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MultiWay Conventional loudspeakers with crossovers 

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12th March 2004, 05:10 PM  #1 
diyAudio Member
Join Date: Feb 2004
Location: Chicago area

Another Impedance Question
OK, I have eight 8 ohm drivers and I want the total load to not be less than 6 ohms.
So... I put 4 drivers in series and the other four in series of 2 each. I then get 3 units of 32 ohms, 16 ohms and 16 ohms. Now I parallel those three units and get 6.4 ohms nominal for the whole shebang. Right?!? Any way to get closer to 8 ohms in this scenario? My brain hurts!  Don't touch that! Its hot! 
12th March 2004, 05:18 PM  #2 
diyAudio Member
Join Date: Jun 2001
Location: Eugene, OR

You're as close as you can get. But, remember the power won't be equally divided between the drivers.

12th March 2004, 06:16 PM  #3 
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

In terms of power 88888888 is better for 6.8 ohms.
But TBH 9 drivers 888888888 for 8 ohms is far more sensible than any combination of 8 drivers. sreten. 
12th March 2004, 06:20 PM  #4  
diyAudio Member
Join Date: Feb 2004
Location: Chicago area

Quote:
 Don't touch that! Its hot! 

12th March 2004, 06:27 PM  #5  
diyAudio Member
Join Date: Feb 2004
Location: Chicago area

Quote:
Well the cabs are done except for finishing and mounting drivers. I should have this knocked out over the weekend! Thanks again. 

12th March 2004, 07:05 PM  #6  
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Join Date: Jan 2004
Location: Virginia, USA

Re: Another Impedance Question
Quote:
Make four groups of 2 in parallel, and put all four groups in series for 16ohms... Peace 

12th March 2004, 11:49 PM  #7  
diyAudio Member
Join Date: Jul 2003
Location: Seattle, WA

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13th March 2004, 12:06 AM  #8  
diyAudio Member
Join Date: Feb 2004
Location: Chicago area

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13th March 2004, 12:22 AM  #9  
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

Quote:
Assuming 88888888 the 8888 drivers will take 1/5 of the power and the 8888 drivers will take 4/5 of the power. This is because the impedance of the 8888 part is 8 ohms, and the 8888 part is 32 ohms. I'd go with sretens suggestion and get a ninth driver to build a 888888888 combination, if possible. Otherwise a 88888888 combination for 16 ohms, or 88888888 combination for 4 ohms might work, depending on the amplifier. 

13th March 2004, 12:47 AM  #10  
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

Quote:
Quote:
Come on chaps get a grip ! All the drivers in a series link if the same will have the same power. The power per parrallel link depends on each links impedance. The power per driver depends on the parallel link impedance and the number of drivers in the parallel link. So for a 100W into 88888888, I is split in the ratio 1/3:1/3:1/2 which normalising = 28.6%: 28.6%: 42.8% so the two 888 handle 28.6% whilst the 88 handles 42.8% So six drivers will handle 9.5 Watts each, and two 21.5 watts each. So simply put the cabinets power handling is halved compared to a proper equal distribution of power between the units. for a 100W into 88888888 current splits in the ratio 2:2:1 i.e. 40%, 40%, 20% so 4 drivers are handling 20w each and 4 drivers 5W each, not good. sreten. 

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