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To sense the capacitance value of a sensor, I found the following circuit.

enter image description here

My knowledge on analogue electronics isn't that great, so I'm not sure how I'd determine the gain of this circuit.

Vo is the output voltage I need, the peak value detector and Vdc can be ignored.

So what is the gain, Vo/Vs, in this circuit?

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If you pick Rf to have a high resistance relative to the impedance of Cf at the input frequency, the gain before the peak detector will be approximately -Cs/Cf.

That's a consequence of the feedback impedance being XCf || Rf ~= XCf and the normal equation for the closed-loop gain of an inverting op-amp amplifier.

One would normally pick the resistor in such a way- typically the purpose of Rf is to provide a path for the bias current of the op-amp. At very low frequencies the gain will be less as the resistor influence manifests, and you can easily write an equation for the gain including the resistor and including the op-amp open-loop gain too, if you want.

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  • \$\begingroup\$ So is the following correct: V_0/V_s = jwC_s/(jwC_f + 1/R_f)? \$\endgroup\$ – Physco111 Jan 7 '20 at 19:14
  • \$\begingroup\$ That's correct, but if you want the magnitude of the gain you'd do a bit more manipulation, which would make it a bit more clear when you can ignore Rf. \$\endgroup\$ – Spehro Pefhany Jan 7 '20 at 20:41

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