To: <towertalk@contesting.com>
>Date: Thu, 30 Jul 1998 11:27:18 -0700 (PDT)
>From: Pat Barthelow <aa6eg@cv.tmx.com>
>
>HI Eric,
>
>From your response to the TA-33 feed question, you may be able
>to help me with a question I have regarding antenna currents
>within a non-ideal coax transmision line..... Consider this
>common situation:
>
>An HF transmitter supplying 100 watts to 10 ft of 50 ohm
>hardline coax, terminated into a Bird, 50 ohm, shielded dummy
>load.
>
>Power formulas (I squared R, and P= I x E) say that there would
>be 1.41 amps of RF flowing up the outside of the center
>conductor, and on the inside of the hardline shield...currents
>exactly matched.....and, neglecting the negligable loss,
>constant anywhere on the 10 ft of hardline. Voltage a tad over
>70 volts. No currents on the outside of the shield...no way to
>"Get Out" from inside the coax.
>
>
>NOW, my own dilemma, and question....
>
>Suppose a PL 259 TEE fitting is inserted partway up the coax.
>That is, a barrel to accept the PL 259s in line, to the dummy
>load, and an unterminated Male fitting sticking out the side.
>The side male pin protrudes just a slight amount out of the
>coupling threaded ring, and the fitting is just left open,
>exposed.
>
>Can RF current flow, (leak) from the inside of the shield, now
>to the outside of the shield, through the Tee fitting, and back
>down or up the coax, and be radiated?
Yes. But at frequencies for which the opening and pin length are
small in terms of wavelength, they would not be large in
magnitude. Whether they radiate from the coax shield depends a
lot on exactly where along the line the tee is placed. There
will certainly be some amount of radiation directly from the pin
in any case.
>
>If so, is the remaining RF current headed to the dummy load on
>the center conductor and matching current on the inside of the
>solid shield still matched (equal and opposite polarity)? Is
>the RF current diminished by the amount of current, leaked to
>the outside of the coax?
The power reaching the load should be diminished by the amount of
the power lost to radiation, to additional line loss from the
mismatch (should be exceedingly small), and to whatever the
reduction is made by the source in response to the amount of
mismatch introduced by the tee. The exact current magnitude and
phase relationship to the voltage in the line between the
generator and the tee depends on a lot of undefined variables
(where is the tee relative to the source?, how big is the tee in
terms of wavelength?, etc). But the currents in the matched line
from the tee to the load should be equal and opposite center
conductor to shield.
73, Eric n7CL
>
>Answers to these questions can help me understand some other
>situations, like RF current flow routes and magnitudes on the
>insides of Chassis' of HF amplifiers, towards the output
>connector if open frame relays, only switching hot RF center
>conductors, are used as antenna switches.....
>
>
>
>
>73, DX de Pat, AA6EG/N6IJ
>"The Contest Station from the Government"
>
>Marina Amateur Radio Contest Station; N6IJ
>599 DX Drive
>Marina CA 93933
>
>aa6eg@tmx.com
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