bi-wire config question
 

Serge Auckland wrote:

You've obviously convinced yourself that it makes a difference. I've
never managed to hear any myself, and when I do the sums, I'm not a
bit surprised. By the way, if you are indeed sure it does make a
difference, have you tried to analyse why and how? What mechanism can
be acting to make the sound better (or even different)?

Not going to start drawing too many ASCII art diagrams at 1am, but a lot
of people think that "bi-wiring" just consists of running two lengths of
cable between amp and speakers. But if you don't remove the bridging
straps at the speaker end, all you're doing is halving the series
resistance of the cable (which should be negligible anyway). Or if you
prefer it as "doing the sums":

C = R / 2

where C is the resistance of the cable run(s) and R is the series
resistance of one run of cable.

Once you take the bridging straps off however, something else is
occurring. And what happens makes an audible difference. Too tired to do
the maths atm, but I'm sure a few ASCII art diagrams will help. (For
those people using brain-dead newsreader software, switch to a
monospaced font at this point.)

The HF driver is drawing a much lower current from the amp than the LF
driver (it doesn't need to move in and out so much, this should be
obvious). If you imagine the cable as a resistor and the driver voice
coil as an inductor, what you've basically got is this (note highly
simplified):

Single wired speaker
====================

+ -----[===]----+-----+
CSR | |
{L} {C}
| |
{D} {D}
CSR | |
- -----[===]----+-----+
LF HF

whe

CSR = Cable Series Resistance
L = Inductor (low pass filter in x-over)
C = Capacitor (high pass filter in x-over)
D = Driver voice coil

Bi-wired speaker
================

+ --+-----[===]----+
| CSR |
| {C}
| |
| {D} HF
\ CSR |
- ---|-+--[===]----+
/ \
+---|-[===]----+
/ CSR |
| {L}
| |
| {D} LF
| CSR |
+--[===]----+

Treat the run of cable between the amp and speaker as if it were a
resistor and it makes it easier to understand what's happening. And
remember that the LF driver can take a hefty current when you're trying
to move a lot of air - and that current is effectively being drawn
through a series resistor (ie the run of speaker cable). By bi-wiring,
you're no longer drawing that high current required by the LF driver
through the same series resistor as the HF driver.

The result - much more linearity from the HF driver, as it's no longer
suffering from current drain via a series resistor when the LF driver
draws current. Therefore no current sag to the HF driver, resulting in a
cleaner and more dynamic HF response.

Again, apologies if this makes little sense, it is 1am and I'm rather
tired...

--
Glenn Richards Tel: (01453) 845735
Squirrel Solutions http://www.squirrelsolutions.co.uk/

IT consultancy, hardware and software support, broadband installation

bi-wire config question
 

"Glenn Richards" wrote in message
. uk...
Treat the run of cable between the amp and speaker as if it were a
resistor and it makes it easier to understand what's happening. And
remember that the LF driver can take a hefty current when you're trying to
move a lot of air - and that current is effectively being drawn through a
series resistor (ie the run of speaker cable). By bi-wiring, you're no
longer drawing that high current required by the LF driver through the
same series resistor as the HF driver.

The result - much more linearity from the HF driver, as it's no longer
suffering from current drain via a series resistor when the LF driver
draws current. Therefore no current sag to the HF driver, resulting in a
cleaner and more dynamic HF response.

Again, apologies if this makes little sense, it is 1am and I'm rather
tired...

--
Glenn Richards Tel: (01453) 845735
Squirrel Solutions http://www.squirrelsolutions.co.uk/

IT consultancy, hardware and software support, broadband installation

But is the wiring between the amp output devices and the speaker terminals
to which you attach the bi-wiring to be considered in this equation or do
you just ignore it and concentrate totally on the speaker cables. If you run
separate wires to the amp output devices (not the amp terminals) are you
then to consider the resistance and other effects of the output devices. Do
we then consider the diameter of wiring between the power supply and the
amplifier circuitry in this equation as it will cause voltage drop at higher
currents. Should we require a power supply (transformer/capacitors) that
suffer no voltage drop at higher current requirements. Are the power supply
capacitor leg wires sufficient to minimise/eliminate voltage drop under
high/heavy current requirements

All you are doing with bi-wiring is placing the terminal that joins the the
LF and HF driver circuit inputs at the amp terminals instead of the speaker
terminals. If the speaker wire is so inadequate that you are able to measure
any significant changes in HF linearity due to the effect of speaker wire
resistance then the wire run is too long or the wire is too light.

bi-wire config question
 

APR wrote:

All you are doing with bi-wiring is placing the terminal that joins the the
LF and HF driver circuit inputs at the amp terminals instead of the speaker
terminals. If the speaker wire is so inadequate that you are able to measure
any significant changes in HF linearity due to the effect of speaker wire
resistance then the wire run is too long or the wire is too light.

