**Re: ribbon....**
AudioGeek said:

** **

can you elaborate on hoe the ribbon conductors should be wired correctly for 8 ohm impedance matching?

Hmmm..that article is not really that clear for what you need.

equations:

LC=1034 * DC...dielectric coeff...teflon=2.7, air=1

Z=sqr(L/C)

if you use L=13.5 nH per foot, C is 206.7 pf per foot.

sqr(13.5/.2067)=8.08 ohms impedance.

So, take a wire pair...use the terman equation...

L=.01016*length*[{2.303*log(D/r)}-{D/length}+{mu*.25}]

length in inches

D is center to center spacing of wires

r is wire radius

mu is 1

with that calculated inductance, divide it by the needed inductance..

Ex: #18 awg zip, .080 spacing, r=.020.... 200 nH per foot..

200/13.5=14.8

use 15 pairs to get the inductance you require...the capacitance will be difficult to calc due to proximity effects..but you get the idea.

For a ribbon, alternate + and -, and give it a whirl..

One of these days, I'll try measuring one, to see how the addition of more pairs changes L,C, and Z...

but not now..

Cheers, John