I have load a 10nF capacitor with a bigger capacitor of 100 uF. Both have the same voltage but when I measure the voltage of my small capacitor I unload it to zero... How can I avoid this effect ? Is it common to unload a capacitor when you measure its voltage with a multimeter ?

  • \$\begingroup\$ Erm, yes. DMMs have a non-infinite resistance. \$\endgroup\$ Jul 21, 2016 at 12:37

1 Answer 1


Some DMMs (or example, the popular Keysight/Agilent 34401A 6.5 digit bench DMM) can be switched from (say) 10M to >10G input impedance.

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This is not a normal feature in handheld meters, and particularly not in cheap ones. You could, however, make a voltage follower buffer with a CMOS-input or JFET-input op-amp that would minimize the input current. It would need a supply voltage higher than the anticipated voltage on the cap, which might not be possible. fA typical input currents are possible, but there will be some input capacitance of a few pF which will also affect your accuracy. The effect of the input capacitance can be minimized by precharging the input to near the expected capacitor voltage.

The time constant (time to mostly discharge the capacitor) of 10M and 10nF is 0.1 second. For 10G it is 100 seconds.

Note that the 34401A meter has an input capacitance of about 700pF so it would instantly decrease the voltage on a 10nF capacitor by a bit if it has not been precharged, and then the discharge (or charge) error due to the input resistance/bias current begins to accumulate.


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