Op Amp Circuit Question

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op amp circuit

Hello,

Does anyone know how this circuit work? I am having some dificulties understanding it. It looks like an integrator. The Vref is fixed at 2.5V and Vchange can be anything from 0-5V.

What is the equation for the output voltage of the op amp (Vout)?

Thanks,
 

circuit question

Is there any circuit diagram with this post ?
 

differential integrator op amp

Sorry I forgot to post the picture. I also figured out the answer between the time I posted and now. But i'll post the pic for discussion purposes.

Thanks,
 

op-amp circuit

With a correction, it's a differential integrator. A shown, R2/C2 are functionless and the transfer function is Vout = 1 + a∫Vin dt.
 

integrating op-amp questions

I think it is a comparator. It would be a differential integrator if both resistors were fed to a different voltage source. It is a comparator that ramps to 0 or Vcc depending on the relative difference between Vref and Vrange. The ramp rate is determined by R1 and C1.

So if:

Vrange < Vref

The output will ramp from Vrange to 0

If: Vrange > Vref,

The output will ramp from Vrange to Vcc.


The circuits ramps up or ramps down, but it does not integrate as the classical integrator.
 

question on opamp circuits

A difference between Vin and Vref is also a difference between two input voltages. But you can regard it as noninvrting integrator as well.

Here's the correction. The circuit is working as integrator if R1•C1 = R2•C2

 

ramp op amp

Yes I agree. The modified picture that you posted is a difference integrator. I am using the original post as a method to slowly turn off a device under over voltage conditions.

Thanks.
 

op amp circuits question and answer

Yes, it looks like an integrator (called BTC integrator, Balanced Time Constants).
However, I am afraid it will not work because of the single supply voltage.
 

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