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?

  • \$\begingroup\$ Or you could charge the capacitor and measure \$ \tau\$ as it discharges through it's internal resistance. \$\endgroup\$ – Chu Jul 14 '18 at 10:41
  • \$\begingroup\$ Any measurement jig will involve some unavoidable inductance, complicating this simple RC case. \$\endgroup\$ – glen_geek Jul 14 '18 at 11:13
  • \$\begingroup\$ Would the effect of inductance be significant on the results? \$\endgroup\$ – Wouter vw Jul 14 '18 at 11:35
  • \$\begingroup\$ @Woutervw The lower your R is (and your R is already pretty low) the faster it discharges and the more L will affect the results. \$\endgroup\$ – DKNguyen Oct 24 '19 at 20:21

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.

  • \$\begingroup\$ Sorry but I'm not too familiar with capacitors, typically what sort of values for ESR are expected? \$\endgroup\$ – Wouter vw Jul 14 '18 at 16:56
  • \$\begingroup\$ Depends on the type and ratings, but typically from a few ohms down to milliohms for the low-Z (eg. polymer) type used in PC motherboards. \$\endgroup\$ – Spehro Pefhany Jul 14 '18 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.