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In Reply to: Re: Biwiring--why do it? posted by Kal Rubinson on March 7, 2007 at 18:01:16:
I think revealing your experiences would be more helpful, obviously you're not big on it but if you were to say 'in my experience bi-wiring is a wast of time and wire' it could be said you didn't hear any difference, and coming from someone as heavily exposed to components as you are, would be valuable input.
Follow Ups:
I did not say anything about your experience, did I? I stated that all the reasons offered for "why" in the post I was responding to are unproven."The "why" for biwiring is by having the bass and the mid/treble signals run through separate cables, the higher current of the bass will not modulate the lower current mid/treble signal."
Unproven and unlikely. Since they come from the same output stage, all biwiring is doing is moving the shunt across the speaker terminals back to the amp terminals."Also, with the two pairs of connectors on biwirable speakers, the crossover inside the speaker can be more easily optimized to take advantage of the arrangement and thus provide a more open, transparent sound."
Unproven and undemonstrated. How does biwiring result in the optimization of the crossover? In order to accommodate biwiring, some crossover topologies, like series filters, are precluded.
Unproven and unlikely. Since they come from the same output stage, all biwiring is doing is moving the shunt across the speaker terminals back to the amp terminals.And by doing that, it is no longer a shunt. In a bi-wire configuation, there are two cable lengths of impedance between the high and low crossover sections. It is not the same circuit.
Dave
"two cable lengths of impedance?" I am not sure what that says but each of those will have a higher series resistance than one cable of twice the thickness. And, by the way, it is always the same circuit since they are connected.
Maybe you need to draw the circuit to see the difference.Suppose, for purpose of argument, that the speaker is a 4 ohm resistive load, with 8 ohms at the low posts and 8 ohms at the high posts, and the speaker cable resistance is 1 ohm. Using a single cable, the total impedance of the cable plus load is 5 ohms (1 + 8/2). When bi-wiring with two such cables, the total impedance is 4.5 ohms ((1 + 8)/2).
If the crossovers were perfect brick walls and there was no overlap in frequency between the high and low networks, then bi-wiring wouldn't make a bit of difference (assuming the cable behaves linearly). But since there is overlap, in the region around the crossover frequency the cable impedance will be different in a bi-wire circuit vs. a single wire circuit. That will affect the amount of current delivered to the speaker in that frequency range, and thus the speaker's output in that frequency range.
Now, what if you use the same two cables to double up a single run (connecting them together at the speaker terminals) instead of bi-wiring. In my example above, the total impedance was also be 4.5 ohms, same as bi-wiring. But there is a difference. With the double run circuit, the impedance difference compared to a single run is the same across the whole frequency range. With the bi-wire circuit, the impedance difference compared to a single run is limited to the frequency range in which currents flow to both high and low crossover networks.
One thing at a time.
"Suppose, for purpose of argument, that the speaker is a 4 ohm resistive load, with 8 ohms at the low posts and 8 ohms at the high posts, and the speaker cable resistance is 1 ohm. Using a single cable, the total impedance of the cable plus load is 5 ohms (1 + 8/2). When bi-wiring with two such cables, the total impedance is 4.5 ohms ((1 + 8)/2)."
Your math is fine but your assumptions are far from real world. An 4ohm system cannot be 8 + 8 across all frequencies (even simplifying it to resistance and not impedance) since each leg has an increasing resistance with frequencies beyond its passband, even at crossover. Also, a cable resistance of 1ohm is HUGE! Using this assumption, unreasonably magnifies all the effects you describe."If the crossovers were perfect brick walls and there was no overlap in frequency between the high and low networks, then bi-wiring wouldn't make a bit of difference (assuming the cable behaves linearly). But since there is overlap, in the region around the crossover frequency the cable impedance will be different in a bi-wire circuit vs. a single wire circuit. That will affect the amount of current delivered to the speaker in that frequency range, and thus the speaker's output in that frequency range."
The cable impedance does not change appreciably with frequency regardless of the crossover. It is the load represented by the speaker that changes. The amp sees that change but, since it is paralleled by a complementary change in the parallel leg, it is minimal."Now, what if you use the same two cables to double up a single run (connecting them together at the speaker terminals) instead of bi-wiring. In my example above, the total impedance was also be 4.5 ohms, same as bi-wiring. But there is a difference. With the double run circuit, the impedance difference compared to a single run is the same across the whole frequency range. With the bi-wire circuit, the impedance difference compared to a single run is limited to the frequency range in which currents flow to both high and low crossover networks."
