electronics testing capacitors

Question

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?

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