# Q formula

#### PeteMcK

##### Member
Hi Guys,

I'm trying to work out the effects of using non-standard values in a 2nd order passive filter, i.e. different from what a standard calculator will give (kind of working backwards);

Is this formula OK?: Q = Z*SQRT(C/L*1000)
(from http://www.passivecrossovers.com/index.htm )

Ideally, I'd like to enter driver Z, L&C into a spreadsheet and have it chunder out the F3, Q, and a nice graph of the predicted freq & phase response...perhaps such a thing exists already??
cheers,
Pete McK

#### Ron E

##### Member
If you have a fairly recent version of excel, you can use the passive crossover designer tool by Jeff Bagby.

Otherwise, If I am thinking correctly today.
Q = C*R/(2*pi*Fc)
Fc = sqrt(1/(L*C))/(2*pi)

H(s) = s^2/(s^2+s/(R*C)+1/(L*C)) for highpass
H(s) = (1/(L*C)/(s^2+s/(R*C)+1/(L*C)) for lowpass

Set s=jw and have at it.... j = sqrt(-1), w=2*pi*f

#### jnb

##### Member
PeteMcK said:
Ideally, I'd like to enter driver Z, L&C into a spreadsheet and have it chunder out the F3, Q, and a nice graph of the predicted freq & phase response...perhaps such a thing exists already??

They do, crossover simulators. Even if you fed one a flat response to begin with it may get the info you're seeking #### PeteMcK

##### Member
Thanks Ron,
I'll play with those formulae;
I had a look at Jeff's 'Passive Crossover Designer' but was overwhelmed by the interface, & it seems to rely on importing files; perhaps I need to search a bit more for a tutorial on it...

JNB, which simulator did you have in mind, if any? (Bear in mind that I'm wanting to work backwards, from the 'result' component values to the F & Q.)

Cheers Guys

#### jnb

##### Member
Working backward from an L and a C and an R, to the F and the Q are each a 10 second calculator job. I assumed you wanted it plotted.

I also assume you got the spreadsheet from the FRD consortium. There is another beside it. Use whichever you like. FRD and ZMA files are easy. The first thing you do is choose a resolution - say, between 30 and 150. E.g. 64 is a nice round number.

Create a text file (notepad will do) and use the following format:

20 84 0
22 84.5 0
24 86 0
27 90 0

etc... where the first column is the frequency (use geometric spacing between frequencies. The second column is the SPL and the third is the phase. Use spaces between data and enter between lines. Decimals are permitted.

Simply create a file with all the same SPL and all 0 phase to use as a default.

The ZMA file is for impedance. The three columns are: F, Z and Zphase.

I have included a flat FRD and ZMA file here to get you started. The ZMA is 8 ohms as you will see if you look at it in a text editor.

P.S. won't let me upload the files right now, I'll try again later.

#### PeteMcK

##### Member
Thanks Jnb, that'll keep me busy for a day or two... #### jnb

##### Member
Same problem attaching the files If you'd care to shoot me a mail I'll send them to you.

#### jnb

##### Member
YGM.

FRD files can also be created using SPL trace fom the FRD consortium so you could use a screenshot or a .jpg of a plot.

#### forr

##### Member
An easy way, using normalisation

~~~~~~~~~~~~~~~~~~
f = 1 Hz, R = 1 Ohm ==> C' = 160000 µF, L' = 160000 µH
. for other frequencies, divide C' et L' by the value of the required frequency
. for other resistive loads, divide C' et L' by the value of the required load

At wo = 2.Pi.fo (fo being the -3 dB frequency),
the impedances of C' and L' are equal to R
ZC' =1/(wo*C') = R
ZL' = wo*L' = R

~~~~~~~~~~~~~~~~~~
f = 1 hz, R = 1 Ohm and Q = 1.00 ==> C" = 160000 µF, L" = 160000 µH
. for other frequencies, divide C" et L" by the value of the required frequency
. for other resistive loads, divide C" et L" by the value of the required load
. for other Qs :
multiply C" by the value of the required Q
divide L" by the value of the required Q

When wo² = 1/L".C" = (2.Pi.fo)² (fo being the resonance frequency resulting from the combination of reactive components C" and L")
ZC" = 1 / (Q*wo*C")
ZL" = (wo*L1") / Q

For Q = 1,
ZC" = ZL" = R, the value of C" and L" are the same as C' and L' of the order 1 filter if the load is the same.

Note
the exact value of 160000 is 159236, the difference is less than 0.5%. It is much more precise than the usual components of passive crossovers and than the driver impedance and frequency (often 5% variation for this last one) .

#### Svante

##### Member
I would like to add to this discussion that while formulas for crossover frequency and Q value are good for understanding, they are not sufficient to design a well behaved crossover filter. This is due to the complex load of the driver, it is far from resistive. There are two ways arounf this. Either make the driver resistive by adding compensatory circuits, or simulate the crossover components until their values generate the desired response with the complex load.

Or use a combination of both.

#### PeteMcK

##### Member
Got 'em Joel, many thanks.

Forr, thanks, that certainly makes it easier.
Svante, yep, I had that thought in the back of my mind, but at this early stage, I'm just trying to get a picture of what changes in the the xover components will do...
now, got my climbing boots on for that learning curve, thanks to you guys already on my way 