So, first, the real world differences are MUCH smaller since 1ohm cables are generally not in play. Second, any impedance variation seen by the amp in one leg is compensated for in the other and has nothing to do with the fixed resistance of the cable.In fact, one might argue that a typical crossover benefits from single-wiring since that assures the intended complementary shunting of out-of-bandwidth impedance rises of one network/driver leg by the other. Introducing real world cables, with their complex impedances, between the legs would certainly complicate that process. Of course, I would think to suggest something as radical as that. :-)
Kal
Your math is fine but your assumptions are far from real world. An 4ohm system cannot be 8 + 8 across all frequencies (even simplifying it to resistance and not impedance) since each leg has an increasing resistance with frequencies beyond its passband, even at crossover.
...
The cable impedance does not change appreciably with frequency regardless of the crossover. It is the load represented by the speaker that changes. The amp sees that change but, since it is paralleled by a complementary change in the parallel leg, it is minimal.
Kal, because the impedance is not constant, my example is only intended to represent the conditions at a single frequency. I chose equal loads on both the high and low end because I was making a point about the effect of bi-wiring around the crossover frequency - where (as you said yourself) out the loads are roughly equal.
Also, a cable resistance of 1ohm is HUGE! Using this assumption, unreasonably magnifies all the effects you describe.It was an example. Make the number as big or small as you wish. My point is that the single wire and bi-wire circuits are NOT equivalent, as you seemed to be claiming.
Second, any impedance variation seen by the amp in one leg is compensated for in the other and has nothing to do with the fixed resistance of the cable.Wrong. At a frequency outside the pass band of the low frequency crossover network, the impedance of that network is very large, so the load impedance is just the impedance of the high frequency network in series with a single cable. At that frequency, it doesn't matter whether both networks share the same cable or not because there is no load from one of the networks. But around the crossover frequency, both the high and low frequency networks present a similar load, and then the bi-wire configuration will result in a smaller impedance. As a result, bi-wiring will produce an increase in speaker output in the frequency range where the crossovers overlap. It may not be big, but it can be explained and measured.
In fact, one might argue that a typical crossover benefits from single-wiring since that assures the intended complementary shunting of out-of-bandwidth impedance rises of one network/driver leg by the other. Introducing real world cables, with their complex impedances, between the legs would certainly complicate that process. Of course, I would think to suggest something as radical as that. :-)One might conversely argue that a speaker which supports bi-wiring may have been tweaked or voiced by the designer to sound its best in a bi-wire configuration. But I have no idea whether this is common.
Well, if you reduce the impedance of the cable to 0.1ohm (for argument), the effects you describe reduce to insignificance (even if I accepted them all).Also, your example of the 4ohm speaker having 2 8ohm legs is unrepresentative. Both legs are usually 4ohms in their respective passbands and each defines the net characteristic impedance in its passband. (Sure, cabinet + resonance effects aside.)
So, let’s look, again, at your analysis:
You say that (keeping your 1ohm cables), “Using a single cable, the total impedance of the cable plus load is 5 ohms (1 + 8/2). When bi-wiring with two such cables, the total impedance is 4.5 ohms ((1 + 8)/2).” Each leg is a 4ohm speaker load and a 1ohm cable load and the result is a 5ohm load either way. The two parenthetical statements should be (1+4) and (1+4). The latter since each leg represents a meaningful load only in its passband. At crossover, there will be some reduction in overall impedance where the two legs are effectively in parallel but see below for comment on that.“As a result, bi-wiring will produce an increase in speaker output in the frequency range where the crossovers overlap. It may not be big, but it can be explained and measured.”
And why, assuming it is true, would that be desirable? Surely, you would want the load and the output to be identical at all frequencies.“One might conversely argue that a speaker which supports bi-wiring may have been tweaked or voiced by the designer to sound its best in a bi-wire configuration. But I have no idea whether this is common.” Well, unless the manufacturer explicitly states that the speakers MUST be used bi-wired, one must assume that bi-wiring is not necessary. I do recall that some Alon speakers were so specified but, since monowiring is almost ubiquitous, it would make sense to for a manufacturer to presume it.
