Mustermann
Well-known member
First have two capacitors in a row. Lets say 1µF each. So you can measure 0.5µF for both.
Next detach C Meter as it will disturb charging. Then charge 1 capacitor to 5V. Second capacitor uncharged.
When removing battery, capacitor stay charged. Then charge the other capacitor with reversed polarity. Both capacitors are charged, 5V and -5V.
So result is 0V
Now remove battery. Both capacitors are still charged. Connect C meter. As Voltage at C meter is 0V it is able to measure without changing the DC Voltage at the capacitors. You are able to measure capacitance of DC charged capacitors.
Because of DC degrading resulting capacitance is different (smaller).
As leaking current is not so high, DC voltage stay for at least minutes (enough time to measure capacitance and verify that capacitors are still charged.
Does this make sense?
Next detach C Meter as it will disturb charging. Then charge 1 capacitor to 5V. Second capacitor uncharged.
When removing battery, capacitor stay charged. Then charge the other capacitor with reversed polarity. Both capacitors are charged, 5V and -5V.
So result is 0V
Now remove battery. Both capacitors are still charged. Connect C meter. As Voltage at C meter is 0V it is able to measure without changing the DC Voltage at the capacitors. You are able to measure capacitance of DC charged capacitors.
Because of DC degrading resulting capacitance is different (smaller).
As leaking current is not so high, DC voltage stay for at least minutes (enough time to measure capacitance and verify that capacitors are still charged.
Does this make sense?
