[Aether Audio]

Friends,

I have considered the biwire question further and have identified a technical issue that may impart an effect on performance.  A technical analysis by J. Lesurf can be found at the link below.

http://www.audioholics.com/education/cables/bi-wiring-part-2-the-cable-conundrum/bi-wiring-from-amplifier-to-loudspeaker

And follow up by jneutron:

http://forums.audioholics.com/forums/showpost.php?p=248924&postcount=99

The above articles represent a thorough analysis and shows that in the case of “perfect” (i.e., lossless) crossover components, there is no net difference in power delivered to the speakers, save that which results from the reduced losses that would result from the increase in total effective wire gauge.  The same result could be had by simply increasing the gauge of a monowire, equivalent to the sum of the biwires.

The fact is though that the real world is a little more complex.  As is pointed out, the analysis does not consider the back emf generated by the drivers themselves or that of the reactive components in the crossover.  It is because of these that I suspect an improvement could be had.  A deeper analysis of the real-world conditions requires a model as a starting point for further examination.  The model I propose is as follows:

The driving amplifier is considered a “perfect” voltage source.  In EE terms this is defined as a source of voltage that has an infinitely low output impedance.  That means that the amplifier can continue to deliver more and more current as the load resistance approaches a dead short-circuit without the output voltage at the terminals collapsing.  Of course, no such circuit exists as it would require infinite current from the wall outlet, on through the amplifier’s power supply and then through to the output devices.

Furthermore, the model requires that the amplifier employ negative feedback for correcting output errors/distortion.  Without the inclusion of this condition the model and hence the analysis collapses.  In fact, this may be the source of any significant effects that are manifest in the real-world scenario.  As such, it represents the cornerstone of my further development.

Next, we are forced to consider that internally, from the amplifier’s output devices to its output terminals, there is some net finite resistance.  After all, wiring and circuit traces must be employed and nobody is making claims of room temperature super conductors being used in their equipment.

As a side note, in the monowire case we could easily transfer all of the speaker cable resistance to the inside of the amplifier, between the output devices and the output terminals.  Then we could (in theory) connect the loudspeaker directly to the output terminals and we would get the same result as if the wires were external to the chassis.

Regardless, there is (and for the time being) always will be some finite resistance between our perfect voltage source amplifier and the loudspeaker load.  Since the amplifier “senses” or “picks off” its negative feedback somewhere in the chain between the output devices and the output terminals, the resistance of the speaker wire will have an effect upon the completed system. 

If the speaker wire exhibits any net reactive term (i.e., capacitance or inductance) then it will impart some complex alteration to the corrective action of the feedback loop.  If it were only a matter of pure resistance being exhibited by the wire, then simply using a larger gauge would be as effective as biwiring.  The effect of cable resistance will be to “decouple” the errors (back emf) generated by the speaker load from the error correcting action of the negative feedback loop.  To the degree this series resistance produces a voltage drop across it, proportionally the error correction at the speaker terminals is reduced.

Albeit it is argued the effect is small, it is often the basis upon which proponents of using large wire gauge cables make their argument.  The effect higher resistance is to reduce amplifier Damping Factor and is most noticeable at low frequencies.  Dividing the spectrum between 2 sets of wires would (theoretically) lower resistance as well, but many argue that simply using a larger gauge monowire arrangement will yield the same result.  But…there is a “catch” to that argument.

In the real world, cables of sufficient reasonable gauge to offer the lowest possible resistance and hence, the highest Damping Factor, are seldom constructed such that they exhibit a net-zero reactance.  Although the reactances they do tend to exhibit can be considered quite small, nevertheless it is unlikely to be zero.  It can be argued that this reactance is so small that for all intents and purposes it can be ignored, and in a general sense this is true.  Our systems still work pretty well using imperfect cables.  Also, it is not impossible to construct such a large gauge cable that exhibits close to perfect resistance, but it’s somewhat difficult and seldom done.

Seeing that the question of audibility is rooted in our quest for ultimate fidelity, we are driven to consider these small reactive components.  Any reactance in our cables will induce a frequency dependant phase shift of the error signal (back emf) generated by the speaker load, as seen by the amplifier’s feedback loop.  At low frequencies it is justifiably arguable that these small reactive effects are inconsequential.  The true question arises when we consider their effects at the highest frequencies of operation.

It is well known that amplifiers employing negative feedback will exhibit a tendency, due to circuit induced phase shifts, toward regenerative (positive feedback) oscillation at some high frequency.  They are designed to position this frequency well above their operational pass-band in order to avoid ultimate self-destruction.  In order to achieve this they are designed such that at the frequency of oscillation, the total loop gain is reduced to less than unity.  Doing so limits the amplifier's ability to correct for errors incurred at or near their upper operational frequency limits.  This is why we often see rising distortion figures as frequency is increased in amplifier test results.

