Let's say I have LM386 or some other audio amplifier. What determines the length, resistance, capacitance and inductance (if L C parameters matter) of the conductor that carries the output of the audio amplifier to the speaker?

  • \$\begingroup\$ You fail to ask about the resistance of the wire which increases with length and is inversely proportional to cross sectional area. Perhaps its the resistance of the wire that is most significant. \$\endgroup\$ – JIm Dearden Jul 19 '16 at 13:09
  • \$\begingroup\$ The resistance will be quite tiny isn't it? Fraction of a ohm. \$\endgroup\$ – quantum231 Jul 19 '16 at 13:14
  • \$\begingroup\$ I have added the resistance add well \$\endgroup\$ – quantum231 Jul 19 '16 at 13:14
  • 5
    \$\begingroup\$ "The resistance will be quite tiny isn't it? Fraction of a ohm." That all depends on the wire. You cannot simply assume that it is insignificant. E.g. 28 AWG twin (copper) wire will have a capacitance of about 15pF/m, 0.43 Ohms/m (allowing for return path 2 x 0.21) and about 1.7 uH /m inductance. A mere 20 metres of 28 awg twin wire will produce a resistance of 8.6 ohms - enough to half the power to an 8 ohm speaker, the capacitance and inductance being insignificant to audio frequencies. \$\endgroup\$ – JIm Dearden Jul 19 '16 at 13:34

The primary limiting factor will be the series resistance of the wire.

Let us work with a speaker of 8 ohms.

If the connecting wire were 18 AWG it has a resistance of 0.021 ohms per meter. If you were to connect the speaker with 50 meters distance from the LM386 (for a total wire path of 100 meters) the wire resistance would be (0.021 * 100) = 2.1 ohms. In such instance over 20% of the signal energy from the amplifier would be spent in the cable as heat and the rest going to the 8 ohm speaker.

(2.1 / (2.1 + 8) = 20.7%

You can use this example to evaluate other wire sizes and understand the effects.

Note: For audio frequencies do not fall victim to the marketing efforts of companies like Monster Cables that would have you believe that you need 8 AWG copper cable with gold plated connectors to connect up an 8 ohm speaker. Conventional low cost 16AWG lamp cord is almost always more than adequate to connect speakers at a reasonable working / listening distance.

|improve this answer|||||
  • 1
    \$\begingroup\$ Nah. Monster Cable is for the poor. You need this to really enjoy an unsurpassed listening experience. \$\endgroup\$ – dim lost faith in SE Jul 19 '16 at 14:35
  • \$\begingroup\$ If the only concern you have is power loss to your speakers, then go with 18AWG wires. If audio quality is your primary criteria, then there are many more considerations, many of which can be dealt with by decreasing cable resistance, and making sure you don't have oxidizing intermetallic contacts (which will soon increase contact resistance and also potentially generate nonlinear galvanic potentials). \$\endgroup\$ – AndyW Jul 19 '16 at 18:35
  • \$\begingroup\$ With an LM386 audio amplifier I doubt there is much consideration of the "audio quality". Then yet the huge amount of hype surrounding what people seem to believe in regarding their perception of "audio quality"!!! \$\endgroup\$ – Michael Karas Jul 19 '16 at 22:46

If you really want to be serious about the subject, EAW do a calculator and Crest have a chart. I haven´t been able to find the calculator on the website, but it´s here, top of the list.

And the Crest table:

Crest Table

And Litz cable actually does work, especially for reducing losses at the higher audio frequencies, but there are other (cheaper) alternatives as well. They´re called graphic equalisers! :)

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.