I was searching some documents and tutorials online for filter design. Then I stumbled upon this TI application report. The following is referring to section 2, figure 7 and figure 8 of the paper. I added a picture of the circuits below for convenience.

Comparing the circuit with this electronics tutorial I am not able to figure out, what the resistors R3 and R4 are supposed to do. And why are R3, R4 and Cin, Cout not needed for the circuit with +/- supply?

Taken from TI Application Report - Filter Desing in Thirty Seconds

  • 2
    \$\begingroup\$ They look like biasing circuitry \$\endgroup\$
    – NickHalden
    Aug 21 '11 at 21:19
  • \$\begingroup\$ Fig 7 is OK for DC feed. Note that as the IC is a "follower" with a gain of unity, you can use a single transistor emitter follower in its place ! :-). Collector to supply, R2/C1 to base, emitter via lowish resistor to ground and C2 to emitter. Magic :-). \$\endgroup\$
    – Russell McMahon
    Aug 22 '11 at 4:57
  • \$\begingroup\$ @Russell McMahon - Novice here, so bear with me. The ideal op amp impedance is infinite. If I supsitute it with a transistor, current has to flow to the transistor. So when I do want to affect the voltage under measurement as less as possible, I have to rely on the op amp? \$\endgroup\$
    – PetPaulsen
    Aug 22 '11 at 9:29
  • \$\begingroup\$ @PetPaulsen - sounds pretty good for a novice :-). Yes. You are correct. If you want best DC accuracy the opamp is defintelt better. The transistor will both load he circuit somewhat and will offset the voltage downwards by a base-emitter voltage drop, so for accurate DC use an op amp. In many cases people who use this sort of ciruit care about the AC components mainly and the transistor solution is effective and cheap. \$\endgroup\$
    – Russell McMahon
    Aug 22 '11 at 9:54
  • \$\begingroup\$ @Russell McMahon - Thank you! Again learned something new :) \$\endgroup\$
    – PetPaulsen
    Aug 22 '11 at 10:02

As JGord says, they are biasing circuitry. The second circuit takes the input through a series capacitor, which removes the DC component of the input level. Since the feedback path is also AC coupled through a cap, there's nothing reliably setting the DC input voltage of the op amp input, other than leakage through the input and the caps. That's why a resistive voltage divider has been added.

In the first circuit, the DC component of the input is provided by the source. Hopefully that is within the range of the op-amp centered on ground. In the second circuit, single supply operation requires that the center voltage be well above ground so AC coupling is pretty much required. Even if it weren't required for that reason though, it would still be useful in for example connectorized audio gear - with AC coupling and internal bias, the circuit takes care of itself without relying on the DC level of the unknown external input. (There is a possible cost though... for example, a sound card with such an input is great for audio, but can't be used for general purpose analog data logging since the bandwidth no longer extends all they way down DC)

  • \$\begingroup\$ I think I understand. Thank you! So in this circuit only the AC component of the signal gets through. My actual goal is to measure a DC voltage, so I have to relay on the external DC level (obviously). In this case I just stick with the bare second-order low pass filter (figure 7). Am I Right? \$\endgroup\$
    – PetPaulsen
    Aug 21 '11 at 22:36
  • \$\begingroup\$ Nitpick: Depending on the op-amp and the capacitor type, the input bias currents of the op-amp could fairly easily dominate the capacitor leakage current. \$\endgroup\$ Nov 16 '14 at 5:12

I'm not really happy with that second circuit.

In the first circuit, you have an Opamp with positive and negative supplies, and an earth on C1.
When do don't have two supplies, a common technique is to put two resistors in series, across the power supply. The middle point, between the two resistors, is called a "virtual earth". Do this with a 12V car battery and you get supplies of +6V an -6V.
Virtual earths aren't particularly good a supplying current, thanks to the resistors used to create them. In your second circuit, they are providing DC bias for R1.
C1 should also be tied to "virtual ground" as in the first circuit. It seems to rely on the AC impedance of a DC power supply being zero: again, thanks to R3 and R4, this is not the case.

  • \$\begingroup\$ You may not be happy with it, but it works. I've done it. \$\endgroup\$
    – Matt Young
    Nov 16 '14 at 4:04
  • \$\begingroup\$ @MattYoung I'm glad to hear it. Making fancy split power supplies is something tackled by those making C-Moy headphone amps (in an Altoids tin). While the fancy ones are considered "better" there's a diminishing return factor. Glad to know this simple cheap version was good enough for your needs. An example of fancier supply design is at: aries1470.blogspot.com.au \$\endgroup\$ Nov 17 '14 at 0:33

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