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Square wave output from op-amp integrator

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go2tariq

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I have a problem I configured my amplification unit as a integrator. As a result I am getting square wave for example see the image file. Could anyone please help me to identify what is hapenning actually.
Thanks in advance.

 

LvW

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Who has told you to place a capacitor in front of the negative input node?
Or do you want to build an S/C integrator?
 

go2tariq

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Actually I use the capacitor in front of negative input because I want to stop any dc current. Without use the dc-coupled capacitor my integrator was saturated in less than 2 seconds. After installing the capacitor infront of negative input the circuit is not saturated anymore.
I didn't use it as a SC integrator.
 

LvW

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Actually I use the capacitor in front of negative input because I want to stop any dc current.
Have you never seen a typical RC integrating device (Miller integrator)?
You told us you are "getting a squarewave" at the output. Perhaps that is correct?
I do not know because I have no information about the input signal.
You see, a lot of information is missing.
 

go2tariq

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Current is an input signal. I simulated this circuit please see the image. Current signal I am using as input and volt signal I am getting as output.
 

LvW

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For my opinion, it looks nice. What is the problem now?
 

go2tariq

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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.
 

LvW

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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?
 

Kral

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You may have a grounding problem. Try to route the ground connections from the input current source and the integrator so that they connect to the power supply at one point only (at the power supply). This is the classic "Star" ground configuration. If the grounds differ in voltage, then you will have a common mode input to the integrator, which would cause the output of the integrator to vary in step with the the ground voltage difference.
 

go2tariq

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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?
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 appreciated
 

LvW

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Further 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?
 

go2tariq

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Further 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?
1. Inverting Input is connected to the sensor and non-inverting input is grounded
2. Relay work on demand, I mean a signal generated on demand.
3. I used a large resistor >300 M ohm but there was leakage of current that I dont want
4. Display no 2 is the result from simulation
5. Drawing no 1 is output of from the amplifier in real. simulation result is drawing no 2
 

LvW

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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.

Recommendation:
*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).
 

go2tariq

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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.

Recommendation:
*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).
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.
 

FvM

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The author has started a number of threads revolving around an integrator with 10 pF feeedback capacitance to measure currents in a fA range. See e.g. https://www.edaboard.com/thread203750.html
Some points didn't get clear for me yet, Up to now, the sensor had been described as a current source, if so, the purpose of the input capacitor can't be understood. If the sensor is a voltage source, then the input capacitor cancels the integrator action.

OK, now it becomes more clear.
Not particularly, I fear.
For the image in post #1, apart from the (minor) differences between two waveforms, many details are missing:
- circuit connected to the input
- respective input current/voltage waveform
- type or properties of the amplifier
- clarification, that the switches are not operated during the measurement, respectively how they have been operated
 

LvW

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Yes, FvM, I agree.
My comment "it becomes more clear" was related to the subject "measurement/simulation" only because I couldn't see any relation between the two plots the author of the thread had presented.
Only now - with posting #14 - we get some information on the input signal (however, just the form -no time information).

It's really a pitty that some people even are not able to describe a problem in such a way that somebody can give a substantial answer.
For my opinion, this should not be a great problem:
Basically we need a description of (a) input signal (form, frequency, amplitude, source properties), (b) required output signal, (c) technical, operational conditions (hardware restrictions, powering, other...)
 

LvW

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Hi go2tariq,

after some complaints (see above) I like to start helping again.

At first, I summarize: You have an unsymmetric input signal from a sensor (current pulse, amplitude:...A) of .... seconds duration.
It is your goal to create the average of this puls. Is this correct?
That's all. The mentioning of the average is very important because of the integrator time constant that is necessary for this purpose. Unfortunately, you did mention this (via drawing) rather late (in posting #14 !!).
As you can see, some information is still missing (input signal amplitude and duration).

I have developped an example for average calculation (see pdf attachement). Please note, that the time constant must me much larger than the signal duration (in my example: factor 100/3=33.3).
Regards
LvW
 

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