# How does the capacitor C1 provide filtering in this second order active low pass filter?

I am looking at this second order active filter:

I can see that the cap C2 provides a short to ground for HF signals; providing filtering, but could someone please tell me how the cap C1 provides any filtering?

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I can see that the cap C2 provides a short to ground for HF signals; providing filtering, but could someone please tell me how the cap C1 provides any filtering?

Consider frequencies much higher than the 3 dB cut-off point - the output is going to be really small (nearly 0V) and it has to be that way because the circuit is a low pass filter. Hence both C1 and C2 are jointly pulling the signal down to 0V. This is now equivalent to this type of filter at way beyond the cut-off: -

Remember, the above circuit is true from frequencies much higher than the 3 dB point of the filter because the output is so small. This proves C1 does do filtering but the effect it has a decade either side of the 3 dB point is much more important than very high up in frequency.

So to answer your question "this is how C1 provides filtering" but, of course if you are looking for a fuller answer that encompasses the 3 dB point then prepare for math because it isn't that obvious what happens just by looking at the picture.

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Ok, that's good, but if there is HF on top of a DC offset - C2 charges to the offset voltage, which is now also at Vout (assuming no amplification from R2 and R1). Now there is no voltage across C1 due to DC because the input and output are the same voltage. Now I don't have a second path to ground. I can see that the HF is passed to Vout, and is now included the the feedback loop (abite half the amplitude) Does the inverting input cancel it out somehow? - lots of maths would be great. :) – Tim Mottram Feb 23 at 21:16
@TimMottram of course you have a 2nd path to ground thru C1 - it doesn't matter what DC offsets are around - a low pass filter will attenuate an AC signal above the cut-off frequency even if there were a hundred volts across the capacitor. It doesn't have to be at ground - it has to be at a voltage that is constant and largely immutable and, of course 0V or 100V fits the bill. Here's the math: daycounter.com/Filters/SallenKeyLP/… – Andy aka Feb 23 at 22:16

At high frequencies C2 moves the Input+ pin near ground so the Input- pin would be also near ground. so the output would also be near ground. so at high frequencies the Vout would be very low and we call it virtual ground. now you can see another low pass filter with R3 on the input and C1 to the ground, interesting :)

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That's a nice answer, but what about a composite signal or a HF with DC offset? – Tim Mottram Feb 23 at 21:10
If you use superposition, you can evaluate DC and HF seperately and sum up the answers in the end. At very low frequencies both caps are open so you would only have a simple gain – Ali80 Feb 23 at 21:13

Well you could always go though a metric ton of algebra and get the exact equation, but I'm guessing you're looking for something a little more intuitive.

If the voltage at Vin is a sinusoid (say sin(t)), then the voltage at V+ will lag by 90 degrees (i.e. -cos(t)) because of the parallel capacitor (to ground). Of course this produces an output signal with the -cos(t) phase. The voltage across C2 will be the difference between these two signals, thus the amplifier injects an out-of phase signal into the input, reducing the amplitude of the resulting signal at V+.

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I don't have enough reputation to comment, but here is maybe answer to your question. This is well known Low Pass filter named Sallen-Key. Here you can see all the equations: Sallen-Key.
Your topology has additional gain with R1 and R2 resistors (for low frequencies).

EDIT: Feedback with C1 is for keeping signal phase non inverting and provides Q enhancement of the signal.

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Thanks for the answer, I have looked at the page, however is gives no indication of why the cap C1 - acting as a short to HF, helps attenuate the frequency. – Tim Mottram Feb 23 at 21:00

First consider the case where C1 were connected to ground, you would have two low pass filters (R3,C1 and R4, C2) in series. The problem is that the second filter loads the first one. The improve this situation the bottom is not connected to ground but to the output of the amplifier.

To see why this helps, imagine a step at the input of the fiter. It will charge C2 through R3 and R4. The voltage that builds up is amplified and increases the voltage at the right side of C1, which is equivalent to charging C1.

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