while experimenting with capacitors, I charged a 15pF cap through a resistor to the voltage of a 1.5v battery. Then when I disconnected the capacitor and tried to measure its voltage with a 10Mohm input impedance multimeter, I found out, to my surprise, that the capacitor voltage (after being charged) is still zero!
Well, to tell whether this a multimeter or a capacitor error, I tried a different capacitor with the same value, and the result is still the same. Then I thought perhaps the capacitor is discharging through the multimeter input impedance, and that's exactly the reason why its effect becomes so strong when measuring low pF caps compared to the usual uF caps, because with low pF values the time constant of the discharging process becomes so small that the voltage reading of the multimeter goes to zero almost instantly.
So, I googled it and found out that actual capacitors discharge through the multimeter input impedance. The reason that I didn't know about it earlier, is because I used to test uF caps, so the voltage seemed almost constant to me.
My question is : How could I slow down the discharging process, when measuring low pF caps, such that I could capture it on the meter screen, without buying a higher input impedance meter?
Best Answer
For a given time constant \$\tau\$, the required value of input resistance is
$$R = \frac{1}{\tau C}$$
For example, for a 1 second time constant and 15pF capacitance, the meter input resistance would be
$$R = \frac{1}{15 \cdot 10^{-12}} = 66.7 \,\mathrm G\Omega$$
Now, that's an enormous input resistance and the capacitor will still discharge in about 5 seconds.
Without further information on what you would like to accomplish, it's difficult to advise you on how to proceed. The calculation above is simply to give you an idea of what you're up against.