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I designed a low pass butterworth filter with the specification to have a 500Hz cut off frequency with diagram as shown:

Low Pass Butterworth Active filter

In the specification it says "Signals need to be attenuated by at least 80dB relative to difference frequency of 400Hz" but I'm not sure what this means?

edit:

the input of filter is a 100mV AC signal, the filtered signal needs to be converted into a 5V signal. I have thought of using this comparator circuit after the output of the filter circuit? Comparator Circuit

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  • \$\begingroup\$ What this means Klaus is that the second order circuit you shown wont be anywhere enough . \$\endgroup\$ – Autistic Jan 6 '17 at 10:31
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    \$\begingroup\$ signals at what frequency need to be attenuated by 80dB with respect to signals at 400Hz? Does the 500Hz 'cutoff' mean -80dB or -3dB? Is the 500Hz the top end of the passband, or the bottom end of the stopband? \$\endgroup\$ – Neil_UK Jan 6 '17 at 12:03
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A 2nd order low-pass butterworth filter as you have shown will have a similar frequency response as shown below in trace 2: -

enter image description here

So, if 500 Hz is re-normalized to coincide with "1" along the x-axis, you will see that at 5000 Hz, the attenuation is 40 dB down and below 500 Hz it's fairly flat all the way to DC (but with some gain dictated by R3 and R4 in your circuit).

In the specification it says "Signals need to be attenuated by at least 80dB relative to difference frequency of 400Hz" but I'm not sure what this means?

It certainly doesn't apply to a 2nd order filter as you have drawn in your question. It might mean that at 900 Hz (500 Hz plus 400 Hz) the attenuation needs to be at least 80 dB and, if that is the case then a fifth order butterworth filter would be required.

It could even imply that what you want is a pass band filter with 80 dB (or greater) attenuation 400 Hz either side of the centre at 500 Hz.

But, at the end of the day, it's down to you understanding the full specification you are given so, maybe publish a little bit more of it.

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  • \$\begingroup\$ by order of the filter, do you mean that I would need a 4 pole rather than 2 pole filter? \$\endgroup\$ – Klaus Jan 6 '17 at 11:33
  • \$\begingroup\$ @Klaus it all depends on how you interpret the specification. As I've said, it could mean that you need a fifth order low pass butterworth filter or maybe a tenth order band pass filter. The spec is unclear so I'd advise you to clarify what is meant. \$\endgroup\$ – Andy aka Jan 6 '17 at 11:56
  • \$\begingroup\$ I have chosen to use a 4 pole butterworth filter; this makes the roll-off 80dB/decade which is what I think I need. The nominal input will be around 100mV but then that needs to be converted to a 5V logic-compatible signal. Would I just need to use a simple op-amp circuit for that? but I understand that the filter would add some gain to the circuit \$\endgroup\$ – Klaus Jan 9 '17 at 11:19
  • \$\begingroup\$ You can convert millivolts (a ten to several hundreds) to a logic signal using a comparator (typically an LM339 for lowish speed ranging up to MAX999 for blistering fast speeds). I wouldn't use an op-amp given what you have said in your question. \$\endgroup\$ – Andy aka Jan 9 '17 at 11:23
  • \$\begingroup\$ I have attached the comparator circuit above, will this be sufficient? \$\endgroup\$ – Klaus Jan 9 '17 at 13:49
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The specification is a bit unclear (because of the term "difference frequency"). It would be much better to say: "The minimum attenuation at (XXX) Hz must be at least (YY) dB.

However, independent on this point (and also indpendent on the required filter order) I am afraid you will never get an attenuation of 80 dB using THIS topology (Sallen-Key). The reason is as follows: It is a known effect that all Sallen-Key lowpass stages suffer from direct signal feedthrough.

That means: The desired filtering action is heavily disturbed in the stop band region because an unwanted part of the input signal is directly coupled to the output through the feedback capacitor C1. A corresponding signal portion is, therefore, produced across the finite opamp output resistance. In many cases, tis effect limits the stopband attenuation at a level of app. (-40...-50) dB only.

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The stop-band is ruined, not only by C1 but by the OpAmp's collapsing gain at the higher frequencies. And the soaring Zout of the OpAmp above UnityGain Bandwidth.

Include some discrete R+C Low Pass Filters scattered amidst the high-order active filter stages, so the R+C passives can take over the attenuation of the very highest frequencies.

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