Location of input for negative feeback bandpass filter

Question

I came across a reference design on a TI application imaged here:

It is clear to me that there is a bandpass filter on the negative loopback branch of the amplier. However, from the few electronics classes I have taken in school, typically the input is seen where the group is placed. Is there any purpose as to why?

From the figure, I deduced the transfer function as:

That is mainly just a sanity check that I can workout the transfer characteristics properly.

This is my first post so don't rip me apart (or do, if I so deserve it.)

Answer 1

In thinking thru such circuits, one approach is to have the INPUT be still, unmoving. Then examine the "noise" response.

In this opamp circuit, if the Vin is unmoving (Vin+), then the opamp strives to make the other input also be unmoving.

If Vin- is unmoving, then the networks around Vin- are DECOUPLED, and you can examine them separately.

The network to Ground is a high_pass filter. This is easy to see, if you realize DC impedance is infinite, making the total gain be Feedback/infinity.

The Feedback network is a low_pass filter, and at very high frequencies the impedance IS zero, so the total gain is Zero/Input_network_impedance = zero.

If you properly set the two corner frequencies, you have a well-controlled passband with gain set by Rfeedback/Rin. In this case, gain is R5/R6. For accurate gain in the "passband" (accurate within a few percent), you need a passband width of several decades (such as 100Hz to 10,000Hz, not quite adequate for audio signals).

Here the low_frequency time_constant (as someone tries to SLOWLY SLOWLY move thru the field_of_view of the PIR, is 33uF * 40Kohm. This is 1.3 second time_constant, thus by moving VERY SLOWLY, you can trespass at will.

The upper end is 100pF and 15Meg ohm. Given 1pF and 1MegOhm is 1 uS, we easily compute a 1.5 millisecond time_constant, or about 100Hz upper corner (F3dB, 45 degree phase shift). Note this DOES NOT attenuate 60Hz energy, thus electro_static power line interference may be a problem; some thinking about shielding might be useful to prevent surprises, or false triggers.

Answer 2

This op-amp appears to be used in a non-inverting configuration, meaning the input comes from the positive terminal and the other side is grounded. However, a feed back network is still useful to control the poles and zeroes of the system.A bandpass filter resembling one of the more complex feedback networks. I am assuming PIR stands for passive infrared sensor and that is what is providing the input.

I think that might answer your question about the location of the input, but more about the how the feedback network actually influences output must me in the TI paper. Maybe you can leave a link in the comments.

Answer 3

Your equation has one big mistake: you’ve added R5 and Zc1 but they are in fact in parallel.

There are 3 independent filters going on here:

The input coupling of C1 with the parallel combo of R3 and R4 yields a .02Hz high pass. This is undoubtedly intended merely as a DC blocker. It is meant to pass all non-zero frequencies.

The R6 and C3 cutoff frequency is not much higher, at just 0.12 Hz. So for any frequency >1Hz or so we can treat C3 as a short and only R6 is in the circuit.

Finally, C2 and R5 form a cutoff at 106Hz. For frequencies below that C2 is pretty much a non-factor, leaving only the huge R5.

This leaves us with a cirucuit that has a gain of R5/R6 (50dB) for frequencies in about the 0.2Hz-50Hz range, dropping off to zero at DC (with the 3dB point at 0.12Hz) and a high end rolloff at 106Hz that drops to a minimum gain of 1.0.

I would call this either a bandpass filter, or just a low pass with a DC blocker. Or I suppose more accurately, a "low shelf filter" as the high end gain doesn't go to zero but to unity.