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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?

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  • \$\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
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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.

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  • \$\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

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