I was wondering if its possible to measure internal resistance of a capacitor from a DC circuit using the below formula and method.

\$\ V=V_0e^{\frac{-t}{\tau}}\$

\$\ lnV=-1/\tau\ *t + ln V_0\$

Using a graph to determine the gradient can I then solve for \$\tau\\\$?

From there I was thinking of using total resistance R and subtracting the resistance of the resistors to find internal resistance.

\$\ R= \frac{\tau}{C}\$

\$\ \frac{1}{R}= \frac{1}{R_{resistor}}+\frac{1}{R_{Internal}}\$

Will this method give me valid results?

## Best Answer

This is not a very good approach because the value of C is very poorly defined (often +80/-20% tolerance) and your external resistor will necessarily be much higher than the ESR of the capacitor, so I don't think you'll have any kind of reliable measurement. You'll be measuring the capacitance mostly, and what's left will be a small fraction of the resistance measurement.

You should run the numbers yourself- determine the sensitivity to each value.

If you measured with two different (say 2:1 or 5:1) relatively low value external resistors over exactly the same voltage change you might be able to get a good reading.