# Electronic – Why does this first-order RC filter with an op-amp display resonance

filteroperational-amplifier

I built this circuit and simulated it in CircuitLab and was trying to make an active low pass filter. However, it shows a frequency response like a resonant low pass filter and I can't figure out why.

Here's the schematic: And the frequency response: The gain of the frequencies below the resonant frequency is set by the ratio of R2/R1 (x10 with these values), as expected for an inverting op-amp circuit. But I can't figure out why there is the resonant peak at 800 Hz for this first-order circuit, or how to calculate the resonant frequency.

And the response of a few different cap values: And the response of different values of R1: Changing the value of R1 has no effect on the resonant frequency, only on the gain of the circuit. I don't understand why this circuit behaves this way. Can anyone explain it to me?

I came up with this equation for the value of \$V_{out}\$:

$$V_{out} = -( \frac{A \times R2}{R1 + R2 + R1R2Cs – A \times R1} ) V_{in}$$

Does this equation look right? It makes sense to me that when \$(R1 + R2 + R1R2Cs) << A \times R1 ,\$ the result is approximately equal to R2 / R1. But I have no idea where the peak at 800 Hz comes from. 