Impedance confusion...

[Cashdollar2009]

Just having trouble wrapping my head around this. On even number calculations I am perfectly fine calculating impedance, but I am just a bit skeptic of working with 6 drivers.

Now, to save myself from drawing it out i'm just going to paste a picture...

3_2ohm_dvc_134ohm.gif

Now, if I were to run six drivers, would I just be able to parallel these two banks of three together to get a final nominal impedance of .67-ohms? And if I were to do that would the power still be theoretically divided among the six drivers?

Thanks.

[will77530]

picture doesnt work

[bkolfo4]

6 drivers with single or dual voice coils? What is the impedance of each coil?

[JR2011bu]

Just having trouble wrapping my head around this. On even number calculations I am perfectly fine calculating impedance, but I am just a bit skeptic of working with 6 drivers. Now, to save myself from drawing it out i'm just going to paste a picture... 3_2ohm_dvc_134ohm.gif Now, if I were to run six drivers, would I just be able to parallel these two banks of three together to get a final nominal impedance of .67-ohms? And if I were to do that would the power still be theoretically divided among the six drivers? Thanks.

6 drivers with single or dual voice coils? What is the impedance of each coil?

from this i'm assuming DVC 2ohm woofers..

and im assuming that the ohm load will be like this

Dual-2 Ohm Subwoofers: 0.7 Ohms

[bkolfo4]

If you assume dual 2 ohm, then series the coils on each sub. Then parallel the 6 subs to the amp. .67 ohms

[Cashdollar2009]

Sorry it took a bit to get back, and your assumptions are correct, dual 2-ohm drivers. The only diagram I could find while my brain wasn't working would be of three D2's. Series'd and then paralleled down to 1.34-ohms. Then I figured I could do each side as such and then parallel the two sides to achieve a .67-ohm load. I mean, it was making sense, and it makes more sense now, I just wanted to double-check that they would still all be seeing the same amount of power per coil.

Thanks guys.

[ChevyBoy95]

if you did this with each pair of 3:

and this with the 2 pairs of 3 (ignoring the ohm loads show below (.5 = .67 / 1 = 1.34)):

this will arrive you at .67 ohms. think of the bottom picture as each lead on the coil is a pos or neg from the groups of three)

easy way to do that diagram above is after you make the groups of 3 to 1.34 ohms, just put both pos in the positive amp terminal and both negs into the negative amp terminal.