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
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.