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Old 16th January 2008, 10:36 PM   #1
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Default Q formula

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
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Impedance varies with frequency, use impedance plots of your drivers and make crossover calculations using the actual impedance of the driver at the crossover frequency
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Old 16th January 2008, 10:54 PM   #2
Ron E is offline Ron E  United States
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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
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Old 16th January 2008, 10:59 PM   #3
jnb is offline jnb  Australia
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Default Re: Q formula

Quote:
Originally posted by PeteMcK
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
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Old 17th January 2008, 02:16 AM   #4
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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
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Impedance varies with frequency, use impedance plots of your drivers and make crossover calculations using the actual impedance of the driver at the crossover frequency
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Old 17th January 2008, 04:15 AM   #5
jnb is offline jnb  Australia
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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.
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Old 17th January 2008, 04:52 AM   #6
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Thanks Jnb, that'll keep me busy for a day or two...:-)
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Old 17th January 2008, 09:29 AM   #7
jnb is offline jnb  Australia
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Same problem attaching the files

If you'd care to shoot me a mail I'll send them to you.
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Old 17th January 2008, 12:00 PM   #8
jnb is offline jnb  Australia
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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.
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Old 17th January 2008, 02:49 PM   #9
forr is offline forr  France
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Default An easy way, using normalisation


Filter order 1, resistive load
~~~~~~~~~~~~~~~~~~
f = 1 Hz, R = 1 Ohm ==> C' = 160000 F, L' = 160000 H
. 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


Filter order 2, resistive load
~~~~~~~~~~~~~~~~~~
f = 1 hz, R = 1 Ohm and Q = 1.00 ==> C" = 160000 F, L" = 160000 H
. 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 wo = 1/L".C" = (2.Pi.fo) (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) .
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Old 17th January 2008, 04:51 PM   #10
Svante is offline Svante  Sweden
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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.
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