Active filter question - Page 1
Forum section: electronics / beginners

author: Cantafford
author_rank: Newbie
timestamp: December 19, 2014, 01:34:25 pm
content: Hello,I have a question about the band pass active filter in the picture that I have uploaded.To my understanding, a band pass filter is supposed to output the frequencies that are within a specified bandwitdh and exclude the ones that are not. So in the uploaded picture, capacitor C1 will limit the signals whose frequencies are too low. Up to this point I understand.Now please correct me if I'm wrong. The signals that have passed C1 and are not 'too high' will NOT GO through C2 but will instead GO INTO THE INVERTING INPUT of the op amp and then there will be an output at Vout. But what if the frequency of the signal is too high and is therefore out of the bandwidth that the filter was supposed to output. What would stop such a signal from reaching the output via the top capacitor C2?Thank you for reading!

author: KM4FER
author_rank: Regular Contributor
timestamp: December 19, 2014, 01:55:55 pm
content: My analysis of this circuit is that the frequencies that are "too high", that is those above the band, WILL pass through the feedback capacitor C2 and arrive at the inverting input of the amp. But, they will have undergone a phase inversion in going through the amp and will therefore be 180 degrees out of phase with the input signal. By superposition then the feedback signal will combine with the input signal and the two will cancel each other out thereby filtering out the high frequencies. Meanwhile C2 provides a higher impedance to the in-band frequencies and they ARE NOT fed back to the inverting input and consequently are not attenuated.earl...

author: Cantafford
author_rank: Newbie
timestamp: December 19, 2014, 02:28:06 pm
content: Thank you, very good explanation I almost got it.If you can tell me why would a signal go through C2 go back to the inverting input and not go through the non-inverting one towards the ground?

author: LvW
author_rank: Frequent Contributor
timestamp: December 19, 2014, 02:32:26 pm
content: Well - let me try a somewhat modified answer (if compared with reply#1).There are different possible approaches to answer your question:1.) It is easy to derive the closed-loop transfer function based onT(w)=-Zf/Zinwith Zf=R2||jwC2 and Zin=R1+jwC1. After proper rearrranging, this will give the classical second-order bandpass function with a second-order denominator.2.) You can explain the principal behaviour of the circuit as a combination of a lowpass and a high pass section.3.) As mentioned in reply#1: Yes - C1 blocks for low frequencies and C2 produces a short for very high frequencies.But this is a rather qualitative sight only.In reality, C1 does not "block" but has an impedance that continuously increases with decreasing frequencies; and C2 produces no "short" but has an impedance 1/jwC2 that decreases continuously with rising frequencies.4.) Of course, you will have feedback for all frequencies - but the amount of feedback (and the phase!) depends on the actual frequency. And this effect leads to the bandpass behaviuor with one single frequency for which the phase contributions from both parts cancel each other.This is the center frequency with a phase shift of -180deg (because of the inverting nature of the amplifier).5.) Forgetting realistic opamp properties, for very high (infinite) frequencies we have a circuit that can be seen as a current-to-voltage inverter having a conversion factor of zero ohms (zero ohms in the feedback path).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 19, 2014, 05:57:14 pm
content: It is assumed the op-amp has an ideal (constant voltage) output.It's also assumed to have infitite gain and phase margin at all frequencies, which is similarly fallacious.Tim