I am trying to come up with the equation for testing electronic capacitors by discharging them with a test pulse using a known resistor value.

A known value resistor is switched in circuit for a test pulse period to discharge the capacitor voltage by 10% over a test period time (T2-T1). Because a known value resistor was used to discharge the capacitor by 10%, it is possible to calculate the actual true capacitance of the capacitor, the equation below is using capacitor voltage and time:

E = 0.5 * C * V^2 ( Energy stored in a capacitor in joules )

Delta_E = ((((((V1 – V2) / 2)+V2) / R) ^2)* R) * (T2-T1) ( Change in capacitor stored energy )

And (0.5 * C * V2^2) + Delta_E = (0.5 * C * V1^2)

C = -1 * (Delta_E / ((0.707*V2+0.707*V1)*(0.707*V2-0.707*V1))

For example, the capacitor was charged up to 50 Volts, a discharge resistor of 475 Ohms was used to reduce the capacitor voltage to 45 Volts, during the test, it took 16.66 milliseconds to discharge the capacitor.

V1 = 50 Volts, V2 = 45 Volts, R = 475 Ohms, T2-T1 = 16.66ms

Delta_E = 0.079135 Joules

C = 333uF

The electronic circuit simulator used a capacitor of 330uF so the result for C is very close.

1) The above equation does not consider that the capacitor discharges exponentially, what is the equation for using exponential voltage decay?

2) The capacitor could have a unknown series resistance called ESR, what would be the equation if ESR is also considered?

  • 2
    \$\begingroup\$ Welcome to EE.SE, Gary. This site uses MathJAX which does a great job with formulas. It also has a schematic editor button on the toolbar. \$\endgroup\$
    – Transistor
    Commented Apr 6, 2016 at 20:00
  • \$\begingroup\$ 1) I can discharge capacitors linearly or exponentially or even slowly depending on what I want. Read any book on electronics basics and you will find the formulas. 2): you need Ohm's law because ESR is caused by a resistor. \$\endgroup\$ Commented Apr 6, 2016 at 20:07
  • 2
    \$\begingroup\$ This smells a bit homeworky. We prefer that homework be identified as such, and are prone to not offer much help if that is not done and a problem smells far too much like "do my homework for me" - which we won't do in any case, but we might guide you in the right direction if you do your part. \$\endgroup\$
    – Ecnerwal
    Commented Apr 6, 2016 at 20:08
  • \$\begingroup\$ Look at the exponential decay equation for RC voltage discharge. You know the beginning and end voltages, the R and t, so just solve for C. If you add an unknown ESR, then the resistance in the RC time constant is now R+ESR. So you have two unknowns and one equation. How do you get a second equation so two unknowns can be solved? Another set of data from a second sample. \$\endgroup\$
    – rioraxe
    Commented Apr 7, 2016 at 2:48
  • \$\begingroup\$ I finished my studies 20 years ago. I am trying to test the power supply capacitors in vintage hi-fi audio amplifiers with ageing capacitors. I want to calculate all of the parasitic resistances and true capacitance, R_leakage, ESR and C. I think to measure ESR i need a first capacitor voltage sample V0, this is taken before the discharge resistor is switched in circuit, then V1 immediately follows when the voltage steps down because of ESR. ESR = -1 * R + ((V0 * R) / V1) (V0 = voltage before R is switched in circuit, R = discharge resistor, V1 = voltage when R is switched in circuit) \$\endgroup\$
    – gary
    Commented Apr 7, 2016 at 10:14

1 Answer 1


ok, i can answer my own question after using LTspice and Mathway tools.

V0 = Voltage before test discharge resistor is in circuit

V1 = Voltage immediately when discharge resistor is in circuit

V2 = Voltage at end of discharge resistor in circuit

t = discharge time

R = discharge resistor

ESR = ?

C = ?

ESR = -1 * R + ((V0 * R) / V1)

C = -t / ( ln(V2/V1) * (R + ESR))

For example

R = 7

V0 = 50.1

V1 = 29.06

V2 = 12.03

t = 0.005

ESR = 5

C = 470uf

  • \$\begingroup\$ I didn't work through your calculations but I think there are much better ways to measure / calculate the ESR. Most use a 100 kHz signal (which is relevant to SMPS, etc.) and monitor the response at the terminals of the capacitor. See 99c ESR meter for example. \$\endgroup\$
    – Transistor
    Commented Apr 7, 2016 at 19:41

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