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Old 12th March 2004, 04:10 PM   #1
Sherman is offline Sherman  United States
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Default 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!

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Old 12th March 2004, 04:18 PM   #2
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You're as close as you can get. But, remember the power won't be equally divided between the drivers.
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Old 12th March 2004, 05:16 PM   #3
sreten is offline sreten  United Kingdom
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In terms of power 8-8-8||8-8-8||8-8 is better for 6.8 ohms.

But TBH 9 drivers 8-8-8||8-8-8||8-8-8 for 8 ohms is
far more sensible than any combination of 8 drivers.

sreten.
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Old 12th March 2004, 05:20 PM   #4
Sherman is offline Sherman  United States
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Quote:
Originally posted by Bill Fitzpatrick
You're as close as you can get. But, remember the power won't be equally divided between the drivers.
Thanks for the help! I figure the four driver unit will have the least power. So if I wire the middle four in that unit (more directly in line with the listener's ear) it might result in more even sound over the entire vertical spread of drivers. What do you think?

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Old 12th March 2004, 05:27 PM   #5
Sherman is offline Sherman  United States
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Quote:
Originally posted by sreten
In terms of power 8-8-8||8-8-8||8-8 is better for 6.8 ohms.

But TBH 9 drivers 8-8-8||8-8-8||8-8-8 for 8 ohms is
far more sensible than any combination of 8 drivers.

sreten.
I like that better than my original! I have to draw it out on paper, I'm a visual type of person, and I didn't draw that combo! I can only blame the mind-numbing effect of too much sawdust!

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

Thanks again.
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Old 12th March 2004, 06:05 PM   #6
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Default Re: Another Impedance Question

Quote:
Originally posted by Sherman
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!

Make four groups of 2 in parallel, and put all four groups in series for 16ohms...

Peace
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Old 12th March 2004, 10:49 PM   #7
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Quote:
the power won't be equally divided between the drivers
Does the greatest power dissipation occur at the first device in series?
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Old 12th March 2004, 11:06 PM   #8
Sherman is offline Sherman  United States
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Quote:
Originally posted by Timn8ter


Does the greatest power dissipation occur at the first device in series?
I'm not sure, my assumption was that the unit with the greatest number of drivers in it would dissipate the most power regardless of its position but I could be totally wrong.
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Old 12th March 2004, 11:22 PM   #9
Svante is offline Svante  Sweden
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Quote:
Originally posted by Timn8ter


Does the greatest power dissipation occur at the first device in series?

Assuming 8-8||8-8||8-8-8-8 the 8-8-8-8 drivers will take 1/5 of the power and the 8-8||8-8 drivers will take 4/5 of the power.

This is because the impedance of the 8-8||8-8 part is 8 ohms, and the 8-8-8-8 part is 32 ohms.

I'd go with sretens suggestion and get a ninth driver to build a 8-8-8||8-8-8||8-8-8 combination, if possible. Otherwise a 8-8-8-8||8-8-8-8 combination for 16 ohms, or 8-8||8-8||8-8||8-8 combination for 4 ohms might work, depending on the amplifier.
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Old 12th March 2004, 11:47 PM   #10
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by Timn8ter


Does the greatest power dissipation occur at the first device in series?

Quote:
Originally posted by Sherman


I'm not sure, my assumption was that the unit with the greatest number of drivers in it would dissipate the most power regardless of its position but I could be totally wrong.

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 8-8-8||8-8-8||8-8,

I is split in the ratio 1/3:1/3:1/2

which normalising

= 28.6%: 28.6%: 42.8%

so the two 8-8-8 handle 28.6% whilst the 8-8 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 8-8||8-8||8-8-8-8

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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