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I'm studying the unity-gain Sallen-Key low pass filter

I can't find any material explaining the rise in knee of the magnitude response at/after the cut-off frequency, other than that the Q factor (and ratio of capacitors) affects this

Is this a type of resonance between the two capacitors?

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  • \$\begingroup\$ Unless there's a gyrator (voltage-to-current and current-to-voltage converter) in between I don't see how two capacitors can resonate with each other. \$\endgroup\$ – Bimpelrekkie Oct 27 '15 at 12:10
  • \$\begingroup\$ Sorry - I am not sure what you mean with "knee". For a reliable answer it is necessary to avoid misunderstandings. Hence my question: Do you refer to the "peaking" of the magnitude response ? This is a typical property of a Chebyshev response. But note that this does NOT happen after the cut-off frequency. In contrary - the peak can be observed below defined the cut-off. \$\endgroup\$ – LvW Oct 27 '15 at 12:50
  • \$\begingroup\$ This graph of a 2nd-order Sallen Key demonstrates a "knee", most life-like in the yellow plot where Q=1. OK, I admit I've never seen anybody with real knees that look like the plots of Q>1, but you get the idea. \$\endgroup\$ – CharlieHanson Oct 27 '15 at 14:34
  • \$\begingroup\$ CharlieHanson-quality values Q>1 are required for higher-order filters. For example, a 4th-order Butterworth filter needs a 2nd-order stage withQ=1.3. But it is very uncommon to use the term "knee" for a simple magnitude peaking. Therefore my question. \$\endgroup\$ – LvW Oct 27 '15 at 14:44
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Forget Sallen/Key for the moment, go back to the equivalent passive implementation of the same filter. (e.g. search "passive butterworth filter" for examples.) You'll see it involves an L and a C per 2nd-order filter section, thus it's basically an L-C resonant circuit, with resistive damping to control the resonance (the height of the knee).

Having understood this, then you can use a C in the feedback path of an amplifier to simulate the L in an L-C resonant circuit - and the Sallen-Key circuit is basically one example of this. As "FakeMoustache" comments, the gyrator is another.

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    \$\begingroup\$ For my opinion, it is somewhat questionable if "a C in the feedback path" is able to "simulate the L". Such a circuit would have an integrating behaviour rather than an inductive characteristics. The Sallen-Key topology is nothing else than a 2nd order ladder RC structure with positice feedback (for producing values Q>0.5. \$\endgroup\$ – LvW Oct 27 '15 at 14:52
  • \$\begingroup\$ @LvW : +1,I plead guilty to over-simplification! The analogy is not exact, and if you can come up with a better one, please do. \$\endgroup\$ – Brian Drummond Oct 27 '15 at 15:06
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    \$\begingroup\$ I rather think, there is no analogy between an LC resonant effect and a RC ladder topology. And, more than that, why should I look for such an analogy? Evrybody knows the classical RC block - and the S+K structure consists of two blocks with positive feedback - and positive feedback can compensate damping effects. I think, this is the most simple explanation. \$\endgroup\$ – LvW Oct 27 '15 at 15:21
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Ideal op-amps are not readily available- real op-amps typically have dominant pole compensation so that the gain drops by 20dB/decade. The output impedance of the amplifier thus appears to increase at a similar rate, and at a high frequency (usually well above the cutoff if the amplifier has been chosen well) the magnitude of the output/input will start to increase with frequency.

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  • \$\begingroup\$ This is true, but in a unity-gain configuration, I don't think that's what the questioner is asking. \$\endgroup\$ – Brian Drummond Oct 27 '15 at 12:09
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    \$\begingroup\$ @BrianDrummond The question would benefit from a schematic and a Bode plot illustrating the 'knee'. \$\endgroup\$ – Spehro Pefhany Oct 27 '15 at 12:15
  • \$\begingroup\$ Yes - what is a "knee"? Amplitude peaking? \$\endgroup\$ – LvW Oct 27 '15 at 14:47
  • \$\begingroup\$ yes it's a common term in engineering for the resonant peak in a filter's frequency response \$\endgroup\$ – Andrew Davis Jan 13 '17 at 11:53
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It's a resonance in the entire circuit. That is, if you change the amplifier gain, either of the resistor values, or either of the capacitor values in the Sallen Key circuit, the response will change.

It's not easy to explain intuitively how the response changes. It's not as easy as it is for instance with an LCR circuit, where the LC controls the frequency, and the R affects the Q. Unfortunately, one has to set up the transfer response for the circuit, and grind through the s algebra.

But you've noticed the capacitor ratio influences mainly Q. The RC product influences mainly the bandwidth. That's about as intuitive as you're going to get.

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