Considering the above, it does not require a great leap of imagination to suggest that phase shifts resulting from cable reactances could negatively influence an amplifier’s feedback loop.  In theory, the amplifiers ability to accurately correct for high frequency errors due to speaker load back emf could be reduced.  In fact, it is altogether possible that such phase shifts would give rise to further errors being generated by the amplifier in its attempt to correct for these phase altered signals presented to its loop.  This is the stuff of TIM and other Inter-modulation Distortion artifacts.

The upshot of all this is that it seems reasonable to suspect that a biwire approach, wherein each cable is optimized for the frequency band it carries, can be argued in favor of.  In theory, this may not be an ultimate requirement as long as a monowire cable that exhibits zero reactance can be found and used.  This represents a major difficulty in that such a cable will only manifest a zero reactance property into a single load impedance.  it is common for loudspeakers to exhibit a changing impedance vs. frequency, often increasing to some degree as frequency is increased.  In practice, it may simply be easier to construct a biwire arrangement that achieves the desired result.

As stated near the beginning, if negative feedback is removed from the equation – all bets are off.  In fact, this whole scenario may be the very reason many advocate the use of amplifiers that make no use of negative feedback.  In the case where the amplifier used does employ negative feedback, it would seem prudent to at least consider the reactive effects imparted by speaker cables…when making our selection thereof – biwiring or not.

-Bob

[jneutron]

D##m..I sent that link you 21 minutes before you posted this..you read fast, you type fast..

There are a few things however..I think you went through it just a tad hastily.

John’s analysis is thorough and shows that in the case of “perfect” (i.e., lossless) crossover components, there is no net difference in power delivered to the speakers, save that which results from the reduced losses that would result from the increase in total effective wire gauge.  The same result could be had by simply increasing the gauge of a monowire, equivalent to the sum of the biwires.

Salient points:

1. The net, or RMS dissipation between biwiring and monowiring is exactly the same if the same guage wire is used, as Bob said.  However, the dissipation profile changes instantaneously.  So, it is not a matter of powerloss per se, but a matter of what signal makes it to the drivers.

2. Regardless of what guage wire is used, monowiring will still show the 2AB modulation of powerloss.  By going to larger guage, the modulation certainly will be reduced, but that is not the "same result".

3. It is unknown at what level the effect is or is not audible.

4. My analysis does not require lossless crossover components, it has nothing to do with them.  In fact, I went to great length to make sure that non perfect amplifier and crossover components were zero'd out of the equation, to eliminate them from the effect under scrutiny.

The fact is though that the real world is a little more complex.  As John points out, his analysis does not consider the back emf generated by the drivers themselves or that of the reactive components in the crossover. 

I do not consider that simply because the initial analysis is already beyond that which is understood...that is no place to start hanging complexities.  For the same reason, I limited the analysis to two simple sines, and first order crossover elements.  We must walk first.  And the first step is independent confirmation.

It is because of these that I suspect an improvement could be had. 

Well, duh..

Seriously, I'll paste some points which are of interest to me...

Next, we are forced to consider that internally, from the amplifier’s output devices to its output terminals, there is some net finite resistance.  After all, wiring and circuit traces must be employed and nobody is making claims of room temperature super conductors being used in their equipment..

I'd be all over anybody who claimed room temp supercons..

Amplifier internal power routing is certainly a big issue, specifically low impedance outputs.  I cringe at current state of the art in terms of internal wiring layout.

As a side note, in the monowire case we could easily transfer all of the speaker cable resistance to the inside of the amplifier, between the output devices and the output terminals.  Then we could (in theory) connect the loudspeaker directly to the output terminals and we would get the same result as if the wires were external to the chassis..

Strangely enough, that was exactly how I configured the first speaker wire I built and send to a guy for audibility evaluation..it was a 10 nanohenry per foot, 288 pf per foot cable with an internal coaxial cable for external node connection to remove wire resistance and reactance from the output equation.  The only issue was the lack of external feedback terminals on this generation amplifier.

Regardless, there is (and for the time being) always will be some finite resistance between our perfect voltage source amplifier and the loudspeaker load.  Since the amplifier “senses” or “picks off” its negative feedback somewhere in the chain between the output devices and the output terminals, the resistance of the speaker wire will have an effect upon the completed system.  .

But it doesn't always have to be like that.

