Biwiring LSI9's. Are they still 4 Ohms?
timswim78
Posts: 38
I understand that the LSI9's should be driven by a 4 ohm amplifier when the jumpers are in place. If the LSI9's are bi-amped do they still require a 4 Ohm amp, or can you get away with something like 6 Ohm or 8 Ohm amps?
Post edited by timswim78 on
Comments
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Bi-wiring a speaker will not affect it's impedance. Bi-amping can.
What do you plan on driving them with?"He who fights with monsters should look to it that he himself does not become a monster. And when you gaze long into an abyss the abyss also gazes into you." Friedrich Nietzsche -
If you bi-amp then you would have different impedances, bi-wiring there is no difference. Think about it, instead of having on wire from the amp to the speakers w/ the jumper in between now you have two wires from the amp to the speaker. Still the same load as the amp sees it since it is still one wire on the amp side.
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Bi-wiring a speaker will not affect it's impedance. Bi-amping can.
What do you plan on driving them with?
I meant bi-amping. I will change my original post. My mistake.
I am using a Denon DRA-395 right now. I do not plan to change, and I am really just asking this question out of curiosity. -
I don't know the actual impedance of each pair of binding posts, but as a whole, the LSi9 is actually closer to a 6ohm speaker."He who fights with monsters should look to it that he himself does not become a monster. And when you gaze long into an abyss the abyss also gazes into you." Friedrich Nietzsche
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http://www.termpro.com/articles/spkrz.html
"Calculating the load impedance for the parallel-wired channel in Figure 2A is a bit more complicated than doing so for speakers wired in series. Using Equation 4, multiply the impedances of each speaker and then divide the result by the sum of the speakers' impedances. You can visualize the result as a single imaginary speaker (Figure 2B), whose impedance is represented by Zt. Zt stands for the equivalent-load impedance, while Za and Zb represent the impedances of speakers A and B, respectively.
Equation 4: Speakers in Parallel
Zt = (Za x Zb) / (Za + Zb)"
The gold connectors act as if they are wired in parallel. So yea, it has to change.Home Theatre: Epson 5020ub, Elite Screen Sable Frame 100", Onkyo 818, Oppo BDP-103, Tivo Series 3, Xbox 360, Sealed Dual Sound Splinter RL-p 15" DIY sub powered by Behringer EP2500 with FBD, QSC RMX1400 powering LSi15, LSiC, LSiFX sides, and Lsi7 for the back, Technics SL-1200M3D -
The crossover divides between the 2 sets of posts, so they are each going to be pretty much equal to the overall impedance by themselves. Remember, if you have a 4 ohm tweeter and a 4 ohm woofer, you don't get 2 ohms when you put them together in parellel with a crossover, this is because the woofer will have an impdance rising above the crossover point and the tweeter rising below the crossover point because of the reactance of the filter....
So, stated simply, if you have a 4 ohm speaker, more than likely one set of posts is an average of 4 ohms and the other is an average of 4 ohms as well, and when they are combined in parellel, they are still 4 ohms because of the crossover... This is pretty general, but when combined, they do not cut in half like you may think they would.... This would only happen if there were no crossover inside the speaker, which is not the case.
The formulas mentioned above are for non reactive circuits and don't relate to your question at all.www.forceaudio.com .... We cut through the BS.
