Can I put two sallen key filters in series without putting a buffer in between them (in order to analyse it mathematically correct)?

I know that when the output impedance of the first sallen key filter is much lower than the input impedance of the first sallen key filter.

  • \$\begingroup\$ Do you have a specific higher order filter response that is required by the chained result and are, instead, asking how to compose that higher order filter into a successive set of Sallen-Key filters of 2nd order to achieve it? \$\endgroup\$ – jonk Feb 28 '18 at 20:02

The output impedance of a Sallen-Key filter is zero, with an ideal op-amp of course, and very close to zero with a real op-amp. One stage can drive another stage without interaction.

Cascading Sallen-Key stages is a standard way to build up a higher order filter.

  • \$\begingroup\$ Thanks for your answer. How do I know/prove that the output impedance is zero? \$\endgroup\$ – sda Feb 28 '18 at 19:43
  • \$\begingroup\$ That comes from op-amp and control theory. It would be best to start another question to get an answer to that. \$\endgroup\$ – John D Feb 28 '18 at 19:47
  • 1
    \$\begingroup\$ How do you know? Because I told you so. BTW, I also told you that with a real op-amp it's not zero, just very close to zero. Perhaps you mean how do I calculate what the output impedance is? The output impedance is change of voltage for change of current. Now the op-amp can supply arbitrary output current change without change at the input pins (an ideal, infinite gain op-amp anyway) so its output voltage doesn't change either. So zero output impedance. A real opamp needs a sniff of deltaV at the input pins, so the output voltage does change, but with very high gain, it changes very little. \$\endgroup\$ – Neil_UK Feb 28 '18 at 19:47
  • \$\begingroup\$ @JohnD I do know control theory. \$\endgroup\$ – sda Feb 28 '18 at 19:47
  • \$\begingroup\$ @Neil_UK Yes indeed how do I calculate it? \$\endgroup\$ – sda Feb 28 '18 at 19:48

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