[August Karlstrom]

On 2010-10-19 17:23, August Karlstrom wrote:
Here is my calculation of the minimum cross-sectional area of speaker
wire. Please tell me if you spot any errors.

***

Unfortunately, some strange characters seem to have creped into the
text. Here is a corrected version:

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The resistance R of a wire can be computed as

R = rho * l / A

where l is the length of the conductor in meters, A is the
cross-sectional area in square meters and rho is the electrical
resistivity in ohm meters.

Now, if we want the resistance of a wire of a specified length to be no
more than a percentage p of the lowest impedance of the speaker R_L we
have the relation

rho * l / A = p * R_L

which is equivalent to

A = rho * l / (p * R_L)

For copper wire the resistivity is 1.68E-8 ohm meters. If we choose p to
be one percent, which according to Arny Krueger ensures that the effect
of the wire is less than 0.1 dB and thus inaudible, we have the relation

A = 1.68E-10 * l / R_L (1)


Example:

Let's say we need two runs of four meter speaker wire. The lowest
impedance of our speaker is 8 ohm. Which cross-sectional area should a
copper wire have for maximum performance? By applying formula (1) we
have that the total area should be at least 8.4E-07 m^2 so a 2x0.50 mm^2
wire should be enough.


/August