Copperfoil speaker cables

[Pan]

There are foil inductors from several makers. These are considered the best aircore coils for serious speaker networks.

They come cheap and are easy to unroll and make a speaker cable of.

You can use the foils without insulation and connectors for the cleanest and fastest fix. If you like it you can insulate with plumbers teflon tape or cotton and put spades/bananas on them (not necessary though).

/Peter

[JonHarrison]

Hi Peter, thank you for your reply.

All,

Does anyone know of an Alphacore supplier (or other foil inductor manufacturer) in the UK. I've had a look on the net and haven't found any. I know Alpha ship internationally but the $35 makes the idea of buying a $15 inductor to experiment with less appealing !

Cheers,

Jon

[protos]

audiocom international or copper foil tape from rs

[promitheus]

Does anybody know, how to calculate the inductance of two foil conductors that laying on one another?

I am going to try to make a cable of 4 foils one on top of the other in a +-+- format. It should have very low inductance but high capacitance. The only problem is that the foil from the Mundorf coils isn´t insulated but there is a thin layer of isolation foil that is a bit bigger in width. My problem is to hold the whole thing together. Maybe running tape across the sides is a good idea.

Does anybody have an idea?

[jneutron]

Originally posted by promitheus
Does anybody know, how to calculate the inductance of two foil conductors that laying on one another?

Does anybody have an idea?

#2...double stick tape.

#1...try this..it is for coaxial construction, so will depart from reality for small width foils or very large spacing..

Cheers, John

[jneutron]

Treat it as a double coaxial construct...

given W, width of the tape...

T, insulation thickness between the tapes..

DC, dielectric coeff of 2.7 (teflon, tefzel)

inner diameter is D_inner =W/3.1415

outer diameter is D_outer =D_inner+(2*W)

inductance is = 60.96*LN(D_outer/D_inner)

Kinda wish I coulda posted the picture instead..

For really far spacings, it reverts to the two wire Terman equation..

Cheers, John

[jneutron]

Ah, figured it out..the jpeg was too high....

Here it is..

While used for dual braid coax, it is applicable to flat conductors, as long as the spacing is small compared to the width..

It is inductance per foot vs diameter (width/pi), for various insulation thickness..so for 1 inch wide ribbons, use 1/pi for the diameter...Bad image, not very clear...let's try this:

<img src="http://forums.audioreview.com/attachment.php?attachmentid=660&stc=1">

Bah...that didn't work either..oh well, it's there if anyone wants a better image..

Cheers, John