Transient time of opamp integrator to reach stability

[zzlong94]

The higher the value of the Rf (feedback resistor), the longer the time it take to reach its stability for it output voltage.
whats the concept of it?
it is because the larger the resistance, it need much time n energy to charge up?

[danadak]

Schematic ? Including driving source....


Regards, Dana.

[Jay_Diddy_B]

Hi,
I true integrator will never settle.
If you are talking about this circuit:



(From your other thread)

The answer output is an exponential described

Vout = -vin x R2/R1 (1 - e^{-t /(R2 x C1)})

It is normally accepted that this type of circuit settles in 5 time constants or 5 x R2 x C.

But, it will continue to settle for ever.

5 time constants gets you (1- e^-5)  equals 0.993 x final value

Regards,

Jay_Diddy_B

[rstofer]

Realistically, you don't have much of an integrator left once you add Rf.  That is, the result of operating on the signal for some period of time is not the integral of the signal over time because the capacitor has been constantly discharged by Rf.  There are simply too many terms in the transfer function.

At best, the circuit is a low pass filter.

http://www.analog.com/en/analog-dialogue/articles/phase-response-in-active-filters-2.html