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I've designed an active bandpass filter based on the one described in this article http://www.bu.edu/eng/courses/ek307/documents/ActiveFilterNotes.pdf (section 13.3.4). However, I want to use a single ended supply, and I want use a non-inverting configuration. I've designed a circuit (below) and run some spice simulations, and everything is working fine.

But then I noticed that R3 is connected to the negative rail (GND) not my virtual ground (Vcc/2). My instinct tells me that R3 should terminate at Vcc/2 but after some experimentation I found that it doesn't matter where it goes - ground, Vcc, Vcc/2, 10 megavolts - no difference.

How can that be? Can someone explain exactly what's going on here? Am I correct that R3 should terminate at Vcc/2?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The point V1 is ac coupled to the rest of your circuit, I don't really know what effect you expect it to have on the rest, maybe you want to elaborate \$\endgroup\$
    – PlasmaHH
    Commented Aug 24, 2016 at 8:10
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    \$\begingroup\$ Any DC voltage on the left side of R3 gets through to C2 and C3 but since these are capacitors, the influence of that DC voltage stops there, it cannot influence the rest of the circuit in any way. C2 and C3 will simply charge to a different DC voltage as you change the DC voltage on R3. \$\endgroup\$
    – Bimpelrekkie
    Commented Aug 24, 2016 at 8:22

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For single supply operation, the DC operating point should be at 50% of the supply voltage. That is what you have done with Vcc/2 connected to the non-inverting input of the opamp circuit which has a positive DC gain of unity. All other nodes remain unaffected from the single supply modus. In particular, connecting R3 to a positive voltage would not change anything because (under staedy-state conditions) the capacitors connected to the other node of R3 cannot allow any DC current.

However, it is to be noted that the classical bandpass response can be obtained only if the signal input is connected to the left node of R3 (instead of ground). In this case, the circuit provides the classical inverting bandpass response (multi-feedback topology). See Fig. 13-15 in the referenced document. Feeding the input signal into the non-inverting input node (as in your case) allows a minimum stop-band gain of unity only (no bandpass response with only a zero in the origin)! Instead, your design gives two negative-real zeros.

If you want a non-inverting configuration, I recommend to use a second-order Sallen-Key topology with a fixed-gain stage (unity gain or gain-of-two).

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    \$\begingroup\$ I suppose it would be helpful for the OP (titch) if the anonymous "downvoter" would mention in a short comment the errors in my above answer. As an alternative, he could give his own (better) answer. \$\endgroup\$
    – LvW
    Commented Aug 24, 2016 at 9:48
  • \$\begingroup\$ Thanks for responding. C2 and C3 blocking the DC makes sense. I've simulated my circuit in TINA TI and I get the same bandpass response as the circuit from fig 13.15. Could you explain a bit more about why my design wouldn't work? \$\endgroup\$
    – titch
    Commented Aug 26, 2016 at 3:05
  • \$\begingroup\$ titch, I do not understand: (1) You claim that your simulation gives the correct results and (2) you ask why it does not work? I repeat: The input must be at the left side of R3 (see Fig. 13.15). In this case, very low frequencies (including DC) are blocked by C2 and the signal output is app. zero (must be the case for a bandpass). In your design (input at he non-inv. input), the gain at DC is "1" (and NOT zero) - and the bandpass does not attenuate at low frequencies. Instead it works as a unity-gain follower. \$\endgroup\$
    – LvW
    Commented Aug 26, 2016 at 7:28
  • \$\begingroup\$ ...and the same applies to very large frequencies (unity gain). Hence, the input at the non-inv. terminal does not provide the classical bandpass response. \$\endgroup\$
    – LvW
    Commented Aug 26, 2016 at 8:23
  • \$\begingroup\$ My electronics is pretty shabby (as you can probably tell!) but I think I understand what you are saying and agree. But it's curious that my simulation says otherwise \$\endgroup\$
    – titch
    Commented Aug 26, 2016 at 22:38

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