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Okay let me try again as nobody seemed to understand my query the first time.

Sallen-Key 2nd-Order Low-Pass Filter

The cutoff frequency f of this Sallen-Key 2nd-Order Low-Pass Filter is f = 1 / 2(pi)RC

For equivalent values of RC, do different combinations of R & C give different signal integrity in practice.

For example R = 1000, C = 5x10-6 has an equivalent cutoff of as R = 100000, C = 5x10-8. But will one be better than the other in practice.

My question is, are there best practices for reducing signal error with regard to the ranges of resistance and capacitance aside from tolerances.

Edit: I am assuming R1=R2 and C1=C2 for a smooth dropoff

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  • \$\begingroup\$ what name did you use the 1st time? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 21 '18 at 21:22
  • \$\begingroup\$ I was asking about this with regard to a linkwitz-riley crossover previously \$\endgroup\$ – Adam Mar 21 '18 at 21:24
  • \$\begingroup\$ I believe Adam is talking about his previous question, Tony. \$\endgroup\$ – KingDuken Mar 21 '18 at 22:38
  • \$\begingroup\$ Consider both Johnson and shot noise. \$\endgroup\$ – Brian Drummond Mar 21 '18 at 23:47
  • \$\begingroup\$ Do you need a successful attenuation at 10X and 1000X and 1,000X the F3dB? How about at 10,000X? Do you require -80db output at 10,000X the F3dB? \$\endgroup\$ – analogsystemsrf Mar 22 '18 at 4:03
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For practical cheap amplifiers, with limited bandwidth, their output impedance begins to rise with frequency. If you've chosen a reasonable amplifier, this will be well into the stopband. At a frequency where the output impedance becomes significant, it fails to control the feedthrough of signal through R1 and C1 directly to the output. This results in a stopband lift, above the expected 40dB/decade fall. Smaller capacitors mitigate this effect.

Keep your Cs well above 100pF or so, to avoid inaccuracy due to strays.

The amplifier input bias current is provided through R. The DC offset across the stage will increase as 2R times the amplifier input bias current. However, for most practical cheap amplifiers, this will be dwarfed by the input offset voltage.

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  • \$\begingroup\$ I agree with everything - however, does this contribution touch the subject of "signal integrity"? \$\endgroup\$ – LvW Mar 22 '18 at 8:28
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Input bias currents

OP AMP input bias current causes some offset without Rf balance and stray E field may induce V into very high impedance e.g. 10M with some applications that don't offer a ground plane. This only becomes significant if your DC error tolerance is very small. thus you would add Rf from out to IN- to match IN+ resistance when Ii produces an offset voltage greater than Vio or your error spec. So 10M is practical some cases 100K in others

Current limit

Since input impedance is R1 +R2 maximum at DC the source must be able to drive this for step voltages and not induce a lower than desired slew rate t=0.35/f The Op AMP must keep up with the input as well so it has to drive R1. CMOS OP AMPS tend to have more load regulation errors from higher open loop output impedance and lower current drive from loading specs, so they are sometimes rated with 10k loads so if using a RRIO type consider this and start with 10K .

Cap

Cap values with stable wide temp NPO/COG are preferred for stable temperature operation, but are lower density so values are limited to smaller ranges requiring larger R values. E-caps can polar if polarity is observed and R values are high. Ceramics in SMD can be microphonic is mechanical shocks are expected. Since stray capacitance can be anywhere from 0.5pF to 5pF your minimum value will depend on it there is a ground plane which adds to capacitance.

Unless you are doing something critical where Resistor noise currents matter I suggest 10k is nominal and 100 to 10M are extremes that can be used but just remember input bias current DC offset errors and input impedance matters.

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The cutoff frequency f of this Sallen-Key 2nd-Order Low-Pass Filter is f = 1 / 2(pi)RC

No, that's not quite true if we use the definition of the cut-off frequency being the point where the signal power at the output has halved (3 dB down). When both resistors and both capacitors are the same value, the filter has 6 dB attenuation at f = 1 / 2(pi)RC.

This is because like all other similar 2nd-order low-pass filters the transfer function magnitude at f = 1 / 2(pi)RC equals Q, the quality factor of the circuit: -

enter image description here

enter image description here

And if you made R1=R2=R and C1=C2=C, Q is 0.5

Hence the TF is 6 dB down at f = 1/2(pi)RC

For equivalent values of RC, do different combinations of R & C give different signal integrity in practice.

Absolutely. If you are prepared to make the capacitor C1:C2 ratio 2:1 you get a Q of 0.7071 and although this doesn't sound much, you get a 3 dB attenuation at the cut-off frequency - this is called the butterworth maximally flat response. If you make the ratio bigger you get higher Q and a bigger resonant peak in the pass-band: -

enter image description here

So, if you are going to use a Sallen key filter but keep resistors and capacitors the same value you are missing a trick to get the best frequency response i.e. the flat butterworth response. In fact you might as well just cascade two RC filters and get rid of the op-amp entirely if you are not going to identify and use the benefits that a controlled peaking circuit can provide. OK using an op-amp gives you good buffering so it's not entirely wasteful!

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