I have the following filter after the DAC chip (currently with unbalanced outputs) working very well, but I need now to change the DAC chip. The new source has differential outputs.

Is there a simple way to convert the filter (maintaining identical performances) using differential inputs, instead of unbalanced like now?

Is there any simple formula for the components? (Output should remain RCA unbalanced). Usually differencial opamp circuitry should have mirroring components on both inputs, correct?

enter image description here

I doesn't matter to still use two opamps or add an extra one.

  • \$\begingroup\$ Components values in the schematic are dummy as I cannot reproduce true ones for copyright reason, but please use these one. \$\endgroup\$ – Gianluca G Sep 24 '18 at 6:24

Yes Diff Amp uses RC feedback on Vin+ to gnd . With low tolerances like 0.1% for 60dB CMRR then use both R diff inputs.


The filter is suboptimal cascaded as a 1+ 3rd order filter =4th order LPF and is better designed with specs for group delay, Q BW, etc., to optimize bandpass range and bandstop rejection, when both Op Amp filters are integrated.

See my other related answers using TI filter design tool. enter image description here

  • \$\begingroup\$ Sorry, but I need to know the schematic by using the same components (same number of zeros and order) . Can you please post a draft? \$\endgroup\$ – Gianluca G Sep 24 '18 at 15:29
  • \$\begingroup\$ I could but you might need to sign an NDA (j/k) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 24 '18 at 15:37
  • \$\begingroup\$ I just realized I cannot do that on an iPad. So imagine same Feedback parts shunted to Vin+ to gnd. Then open ground and use differential input R’s \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 24 '18 at 16:00
  • \$\begingroup\$ Disregard 2nd diff output sht cct. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 24 '18 at 16:53
  • \$\begingroup\$ Sorry, but this doesn't reply to my initial question: need exactly the same components values and slopes which I'm using in my attached circuit. \$\endgroup\$ – Gianluca G Sep 24 '18 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.