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If I have a differential signal on a pair of wires, then how would I do a simple RC low pass filter on this differential signal?

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    \$\begingroup\$ Like any other signal, just duplicate the filters on both differentials? \$\endgroup\$
    – Lundin
    Commented Aug 31, 2020 at 13:17
  • \$\begingroup\$ Following up on what Andy aka provided, the + Source impedance and mismatches + the load impedance and mismatches + the parasitic capacitances and imbalances ALL matter, in achieving good/useful differential and common mode filtering. \$\endgroup\$ Commented Aug 31, 2020 at 13:42
  • \$\begingroup\$ Alternatively, an R in series in each leg, and C/2 from one leg to the other. Or combine them to get a capacitive delta (the best approach in noisy environments) \$\endgroup\$
    – user16324
    Commented Aug 31, 2020 at 13:45

1 Answer 1

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how would I do a simple RC low pass filter on this differential signal?

The simplest method: -

enter image description here

Source. This circuit only filters differential signals; if there is an in-phase signal (or interference) on both lines (called a common-mode signal/noise) it provides no filtering to it.

A more complex but effective method: -

enter image description here

Modified from this Source. This filter provides differential and common mode filtering. C2 and C3 are the extras but, be aware that if the differential signal is required to remain highly "balanced" in nature, tolerances on C2 and C3 can displace the perfect symmetry at higher frequencies.

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  • \$\begingroup\$ I'm trying to calculate a modification of the 2nd circuit. It has additional resistors between C1 and C2\C3. Could you elaborate on a purpose of those resistors? Is it still a 1st or 2nd order LPF? Would you mind adding some cut-off frequency calculation formulas for this circuit? \$\endgroup\$ Commented Jun 30 at 5:35
  • \$\begingroup\$ Please raise a new post on this theme. \$\endgroup\$
    – Andy aka
    Commented Jun 30 at 9:02
  • \$\begingroup\$ Thank you for the response! I've posted it here. \$\endgroup\$ Commented Jun 30 at 21:29

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