
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
MultiWay Conventional loudspeakers with crossovers 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
16th January 2008, 10:36 PM  #1 
diyAudio Member
Join Date: Feb 2002
Location: Western Sydney

Q formula
Hi Guys,
I'm trying to work out the effects of using nonstandard 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
__________________
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 
16th January 2008, 10:54 PM  #2 
diyAudio Member
Join Date: Jun 2002
Location: USA, MN

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
__________________
Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. Carl Sagan Armaments, universal debt, and planned obsolescencethose are the three pillars of Western prosperity. —Aldous Huxley 
16th January 2008, 10:59 PM  #3  
diyAudio Member
Join Date: Dec 2006

Re: Q formula
Quote:


17th January 2008, 02:16 AM  #4 
diyAudio Member
Join Date: Feb 2002
Location: Western Sydney

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
__________________
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 
17th January 2008, 04:15 AM  #5 
diyAudio Member
Join Date: Dec 2006

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. 
17th January 2008, 04:52 AM  #6 
diyAudio Member
Join Date: Feb 2002
Location: Western Sydney

Thanks Jnb, that'll keep me busy for a day or two...:)
__________________
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 
17th January 2008, 09:29 AM  #7 
diyAudio Member
Join Date: Dec 2006

Same problem attaching the files
If you'd care to shoot me a mail I'll send them to you. 
17th January 2008, 12:00 PM  #8 
diyAudio Member
Join Date: Dec 2006

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. 
17th January 2008, 02:49 PM  #9 
diyAudio Member
Join Date: Dec 2004
Location: Next door

An easy way, using normalisation
Filter order 1, resistive load ~~~~~~~~~~~~~~~~~~ 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 Filter order 2, resistive load ~~~~~~~~~~~~~~~~~~ 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) . 
17th January 2008, 04:51 PM  #10 
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

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. 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
box formula  crs1  MultiWay  29  23rd January 2008 09:53 AM 
Horn formula ?  Bernhard  MultiWay  2  23rd May 2004 07:47 PM 
gain formula  xplod1236  Chip Amps  1  6th April 2004 04:47 PM 
Need help on xover formula  Jay  MultiWay  3  12th August 2003 05:28 PM 
Anyone have the formula for...  Rino odorico  MultiWay  4  22nd January 2003 04:07 PM 
New To Site?  Need Help? 