Glenn said recently:
"(HF needs less current so a thinner cable will suffice)."
Which shows that there is no point in discussing technical stuff with him,
unless it is to give the rest of us a laugh.

--
Eiron

No good deed ever goes unpunished.

bi-wire config question
 

Eiron wrote:

Glenn said recently:
"(HF needs less current so a thinner cable will suffice)."
Which shows that there is no point in discussing technical stuff with
him, unless it is to give the rest of us a laugh.

It's called "speaking in layman's terms". Some people aren't technically
minded, and don't really care about the how and the why, just the what.

But since you're being pedantic, let's put it another way. A loudspeaker
voice coil is basically a linear motor. The LF driver is going to draw
more current than the HF driver - which should be blindingly obvious as
it has to move in and out much more (a throw of anything up to an inch
or so). The HF driver only moves by a few microns, as the frequencies it
is reproducing are far higher.

Therefore the HF driver is going to draw far less current than the LF
driver - and therefore is less likely to be affected by a slightly
higher series resistance (eg a thinner cable) than the LF driver.

--
Glenn Richards Tel: (01453) 845735
Squirrel Solutions http://www.squirrelsolutions.co.uk/

IT consultancy, hardware and software support, broadband installation

bi-wire config question
 

APR wrote:

All you are doing with bi-wiring is placing the terminal that joins
the the LF and HF driver circuit inputs at the amp terminals instead
of the speaker terminals. If the speaker wire is so inadequate that
you are able to measure any significant changes in HF linearity due
to the effect of speaker wire resistance then the wire run is too
long or the wire is too light.

Ok, for those people that still don't "get it", here's an experiment
that will demonstrate visually what's going on:

Take a 12V DC power supply (one of those bench PSUs capable of supplying
several amps will suffice) and a 12V 1W bulb (any type will do, it's
easier to perform the experiment using a MES bulb and batten holder with
screw terminals though).

Connect the bulb to the power supply using 5 metres of 13-strand zip
wire and power up. Observe the brightness of the bulb.

Now take a second bulb and holder, and attach a few inches of the same
wire to the second holder. Connect the second bulb in parallel with the
first so that it is "chained" from the first bulb, ie you've got 5m of
cable from PSU to first bulb, then a few inches from the first bulb to
the second bulb. Make this connection with the power turned on.

As you connect the second bulb, you'll see the first bulb's brightness
decrease. This is caused by a voltage drop in the cable. Disconnect the
second bulb and the brightness of the first will increase again.

Now replace the short piece of wire on the second bulb with another 5
metre length, and connect the two bulbs in parallel by attaching two
sets of wire to the power supply. This time when you connect the second
bulb the first one won't dim, and the second will light up at full
brightness.

This is exactly what is happening with your speaker cables. The cable
run acts like a series resistor, and the load (in this case the driver
voice coil) causes a voltage drop across that series resistor. By
bi-wiring you are avoiding the load from the LF driver causing a voltage
drop to the HF driver. (Again this is a greatly simplified description
of what's going on.)

I've specified 13-strand zip wire in the experiment above as it makes
the results more obvious, but the effect will still stand with thicker
cable, it'll just be harder to see. And eventually you'll get to a
thickness of cable that will have a low enough series resistance that
the effect will no longer occur - but in the case of speakers that cable
will either be so expensive or so unwieldy that it makes more sense to
use two thinner runs of cable.

--
Glenn Richards Tel: (01453) 845735
Squirrel Solutions http://www.squirrelsolutions.co.uk/

IT consultancy, hardware and software support, broadband installation

bi-wire config question
 

On Sun, 18 Jun 2006 07:45:45 +0100, Glenn Richards
wrote:

Eiron wrote:

Glenn said recently:
"(HF needs less current so a thinner cable will suffice)."
Which shows that there is no point in discussing technical stuff with
him, unless it is to give the rest of us a laugh.

It's called "speaking in layman's terms". Some people aren't technically
minded, and don't really care about the how and the why, just the what.

But since you're being pedantic, let's put it another way. A loudspeaker
voice coil is basically a linear motor. The LF driver is going to draw
more current than the HF driver - which should be blindingly obvious as
it has to move in and out much more (a throw of anything up to an inch
or so). The HF driver only moves by a few microns, as the frequencies it
is reproducing are far higher.

Therefore the HF driver is going to draw far less current than the LF
driver - and therefore is less likely to be affected by a slightly
higher series resistance (eg a thinner cable) than the LF driver.

Glenn. Do yourself a favour and stop posting right now. Go read up on
Ohms law, apply it to this situation, then come back and say sorry.
This was one of the funniest posts that has appeared here in ages.

