Geofex parametric EQ "Q" question

[Heemis]

So I'm looking at the schematic for the parametric EQ circuit at geofex... the article is very enlightening!  I just have one question that the page doesn't answer.  I'm interested in knowing the maximum and minimum Q values for this circuit:  http://www.geofex.com/Article_Folders/EQs/parmet.gif

If anyone knows, or could even just tell me how to calculate the Q I would really appreciate it!

[fuzzo]

http://www.muzique.com/lab/gyrator.htm

you can see the Q value but I don't know what that value means.

[Heemis]

Thanks very much for the reply... I've punched in the values in the calc, and I'm finding as the resistance of R1 changes to alter the frequency, the Q also changes... this would make this circuit a bit less useful for me.  Anyone have any ideas as to how to avoid this?  I'm basically looking for a notch filter with variable boost/cut, frequency, and Q controls, where the Q doesn't change as frequency changes.

[jacobyjd]

Perhaps you could use a dual-ganged pot to counteract the change in Q as you change the frequency. This would keep the Q 'centered' on the setting that your Q control pot would be set to. Maybe.

[Heemis]

Thanks for all the ideas guys... I'm still not so sure if i'm going to get this circuit to do what I want.  Anyone have any ideas/links to similar circuits that will have a constant Q over change in frequency?

[Heemis]

Just a friendly bump for more info.

[fuzzo]

Isn't really about "Q" but what's the gain of the final AOP stage (the one you mix dry signal/inductor) ? I wonder if I need to put a volume control or not at the end (I'm working on a Overdrive +  EQ project)

[Heemis]

The gain of the final amplification stage could be tweaked by the 2k7 resistor from - in to out... I am assuming the circuit will be at unity gain if the EQs boost/cuts controls are left flat, is this correct?

[fuzzo]

If I don't say mistakes, the gain depends of the position of cut/boost pot that increase or reduce (with 2K7 resitor placed before it). So, with the cut/boost to 10K , we've 12.7/2.7 = 0.21 and with the pot at min we've (2.7/2.7) 1. So any boost at the end, and even a lower signal.

But it's little weird, I think I'm wrong or I forgot something else.

[Heemis]

Well, the two 10k pots are in parallel, so that changes the total resistance.  it's not actually 12.7 k with the both at full resistance.  Also, the signal feeds into both the + and - inputs of the op amp... I don't know how this effects the gain, but it seems like it would... can anyone shed some light?

[Heemis]

Still haven't been able to figure this out... any Q wizards out there?

[Lurco]

Still haven't been able to figure this out... any Q wizards out there?

RANE