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Calculate Impedance for BiWired

post #1 of 9
Thread Starter 
As the title says:

How do you calculate impedance for BiWired speakers?
I have speakers rated at 8 ohms and Biwired from the Speaker A and Speaker B outputs from my amp.
What does the amp see at each speaker output?


Edited by billbartuska - 10/22/13 at 5:08am
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post #2 of 9
Wiring from one terminal of "Speaker A" output and one terminal of "speaker B" output on your amp is not bi-wiring, it is more of a bridge mono mode in the amp. Sort of. But not exactly. IDK what you are doing really with that picture because it doesnt make sense. Your kinda halfway between bi-amping and bridge mono mode, but not exactly proper into either camp... My guess would be all that you are doing right now is screwing up your signal pretty hardcore and sending part of the left and part of the right signal to both speakers




here is an image I made of how to do Bi-wire, bi-amp, and bridge mono modes. For bridge mono, the amp must be set into that mode you cannot just wire it up and hope for the best. Most home amps do not have this capability from what I have seen. You will also need a second amp for the right speaker in bridge mono mode since both channels are used to drive the left speaker. Bi-amp mode you will need another amp as well to power the right speaker since both channels are already used to drive the left speaker.






As for impedance, someone else can answer that. It depends on which mode you are using to wire things.
Edited by EniGma1987 - 10/22/13 at 8:35am
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post #3 of 9
Quote:
Originally Posted by billbartuska View Post


Where the heck did you get that image?

Its really confusing, but both images are electrically equivalent- the A and B speaker outputs on an amplifier are wired in parallel, so they're the same.

If you have Bi-wired speakers, you need an amplifier setup with the crossovers needed to output the correct frequencies to each side. Each speaker will require 2 amplifiers, (Hence the bi part of bi-amp) for a total of 4 amplifiers for stereo.


EniGma1987, look at the image more carefully- the Left and Rights on both images are actually kept separate. It took me a few minutes of staring to get it.
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post #4 of 9
Thread Starter 
Thanks for the responses EniGma1987 and kevmatic.!
Have another look if you will. The BiWiring just begins farther up the circuits.



So, again, anybody have an idea how to calculate impedance if the speakers are 8 ohm?
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post #5 of 9
The amp should still be seeing the same load, just through two sets of cables instead of one. That means impedance should be the same overall, but cable impedance will be half. However cable impedance is probably next to nothing anyway so it wouldnt really make a difference. At least that is my take on it.
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post #6 of 9
Quote:
Originally Posted by billbartuska View Post

Thanks for the responses EniGma1987 and kevmatic.!
Have another look if you will. The BiWiring just begins farther up the circuits.

So, again, anybody have an idea how to calculate impedance if the speakers are 8 ohm?

Yes, the image is correct. The impedance for loads in parallel that are the same resistance, its simply the resistance of one divided by the numer of them. Two 8 ohm speakers is 8/2 = 4 ohm.

Quote:
Originally Posted by EniGma1987 View Post

The amp should still be seeing the same load, just through two sets of cables instead of one. That means impedance should be the same overall, but cable impedance will be half. However cable impedance is probably next to nothing anyway so it wouldnt really make a difference. At least that is my take on it.

You are correct.
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post #7 of 9
Thread Starter 
Thanks. That makes it 195/195 Watts.
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post #8 of 9
If you don't have an active crossover don't bother. EDIT: I thought you were bi-amping. Biwiring is one of those topics I don't touch.
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post #9 of 9
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