bi-wire technical discussion

Garmt van der Zel

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Jul 18, 1996, 3:00:00 AM7/18/96

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> So there you have it: a mechanism for distortion in a non bi-wired
> loudspeaker.
> Is it the main mechanism? I think it is likely.
> Is it audible? I suggest that this would depend on the amount, which
> depends on the values of parameters 1 to 4 above.
>
> Yours comments are awaited!
>
> John Harding

This is the only correct theory regarding the technical merits of
bi-wiring! I've read a German magazine that has already measured the
quantities of distortion and indeed they found differences between
bi-wiring, single-wiring and bi-amping (because of total seperation
between bass and treble with bi-amping, this method turned out to be
best, technically...), on the basis of this theory. Whether it is purely
harmonic distortion that makes the difference I'm not sure, but you
already mentioned that you haven't taken this in account.

Try to move the cone of your speaker with and without shorting the
connections. You will feel the difference! This is purely because of
back-emf's.

Garmt van der Zel
Delft University of Techology
Faculty of Applied Physics
The Netherlands

Bob Myers

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Jul 18, 1996, 3:00:00 AM7/18/96

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harding (john.h...@pt.teradyne.com) wrote:

> 3. The cable reistance is 0.5 ohms (this is quite a long cable)

Yes, it is. If it's 12 AWG, then this cable (assuming the total resistance
of two conductors is what is intended above) is 157.4 feet long. The question
now is whether or not this analysis corresponds to a realistic situation,
and why the effect of biwiring could not be duplicated simply by using a
larger single conductor.

> emf is 10 millivolts. This is split between voice coil resistance and
> cable resistance in the ratio 4.5 to 0.5. Therefore 1 millivolt of
> distortion is applied to the tweeter, in its pass band. This corresponds
> to 0.1% distortion.

And given that most of us are probably using cables about a tenth as long
as you assumed, twe'd be seeing 0.01% distortion - assuming that we accept
your other parameters.

> If the cable is not shared between bass unit and tweeter, this mechanism
> will not take effect, and this distortion will be zero. If the cable is
> shared, but is simply made thicker, the distortion will be reduced, but
> not eliminated.

This is incorrect. The path from "bass unit" to tweeter is longer in the
biwired case, but still exists. If the mechanism shown here is really
a problem, it is still non-zero in this case. Making the cable thicker
DOES work, as advertised.

Bob Myers KC0EW Hewlett-Packard Co. |Opinions expressed here are not
Workstations Systems Div.|those of my employer or any other
my...@fc.hp.com Fort Collins, Colorado |sentient life-form on this planet.

Mike Maloney

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Jul 19, 1996, 3:00:00 AM7/19/96

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I want to preface this by saying - this is not an area of expertise
for me, and there may be some obvious flaws in the reasoning here that
I have missed. But a couple of assumptions in this discussion
bothered me.

The original post has been edited for brevity-
harding <john.h...@pt.teradyne.com> wrote:

>How Bi Wiring could affect distortion levels

>Consider first that a bass loudspeaker is not well modelled as a
>resistor. A more acurate model is perhaps 5 ohms of resistance (in the
>voice coil and the crossover inductor-if any) in series with a voltage
>generator representing the back emf of the voice coil. This voltage is
>directly proportional to the velocity of the voice coil.

Isn't a more accurate model that of an inductor? Or more accurately, a
network of inductors? A resistive element has no time-domain
characteristics. Speakers are innately reactive.

>Consider next that most loudspeakers generate some distortion. This is
>both harmonic and intermodulation distortion. For simplicity's sake I
>will consider only harmonic effects here. As the voice coil is now
>moving in a manner which is not a perfect analog of the input waveform,
>the back emf which it generates will contain some distortion. I accept
>that some forms of distortion will not feed back into the voice coil,
>but I suggest that a certain amount will.

>Now, if the amplifier output impedance is effectively zero, the bass
>unit voice coil back emf will appear across the voice coil resistance,
>the cross over inductor resistance and inductance (if any), and the
>loudspeaker cable resistance.

What is "damping factor" and how does it apply to your model?

>Now consider a signal with frequency just below the cross over
>frequency. This will generate harmonic distortion which is above the
>cross over frequency. Some proportion of this distortion will appear
>across the loudspeaker cables, and be fed into the tweeter.

>Conclusion: There will be a small but measureable amount of distorted
>signal applied to the tweeter when the speaker is not bi-wired.

Your bi-wire example describes using one amplifier channel connected
in parallel to 1) a high-pass filter connected to a high-frequency
driver; and 2) [a low-pass filter connected to] a low-frequency
driver - we'll make the low-pass section optional to maintain symmetry
with your post.
Why, then, do you assert that there is no connection between the HF
and LF sections? They are connected in parallel to the same amplifier,
right? This parallel link does have the added resistance of cable
resistance x 2, but is that in fact significant? And if you choose to
strain at this particular gnat, then when are you going to start
looking at the reactive properties of the cabling (which you just
doubled) and compromise between the effects?

