Thread: Coax cable calculation View Single Post
#6
Posted to sci.electronics.repair
 Dave Platt[_2_] external usenet poster Posts: 201
Coax cable calculation

In article ,
Trevor Wilson wrote:

Here's a question that stumped me back in the old slide rule days:

A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using
an insulator with a dielectric constant of 2.56, is yo have a
characteristic impedance of 2,000 Ohms. What must be the outer conductor
diameter?

Almost sounds like an OLD ("slide rule") homework problem....

**Correct. It was.

I thought someone would have come up with an answer before now. It's not
an overly difficult calculation.

Well, just for grins... using the formula

Z0 = 138 log10 ((Dd/Dc)*(1/sqrt(eR)))

it seems to transform to

Dd = 10^(Z0/138) * sqrt(eR) * Dc

with Z0 = 2000, eR = 2.56, and Dc = .001, we'd get approximately

Dd = 10^14.5 * 1.6 * .001

Dd = 1.6 * 10^11.5

Dd = 5.1 * 10^11

or about half a trillion inches. That's if I didn't screw up
something using my slide rule (yes, I have a nice 10" Hemmi here at
my desk) or slip a decimal point somewhere along the way.

This seems a bit impractical for a coaxial cable. The
transmission-line cutoff frequency would be ridiculously low, and
construction of a matching section would be, well, interesting, to say
the least. And, I rather doubt that the classic formula is actually
applicable for impedances (and diameters) this large.

A solid cylinder of dielectric material this large would probably
insist on collapsing under its own gravitational self-attraction,
leaving you with a neutron star or a black hole.