I have been told, face to face, by several speaker manufacturers who ‘recommend’ bi-wiring in their user’s manuals that they provide such a facility to accommodate their dealers (who want to sell more cable) and some of their clients who wish to do it.
Now, I do not see any disadvantage to bi-wiring (heck, I use it) but I have not seen any convincing demonstration of “why” it is advantageous. And that “why” is what I responded to, not any issues of perception or preference.
So, let’s look, again, at your analysis:
You say that (keeping your 1ohm cables), “Using a single cable, the total impedance of the cable plus load is 5 ohms (1 + 8/2). When bi-wiring with two such cables, the total impedance is 4.5 ohms ((1 + 8)/2).” Each leg is a 4ohm speaker load and a 1ohm cable load and the result is a 5ohm load either way. The two parenthetical statements should be (1+4) and (1+4). The latter since each leg represents a meaningful load only in its passband. At crossover, there will be some reduction in overall impedance where the two legs are effectively in parallel but see below for comment on that.I think you're still missing it. (1+4) is never in parallel with (1+4). As I stated before, I assumed both loads are roughly equal at the crossover point. If the individual loads were 4 ohms in their respective passbands, they are going to be higher than that at the crossover point. That is where the 8 ohms comes from - an assumption that the crossovers are designed such that the speaker output and impedance remain consistent through the crossover point.
To make it clearer, let's talk about three frequencies: f_L, f_H, and f_C. f_C is the crossover frequency. f_L is a lower frequency within the passband of the low network and well out of the pass band of the high network. Vice versa for for f_H. In my example, at f_H the impedance of the high network is 4 ohms and the impedance of the low network is much higher, say 1000 ohms. At this frequency, for the single wire case the load impedance is 1+1/(1/4+1/1000) = 4.984 ohms, and for the bi-wire case the load impedance is 1/(1/(1+4) + 1/(1+1000)) = 4.975 ohms. The difference is 0.009 ohms, or about 1% of the cable impedance. Same situation at f_L.
At f_C, in my example the low and high networks are producing half the output and have twice the impedance they have in their passband, so that in combination their output and impedance remains roughly constant through the crossover region. At this frequency, as my example previously showed, the impedance for the single wire case is 5 ohms and the impedance for the bi-wire case is 4.5 ohms.
If that doesn't make it clear why bi-wiring produces an impedance change near the crossover reason, I give up.
Well, unless the manufacturer explicitly states that the speakers MUST be used bi-wired, one must assume that bi-wiring is not necessary. I do recall that some Alon speakers were so specified but, since monowiring is almost ubiquitous, it would make sense to for a manufacturer to presume it.I have been told, face to face, by several speaker manufacturers who ‘recommend’ bi-wiring in their user’s manuals that they provide such a facility to accommodate their dealers (who want to sell more cable) and some of their clients who wish to do it.
Honestly, I don't think bi-wiring makes enough difference to ever be necessary. It's more of a tweak thing in my view.
What really bothers me is this: Bi-wiring is conceptually simple; just parallel vs. series impedances. Anybody who made it through the first few weeks of an EE 101 course can do the circuit analysis. But even though this hobby is full of people with EE degrees, there still seems to be widespread misunderstanding. For example, if I hear one more voodoo explanation, or even another bi-wiring argument based on the supposed interaction of frequencies which ignores superposition, I'm going to be sick.
"If that doesn't make it clear why bi-wiring produces an impedance change near the crossover reason, I give up."
Sure it does but, as I stated, a 1ohm cable is a poor example and, as you estimate for the 1ohm situation, the difference is less than half the cable the impedance of the cable, even at crossover. Thus:(1) The difference shrinks to (greater?) insignificance with real world high-quality cables.
(2) A change in impedance at crossover compared to the rest of the spectrum is not a generally desirable outcome."Honestly, I don't think bi-wiring makes enough difference to ever be necessary. It's more of a tweak thing in my view."
Whew! That is my position, as well.
A similar discussion came up many years ago at the Madisound board. It prompted my first attempt at a web page. The link here goes to a page showing various incarnations of a 3-way with impedance and SPL for compensated vs. uncompensated and with a 1-ohm resistive cable, exaggerated for visualization. Of course every system will have a different impact from the cable and will likely not be restricted to just the crossover area.
Bravo!
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