Capacitor – Why Measured Leakage Current is Lower Than Calculated Value

You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric, and is ideally infinity.

The formula you use is correct, but the value you come out with is NOT the ESR, is the leakage resistance. Once the capacitor is charged, if you leave it it slowly discharges trough the leakage resistor with a time constant \$R_{leak}\cdot C\$, so \$R_{leak}\$ is what you calculated, approximately \$50M\Omega\$, that is plausible.

To calculate the ESR you need to measure how long does it take the capacitor to discharge through a much smaller resistor, let's call it \$R_{dis}\$. When you discharghe the capacitor through \$R_{dis}\$ the total resistance through which it discharges is actually \$R_{dis}+R_{ESR}\$, so using the very same formula you used for the leakage resistance you can calculate the ESR.

But is it really that easy? Of course not.

The ESR is hopefully quite small, tenths of milliohms if you have a very good capacitor up to a few ohms. Since in the formula you have \$R_{dis}+R_{ESR}\$ you don't want an eccessive \$R_{dis}\$ to mask \$R_{ESR}\$. Ok then! Why don't we choose \$R_{dis}=0\Omega\$? Easy question:

• \$0\Omega\$ resistance does not exist. But i can make it small!
• Time. You need to be capable to measure how long does it take to the capacitor to discharge.

If you charge the capacitor to a certain voltage it will take \$\tau\ln{2}\approx0.7\cdot\tau\$ where \$\tau=RC\$. If \$R=R_{ESR}+R_{dis}=1\Omega+1\Omega=2\Omega\$ and \$C=680\mu F\$ that's less than 1ms. Without proper equipment, that is a properly set oscilloscope, you can't easily measure the ESR.

Last but not least, keep in mind that electrolytic capacitors values have a tolerance of \$\pm10\%\$, that leads to: $$ R_{ESR}=\frac{t_{dis}}{\left( C\pm C/10\right)\ln{2}} - R_{dis} $$ with the above numbers, t=1ms, C=\$680\mu F\$, \$R_{dis}=1\Omega\$, this translates to: $$ R_{ESR}\in\left[0.91,1.33\right]\Omega $$ That's 10% down and over 30% up.

Gate of a MOSFET cannot drain 5mA current. If you said 5pA I could understand, but 5mA is practically and theoretically impossible.

Either;

• There is a leakage somewhere across the gate and source pins of the MOSFET.
• Your multimeter has a bias.
• The MOSFET is broken.
• There is a wrongly done connection around the gate.

Suggestions:

• Connect a gate resistor of 10\$\Omega\$ or 100\$\Omega\$. Send 1kHz, 0-5V pulses to the gate. If you can, amplify the voltage on the resistor by using an opamp difference amplifier. Observe the voltage drop on the resistor by using an oscilloscope (not with that multimeter).
• If you constructed your circuit on a breadboard, reconstruct it in a different region.

The circuit is indeed constructed on a breadboard. Which effects would show this behaviour?

Dirt inside the breadboard structure my cause leakage current.

I’m still open for blaming the multimeter though

You should. That's probably the only culprit.

The FET seems to do what it is supposed to. Is there a failure mode which introduces such a large leak current but makes it still work correctly?

The system will still work if there is a ghost resistance between gate and source pins which will draw 5mA current. The system will still work even if that resister were very small in value as long as the power supply can supply enough current.