In the real world, cables of sufficient reasonable gauge to offer the lowest possible resistance and hence, the highest Damping Factor, are seldom constructed such that they exhibit a net-zero reactance. .

A net zero reactance requires infinite propagation velocity.  Wire reactance follows this rule:

L C = 1034 EDC.   L in nH per foot, C in pf per foot, EDC is the effective dielectric constant.

For a coaxial cable, EDC will be exactly the dielectric coefficient (air = 1).  Regardless of what the coax looks like, if you use the same dielectric, the product LC will always be the same.

For ALL OTHER cable geometries, the LC product will at best equal 1034 EDC, or it will be higher.  It can never be lower.

For ribbon sets, EDC is very close to the dielectric coefficient when the aspect ratio is over 20 or so...a 2 mil kapton dielectric film sandwiched between two half inch wide strips gives 1200 pf per foot, 2.3 nanohenries per foot, and a cable Z of 1.37 ohms.  (built and tested, it worked).

I've developed the equations sufficiently, and popped them into an excel spreadsheet, so that I can design a coaxial cable of any impedance, and any L or C (pick one, the other pops out from the equation).

Parallel wire pairs use terman's equation..gotta sheet for that also.  But I digress.

The upshot of all this is that it seems reasonable to suspect that a biwire approach, wherein each cable is optimized for the frequency band it carries, can be argued in favor of..

That is the crux.  However, it is necessary to make sure that if the effect is audible, the speakers and their design must play a big role.  Remember, if biwiring does indeed present as audible, the speaker MUST be designed for it.  Otherwise, your changing something in front of an ALREADY optimized system..

Cheers, John

[Aether Audio]

John,

D##m..I sent that link you 21 minutes before you posted this..you read fast, you type fast..

Not actually.  I was working on the post for some time before you sent me the link.  I want to apologize too.  I did a search on my own at AH and followed the links.  In so doing I initially mistook JL's paper for yours and had to go back and correct it - unfortunately after I had already posted.  In that, I also skimmed things a little too quickly and ended up a little mixed up on who said what.  Sorry about that.

I was also under the impression that the power loss modulation was only internal to the cable.  After closer inspection it appears that may not be so.   Well...another reason to consider biwiring.

Strangely enough, that was exactly how I configured the first speaker wire I built and send to a guy for audibility evaluation..it was a 10 nanohenry per foot, 288 pf per foot cable with an internal coaxial cable for external node connection to remove wire resistance and reactance from the output equation. 

I did the a similar thing once with a Crown Macro-Reference amp except the coax was external to the speaker wires.  it still worked though and was clearly audible.

A net zero reactance requires infinite propagation velocity.  Wire reactance follows this rule:

L C = 1034 EDC.   L in nH per foot, C in pf per foot, EDC is the effective dielectric constant.

As if I already knew that.   That's why "you da man" and I'm a "hack."  Thanks though...now I'm armed to the teeth!

Parallel wire pairs use terman's equation..gotta sheet for that also.

Does it work for ribbons too?  How much $$ you want for it? aa

Remember, if biwiring does indeed present as audible, the speaker MUST be designed for it.  Otherwise, your changing something in front of an ALREADY optimized system..

OK Mr. Wise Guy  ...tell me how you do that?  We're stuck with the crossover components that the driver's load impedance calls for.  Most of us don't make our own drivers and what difference would it make anyway?  Cable reactances are orders of magnitude lower than those of the crossover parts and/or drivers.  It seems to me they would swamp any cable reactances - whatever they were.  Obviously I'm missing something.  Be gentle.

Thanks my friend!
-Bob

[jneutron]

I was also under the impression that the power loss modulation was only internal to the cable.  After closer inspection it appears that may not be so.   Well...another reason to consider biwiring.

Actually, it's entirely within the cable.  But it is caused by the branches at the crossover.

Does it work for ribbons too?  How much $$ you want for it? aa

Ribbons:

Step 1.  Choose the ribbon width.
2.  Choose the dielectric and thickness.

3. Calculate the flat plate capacitance.

4. Use LC = 1034 DC to calculate the inductance.

5.  Use Z = sqr(L/C) to calculate impedance.

Example:  Copper 1/2 inch wide, kapton 1 mil thick, DC = 2.7

Capacitance equation:

C = (epsilon r times epsilon o times area) / thickness (sorry about no html scripted variables..)

Epsilon r is 2.7
Epsilon o is 8.854 times 10 e-12 farads/meter
area in square meters, thickness in meters.

Plug in, and get 94 pf per square cm.  At 1 foot, half inch wide, 3,637 pf per foot.

If you use 2 mil thick kapton, that is 1821 pf per foot.