I see a major dispute brewing between these two areas of concern. How
do you wish to resolve this? Do you wish to state categorically that a
speaker cable is non-reactive? This would help solidify your model.
By virtue of your example, in a bi-wiring situation, higher cable
resistance would provide more isolation between components, and hence,
lower distortion. How do you propose to sell that notion to the
community? [ I can see it now... Mark Levinson introduces "Hi-R Pro",
Monster "MultiPhase Resistive Speaker Linkages - performs like a cable
half its size!"; sorry, I couldn't help myself]

So - I will not deny that electronically, there is a difference
between bi-wiring and single wiring. But as far as I can see, this
same effect would be rendered by inserting two resistors as a voltage
divider in the crossover at the first distribution node between
high-pass and low-pass sections. The value of the resistor would be
equivalent to - what? The equivalent of 150' of 12 gauge wire? Or some
lower figure?

In an ideal world, the listener will hear precisely what the recordist
put on tape. In the real world this is limited by similarity of
technology, re-creation of playback levels, and environment. Any
attempt to improve any individual element of this equation without
equal attention to the others is pointless.

Aesthetically, the recording engineer is the final link in the
performance before it reaches the listener. His/her technical
judgments are based on the particular monitor/room arrangement in
which he is situated. The only guaranteed way that the listener will
experience this mix in the way intended is to be there. Anything else
is a latently flawed experience-from a purely philosophical
standpoint. (I assume that someone out there might think the ability
to reproduce a recording accurately is important. Perhaps I am
wrong.)

Conclusion:
I am bewildered by the priorities of anyone who is enthusiastic about
technical issues that are generally ignored by recording
professionals. If there any evidence that more than 1% of
self-proclaimed audiophiles spent as much time or energy on their
acoustical environment as their equipment, it would be much easier to
give creedence to such arcane practices as bi-wiring.

Mike Maloney

Richard D Pierce

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Jul 19, 1996, 3:00:00 AM7/19/96

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In article <4sl38l$p...@steadfast.teradyne.com>,

harding <john.h...@pt.teradyne.com> wrote:
>How Bi Wiring could affect distortion levels
>
>Consider first that a bass loudspeaker is not well modelled as a
>resistor.

That's fine, but I know of no one who ever seriously made such an
assertion.

>A more acurate model is perhaps 5 ohms of resistance (in the
>voice coil and the crossover inductor-if any) in series with a voltage
>generator representing the back emf of the voice coil. This voltage is
>directly proportional to the velocity of the voice coil.

And this model is, in fact, far less accurate than the simple resistor
model. It ignores, for example, the fact that the back-emf generator many
are so fond as quoting as the evil agent is nothing more than the
electrical equivalent of the motional impedance of the loudspeaker. A
simple voltage generator doesn't come close to modeling this and will
generate all sorts of other non-existant problems because of the
inaccuracy of the model. For instance, your model presupposes that the
"back-emf" is independent of frequency: it is moast assuredly not. For
example, above the mid-band, it is effectively inactive.

A FAR more accurate model is the one that models the physical reality
more closely. The motional impedance of the driver is a direct result of
the fact that you have a mechanical resonance resulting from the
interaction of the drivers moving mass and the system compliance with
frictional, acoustical and electrical losses. No magic back-emf
generator is needed. The speaker responds simply because the mechanical
resonance is storing energy. As such, one can FULLY model this behaviour
electrically as (in the case of generalized direct-radiator systems)

where

Electrical portion of driver:
Re = voice coil DC resistance
Lvc = voice coil inductance

Mechanical portion of driver:
Lces = inductive equivalent of driver compliance
Res = resistive equivalent of suspension losses
Cmes = capacitive equivalent of driver moving mass

Acoustical portion of enclosure
Lceb = inductive portion of enclosure compliance
Reb = lumped absorption, leakage and port losses
Cmep = capacitive equivalent of port acoustical inertance

Use such a model, even physically build it, if you so desire, use the
right parts values, and it will behave PRECISELY like a loudspeaker. And
look: no voltage source! The back-EMF IS NOT a coltage source. The whole
phenomenon is simply the energyu storage in a resonance, nothing more. No
mysterious voltage source is needed, or even desirable.

>Consider next that most loudspeakers generate some distortion. This is
>both harmonic and intermodulation distortion. For simplicity's sake I
>will consider only harmonic effects here. As the voice coil is now
>moving in a manner which is not a perfect analog of the input waveform,
>the back emf which it generates will contain some distortion. I accept
>that some forms of distortion will not feed back into the voice coil,
>but I suggest that a certain amount will.

Fine, substitute non-linear values for the compoenents in the circuit
above. Works just fine.

