Shelving LPF technical question.

[Pano]

Here is a rather technical question about active shelving low pass filters. I didn’t know if this should go in here or in “Solid State”.

Here is the question:

I want to build a shelving low pass filter to compensate for the Open Baffle roll-off. This is an active filter built around an opamp. The filter should have a gain difference of 6dB from point F1 to point F2. Gain above and below should be flat. F2 is normally the F-equal point where the back wave of the Open Baffle wraps around and begins to cancel out the front wave. F1 is often chosen as 20Hz, the lower limit of the audio band, though F1 can be chosen lower.

OK, no new ground here, we can see this put in place on Siegfried Linkwitz’s site for his open baffle speakers. But here is the question:

Linkwitz gives a formula to determine the F1 and F2 frequencies, as well as the gain at these two points. His example here

shows both an inverting and a non inverting version. I am interested in the inverting version.

You can see on the page above his formula for determining F1, F2 and the gain at those points. But this formula does not make sense to me. There is no mention of C in the gain formulas.

The normal gain formula for an inverting opamp is G=R2/R1. That is what Linkwitz gives as G1. But how can that be for this circuit? What about C? If we ignore C for the moment, then just having R3 in place will change the effective value of R2. R2 and R3 must be taken together with the resulting value being RT = (R1R2)/(R1 + R2). We could call the result RT (total). Obviously adding R3 in parallel with R2 is going to make an RT lower than R2. Our new gain formula becomes G=RT/R1. This is accounted for in Linkwitz’s formula for G2. (though it doesn’t seems to be written that way)

Now we add in C in series with R3. The impedance of this RC circuit will change with frequency, becoming lower as the frequency increases. So RT becomes lower as the frequency decreases. That changes the ratio of RT/R1 dependant on frequency. And that’s the while point of the circuit. Gain goes down as the frequency goes up.

But the formulas shown for G2 refer ONLY to R1-R2-R3 – C is ignored in the formulas. Why? It will certainly make a difference. And in this circuit we can’t really calculate gain on the basis of R1/R2 alone. R3 and C have to be taken into account, right? They will change the value of R2 and become what I have called RT.

Does anyone have any ideas about this? What is Gain 2? How is the gain calculated with R3 and C in place? Or are the formulas given correct? They don’t make sense to me, but perhaps I’m missing something here. Is the answer in the top formula V2/V1?

Any help on this will be very welcome.

[pinkmouse]

R3 only comes into effect when the circuit is working at higher frquencies, otherwise it is invisible to the rest of the circuit because of C, so G1 applies. Once C starts working, R3 is seen by the rest of the circuit, and the gain changes as per G2. The only thing that C changes is the frequency of the rolloff, it does not affect the relative levels of G1 and G2 at all.

[AndrewT]

Hi,
imagine C as either a short circuit or as an open circuit.

This allows one to calculate the gains on the flat portions of the curves. This is not the gain @ F1 and F2.
The gain at these respective points are 3db above or below the flat portions.

If you want 6db between the points F1 & F2 then you need 12db between the flat portions.

Now select C to move the 6db/octave slope to the correct frequency location.

[pinkmouse]

Well done Andrew, much clearer than my attempt!

[Pano]

A couple of typos in my post above.

Value of RT should be: RT = (R2*R3)/(R2 + R3) ignoring C.

SHould have said "So RT becomes higher as the frequency decreases". With C in series.

[Pano]

Originally posted by pinkmouse
R3 only comes into effect when the circuit is working at higher frquencies, otherwise it is invisible to the rest of the circuit because of C

Originally posted by AndrewT
imagine C as either a short circuit or as an open circuit.

OK, these both make sense to me. The combined impedance of R3+C becomes very high at low frequencies, so it has little effect on R2. At higher frequencies the impedance of C is very low, so R3 now has more effect on R2. Either a short or an open circuit (almost).

Thanks for the help guys, it seems simpler now. Sometimes you just have to think it thru. I'll do some more number crunching a report back.