Have you never seen a typical RC integrating device (Miller integrator)?Actually I use the capacitor in front of negative input because I want to stop any dc current.
I am afraid, I cannot answer because important information are still missing..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.
Thanks for your reply, in the first displayed image both are the output at different time, means the results are consistent, Reed relay switches are installed for zeroing the capacitor especially the feedback capacitor. Yes supply voltage is connected to the opamp. Finally these results are real. Circuit is configured and in working condition. As I explained before my project require a pure integrator but the problem with pure integrator was saturation. It took only couple of seconds. After installing de-coupling capacitor the problem of saturation is solved but 2hz square wave come as problem. Your help is much appreciatedI 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. Inverting Input is connected to the sensor and non-inverting input is groundedFurther questions/comments:
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?
I tried to use the resistor across the feed back capacitor to discharge the capacitor but the output results having a charge leakage. I am sending you one example that how the input goes in to amplifier and the output result I am expecting.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.