Actually I use the capacitor in front of negative input because I want to stop any dc current.
The problem is the results I am getting as I posted earlier. The 2 square wave per second. The image contains both the circuit and output result if you can see then these square wave I have no idea where they are coming from. I want to eleminate these square waves infact.
I am afraid, I cannot answer because important information are still missing..
The display shows two functions. What is input and what is output? Or two outputs?
What is the purpose of the switches and are they activated or not? If yes, which takt rate?
Are supply voltages connected to the opamp?
Is it a real model or an ideal model?
1.) Are the squarewas at the output created without any input signal? Input grounded or open?
2.) What about the working frequency of the relay? In any case, remove the series capacitor and the relay across it. Otherwise the opamp gets no dc bias current.
3.) Replace the relay across the feedback capacitor by a large resistor (value depends on open loop gain and the input operating frequency)
4.) Why the display in posting#5 looks so good? There seems to be no problem. Why?
5.) Comes just into my mind: Is drawing #1 a measurement rather than simulation?
OK, now it becomes more clear. There was a confusion (on my side) between measurement and simulation.
But still I know nothing about the sensor output signal that is to be integrated.
*Connect the sensor directly to the inverting input.
*Place a resistor Rp kohms (or less) across the feedback capacitor.
*Choose C for T=RpC according to your integration requirements (integration time constant)
*Determine sensor output resistance R
*DC gain -Rp/R should be large enough (at least 40 dB).
Not particularly, I fear.OK, now it becomes more clear.