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opamp integrator- doubts about equation

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Hi everyone. I was reading this:

https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/differentiator-integrator-circuits/

and came across with this:
"However, if we apply a constant, positive voltage to the input, the op-amp output will fall negative at a linear rate, in an attempt to produce the changing voltage across the capacitor necessary to maintain the current established by the voltage difference across the resistor."

Can't understand why a positive voltage to the input will make the output go negative and the current bit on the explanation. Has anyone analysed the integrator from this point of view? Can anyone explain in another word?

Another thing that can't understand is why, in the expression, we have -Vin/(RC). Why it is negative? If we use nodal analysis this is not held.

Thanks.
 

The opamp is in an inverting circuit. When the input to the resistor is positive then the output of the opamp is negative. If the input to the resistor is negative then the output of the opamp is positive.
 

Hi,

If we use nodal analysis this is not held.
Click to expand...
I assume your analysis is wrong.
Did you include the OPAMP as a regulator that regulates the center node to be "virtual ground"?

Klaus
 

No Klaus, I haven't. But I think you are right. I missed a signal.

However there is another thing. The Vin that they are showing in the website is the differential input?

Taking opportunity another question. If instead of 0V at one of the inputs (+ve) I had 1V or so, the feedback would again try to put on the -ve input 1V correct?
 

Hi,

The Vin that they are showing in the website is the differential input?
Click to expand...
Vin is a single ended input. It is referenced to GND because non_inverting_input is GND.

If you put any other signal to the non_inverting_input, then Vin is referenced to this signal.
So as global description: Vin is referenced to non_inverting_input.

To your example:
If non_inverting_input is 1V, then
- if Vin is 1V, too --> Vout won´t change.

Klaus
 

Yes, that's the point. So if you were to have the non_inverting_input connected to a reference in this case say 1V, the inverting_input would be forced to that value right? The output would only react then if the input voltage was to go above or below this voltage, 1V.

Can this integrator be seen as something that does an average? If it can, the average is being done where, at the output?
 

The simple integrator shown has no DC negative feedback then the average input voltage must be extremely close to the non-inverting input voltage of the opamp, and even then the DC input offset voltage of the opamp will be amplified a few hundred thousand times to latch up the output of the opamp.
 

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