jonnythan said:
"the current it needs for the woofers (and they need lots of current) through one speaker cable, while the midrange tweeter section would draw less current (it doesn't need as much) through its own speaker cable."
Sure, they're coming through separate *speaker cables*, but they're coming from the exact same binding post at the amplifier, and the exact same internal wiring from the amplifier. Do you get that? All of the current is passing through a single binding post on the amp - it doesn't matter if it splits into two wires immediately after that gold post or ten feet after that gold post.
Both the "low" frequencies and the "high" frequencies are passing through the single wire that connects to the single binding post on the back of the amp.
Drop an ohmmeter in between the two "separate" red posts or black posts on the back of the speaker and you will measure a resistance of a big fat 0.
How about if you use one wire halfway to the speaker, then break it into two wires?
What if you use one wire all the way to the speaker, but break it into two wires four inches before the terminals?
What if you use one wire an inch long that then breaks into two wires for the rest of the run?
These are all 100% identical electrically.
PENG, think about it for a second. Look at the red binding post on the back of the amp. See how it goes into the amplifier? All of the current for both drivers (assuming a two driver speaker) is going through that single gold post. It's got to split to two cables somewhere between that gold post and the crossover network. When you have two wires between that post and the speaker, how is that different than having what is essentially an extension of the post that doesn't split until the speaker binding post?
Johnny, I did think about the theory behind this a lot before I posted. I actually measured the current in each pair of the bi-wire cables. The pair that goes to the low frequency crossover/drivers binding posts did draw much higher current as expected. If I had a scope and a spectrum analyser, I have no doubt it would have shown that the frequencies in one pair would be mainly high frequencies and in the other mainly low frequencies. You can do it yourself if you have bi-wireable speakers.
I have never said this would mean an audible difference in sound, all I am saying, is that those who claim bi-wiring allows the amp to send signals of different frequency bands to each pair of wires (including What*hi*fi magazine, Axiom, and many other speaker manufacturers) are valid and are based on basic electrical theory.
Let’s consider the following (just one channel):
1) We agree both pair of wires (bi-wire) is connected to the same pair output binding posts at the amp.
2) We agree that the two pair of cables have identical impedance but with the link at the speaker removed, one pair will be connected to the low frequency XO/drivers binding posts and the other to the high freq ones.
3) Ohm's law say V=IZ, or I=V/Z, Z being the impedance.
Let’s call the path from the amp to the speaker’s internal L.F. XO terminal (via the unlinked binding posts) path A and the one from the SAME terminals at the amp to the speaker’s internal H.F. XO path B.
Impedance of path A= Impedance of cable A (negligible for short 12 gad. runs)+Impedance of L.F. XO
Impedance of path B=Impedance of cable B (same as A’s) Impedance of H.F. XO
The above is true because the each path consists of basically two components in series. The overall impedance of each path is predominantly dependent on the XO/drivers as the impedance of the cables are very low. I am excluding the impedance of the drivers for now to make things less complicated.
Assuming we do agree the impedance of cable A and B are the same and are negligible due to the same gauge, length etc., and are negligible for short runs of 12 gauge wires, we still end up with path A having a different impedance than path B.
Path A’s overall impedance would be much higher for high frequency signals and much lower for low frequency signals. Path B will behave in an opposite way. Remember the low frequency XO network/drivers and the high frequency XO network/drivers are no longer connected together inside the speaker box once the links are removed.
I know you keep making the point about the two bi-wired pairs are still connected to the same terminals at the amp, but there is no argument here. Only that, just because they both originate from the same point does not negate the bi-wire effect from a purely electrical stand point. May be you can think for a second too, everything in you house are ultimately connected to the same incoming cable from the utility company, are you saying then the currents in all of your circuits are the same, whether they are terminated at your amp, TV set of light bulbs?
This is probably my last attempt to try and dispel the myth (exist over the web) that I quoted in my first post and I would like to say one more time: I am NOT saying bi-wiring would or would not make any audible sound difference. I am NOT suggesting there is any different in impedance between those red binding posts. I totally agree that the impedance of each pair of the cables used in the bi-wire configuration would be the same as long as the lengths are roughly the same. But I am saying that, due to the different impedance vs frequency characteristics of the high and low frequency networks (filters), the H.F. cable will end up carrying mostly H.F. signals and the L.F. cable will end up carrying mostly L.F. signals, regardless of the same negligible impedance of the cables themselves. It’s the XO’s that draw the signal current they prefer, from the amp (yes, the same terminals, already agree). Thank you Johnny, and everyone else…..