The implication of this post is that cables are nonlinear - they
distort.

d

--
Pearce Consulting
http://www.pearce.uk.com

bi-wire config question
 

On Sun, 18 Jun 2006 08:00:27 +0100, Glenn Richards
wrote:

APR wrote:

All you are doing with bi-wiring is placing the terminal that joins
the the LF and HF driver circuit inputs at the amp terminals instead
of the speaker terminals. If the speaker wire is so inadequate that
you are able to measure any significant changes in HF linearity due
to the effect of speaker wire resistance then the wire run is too
long or the wire is too light.

Ok, for those people that still don't "get it", here's an experiment
that will demonstrate visually what's going on:

Take a 12V DC power supply (one of those bench PSUs capable of supplying
several amps will suffice) and a 12V 1W bulb (any type will do, it's
easier to perform the experiment using a MES bulb and batten holder with
screw terminals though).

Connect the bulb to the power supply using 5 metres of 13-strand zip
wire and power up. Observe the brightness of the bulb.

Now take a second bulb and holder, and attach a few inches of the same
wire to the second holder. Connect the second bulb in parallel with the
first so that it is "chained" from the first bulb, ie you've got 5m of
cable from PSU to first bulb, then a few inches from the first bulb to
the second bulb. Make this connection with the power turned on.

As you connect the second bulb, you'll see the first bulb's brightness
decrease. This is caused by a voltage drop in the cable. Disconnect the
second bulb and the brightness of the first will increase again.

Now replace the short piece of wire on the second bulb with another 5
metre length, and connect the two bulbs in parallel by attaching two
sets of wire to the power supply. This time when you connect the second
bulb the first one won't dim, and the second will light up at full
brightness.

This is exactly what is happening with your speaker cables. The cable
run acts like a series resistor, and the load (in this case the driver
voice coil) causes a voltage drop across that series resistor. By
bi-wiring you are avoiding the load from the LF driver causing a voltage
drop to the HF driver. (Again this is a greatly simplified description
of what's going on.)

I've specified 13-strand zip wire in the experiment above as it makes
the results more obvious, but the effect will still stand with thicker
cable, it'll just be harder to see. And eventually you'll get to a
thickness of cable that will have a low enough series resistance that
the effect will no longer occur - but in the case of speakers that cable
will either be so expensive or so unwieldy that it makes more sense to
use two thinner runs of cable.

OK, Glenn - two things. First lets assume that your experiment is
applicable. Now complete it. Use you second case - with the two wires
run separately right back to the source ( for maximum brightness). Now
take your short piece of wire and join the two bulbs together. If
biwiring made a difference, there should be a change in brightness as
you do this. There is no change. All you have shown is that it is
usually better to have thicker wire.

The second point is that the tweeter and woofer are not in parallel.
Does that surprise you? This because we are dealing with signals in
defined frequency bands, and we have a crossover, which presents a
high impedance to the cable in the stopband of each driver. This means
that low frequency signals - no matter how big - do not suck voltage
away from the tweeter. In fact the tweeter doesn't even know there is
a woofer there. There are of course electrical devices that have the
effect you describe; they are called modulators and rely on controlled
non-linearity to achieve a multiplying function. If you can show that
cables act as modulators, you will have saved the broadcasting
industry a fortune overnight and your fame will be assured.

Glenn, as I said in another post, please stop this. You have a
minuscule technical knowledge which ends far short of an ability to
understand and apply Ohm's law, but you persist in airing your
howlers. I know we are all amused when we have nothing better to do,
but you really are doing yourself no favours.

d

--
Pearce Consulting
http://www.pearce.uk.com

bi-wire config question
 

"Glenn Richards" wrote in message
...
But since you're being pedantic, let's put it another way. A loudspeaker
voice coil is basically a linear motor. The LF driver is going to draw
more current than the HF driver - which should be blindingly obvious as it
has to move in and out much more (a throw of anything up to an inch or
so). The HF driver only moves by a few microns, as the frequencies it is
reproducing are far higher.

Glen, a LF driver with 25mm travel at 50 hertz travels the same distance as
a HF driver travelling 0.01mm at 12,500 hertz. Admittedly the energy
required to accelerate the lower mass of a HF driver will be significantly
less, however, depending on the crossover frequency the energy to a HF
circuit can be greater then that going to a LF unit, especially in 3 way and
up systems where the LF crossover can be well below 300 hertz.

bi-wire config question
 

In article , Glenn Richards
wrote:
Serge Auckland wrote:

Are you seriously trying to tell us that tri or bi-wiring makes (or
even can make) a difference? Have you done the sums to see what effect
bi or tri wiring makes?