>Now, if the amplifier output impedance is effectively zero, the bass
>unit voice coil back emf will appear across the voice coil resistance,
>the cross over inductor resistance and inductance (if any), and the
>loudspeaker cable resistance.

WRONG! It will appear, even in your model, IN SERIES, NOT across, or in
parallel. Let's look at your model:

+------ Re -----+
| |
Rg Vemf
| |
+---------------+

where Rg is the amplifier output resistance. Now, assume YOUR figure that
the amplifier output resistance Rg is 0. What is the magntidue of the
voltage that appears across it? Well, it is:

Rg
Vrg = Vemf --------
Rg + Re

And, if Rg is 0, then Vrg MUST be 0, even if Vemf is 1,000,000 volts.

>Now consider a signal with frequency just below the cross over
>frequency.

What signal? From the driver's "back-emf?" This is where your model falls
completely apart. By ignoring completely the fact that there is no
:back-emf" generator and, instead, using a simple voltage source, you now
have made an assumption that simply has no analog whatsoever ion real
speakers. The motional impedance at the frequency you're talkig about is
effectively 0. Instead, you now have not only the DC resistance between
the amplifier and the motional impedance, but you also have the increased
impedance due to the voice coil inductance, the greater eddy current
losses in the pole piece, etc.

>This will generate harmonic distortion which is above the
>cross over frequency. Some proportion of this distortion will appear
>across the loudspeaker cables, and be fed into the tweeter.

>Now, the question is, how large is this signal?

Do tell.

>I will make the following assumptions which are probably worst case:
>
>1. The voice coil resistance is 4.5 ohms
>2. A crossover inductor is not fitted (this is true of several high
>quality loudspeakers)
>
>3. The cable resistance is 0.5 ohms (this is quite a long cable)
>
>4. The bass unit generates 1% harmonic distortion in the voice coil in
>response to a 1 volt applied signal just below the cross over frequency.

And this value is orders of magnitude off the reality. For a moderate
quality 8" driver at resonance (where the motional impedance will be
maximum, the excursion will be at a maximum, and thus the distortion will
be at a maximum), this frequency might be 40-50 Hz in a system. But at,
say 1 kHz, the total excursio of the driver is now fractions of a
millimeter for very high levels and this level of distortion simply does
not exist. Try a value of 10 mV for your "back-EMF" instead.

>In this case, the distortion that is generated in the voice coil back

>emf is 10 millivolts. This is split between voice coil resistance and
>cable resistance in the ratio 4.5 to 0.5.

It's actually:

0.5
Vdt = Vd---------
4.5 + 0.5

but you rounded to get to the right number.

>Therefore 1 millivolt of
>distortion is applied to the tweeter, in its pass band. This corresponds
>to 0.1% distortion.

But your basic assumptions are WAY off. Given figure that are more in
line with reality, your a couple of orders of magnitude too high.

>Clearly, in the case where a bass unit crossover inductor is fitted, the
>distortion will be attenuated by the inductor, and will be less. It will
>not be zero.

Yes, it's clear if you ignore what the real figures are.

>So there you have it: a mechanism for distortion in a non bi-wired
>loudspeaker.

One which is, unfortunately, based on a model for loudspeakers which does
not come close to reflecting how they actually work.

>Is it the main mechanism? I think it is likely.

Not when you actually realize the way drivers really work, it isn't.

>Is it audible? I suggest that this would depend on the amount, which
>depends on the values of parameters 1 to 4 above.

Well, since both the model itself and the assumptions for the magntidue
of the effects are way off, well...

Paul Pavluk

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Jul 20, 1996, 3:00:00 AM7/20/96

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In article <4slt79$4...@fcnews.fc.hp.com>, my...@fc.hp.com says...
[snip]

>> If the cable is not shared between bass unit and tweeter, this
mechanism
>> will not take effect, and this distortion will be zero. If the cable
is
>> shared, but is simply made thicker, the distortion will be reduced,
but
>> not eliminated.
>
>This is incorrect. The path from "bass unit" to tweeter is longer in
the
>biwired case, but still exists. If the mechanism shown here is really
>a problem, it is still non-zero in this case. Making the cable thicker
>DOES work, as advertised.
>

Bob, I wonder if the amplifier will remove this affect by virtue of its
negative feedback? This would then remove the back emf from affecting the
other drivers. (If this is the cause of the distortion)

Tim Brown

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Jul 22, 1996, 3:00:00 AM7/22/96

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In article <4sqf34$9...@sinus.seqeb.gov.au>,
pp...@un.seqeb.gov.au (Paul Pavluk) wrote:

>Bob, I wonder if the amplifier will remove this affect by virtue of its
>negative feedback? This would then remove the back emf from affecting the
>other drivers. (If this is the cause of the distortion)

An amp can have negative and still have a finite or even low damping factor.
Tube amps are a whole different story, and high-enders often prefer them.

Tim