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For sensing capacitance value of sensors below circuit has common usage in literature.

reference picture

I want to continue talking about simulated circuit in bottom.
Could be possible writing an equation similar to Vout/Vin=XC2/XC1 using virtual ground?(2nd pin of OPAMP)
XC1= 723ohm, XC2=723 ohm
Without R1 in the feedback path, output voltage negative rail saturated without R1 With R1, gain is -1 and output with R1 value of 68k What is the role R1? What is the transfer function of this circuit?

The simulated circuit enter image description here

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As you've found out, without R1, there's no DC path to the amplifier inverting input. Input leakage current will take it off to rail (+ve or -ve depending on the sign of the current) and the amplifier will stop working.

There are two principal options for the value of R1

a) R1.Cf >> input period. Then the gain is dominated by the value of Cf, and you don't need to know the exact frequency. R1 must be small enough to source all of the input leakage current, but that's generally fairly easy for most opamps with a 1M or 10M resistor.

b) R1.Cf is not >> input period. Then you need to allow for R1 and know the frequency if you are calculating the gain. If you are simply making a capacitance meter, and you calibrate the scale with a known capacitor, and keep the frequency constant, then you can ignore their actual values.

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  • \$\begingroup\$ Thanks Neil, I understand now. If R1 removed DC voltage on opamp's input is 128mV.But with R1 this value simulated as -28uV. This offset voltage causes opamp's output be saturated. DC path of inputs is critical. \$\endgroup\$ – Berker Işık Jan 16 '19 at 16:53

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