Re: 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

where:

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.
.

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