Inductance falls out of LC =1034 DC, as DC is 2.7

1 mil kapton, 3643 pf, .766 nH, Z=.458
2 mil kapton, 1821 pf, 1.53 nH, Z=.917
3 mil kapton, 1214 pf, 2.29 nH, Z=1.37

Note, previous post said 2 mil kapton, but I forgot I have 1 mil of transfer tape adhesive in there also..

The only issue with the construct is the fact that the copper will kink if you bend it too much.  What really should be used is copper braid.  Two flat braids bonded to a center insulator.  Now that'll work.

(send cash)

OK Mr. Wise Guy  ...tell me how you do that?  We're stuck with the crossover components that the driver's load impedance calls for.  Most of us don't make our own drivers and what difference would it make anyway?  Cable reactances are orders of magnitude lower than those of the crossover parts and/or drivers.  It seems to me they would swamp any cable reactances - whatever they were.  Obviously I'm missing something.  Be gentle.

Thanks my friend!
-Bob

I'd start with a biwire set using two cables, each with about 20 ohms characteristic impedance.  That way, you're only presenting about 300 pf per foot loading to the amp.

Then optimize the speakers to sound best with those cables.  IFF biwiring makes an audible difference with a speaker, then the speakers have to be optimized to them..that is only logical sense.  (note, I do not know if biwiring makes an audible difference).  Edit:  (note IFF is not a typo, it is "if and only if".)

I also do not know how bad the capacitive load will actually be, because the cables are presenting as a transmission line would.  If you atttach a thousand mile cable to an amplifier, with 288 pf per foot and 10 nH per foot, the cable Z is very close to 8 ohm, will the amp oscillate?

Cheers, John

[jneutron]

th  dc     capacitance   .5 inch wide      inductance     impedance
               per cm sq    cap per foot                           ohms
1   2.7   94.11732283   3643.24392      0.76629511   0.458620964
2   2.7   47.05866142   1821.62196      1.53259022   0.917241928
3   2.7   31.37244094   1214.41464      2.29888533   1.375862891
4   2.7   23.52933071   910.81098      3.06518044   1.834483855
5   2.7   18.82346457   728.648784      3.831475549   2.293104819
6   2.7   15.68622047   607.20732      4.597770659   2.751725783
7   2.7   13.44533183   520.4634171      5.364065769   3.210346746
8   2.7   11.76466535   455.40549      6.130360879   3.66896771
9   2.7   10.45748031   404.80488      6.896655989   4.127588674
10   2.7   9.411732283   364.324392      7.662951099   4.586209638
11   2.7   8.556120258   331.2039927      8.429246209   5.044830602
12   2.7   7.843110236   303.60366      9.195541319   5.503451565
13   2.7   7.239794064   280.2495323      9.961836428   5.962072529
14   2.7   6.722665917   260.2317086      10.72813154   6.420693493
15   2.7   6.274488189   242.882928      11.49442665   6.879314457
16   2.7   5.882332677   227.702745      12.26072176   7.33793542
17   2.7   5.536313108   214.3084659      13.02701687   7.796556384
18   2.7   5.228740157   202.40244      13.79331198   8.255177348
19   2.7   4.953543307   191.74968      14.55960709   8.713798312
20   2.7   4.705866142   182.162196      15.3259022   9.172419276
21   2.7   4.481777278   173.4878057      16.09219731   9.631040239
22   2.7   4.278060129   165.6019964      16.85849242   10.0896612
23   2.7   4.092057515   158.4019096      17.62478753   10.54828217
24   2.7   3.921555118   151.80183      18.39108264   11.00690313
25   2.7   3.764692913   145.7297568      19.15737775   11.46552409


Sheesh, that only took about 19 edits...guess I'll have to set the excel cells to a simpler number format, like ####.##.

If you double the width, you double the capacitance and halve the inductance.  Z will also halve.

Cheers, John

[brj]

Bob, are you sure a fully active speaker system wouldn't be less complicated after all?

Thanks for all of the great reading, guys!

[jneutron]

Bob, are you sure a fully active speaker system wouldn't be less complicated after all?

Hey...we do the math.......so you don't have to..

Thanks for all of the great reading, guys!

Great reading???   You better get a life...you don't wanna be like me, does ya??

Cheers, John

[Aether Audio]

WOW John!!!...you're over the top

Thanks man!  Heck, you wanna build 'em for me too?

Take care,
-Bob

[jneutron]

Heck, you wanna build 'em for me too?

Take care,
-Bob

Building the wire is the easy part. I build many things, using many different materials.. Whaddya want?

Cheers, John