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I am in the process of designing a Butterworth low pass filter in Sallen-Key topology. I use the Analogue Devices Filter design tool.

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After reading the TI application report I have calculated all the values base on the formulas provided. Fc Cut-off frequency (frequency at which phase shift is equal to 90°). Q quality factor by sliding between Chebyshev to Bessel. K is gain converted from dB. I have noticed that the capacitor ratio is fixed as 10. Sensible to select the ratio of caps rather than resistors. Still unsure why 10 was selected. Perhaps getting any series of caps in the one-decade ratio is easy.

https://www.ti.com/lit/an/sloa024b/sloa024b.pdf

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I simulated the circuit in LTSpice.

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I have noticed about a 70Hz difference between the point at which the phase shifts by 90° and the -3dB point. Therefore I am interested in how to calculate the relation between both. When Filter parameters are selected in Filter Design Tool the frequency, in this case, 2KHz refers to -3dB. However, LTSpice simulation shows that at 2KHz the phase shift is equal to 90° and the signal is attenuated by 2.7dB. I also used an online calculator to verify. (Please note that C1 and C2 notation is inversed in the TI document compared to the Okawa-denshi calculator). Sallen-Key Low-pass Filter Design Tool - Result - (okawa-denshi.jp)

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Lastly in the TI application report on page 4, the formula for system gain at f=fc is simplified to: H(lp)=-jKQ in case of Bessel filter Q=0.707 and K=2. Therefore, H(lp) = 1.414 = 3dB.

I am a bit puzzled as to why Filter Design Tool asks for passband frequency and then uses it as cut-off frequency.

Additionally, the formula on page 4 refers to gain not phase.

Case with ideal op-amp and greater accuracy of the components.

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Simulating with an ideal op-amp and much greater precision of the components decreased the difference between -3dB and 90° phase shift point (pole of my system). I begin to suspect that my initial mismatch might be caused by rounding the components to E series components. Before I accept this as an answer perhaps someone could confirm or deny this hypothesis.

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    \$\begingroup\$ Your title says "Butterworth", yet you keep talking about a Bessel. Which one is it? Also, the -3 dB for Bessel is not the same as the -90 deg. \$\endgroup\$ Apr 6 at 12:08
  • \$\begingroup\$ @aconcernedcitizen Thank you for pointing out it is a Butterworth filter. \$\endgroup\$
    – Wintermute
    Apr 6 at 12:52

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I have noticed that the capacitor ratio is fixed as 10. Sensible to select the ratio of caps rather than resistors. Still unsure why 10 was selected. Perhaps getting any series of caps in the one-decade ratio is easy.

One decade or more. A common rule of thumb approximation of when one thing starts to affect another thing is 10:1. For example, if you have a 10K ohm resistor in series with a 10 ohm resistor, you pretty much can ignore the 10 value and just use 10K in circuit calculations. However, if the 10K is in series with 5K, the difference between 10K and 15 K in circuit calculations is very large. The crossover between "worry about it" and "ignore it" is approx. 10:1. Of course, this kind of estimating varies with the circuit's specific details.

In your circuit, the second R-C stage is connected to the middle of the first R-C stage. This means that the second R-C impedance can be seen as a complex impedance load to ground. From an AC point of view, it is in parallel with the 2 nF feedback capacitor. If this load is a significant percentage the first circuit's impedance, it has a direct affect in shifting the cutoff frequency of the filter and lessening the sharpness of the transition between the passband and the stopband. How large a percentage is significant? At least (wait for it . . . .) 10%. Again, this is an approximation, but without knowing more about the article you read and the calculations built into the application, what I see is the 10% rule in action. The 2nd R wants to be at least 10x the first one (in this case, 35x). Beyond that, increase it until its C becomes a convenient value.

Even at 35x, the second network affects the first. This might be built into the equations inside the app, so that the loading affect is accounted for no matter how close or far apart the two networks are in impedance. As this is Analog Devices, that probably is a safe assumption. Select the first R-C values based on the equations for the filter type, minimize the loading by selecting two caps 10:1 apart in value, run the math to get the second resistor value, adjust it for the closest commercially available value. Or something like that.

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  • \$\begingroup\$ I linked the TI material above. Isn’t the loading compensated by positive feedback? Perhaps you have a point in terms of impedances and the goal is to make op-amp work less hard. Regarding the impedance rule of thumb. My understanding is that if we cascade two passive RC networks and make the impedance of the second much higher to the point we do not consider drawing current from the first we can achieve Q of max 0.5. Therefore isn’t the whole point of the active component is to boost Q and compensate for loading? \$\endgroup\$
    – Wintermute
    Apr 6 at 14:10
  • \$\begingroup\$ I usually make a 2nd-order Sallen-Key Butterworth lowpass filter with a gain of 1 and equal-value resistors and double the capacitor value for the positive feedback. I also use equal-values for capacitors and resistors with a gain of about 1.6 times. \$\endgroup\$
    – Audioguru
    Apr 6 at 14:32
  • \$\begingroup\$ @Audioguru The first approach will not work as I require only one stage and flexibility in voltage gain selection without needing to redesign the Rs and Cs (apart pf R3 and R4 obviously). This is why the circuit I selected is so versatile. Making Rs or Cs the same value could simplify some calculations. I would prefer to equalise the Cs as getting any ratio of Rs is much easier. Nevertheless, once you type all the formulas into excel equalising Cs does not make much change in terms of time required for design. TI paper I referenced describes this method in subsection 3.3 \$\endgroup\$
    – Wintermute
    Apr 6 at 15:59
  • \$\begingroup\$ @Audioguru The second approach suggested is not suitable as Q is fully controlled by gain. Therefore I would have to employ a second op-amp to control the magnitude of the passband signal. The second approach is also described in the TI paper referenced in section 3.4 \$\endgroup\$
    – Wintermute
    Apr 6 at 15:59

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