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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?

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    \$\begingroup\$ What are you actually trying to measure? What information are you trying to obtain with your experiment? \$\endgroup\$ – Daniel May 23 '14 at 19:38
  • \$\begingroup\$ You can connect the capacitor to a MOSFET and measure its output, but, as @Daniel mentioned, better to know what you'll use it for. Then it could be said this would be an acceptable solution. \$\endgroup\$ – fceconel May 23 '14 at 19:49
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    \$\begingroup\$ Actually, no. The gate capacitance is typically much greater than 15 pf, so if you just hooked up a MOSFET you'd get a voltage reduction roughly equal to the ratio of the two capacitors as the gate charged up. \$\endgroup\$ – WhatRoughBeast May 23 '14 at 21:00
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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.

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  • \$\begingroup\$ Although the OP specifically stated that he's not interested in buying a new DMM, your 67 Gohm figure is really very easy to come by. It only takes money. www3.imperial.ac.uk/pls/portallive/docs/1/7293196.PDF shows what you can get - in this case an input impedance of 200 T ohm. That's right: 200,000 Gohm. And available on eBay for $3k. \$\endgroup\$ – WhatRoughBeast May 24 '14 at 2:40
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You could just make a voltage follower from an op-amp. A suitable op-amp follower might have a drift of +/-50~+/-500mV/minute with a 15pF cap on the input and an input of a volt or so.

One problem is that the op-amp itself might have 5pF of input capacitance so you'd change the voltage by connecting it to the capacitor. It would also need careful construction to keep leakages low, and would not do well at elevated temperatures or in less-than-benign environments.

schematic

simulate this circuit – Schematic created using CircuitLab

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