samiran_dam
Full Member level 2
Hi all,
If I make a circuit as shown using an ideal op-amp (i.e. open-loop gain ≈ ∞, unity-gain-frequency ≈ ∞ and output resistance ≈ 0), then it will behave as perfect integrator.
But, as the op-amps can not be ideal, what and all non-idealities are expected in the integrator? As per my calculations, I found that there will be zeros in the transfer function along with the pole (which is supposed to come at '0' frequency) coming at a finite non-zero frequency. And there is one more pole coming at high frequency.
If anybody can intuitively explain the behavior of the op-amp based integrator it will be very helpful for me.
Thanks & Regards,
Samiran
If I make a circuit as shown using an ideal op-amp (i.e. open-loop gain ≈ ∞, unity-gain-frequency ≈ ∞ and output resistance ≈ 0), then it will behave as perfect integrator.
But, as the op-amps can not be ideal, what and all non-idealities are expected in the integrator? As per my calculations, I found that there will be zeros in the transfer function along with the pole (which is supposed to come at '0' frequency) coming at a finite non-zero frequency. And there is one more pole coming at high frequency.
If anybody can intuitively explain the behavior of the op-amp based integrator it will be very helpful for me.
Thanks & Regards,
Samiran