You can model a leaky capacitor as a good capacitor C with an internal resistance Rp in parallel with it. The goal is here is to measure Rp. The smaller Rp, the bigger the leakage current, and the faster the capacitor discharges itself. If Rp is very large, ideally infinite, then basically the capacitor does not leak. A capacitor blocks DC, but a resistor does not. So a way to measure the size of Rp is to put this capacitor-to-be-tested in series with a voltage source and a small current sensing resistor Ra. Then wait enough time for the capacitor part C, to charge. If a current still flows (i.e. there is a nonzero voltage across Ra) after sufficient time has passed to completely charge C, then that current is the current flowing flowing through Rp. A diagram of this setup is shown in the attached picture. As an example, I'm using a capacitor with a nominal value of 100 uF, and a 1K resistor for Ra, my current sensing resistor. After 5*Ra*C seconds have passed, the capacitor should be 99% charged (since 1-exp(-5) = 0.99326 ). For Ra=1K and C = 100 uF, 5 time constants is 5*Ra*C = 5*(10^3ohm)*(10^-4F)= 2.5 seconds. After this amount of time has passed, if there is still a voltage across across Ra, and that voltage is steady, not falling to (1/e) its previous value every time constant, then that's the current flowing through Ra, and Rp. And I can calculate Rp, from Ohm's law: V = I*(Rp +Ra) ==> Rp = (V/I) - Ra A picture of this setup is attached. Large version here:https://www.instructables.com/file/F8YABNNGWT2MKAT/
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Do you mean leakage current ?
yes,and i want to identify with non-visual inspection,but how?
MEASURE it !! Charge it to a known voltage, and observe the decay with time. Voltage = C/Q dq/dt = current, so dV/dt is proportional to leakage current. You could charge the cap from a constant current source, timing how long it takes to charge to your known voltage. Make the source current >>> leakage current. Now you know the capacitance, since you know the charge time. With that you can characterise the discharge properly.Steve