Go Back   Home > Forums > Loudspeakers > Multi-Way

Multi-Way 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
Reply
 
Thread Tools Search this Thread
Old 13th May 2013, 09:47 AM   #11
diyAudio Member
 
Join Date: Jul 2010
Location: Western China
Those calculators are great, but the OP was about finding the solution for a 1-cap, no coil xo and the 1st Order BW has a coil in it.

I am trying to find the formula WITHOUT an inductor or resistor, just two drivers and a cap. Reverse engineering my S.0 to figure out what they were doing.

Quote:
High Pass filters allow high frequencies above a selected crossover frequency to pass, filtering out all frequencies below it. In a first order (6dB per octave) filter, this consist of a capacitor in series with a loudspeaker. Just above the crossover frequency, the capacitor begins to add resistance to the circuit. At the crossover frequency enough resistance has been added to equal the resistance of the loudspeaker and reduce the power by 3 dB or 50 %. One octave below the crossover frequency, power has been reduced by 6 dB or 75%. Each octave lower reduces the power by an additional 6 dB. The size of the capacitor will be determined by the impedance of the loudspeaker(s) and the desired crossover point. The smaller the size or value of a capacitor (microfarads, fd or mfd) is, not physical size, the higher the high pass frequency will be.
This is very clear and I understand it, but there is no formula to calculate values? I understand that the smaller the value of the cap the higher the pass frequency, but HOW DO I CALCULATE THIS when there is no coil in the "crossover"?
What is the formula? Something like Fxo = K * [1 / (c)] where c is in farads and K is a constant.

":Just above the crossover frequency, the capacitor begins to add resistance to the circuit."

How do I calculate this xo frequency in a one capacitor filter with no coil or inductor or resistor in the circuit?

Last edited by miragem3i; 13th May 2013 at 09:55 AM.
  Reply With Quote
Old 13th May 2013, 09:57 AM   #12
diyAudio Member
 
Join Date: Jul 2010
Location: Western China
Quote:
Originally Posted by sofaspud View Post
C = 1/2πfR
where R is the driver impedance
or f = 1/2πCR
Which driver impedance? There are two drivers in the circuit. The tweeter impedance, the mid impedance or the combined series impedance of both?

I assume you intend this:
f = 1 / [2πCR]

The original set up was one 4.7 uF capacitor, a very common part, two drivers as DC resistance measured below, and no other parts. No coil no resistors.
The value of the tweeter impedance *seems* to be the value to use, but since this circuit includes a coil and mine had no coil I cannot be certain.

I used the calculators above at http://www.diyaudioandvideo.com/Calculator/XOver/

1st Order Butterworth
xo frequency: 5551 Hertz [trial and error using goal seek]
6.1 Ohm Tweeter / 5.4 Ohm Woofer
Parts List
Capacitors
C1 = 4.7 uF
Inductors
L1 = 0.15 mH

But the xo had NO COIL INDUCTOR. How much error would be in the calculations? Or did they just fudge it and include the inductance of the speaker coils and wire in the circuit and call it a day?

Last edited by miragem3i; 13th May 2013 at 10:18 AM.
  Reply With Quote
Old 13th May 2013, 12:06 PM   #13
Pano is offline Pano  United States
diyAudio Moderator
 
Pano's Avatar
 
Join Date: Oct 2004
Location: Milliways
Blog Entries: 4
Quote:
Originally Posted by miragem3i View Post
Those calculators are great, but the OP was about finding the solution for a 1-cap, no coil xo and the 1st Order BW has a coil in it.
Did you look carefully at the calculator I linked?
2-Way Crossover Designer / Calculator

You'll see that the 1st order low pass section is just a cap. No coil. The coil is for the low pass section. Please look again. The formula is on the help page, if you want to do it that way.
2-Way Crossover Design / Calculator Help
__________________
Take the Speaker Voltage Test!
  Reply With Quote
Old 13th May 2013, 12:45 PM   #14
diyAudio Member
 
Join Date: Apr 2011
The single capacitor is the high pass tot the tweet. The inductor is the low pass to the woofer. I am not aware that you can create both high and low pass with one capacitor. You will note that the calculator that Pano linked asked for the impedence of each driver.

It may be that your system was running the woofer full range.
  Reply With Quote
Old 13th May 2013, 01:16 PM   #15
AndrewT is offline AndrewT  Scotland
diyAudio Member
 
Join Date: Jul 2004
Location: Scottish Borders
A 2way crossover is TWO filters combined,
One Filter is a Low Pass that feeds the Bass driver.
The other Filter is a High Pass that feeds the Treble driver.

If you have ONE driver and want a high pass filter then add a capacitor in series.
If you want a low pass filter then add an inductor in series.

A Capacitor in series with a driver can only be a High Pass Filter.

