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MultiWay Conventional loudspeakers with crossovers 

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20th June 2005, 05:43 AM  #1 
diyAudio Member
Join Date: Sep 2004
Location: California

damping of series vs parallel
Would (Can) someone please explain why the electrical damping is the same for drivers in parallel as in series. Assuming the drivers are the same in the same cabinet.
In open loop ( no amp and drivers resonating) in series we would have twice the voltage as in parallel and in closed loop (accross a minimal impeadance, Zo) we will have twice the current in parallel as in series. But if Zo is small as it is then in both case we will crrent limit and voltage generated at the drivers will be I*Zo and power dissapated is I*I*Zo and therefore in parallel power dissapation would be 4 fold. I know I am making a mistake but I am not seeing it. Your insight is appreciated as is any equations. Cheers 
20th June 2005, 08:07 PM  #2 
diyAudio Member
Join Date: Jun 2005
Location: Kent, UK

I'm unconvinced the electrical damping is the same. JBL don't think so when discussing twin woofers.........Quote.........
"We recommend that you avoid connecting separate woofers in series. Because the amplifierdamping factor (the amplifier’s ability to control the motion of the woofer) is expressed as a ratio of terminal impedance (the sum of speaker impedance, wire resistance and the D.C. resistance of any crossover coil connected to the woofer) to amplifieroutput impedance, connecting separate woofers in series reduces the damping factor of the amplifier to a value less than 1. This will result in poor transient response." I tend to agree unless someone can show otherwise. 
20th June 2005, 08:18 PM  #3 
diyAudio Moderator

If the drivers are identical, I think that symmetry forces the damping to be the same.
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Remember: life is ten per cent what happens to you, ten per cent how you respond to it, and eighty per cent how good your reflexes are when the Tall Ones come at your throat with their pincers. 
20th June 2005, 09:42 PM  #4  
diyAudio Member
Join Date: Mar 2005
Location: Deep South

Quote:


21st June 2005, 04:16 AM  #5 
diyAudio Member
Join Date: Sep 2004
Location: California

thanks but still no answer
The quote may be an accurate qoute but who ever said it is very wrong. Damping for a single driver is the speaker impedance versus all the other impeadances in the circuit (line + filter + amp) not summed with some of them.
And as stated above series would if anything increase the damping though I don't belive it changes it. But I don't have a succint explanation why. The simple questions are the hardest and this one is of paramount importance and we don't know. John K writes "The two driver is series will generate twice the back EMF but that voltage will be applied across the two VC's impedance. Twice the back EMF across twice the resistance yoields the same reverse current and same damping. " I have a lot more to say but i have to go for a while Take care 
21st June 2005, 11:37 AM  #6  
diyAudio Moderator

Quote:
Now, double that voltage and series two Zs. The voltage across each Z is still V. If that is the case, the damping has remained unchanged. A quick simulation in Calsod showed exactly that the frequency response did not change a bit, therefore damping stays the same. A reasonable question, and one that John K raises, is what happens when nonidentical drivers are put in series...
__________________
Remember: life is ten per cent what happens to you, ten per cent how you respond to it, and eighty per cent how good your reflexes are when the Tall Ones come at your throat with their pincers. 

21st June 2005, 07:44 PM  #7 
diyAudio Member
Join Date: Jun 2005
Location: Kent, UK

The electrical damping factor on the back EMF may be the same for nice steady tones driving both identical series woofers, but it's unlikely to be the same in the real world because of one experimental observation.....
If you tap your finger on a woofer cone you'll probably hear a resonant thud, but short the terminals out and you get a shorter more damped and quieter sound. Put 2 in series shorted out and tap one cone and it will be resonant again but not as bad the open circuit version. This is because the generated current flowing is reduced by the other coil which in this situation just look like an approx 6R series resistor. Therefore any resonances etc that are different between the 2 woofers won't get the full damping. No problem though if everything is identical between the 2 woofers, but what chance of that is there in a real speaker cabinet with real woofers? 
21st June 2005, 11:15 PM  #8  
diyAudio Member
Join Date: Sep 2004
Location: California

good thoughts but still no clear concise answer
Quote:
Take care Quote:
Heres my issue my mids are very stable through therie operating rage at 4.9 to 5.1 ohms therfore in parallel 2.5 ohms and series 10 ohms. the amps are meitners w/67 volt mains and max current output of 35 amps (not continuous probably more like 18 amps. refering to Otawa's (probably misspelled) contention that impedance drops to 1/6th of nominal impedance we get this Parallel max i = (672)/2.5 * 6 =~ 170 amps Series max i = (672)/10 * 6 = 39 amps parallel seems like a better option presently the system is parallel and it does bleed out well below pushing the preamp (also meitner). But to generate a comparable crossover to just see will cost about 300 buckc it is all 8 gauge northcreek coils and their caps not to mention tearing the speakers apart and rebalancing the tweeters. Thanks for your thoughts Take care 

21st June 2005, 11:31 PM  #9  
diyAudio Moderator

Quote:
__________________
Remember: life is ten per cent what happens to you, ten per cent how you respond to it, and eighty per cent how good your reflexes are when the Tall Ones come at your throat with their pincers. 

22nd June 2005, 05:30 AM  #10 
diyAudio Member
Join Date: Sep 2004
Location: California

hmmm
I am onn my way to the gym but I am not sure. I will have to think about it but if the amps are taken out of it there is no damping (electrical)  the damping characteristic is due to i not v. And we must be able to define the power dissapation in an equation such I squared R, where R is the output resistance of the amp.
I appreciate your input Take care 
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