At HF the losses in most transmission lines are copper losses, not
dielectric losses; so the losses are directly proportional to the square
of the current.
Picture the current standing wave along a mismatched TL with an SWR of,
say, 4:1. At the current maxima the current will be twice what it would
be in the matched case for the same power, and at the current minima it
will be half what it would be in the matched case. So the cable losses
at the current maxima will be 4 times what they would be in the matched
case, and at the current minima they will be one quarter what they would
be in the matched case.
Integrated over a long length of line the net effect is more loss than
in the matched case; but for short lengths of line that isn't
necessarily true.
Take as an example calculating the losses in 2ft of RG58 wound on a
ferrite core to form a 1:1 balun. Feeding a matched 50 Ohm load with
100W the differential-mode current in the coax will be 1.4A; But if we
now swap the 50 Ohm load for a 200 Ohm load (SWR=4:1), the current at
the load end of the coax will drop to 0.7A and the power loss at that
point in the cable will be one quarter what it was in the matched case.
Further back from the load, the current in the cable will rise due to
the standing wave pattern, and the difference reduces; however for short
lengths of cable the integrated loss can be less than the matched loss.
At, say, 21MHz the matched loss for 2ft of RG58 is 0.036dB, but with a
200 Ohm termination it drops to 0.022dB - about 40% less.
The converse happens if the termination is a low resistance, because the
current is then higher than in the matched case. With a 12.5 Ohm load
the loss rises to 0.131dB - significantly more that in the matched case.
So there we have an example where the cable losses for the same 4:1 SWR
are very different - in one case higher than the matched loss and in the
other case lower than the matched loss. Therefore any loss calculator
which asks for SWR as an input parameter rather than actual load
impedance is making assumptions which might not be valid! And any table
that simply shows "additional loss" due to some particular SWR should be
treated with caution.
Hope that's clear,
Steve G3TXQ
On 15/09/2014 22:13, Bob McGraw - K4TAX wrote:
OK, I'll fall for it, how does a cable have a negative dielectric loss
figure?
73
Bob, K4TAX
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