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Every capacitor has a certain series resistance. This resistance is not only caused by the leads, but also by the metal plates and the dielectric the capacitor is made of. The sum of these resistances is called ESR, Equivalent Series Resistance. This resistance will not always remain the same, but may increase due to aging.

When will the ESR bother us? Of couse this depends on how large the ESR is and the application in which the capacitor is used. Assume the ESR of capacitor C in the filter above is 10Ω. At very high frequencies the output voltage will not be 0V, but 1V∙(10/1010) = 10mV. In most cases, this will not be any problem. However, if resistor R were also 10Ω, the output voltage would have been 0.5V!

We can also expect ESR problems when large charge and discharge currents flow though the capacitor. Remember, a large current means a large voltage across the series resistance. This may even heat up the capacitor. If a capacitor heats up, the ESR may increase. This will heat up the capacitor even more, and so on. Eventually (and this may take months) the capacitor will be ready for the dumpster. Troubleshooting can be a pain; a simple capacitance meter uses small currents and will therefore not notice that the ESR has increased.

How can we measusre the ESR? The are special ESR meters available for this purpose, but these are pretty expensive. Most of the time we only need an indication. We can connect the capacitor to a power supply via a known resistor R and a switch. If the switch is open, the voltage across the capacitor and ESR will be 0V. On the momen the switch is closed, the capacitor is still empty. The voltage we measure across the capacitor is therefore equal to the voltage across its ESR. If that voltage is equal to half the supply voltage, the ESR must be equal to the known resistor R. Of course: the lower the voltage, the lower the ESR must be. The disadvantage of this method is that the power supply must be able to deliver the current peak. Moreover, we must also include the internal resistance of the power supply in our calculations. That's why we often use the opposite method: we charge a capacitor to a certain voltage and then discharge it via a known resistor. Of couse: the higher the voltage at the moment of discharge, the lower the ESR must be. Please find below a picture of both methods. Resistor R is 10Ω. The supply voltage is 1V.

At t=0, the voltage across the ESR is about 0.34V. So ESR/(R+ESR)=0.34 => ESR=R(0.34/(1-0.34)) = 10(0.34/0.66) = 5.2Ω.

At t=100us, the capacitor is discharged via the same resistor R. The voltage immediately drops to 0.66V.

So R/(ESR+R)=0.66 => ESR=R((1-0.66)/0.66) = 10(0.34/0.66) = 5.2Ω.