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Hi,
Hope everyone is doing great.
I am building a circuit in LTSpice using the inverting configuration of the LT1028 operational amplifier. The input signal provided is 0.7 Vp, which is applied to R1 (5M ohm). For the feedback resistor (Rf), I am using my biosensor, whose impedance varies from 10M ohm to 25M ohm. The equivalent circuit of my biosensor is attached in the picture. I want to measure the value of Rf using the formula:
Rf= R1 x ( −Vout/Vin)
After R1, the voltage drops from 0.7 Vp to 60 uV before reaching Rf (the biosensor).
I have created a separate circuit in LTSpice representing the equivalent circuit of my biosensor. The values I used for R and C in this LTSpice circuit are based on the impedance analyzer readings. The actual impedance of my sensor is 15M ohms, but I used the values of R and C provided by the impedance analyzer, where R and C are in parallel. The voltage applied to this circuit is 60 uV, but the output I obtained from the circuit is very low (around 0.8 uV), and the waveform after the circuit is not a sine wave.
Given this, the impedance value for Rf that I calculated is incorrect.
I have a few questions:

Did I correctly build the equivalent circuit of my biosensor in LTSpice? If not, could someone suggest the necessary changes to the circuit?

Why am I not getting the correct value for Rf when I use the equivalent LTSpice model? When I use a simple resistor for Rf, I can calculate the correct value using the above formula.

Can someone explain why the LTSpice model of my equivalent circuit alters Vout and why the parallel capacitor significantly affects Vout?
For reference, I will attach the picture and the LTSpice ASC file.

The LT1022 is a splendid operational amplifier with high impedance JFET inputs. However, it still has leakage current in the input pins. This can be significant when the input and the feedback resistors are in the MΩ range.

You may obtain a slight improvement if you incorporate a unit gain buffer amplifier. Use two op amps. The first is a non-inverting op amp with input directly to the positive terminal. The second is a traditional op amp to provide the required gain. You will still have some leakage current, but your input no longer needs to “fight” against the feedback resistor. You may or may not be able to implement a temperature-based calibration scheme to compensate for the input bias.

Thank you very much for the reply. I will look into the 1022.

First, the equivalent circuit I made yesterday was incorrect. I am attaching a picture of the correct equivalent circuit.

When I use this equivalent circuit, which represents my biosensor in my complete circuit

having inverting amplifier configuration, I get a very high voltage. Based on that, the impedance of my biosensor, as per the calculations, is 35MΩ, whereas it should be 15MΩ.

Hi @Kamran ,
Your circuit seems to be an integrator, as the feedback path is purely capacitive. As there is no DC-path at the feedback loop, the output drifts either to the positive or to the negative supply. You could test swapping your sensor and the R1. Then the OpAmp would be DC-biased properly.
Cheers, heke