On May 2, 3:16 pm, Chris Jewell wrote:
Someone asked in a Technician license class *why* 450 ohm window line
has much lower loss than an equal length of 50 ohm coax does at the
same frequency. The instructor knew that it is true, but could not
say why. I have an idea about the answer, and would like to know how
I'm doing. I'm one of those hams who is NOT a EE, so I'm trying to
work this out based on high school physics, which at least back in
1964, covered AC but not RF.

I observe that at higher Z, the voltage is higher while the current is
lower for a given power level. Ohmic losses are proportional to the
square of the current. This is the same reason that long distance
power transmission is done at high voltage.

For example, 100 W through 50 ohms is 1.4A @ 70V, while 100W through
450 ohms is 0.47A @ 212V. That is, 9 times the impedance results in
1/3 the current, which results in 1/9 the ohmic loss through the
resistance of the transmission line.

I expect that we also need to account for the difference in R
resulting from different conductor diameter and skin effect, and
probably difference in the dialectrics, neither of which I yet know
how to calculate. Apart from those factors, is my explanation based
on the current vs. impendance:

1. Basically correct?

2. On the right track, but oversimplified, and thus not much use?

3. Completely out in left field?

Thank you.

--
73 DE KW6H Chris Jewell Gualala CA USA


Owen seems to have covered it pretty well. Here are a few more
related observations:

To a pretty good approximation, the matched-line loss in dB/100 feet
is given by:

A100 = 4.34*Rt/Zo + 2.78*f*sqrt(r.d.c.)*Fp

where A100 is loss, dB/100ft
Rt is the RF resistance of the conductors in ohms/100 feet (see
below)
Zo is the nominal line impedance, ohms
f is the operating frequency, MHz
r.d.c. is the relative dielectric constant of the dielectric between
the conductors
Fp is the dielectric's power factor at frequency f

Rt for round copper conductors is given for coax by
Rt = 0.1*(1/d + 1/D)*sqrt(f)
where d and D are the inner conductor diameter and the inside diameter
of the
outer conductor, both in inches
and for two-wire balanced line,
Rt = 0.2*sqrt(f)/d
where d is the diameter of either conductor, inches. That doesn't
account for the proximity effect Owen mentioned.

Conductor loss is increased by stranding and by brading. Smooth solid
conductors give lowest loss in almost all circumstances.

It turns out that if there is negligible loss in the dielectric, the
coaxial line loss for a given D will be minimized for D/d = 3.59,
independent of the dielectric you put in. If it's air dielectric, the
impedance for minimum loss is about 76 ohms, but if you fill the line
with solid polyethylene, the impedance is a little above 50 ohms. In
cases where the power handling capacity of the line is limited by
power dissipation and not by arc-over (which is almost certainly true
unless you're dealing with short, high-power pulses), the lowest line
loss will give you very close to the coolest conductors, so if you use
solid-dielectric coax, 50 ohms is a good value with respect to line
loss. The loss-vs-impedance has a rather broad peak, though, so it's
not critical.

If the line is operating into a load not close to the Zo of the line,
there is additional loss. In a qualitative way, you can see that by
realizing that standing waves create areas of high current, and as you
noted, the loss goes up as the square of the current. The squared
effect means that the low loss in the areas of low current don't
cancel out the high loss in the areas of high current. To be accurate
about it, you need to integrate the net current squared times the
resistance per unit length, along the length of the line.

Cheers,
Tom