Dumb parallel question - diyAudio
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 January 2006, 01:43 AM   #1
diyAudio Member
 
Cloth Ears's Avatar
 
Join Date: Dec 2001
Location: Melbourne, Australia
Default Dumb parallel question

Hi all,

I couldn't find the answer to this (not exactly sure how to search for it either, although I tried), so here it is:

When you wire 2 8ohm woofers in parallel, the load is perceived as 4ohms, but when you wire an 8ohm woofer and an 8ohm tweeter in parallel (with a passive xover), the load is still perceived as 8ohms. Is this because they're not really in parallel due to the signal only really going to one or the other?

Please answer in simple words, as my knowledge of the mysterious ways of electrons would just about fill a thimble...
__________________
Jont.
"It is impossible to build a fool proof system; because fools are so ingenious."
  Reply With Quote
Old 13th January 2006, 02:26 AM   #2
Speakerholic
diyAudio Moderator
 
Cal Weldon's Avatar
 
Join Date: Jan 2004
Location: Near Vancouver
Default Re: Dumb parallel question

Quote:
Originally posted by Cloth Ears
Is this because they're not really in parallel due to the signal only really going to one or the other?
Yes

Quote:
[i]Please answer in simple words [/B]
OK
__________________
Next stop: Margaritaville
Some of Cal's stuff | Cal Weldon Consulting
  Reply With Quote
Old 13th January 2006, 03:27 AM   #3
simon5 is offline simon5  Canada
diyAudio Member
 
Join Date: Nov 2004
Location: Québec, Québec
It's like over the crossover point, the tweeter impedance is 8 ohms but the woofer impedance is 1000 ohms for example. So 8 ohms in parallel with 1000 ohms = ~8 ohms.

Same thing under the crossover point, the woofer impedance is 8 ohms but the tweeter impedance is 1000 ohms for example. So 8 ohms in parallel with 1000 ohms = ~8 ohms.

So like you said, it's like they are not really in parallel because one is negligeable versus the other.

Sorry I wasn't as simple as Mr. Weldon hehe!
__________________
DIYaudio for President !
  Reply With Quote
Old 13th January 2006, 04:02 AM   #4
diyAudio Member
 
Join Date: Oct 2004
Location: KL
i remember reading an article earlier about this... the explanation goes like this if i remember

assumption xover 3.5KHz

so 3.5KHz below will be loaded to ur woofer (8ohm) only cause the tweeter has no/very little energy... and freq. 3.5kHz and above will be loaded to ur tweeter which is also 8ohm, and very little energy goes to ur woofer... so basically u r 'cheating' the amp into believing that the range it is outputing 20Hz - 20kHz is actually a full range 8ohm....

this explain why u normally get lower impedance when it's near/ at xover point... this is because the energy goes to both the tweet and woofer and the load is calculated as a parallel load 4 Ohm...

some high end speaker (e.g. focal-jmlab) will list down nominal impedance and min impedance

parallel calculation only work if all the energy goes to both the drivers (or more)... if it only goes to 1 driver then we only calculate the load of that driver at that freq.
  Reply With Quote
Old 13th January 2006, 06:01 AM   #5
diyAudio Member
 
David Gatti's Avatar
 
Join Date: Mar 2004
Location: Melbourne, Australia
G'day Clothears

I'll put it another way.
A woofer usually has an inductor (coil) in series with it. An inductor's resistance increases with increasing frequency (actually called impedance). At low frequencies it gradually starts to look like a short, but at high frequencies it starts to look like an open circuit, gradually reducing the amoount of signal the woofer sees.
A tweeter has a capacitor in series . A capacitor's impedance increases as the frequency decreases, so at high frequencies it just looks like a short and at low frequencies it starts to look like an open circuit (at DC it is an open circuit). Hence it rolls off the low frequencies reaching the tweeter.
The upshot is that the parallel combination of the inductor+woofer and capacitor+tweeter has one impedance rising while the other is falling and the result is more or less that they cancel each other and you get back to your nominal 8 ohms.
  Reply With Quote
Old 13th January 2006, 06:12 AM   #6
diyAudio Member
 
Cloth Ears's Avatar
 
Join Date: Dec 2001
Location: Melbourne, Australia
Thanks everyone, I'd guessed, but I wasn't sure. It never stopped me, but it's a nice thing to know.

simon5 - does the xover really effectively raise the impedance of the '-passed driver (or was this for simplicity)?

Cal - - sorry I couldn't word the question for a 'no' answer. I made you type 5 letters instead of 4...
__________________
Jont.
"It is impossible to build a fool proof system; because fools are so ingenious."
  Reply With Quote
Old 13th January 2006, 06:59 AM   #7
simon5 is offline simon5  Canada
diyAudio Member
 
Join Date: Nov 2004
Location: Québec, Québec
No, the crossover doesn't really raise the driver impedance, but it raises its impedance, and since it's in series with the driver, for simplicity I said it raises the driver impedance.

Well, the "ideal" driver is always at the same impedance (8 ohms) but the x-over component in series with it will change the total impedance as David Gatti said because the x-over component impedance changes with frequency. You need to sum the driver impedance with the impedance of the x-over component in series with it at the frequency you're looking at.

Like if you use a Linkwitz-Riley x-over for flat FR...
At the crossover point, tweeter impedance = 8 ohms, capacitor impedance = 8 ohms, inductance impedance = 8 ohms, woofer impedance = 8 ohms.

So both parts of the circuit are at 16 ohms, so in parallel you get 8 ohms.

Like David Gatti said, over the x-over point, the capacitor impedance is very low (let's say 0.001 ohms) and inductance impedance is very high (let's say 1000 ohms).

So woofer+inductance part is at 1000 ohms (inductance) + 8 ohms (woofer) = 1008 ohms total and tweeter+capacitor part is at 0.001 ohms (capacitor) + 8 ohms (tweeter) = 8.001 ohms, so in parallel you get :
( 1/1008 + 1/8.001 )^-1 = 7.9379921870155639610590934457742 ohms

I said the "ideal" driver, because in reality drivers have inductance as you probably know. The famous Le parameter in Thiele-Small parameters. If you look at most subwoofer impedance curve, impedance get really high when frequency is high. That's why we use impedance compensation circuits in crossovers...
__________________
DIYaudio for President !
  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
Dumb question, but... johnm Tubes / Valves 3 13th June 2008 11:14 PM
Parallel inductors (dumb question) w/ Zobel followup arc2v Multi-Way 11 19th November 2007 09:59 PM
maybe this is too dumb a question... dannyspencer Solid State 6 18th April 2004 05:57 PM
Thi is probably a dumb question.... Konnan101 Multi-Way 4 11th January 2004 02:40 PM
Dumb Zen V4 Question Panelhead Pass Labs 4 26th April 2003 05:15 PM


New To Site? Need Help?

All times are GMT. The time now is 12:28 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