Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

Owen's explicit method should get it right in all cases. If you select
say 0.125 wavelengths of RG213 in Owen's online calculator, the
predicted loss with a 100 ohm load resistance is *less* than the matched
loss. If you change the load to 25 ohms, the predicted loss is *greater*
than the matched loss.

Both of these results make perfect physical sense because the largest
part of the loss is proportional to the square of the current, which
will be greater with the lower-resistance load. The two different
resistances correctly give different results, yet they both have a VSWR
of 2 (based on the 50-ohm system impedance). This shows that VSWR does
not contain sufficient information to give an explicit single-valued
result.

Owen's program will accept a VSWR input, but it correctly posts a bold
red warning that the result is an estimate. If you let the program
select the worst-case load impedance for the supplied value of VSWR,
you're back on track and it can calculate an explicit result.


Although we're debating fractions of a milliBel here, the debate has
shown how often the terms "VSWR" and "load impedance" are used
interchangeably - which they aren't. It isn't a big mistake here, but it
can be in other applications. For example, a solid-state PA designed
for a 50 ohm load will respond very differently to load *impedances* of
100 or 25 ohms, yet the load *VSWR* is the same in both cases.



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73 from Ian GM3SEK