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I was wondering about this question. What if the only design requirement is having the highest steepness of the roll-off but using only one op-amp? I know it's possible to construct approximations of second order low-pass filters with one but what about higher orders?

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  • \$\begingroup\$ I am a bit confused about some answers and their up-voting. I suppose you are asking for an active filter with one single opamp only? Or not? Or is a complete 10th-order chip allowed? Or a cascade of 5 stages ? Or even a passive filter? What do you need? \$\endgroup\$ – LvW Feb 25 '15 at 20:12
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You can make a high-order low-pass filter with no op-amps at all. In principle you could obtain any roll-off rate you like, given enough pi or T sections. You could then add an op-amp output buffer to make your filter a "one op-amp" filter if for some reason you'd like to do that.

But high-order filters tend to be very sensitive to small changes in component values, so they often require hand-tuning each circuit to achieve good performance, and the performance could drift as temperature changes.

The benefit of using active filters is that the op-amp allows low output impedance and high input impedance, reducing the loading effect of one stage on the next. This means you can cascade several first or second order filters without much interaction from stage to stage, and reduce the need for tuning. If you try to make a high-order active filter in one stage, you're giving up a key advantage of using active filters.

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    \$\begingroup\$ Excuse my lack of experience, but what is "good performance"? If the component values change, does the roll-off get worse, or is it just the response around the corner frequencies that gets ugly? \$\endgroup\$ – Greg d'Eon Feb 25 '15 at 19:53
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    \$\begingroup\$ The bandwidth might change, the gain might not be constant in the passband, the shape of the transition might change, and maybe ripples of lower loss could appear in the stop band. \$\endgroup\$ – The Photon Feb 25 '15 at 21:16
  • \$\begingroup\$ Another advantage of using op amps is that they eliminate the need to use inductors. \$\endgroup\$ – supercat Feb 25 '15 at 21:25
  • \$\begingroup\$ Instead of an opamp output buffer, wouldn't it be desirable to put the Pi / T filter sections in the feedback path, so that both input and output impedance will be improved? Or do multiple stages kill phase margin and cause instability? \$\endgroup\$ – Ben Voigt Feb 25 '15 at 23:21
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You can make a 10th order filter using an 8 pin SOIC chip like the LTC1569-6

enter image description here

This doesn't use any conventional op-amps but, keeping in with the spirit of the question and @thephoton's answer you might regard this chip as counting as at least one op-amp (equivalent).

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  • \$\begingroup\$ Right where I was going to go! I'll just add the warning that there are some design considerations for using switched capacitor filters. The biggest is of course the possibility of aliasing. Your clock is your sample frequency, and you need to pay attention to Mr Nyquist. The second is clock bleedthrough, and the distinct possibility that you may need to post-filter to knock this down (though an RC is usually enough) \$\endgroup\$ – Scott Seidman Feb 25 '15 at 19:46
  • \$\begingroup\$ @ScottSeidman I couldn't have said it better \$\endgroup\$ – Andy aka Feb 25 '15 at 20:32
  • \$\begingroup\$ @ScottSeidman care to explain a little bit more about clock bleedthrough ? Not sure if I should open up a question for this. \$\endgroup\$ – efox29 Feb 26 '15 at 6:02
  • \$\begingroup\$ @efox29 I believe Scott is referring to the switching clock used for these types of filters. inevitably a little bit of the clock signal finds its way to the output just as with a DAC you get small glitches on the analogue output when updating a DAC value from the digital system. Fairly easy to get rid of. \$\endgroup\$ – Andy aka Feb 26 '15 at 8:17
  • \$\begingroup\$ @Andyaka ah, yes I have experienced something like that with the switched cap filters - i just thought it was just how it works. Didn't realize the cause. Good to know! Would an RC be sufficient ? Fc ? \$\endgroup\$ – efox29 Feb 26 '15 at 16:56

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