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MultiWay Conventional loudspeakers with crossovers 

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13th January 2006, 12:43 AM  #1 
diyAudio Member
Join Date: Dec 2001
Location: Melbourne, Australia

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...
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13th January 2006, 01:26 AM  #2  
Speakerholic
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Re: Dumb parallel question
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13th January 2006, 02:27 AM  #3 
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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!
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13th January 2006, 03:02 AM  #4 
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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. focaljmlab) 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. 
13th January 2006, 05:01 AM  #5 
diyAudio Member
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. 
13th January 2006, 05:12 AM  #6 
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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...
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13th January 2006, 05:59 AM  #7 
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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 xover component in series with it will change the total impedance as David Gatti said because the xover component impedance changes with frequency. You need to sum the driver impedance with the impedance of the xover component in series with it at the frequency you're looking at. Like if you use a LinkwitzRiley xover 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 xover 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 ThieleSmall 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...
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