I was under the impression that the differences between a Chebyshev, Butterworth, and Bessel filter was that they represented the cases of an underdamped, critically damped, and overdamped systems, respectively. However, after revisiting the transfer function for a second-order Sallen-Key filter, it seems I am mistaken as a Butterworth filter is defined with \$\mathrm{Q} = \frac{1}{\sqrt{2}}\$ or 0.707 placing it in the underdamped category. For a critically-damped filter, Q must be equal to 0.5. Why then doesn't an ideal Butterworth filter seem to exhibit ringing as can be seen with a Chebyshev filter?
Electronic – Why doesn’t a Sallen-Key Butterworth Filter ring
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One ting that jumps out is you have not selected a model for the diode - right click and select one from the list. Otherwise it may not simulate correctly.
I think what you are seeing may be to do with the rectifier not being able to sink current (only a small amount will flow through the 200k resistor), so the input signal will not look as it would unloaded, it will "charge" to it's peak. Then the larger the cap, the longer the fall time when the signal stops. Try putting a non-inverting buffer in between the rectifier and Sallen Key input. This should make both filters respond in the same way.
The Sallen-Key filter can be realized with gain but component value adjustments are necessary to reduce Q. Q rises as gain increases from unity and the feedback capacitor is usually the component to lower. Another way of looking at this is by thinking about the feedback capacitor - it's the only component that is affected by gain - if the gain were (say) 10 and the feedback signal to the cap were reduced by 10 the circuit would perform the same theoretically. Reducing the cap or potting down the input to the cap has the same effect.
Here is a diagram from TI source: -
As you can see, the frequency cut-off is exactly the same as the unity gain circuit but Q is now dependent of gain 1+R4/R3.
The source (TI) linked above should allow you to derive the high pass version
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Best Answer
A Butterworth filter has a flat pass band (spectral response): -
But it will still ring when a transient comes along (time response): -
Only a critically damped or over-damped filter will not produce overshoot (ringing).
EDIED to show that a 2nd order Bessel filter does produce 0.433 % overshoot: -
Calculator source
The resistor value is manipulated to give a zeta of precisely 0.866 with Fd of 0.5*Fn (aka Bessel).