I have a capacitor C which changes with time. Now I want to measure the varying capacitance as a voltage function, so I need to design a circuit for that. My motivation around the problem is that, I want to apply a constant voltage to the capacitor always, so if I have a change in capacitance, I will have a change in the current of the circuit.

Q=CV, so for constant voltages dQ/dt=VdC/dt; so if I take that current and pass it through a current to voltage changer and then pass it through an integrator, I might get my output voltage as a linear function of capacitance, so I have set up the following circuit for the same.


simulate this circuit – Schematic created using CircuitLab

My only apprehension is that since the output impedance of op-amps are very low, hence my initial voltage follower circuit might sink my information current completely and I might be left with no current to process in the 2nd and third stages. Am I being logical here or am I being mislead somewhere, if I am correct in my assumption, please give me some suggestions to work around the process or suggest a different circuit which I might find useful.

  • \$\begingroup\$ Yes, you're right to worry. The first opamp will supply all of the current to/from the capacitor as it changes value. But before offering suggestions, what is the nominal value of the capacitance, and how big are the variations you expect to see? \$\endgroup\$ – Dave Tweed Nov 22 '14 at 5:10
  • \$\begingroup\$ What Dave said, also the frequency range (particularly the low end). \$\endgroup\$ – Spehro Pefhany Nov 22 '14 at 5:14
  • \$\begingroup\$ I have 2 such cases, firstly I have a macro capacitor, hence a capacitance value of an odd 20uF but in my second case, I am using an MEMS capacitor, hence the value which I would use would generally be extremely low (C<0.1pF). I want to use dc voltage to measure my changes so no frequency issue there. But if you want to know the frequency at which my capacitance changes that would be a difficult one because it would be used to measure on road frequency stimulations, hence my variation generally would be limited to 100Hz \$\endgroup\$ – ubuntu_noob Nov 22 '14 at 5:30
  • \$\begingroup\$ Do you understand how a condensor (capacitance) microphone works? \$\endgroup\$ – Dave Tweed Nov 22 '14 at 12:01
  • \$\begingroup\$ To measure your V* dC/dt current you would need the cap (C1) to be in series with the inverting input (OA2). If you are trying to measure variations in 0.1 pF that will be hard. Another way to measure C is to use an AC voltage and measure the current. C * dV/dt... you can then pick your frequency, to be much higher than the 100 Hz. \$\endgroup\$ – George Herold Nov 22 '14 at 12:50

maybe it is simpler than you immangine :)

the circuit you draw is not a good one because OA2 will have its output saturated

maybe this can be a solution:


simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Could you please explain the circuit? How does taking an output voltage across the non-inverting pin help anyways? I think if we can have a prefixed dc voltage across non inverting pin and the capacitance across the inverting pin, it would be of considerable significance since I would have a high impedance voltage source, something which I am lacking. I figured another way of doing it yesterday, I'll wait for your comment and then post my way of doing it if necessary \$\endgroup\$ – ubuntu_noob Nov 23 '14 at 13:17
  • \$\begingroup\$ DC analysys: C1 is an open circuit. The inverting pin of the OPAMP follows the non inverting pin. This means that you have V1 accross C1. AC analysis: the circuit works as a trans impedence \$\endgroup\$ – 3NZ0 Nov 24 '14 at 14:28
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    \$\begingroup\$ I'm sorry, you misunderstood my answer. V1 is not the signal you want to read but the voltage you want to use to bias the capacitance C1 :) the signal you need is of course at the output of the OPAMP \$\endgroup\$ – 3NZ0 Nov 24 '14 at 14:38
  • \$\begingroup\$ Yes, in that case you have the right solution, I was looking for a voltage source that would have a very high impedance, this arrangement didn't strike me. Thanks a lot :) \$\endgroup\$ – ubuntu_noob Nov 25 '14 at 0:27

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