I'm designing an active low-pass filter for a differential input. The input is on a ±100V scale and needs to be scaled down to the standard ±10V scale.

I found this application note which describes how to design a 2nd order multiple feedback filter and convert it to differential inputs:


However, they state in their first step that the magnitude of the amplification factor H0 needs to be at least one, without giving a reason (in our case H_0 = -0.1).

I don't see a reason for such a restriction, so I ignored it and designed such a filter (Bessel, with about 9kHz corner frequency) and it seems to work as expected, both when simulated with LT-Spice and in hardware.

Does anybody know where this condition originates? Is it a real restriction, or is it just arbitrary, maybe copied from somewhere without understanding the math behind it? Is it maybe because you would tend to use a different topology for gain less than one, though there is nothing really wrong in using multiple feedback?

Curiously, TI's Filter Pro also didn't let me enter anything less than 1 for the amplification.

  • \$\begingroup\$ It would help if you posted your circuit with the attenuation of 10 built in. I'm not sure how you could have implemented it from the circuit you linked. \$\endgroup\$ – Andy aka Nov 7 '13 at 20:31
  • \$\begingroup\$ The values are R1=100kΩ, R4=10kΩ (hence the attenuation of 10), R3=1.5kΩ, C5=2.2nF. Instead of two C2=6.6nF there is a single 3.3nF resulting from putting them in series and removing the connection to ground. I think that single C gives better common mode rejection. It is also done like that in a document by TI: ti.com/general/docs/lit/… (page 37). \$\endgroup\$ – starblue Nov 7 '13 at 21:20
  • \$\begingroup\$ OK I get it now. Seems very reasonable what you've done but does the filter shape still look decent with the 10:1 attenuation? I mean, it doesn't turn into a very sloppy low pass with boring old gradual roll off from dc or something like that? \$\endgroup\$ – Andy aka Nov 7 '13 at 22:31
  • \$\begingroup\$ @Ansy aka Good question. The second-order certainly shows, above ca 36kHz it is better than the ca 3kHz single-order low-pass we currently have. How it compares to two 9 kHz low-passes in series is difficult to see. \$\endgroup\$ – starblue Nov 8 '13 at 7:43

I don't see either why it is a requirement. However, you can achieve a 2nd order LPF with attenuation without an an op-amp, and without introducing power supply noise, dealing with input bias current noise, and the other few limitations of op amps.

From TI's Design Methodology for MFB Filters

Using an active filter for a Q < 0.5 (two real poles) and/or gain < 1 is unlikely because a simple passive circuit can easily provide attenuation and two real poles without requiring an active element.

At this point active vs passive really depends on the application requirements such as cutoff frequency, Zin, Zout, area, cost. Additionally, the typical differential passive LPF will have a differential output, while the active circuit in question has the output referenced to ground.

  • \$\begingroup\$ I think we need that opamp to get from differential to single-ended. And the Bessel filter characteristic should be somewhat better than what can be achieved with two real poles. \$\endgroup\$ – starblue Nov 8 '13 at 7:46

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