Forsell EQ math, I'm not understanding Q

Hi all,
For fun and maybe for a build down the line I've been messing around with the Forsell EQ guide, and this calculator...

http://www.pronine.ca/lcf.htm

There's some good discussion here...
http://www.groupdiy.com/index.php?topic=14745.msg172992#msg172992

Plugging in different values into the calculator that I'm grabbing from NYD's schematic, the g-pultec, the available inductors from carnhill etc I'm getting wildly different values for the impedance of the different filters.  What's an acceptable range to shoot for?

I can follow the equation for determining Q, but I don't understand how it relates to the impedance of an LC filter.  Also in the white paper Forsell sort of leaves it at the Q being determined by the quality of available inductors.  But like what's a real world, Q?

Thanks 

It is quite simple. There are many pairs of LC values that will resonate at a given frequency. 1000mH and 100nF resonate at close to 503Hz. So do 500mH and 200nF and also 2000mH and 50nF. At resonance the reactance of the capacitor and inductor are equal and opposite and cancel out but the value of the two  impedances at resonance is different for each of the above pairs of values that all resonate at 503Hz. For 1000mH/100nF it is 3162 ohms, for 500mH/200nF it is 1581 ohms and for 2000mH/50nF it is 6324 ohms. This impedance is known as the characteristic impedance of the LC resonant circuit and is equal to 2*pi*f*L or 1/2*pi*f*C and is often denoted by Zo.

The Q of a an LC circuit at resonance depends on the residual resistance in the circuit after the reactances have cancelled each other out. Some of this residual resistance will be in the inductor itself and the rest is in the circuit connected to the LC. If the total residual resistance is R then the Q of the LC circuit is Zo/R.

Now since Q is proportional to Zo and Zo is proportional to L it follows that for a higher Q resonance at a given frequency you need to use a larger inductance and a correspondingly smaller capacitance. So in our example above, the 2000mH/50nF combination would have the highest Q.

The really hard bit is working out exactly what is the value of R because in most filters this depends on circuit values outside the LC itself e.g. pot and resistor values. In some passive Es it is compicated by the fact that part of the R is reflected impedance from an input transformer as in the case of the Pultec EQP1A. I still find this difficult to do by inspection and I usually resort to simulation to determine actual Q values.

Cheers

Ian

> wildly different values for the impedance of the different filters.  What's an acceptable range to shoot for?

Ask your circuit designer.

In hi-fi it is most convenient to aim for 1K source 10K load.

However inductors to give 10K impedance at low-frequency are costly.

OTOH if the load can be 1Meg the capacitors may be much smaller.

Low Z means cheaper chokes and costlier caps. High-Z means cheap caps and impossible chokes. The balance shifts from bass to treble. The high/low-Z specification also impacts amplifier costs, so a fully optimized product has the choke/cap guy and the amplifier guy sit together. "Milli-Henry is cheap!" "Ah, sub-ohm impedance means massive drivers!"

Great explanations.

the 2000mH/50nF combination would have the highest Q

Click to expand...

Does a higher value mean a wider or narrower peak?  I'm not really used to Q being expressed as a numeric value.

If the total residual resistance is R then the Q of the LC circuit is Zo/R.

Click to expand...

OK, for instance on NYD's schematic he specifies certain DCR values for the inductors.  Ignoring the other impedance factors you guys described, is DCR a reasonable value to use in the Zo/R equation?

substitute said:

Great explanations.

the 2000mH/50nF combination would have the highest Q

Click to expand...

Does a higher value mean a wider or narrower peak?  I'm not really used to Q being expressed as a numeric value.

Click to expand...

The higher the Q the narrower or sharper the peak. Q is also the the bandwidth of the peak divided by the centre frequency. So if you have a plot of the frequency response you can estimate Q by finding the -3dB points on either side of the peak. The peak frequency divided by the difference between the -3dB frequencies gives you the Q. If you check out the published curves for the Pultec EQP1A you will see the high boost has a Q of about 3. Notice that the circuit actually includes a series pot to alter the residual resistance and hence the Q

If the total residual resistance is R then the Q of the LC circuit is Zo/R.

Click to expand...

OK, for instance on NYD's schematic he specifies certain DCR values for the inductors.  Ignoring the other impedance factors you guys described, is DCR a reasonable value to use in the Zo/R equation?

Click to expand...

No. I suspect NYD specifies them so they can be included along with the other circuit resistances in defining the overall residual resistance and hence the Q. The thing to remember is that just about all passive |LC audio EQ uses a series resonant circuit. That is an L in series with a C in series with an R. R is what I call the residual resistance and includes the resistance of the inductor in series with the other resistances. You need to go through the circuit in a loop from one end of the series LC passing all the circuit resistances and back to the other end of the series LC to estimate the residual R. As I said before this is often not easy . I am not sure what this involves in NYD's schematic because I don't have his schematic to hand. Can you post a link to it please?

Cheers

Ian

Thanks Ian!
I woke up this morning thinking DUH, of course the larger number represents a narrower peak.

Here's a link to the NYD passive EQ....

http://www.twin-x.com/groupdiy/albums/userpics/NYD-EQ1revA.pdf

It uses hard to find inductor values, so I thought it be "fun" to try and rework the circuit using the multi taps available at carnhill or cinemag.  That's what got me thinking down this path, then I read the forsell paper, then I read it like another ten times.

You need to go through the circuit in a loop from one end of the series LC passing all the circuit resistances and back to the other end of the series LC to estimate the residual R. As I said before this is often not easy

Click to expand...

Ah, no short cuts, genuine circuit design is not for the faint of heart. 

I also found this, it's for gyrator filters, but an elegant piece of coding I feel...

http://awasteofsalt.com/gyrator/

substitute said:

Thanks Ian!
I woke up this morning thinking DUH, of course the larger number represents a narrower peak.

Here's a link to the NYD passive EQ....

http://www.twin-x.com/groupdiy/albums/userpics/NYD-EQ1revA.pdf

It uses hard to find inductor values, so I thought it be "fun" to try and rework the circuit using the multi taps available at carnhill or cinemag.  That's what got me thinking down this path, then I read the forsell paper, then I read it like another ten times.

Click to expand...

The easy way to tweak the NYD EQ is to work out the characteristic impedance he is using.  So his 40Hz setting uses an 820mH inductor which at 40Hz has an impedance of 206 ohms. Similarly for the 100Hz setting he uses 330mH which at 100Hz has an impedance of 207 ohms - however, these two setting are not resonant but simple hi/lo pass using an inductor instead of a capacitor. Even so, you can uses 206 ohms as the characteristic impedance if you want to calculate inductors for other frequencies or the frequencies you would get with other inductor values. The third position is 300Hz which has a series 1uF capacitor so it is resonant. At 300Hz, the reactance of 330mH is 622 ohms and the reactance of 1uF at this frequency is 530 so it is close to resonance at 300Hz (actual resonance is 277Hz). So to try different inductor values I would recommend assuming the characteristic impedance is 575 ohms (the reactance of 330mH at 277Hz) and you should get similar Q values to the original circuit.

You can do similar calculations for the mid band all of which are resonant. And lastly you can do the same for the hi band but note that the 15KHz cut just uses a capacitor.

Hope that helps.

Cheers

Ian

Thank you, it does.  If I get beyond the stage of playing with algebra and actually start tacking something together I'll be sure to post back with more questions.