor is another approach, which may still be a rough model... it decays exponentially. In the usual application, a decaying voltage likely causes a circuit to malfunction before one time-constant expires. So you should be more interested in initial discharge rate - the experimental method described does so.

• Yes, the voltmeter was only attached momentarily. Can you explain where you got the formula I=C(dV/dt)? I haven’t seen that before. Oct 4, 2021 at 20:08
• Charge on a capacitor is given by $Q = CV$. Current is the rate of flow of charge with respect to time so differentiate both sides remembering that $C$ is a constant: $I = \frac {dQ}{dt} = C \frac {dV}{dt}$. Oct 4, 2021 at 20:11