The standard formula applies.
F-3dB = 1 / 2 / Pi / C / Z where Pi = 3.14159, C = Capacitance in Farads, Z = Driver impedance AT THE CROSSOVER FREQUENCY. This impedance may not be the same as the "nominal driver impedance"
__________________
regards Andrew T.
  Reply With Quote
Old 13th May 2013, 01:51 PM   #16
diyAudio Member
 
Join Date: Sep 2004
Location: Minneapolis, Minnesota, USA
Hi miragem3i,

One of my favorite all-time quotes is from fairly wise individual named Albert Einstine; "Everything should be made as simple as possible, but not simpler." What you're trying to do is "to simple" and it is going to be extremely difficult to nearly impossible to get it done right. You can't apply any math without some well done measurements of the drivers, and if you don't happen to have a lot of expensive (lab grade) test equipment and maybe an anechoic chamber handy, it's going to be difficult to say the least.
In order to build a 2-way speaker with only one cap for the tweeter high pass will require the bass driver to have a very smooth upper roll-off with minimal cone break up, finding one of those is going to be hard, they are rare. Next you'll have to measure the frequency response and impedance curve of the bass driver to determine the cap value for the tweeter high pass. And the two drivers selected have to be matched for SPL if you insist on "no resistors", yet another constraint on driver selection. And after going through all of that you're still faced with the reality of driver offset causing frequency response problems in the cross over region.
If you want to build a well performing speaker system, I suggest you avoid "to simple" and stick with a more traditional approach. Check out 2nd order linkwitz, they are relatively easy to design and build, and give pretty good results.

Mike
  Reply With Quote
Old 13th May 2013, 07:02 PM   #17
diyAudio Member
 
Join Date: Jan 2008
Quote:
Originally Posted by miragem3i View Post
I am trying to find the formula WITHOUT an inductor or resistor, just two drivers and a cap. Reverse engineering my S.0 to figure out what they were doing.
The formulas and calculators are almost worthless, as they assume the speaker is a fixed impedance, like a resistor.

A speaker is not like a simple resistor, it has impedance and reactance that vary with frequency.

Look at the example below, the calculator says an 18 uf capacitor will provide a 550 Hz 1st order Butterworth crossover.

The chart shows the actual measurement of an 18 uf capacitor in series with a 16 ohm horn driver, and the same unit without the capacitor.

As you can see, the response with the capacitor looks nothing like the calculator suggests, in fact the capacitor causes more attenuation above 550 Hz than some frequencies below. The single capacitor is useless as a crossover with this speaker, though it might have the desired effect with another driver that has a different impedance curve.

Calculation of what the capacitor in your S.0 does is impossible unless you had an impedance chart of the driver.
Attached Images
File Type: png Crossover Calculator.png (29.0 KB, 68 views)
File Type: png With &Without 18MFD cap.png (66.8 KB, 67 views)
  Reply With Quote
Old 13th May 2013, 10:19 PM   #18
Pano is offline Pano  United States
diyAudio Moderator
 
Pano's Avatar
 
Join Date: Oct 2004
Location: Milliways
Blog Entries: 4
Quote:
Originally Posted by Pano View Post
You'll see that the 1st order low pass section is just a cap.
Oops, sorry. In the above I meant HIGH pass. High pass is a single cap. Sorry.
  Reply With Quote
Old 14th May 2013, 07:23 AM   #19
diyAudio Member
 
Join Date: Jul 2010
Location: Western China
Quote:
Originally Posted by Pano View Post
Did you look carefully at the calculator I linked?
2-Way Crossover Designer / Calculator

You'll see that the 1st order low pass section is just a cap. No coil. The coil is for the low pass section. Please look again. The formula is on the help page, if you want to do it that way.
2-Way Crossover Design / Calculator Help
Why would the calculations be identical if there is a coil removed from the circuit? The tweeter 'sees' the coil across the terminals of the mid as part of that driver.

Quote:
The inductors (coils) in a LPF have resistance. This resistance affects the impedance of the entire circuit.
If you are claiming that it's too small to matter or close enough or the mH are overwhelmed by the larger circuit impedance, then just say so.

Otherwise why does the omission of this component not matter?::
Inductors
L1 = 0.15 mH

Last edited by miragem3i; 14th May 2013 at 07:26 AM.
  Reply With Quote
Old 14th May 2013, 07:46 AM   #20
diyAudio Member
 
sofaspud's Avatar
 
Join Date: Nov 2005
Location: San Antonio
Quote:
Originally Posted by miragem3i View Post
Why would the calculations be identical if there is a coil removed from the circuit? The tweeter 'sees' the coil across the terminals of the mid as part of that driver.
It doesn't really see the coil, since it is a parallel circuit (and therefore voltage is the same in each branch). The cap blocks lows to the tweeters, while the coil blocks highs to the woofer. And the cap will block those lows whether the coil is there or not, so the calculations are identical.
__________________
It is error only, and not truth, that shrinks from enquiry. - Thomas Paine
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
multiple feedback low pass filter question graygem Analog Line Level 20 14th July 2011 06:01 PM
DIY One fader mixer with Filter Low Pass, Hi Pass, Notch Pass Filter PGM stevep314 Analog Line Level 0 10th August 2010 09:15 PM
Extremely basic ward cleaver Construction Tips 3 29th December 2009 11:16 PM
Low Pass Filter question synthius Chip Amps 18 28th February 2008 12:13 PM
simple low pass RC filter noob question mr.duck Parts 13 22nd July 2006 02:16 AM


New To Site? Need Help?

All times are GMT. The time now is 02:17 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright 1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2