I've done several comparisons, using one then two runs of identical
cable (originally tested it with Gale XL-105). When I switched to
bi-wiring the sound was clearer, switching back to single wiring made it
sound muddy.

However, IIUC from previous discussions, none of the "comparisions" you
have reported here have been carried out in the ways we have discussed and
which would then allow the 'results' to reliable as evidence.

[snip]

More recently I did some tests using my own setup, single-wiring using
two of the four cores of Chord Rumour 4, then bi-wiring, then bi-amping.
Could hear a significant difference in clarity between single and
bi-wired, but couldn't hear any difference between bi-wired and bi-amped
[2]. So the cables were identical on both tests, just with bi-wiring
two runs of the same type of cable (or in this case a 4-core bi-wire
cable) were in use.

This prompts the following questions (amongst others)...

What test protocols did you use? How many times did you repeat the
comparisons? How did you 'randomise' the order of listening? How did you
ensure level matching, etc? How did you ensure that you had no knowledge of
which arrangement was in use at a given moment *other* than the sounds
produced? How did you deal with various other factors - e.g. variations
in hearing physiology with time and exposure to sounds? How did you do a
statistical analysis, and what (numerical) level of confidence did this
return as an estimate?

My recollection from previous reports you have made was that you did not
deal with the above experimental/protocol/assessment issues. The result
being that your conclusions were of little use as 'evidence'. However if
what you are reporting here was a more carefully run test I'd be interested
in the details.

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html

bi-wire config question
 

Glenn Richards wrote:
Serge Auckland wrote:

You've obviously convinced yourself that it makes a difference. I've
never managed to hear any myself, and when I do the sums, I'm not a
bit surprised. By the way, if you are indeed sure it does make a
difference, have you tried to analyse why and how? What mechanism can
be acting to make the sound better (or even different)?

Not going to start drawing too many ASCII art diagrams at 1am, but a lot
of people think that "bi-wiring" just consists of running two lengths of
cable between amp and speakers. But if you don't remove the bridging
straps at the speaker end, all you're doing is halving the series
resistance of the cable (which should be negligible anyway). Or if you
prefer it as "doing the sums":

C = R / 2

where C is the resistance of the cable run(s) and R is the series
resistance of one run of cable.

Once you take the bridging straps off however, something else is
occurring. And what happens makes an audible difference. Too tired to do
the maths atm, but I'm sure a few ASCII art diagrams will help. (For
those people using brain-dead newsreader software, switch to a
monospaced font at this point.)

The HF driver is drawing a much lower current from the amp than the LF
driver (it doesn't need to move in and out so much, this should be
obvious). If you imagine the cable as a resistor and the driver voice
coil as an inductor, what you've basically got is this (note highly
simplified):

Single wired speaker
====================

+ -----[===]----+-----+
CSR | |
{L} {C}
| |
{D} {D}
CSR | |
- -----[===]----+-----+
LF HF

whe

CSR = Cable Series Resistance
L = Inductor (low pass filter in x-over)
C = Capacitor (high pass filter in x-over)
D = Driver voice coil

Bi-wired speaker
================

+ --+-----[===]----+
| CSR |
| {C}
| |
| {D} HF
\ CSR |
- ---|-+--[===]----+
/ \
+---|-[===]----+
/ CSR |
| {L}
| |
| {D} LF
| CSR |
+--[===]----+

Treat the run of cable between the amp and speaker as if it were a
resistor and it makes it easier to understand what's happening. And
remember that the LF driver can take a hefty current when you're trying
to move a lot of air - and that current is effectively being drawn
through a series resistor (ie the run of speaker cable). By bi-wiring,
you're no longer drawing that high current required by the LF driver
through the same series resistor as the HF driver.

Yes, but that doesn't matter at all, as the current drawn by the LF
driver is isolated from the current drawn by the HF driver by virtue of
being at different frequencies, and the current is being drawn through a
frequency-sensitive network, the crossover. The only exception is at or
close to the cross-over frequency itself. When you substitute real-world
values for the cable resistance, you will find that the drop in level at
the crossover frequency is so small (fractions of a dB) that it won't
affect the sound heard. Outside the crossover frequency, it could matter
if the cable were to be non-linear, that is, exhibited intermodulation
distortion, but cables are linear to a huge degree, so that can't
account for the change in sound from bi-wiring.

The result - much more linearity from the HF driver, as it's no longer
suffering from current drain via a series resistor when the LF driver
draws current. Therefore no current sag to the HF driver, resulting in a
cleaner and more dynamic HF response.

No, for the reasons mentioned above.

Again, apologies if this makes little sense, it is 1am and I'm rather
tired..

No apology needed, it was well argued, but I feel it is wrong, unless
I've missed something.

S.