> >I don't agree with your ending statement. In a coaxial cable the RF does
> >not travel down the center conductor - at least it shouldn't.
>
> If RF does not travel down the center conductor, how does current get to
> the load?
I did't speak clear enough. The RF current flows on the SURFACE (depending
on skin depth) of the center conductor and the inner surface of the shield.
When I spoke of "RF energy" I mean an an RF field. The RF field is in the
dielectric. Since the dielectric has losses that is where the majority of
the heating will occur.
Quoting from "Applied Electromagnetism" by Shen and Kong:
" The total current on the inner surface of the outer conductor is equal to
that on the inner conductor with the opposite sign. Because the electric
and the magnetic fields are both transverse to the direction of wave
propagation, the particular set of fields is called the transverse
electromagnetic (TEM) mode of the coaxial line."
So the RF energy propagates down the transmission line as a wavefront. That
is what I was saying.
Continuing later regarding losses:
"The E field of the TEM mode in the coaxial line has a slight longitudinal
component. Some of the power leaks into the conducting surfaces and is
dissipated as heat. Thus, the signal in the line is attenuated as it
propagates."
Now, I don't think this disagrees with anything I have said. Perhaps I
should make another clarification. The loss and heating in the dielectric
will certainly depend on the type of dielectric used. In an air dielectric
coax (such as the larger heliax cables) perhaps most of the attenuation
comes from the conductors. However, in a standard, RG-8 type cable, I would
bet that the majority of the attenuation comes from dielectric losses. It
would be interesting to determine the percentages.
> >material. RF energy should ideally travel only on the surface of the
> >conductor. Since no conductor is a perfect conductor,
>
> So there is really and truly no such thing as a super-conductor?
Even a super-conductor has loss, Rich. Just like there is truly no such
thing as a non-inductive resistor.
73,
Jon
NA9D
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