ESR Calculation of Capacitor

R is the resistance of the load connected across the capacitor ternimals, Guv. And T is a time-constant; the decay is exponential.

ESR of cap should be in the data sheet.
Del

Internal leakage resistance and equivalent series resistance (ESR) are 2 different parameters. Which one do you really want?

Leakage measurement may require a very high impedance voltmeter, but is otherwise pretty simple.

DC and low frequency ESR may be measured and calculated:

Ri = internal resistance or ESR
RL1 = test load resister 1
RL2 = test load resister 2
C = device capacitance
t1 = time constant 1
t2 = time constant 2

Measure:

t1 = (Ri+RL1)*C
t2 = (Ri+RL2)*C

Solve for unknown value Ri and calculate ESR:

Ri = (t1*RL2 - t2*RL1)/(t2-t1)

or review the device data sheet ;-)

Normally capacitors are discharged with external parallel resistors, R in the drawing below, with a time constant, t = R C. The formula is V(t) = V_initial e^-t/RC, and as time passes, for every amount of time, t, the voltage on the capacitor will drop by an additional 63%, or 37% will be left. After five time constants, it'll be under 0.7% of the initial voltage. For many purposes, the capacitor can be considered discharged.

The drawing also show an internal resistor, or leakage resistance, which may be what you have in mind. This leads to a self-discharge time. Normally we'd like the leakage resistance to be as high as possible. Often we consider it to be infinite, assuming the capacitor doesn't discharge on its own.

Manufacturers often have a leakage spec, in units called megohm-microfarads, which as you can see is simply the self-discharge time, in seconds. This terminology helps to reinforce the concept that a larger capacitor will have more leakage paths, and hence a proportionally smaller leakage resistance. A typical value for an electrolytic capacitor might be 100 MΩ-μF. In this case, five time constants for self-discharge would be about 10 minutes (but most of us would still use a screwdriver to short the leads before we put our fingers across it, even if we had waited an hour!).

Manufacturers also have other ways of specifying leakage resistance, for example Panasonic says "I <= 0.06 CV +10 (μA) After 2 minutes application of rated working voltage at +20°C", for many of their electrolytic types. Notice the fixed 10uA term, independent of voltage; hah, they're telling you this part is going to self discharge sooner or later! They're also giving us a very conservative spec, where the capacitor has a 1/0.06 = 17-second discharge time constant. Hey, we hope and it likely is, much longer than that!

In your subject heading you mention ESR. That's the capacitor's series resistance shown in the drawing. It's normally a very small value, under an ohm, and plays no significant role in the capacitor's discharge, unless you are shorting the capacitor or discharging it VERY fast. The value is so small it's impractical to measure it using a capacitor-discharge approach, as mjb1962853 suggested. One way is to put a high-current step load on the capacitor and measure a step-change on the terminal voltage. Another is to charge an inductor to several amps, with a MOSFET switch to ground, and open the switch letting the inductor flyback with its current going through a diode into the capacitor, and observe a step-voltage change, R_esr = ΔV / I_step.

There are low-cost esr meters available, with limited accuracy, so far as I've observed, although they may be useful for identifying a defective capacitor with an excessively-high esr. An electroltic capacitor's esr value changes with frequency, and an impedance analyzer is often used to observe this. You may find the answer given on the datasheet in units of tan δ, at 120Hz, etc., which we won't go into now.

"The value is so small it's impractical to measure it using a capacitor-discharge approach,..."

I agree the ESR value IS typically small, but it may be significant depending on the application.

I deal primarily with Pulsed Power applications & devices. When initially charged to 300V, one such device had the following discharge characteristics:

RL1 = 0.03 (ohms)
RL2 = 0.06 (ohms)
C = 0.099 (Farads)
t1 = 0.00316 (s)
t2 = 0.00613 (s)

Ri=(t1*RL2-t2*RL1)/(t2-t1)

Ri = 0.002 (ohms)



The OP asked for a way to "calculate" the ESR of a capacitor. The value obtained by the measurement & calculation shown above was accurate for this particular application. Although the value was actually a combination of the capacitor ESR and the high current copper bus bar resistance, it was still a realistic ESR for the complete discharge circuit.