Jim Reid wrote:
>
> At 06:32 PM 8/1/97 -0400, you wrote:
> >Hi Jim,
> > I vote for the loss being identical to that of a single run of
> the
> >same coax.
>
> Well if the two transmission lines are identical
> and everything is matched, then one half of the
> RF current is going to flow in each line. Note that
no, power is i-squared*Z0 so 1/2 the power is .707 times
the current.
>
> Power lost is current squared times the loss resistance.
> So if half the current is in each identical line,
again, .707 of the original current
> then one half squared is one fourth. Add the two fourths
.707 squared is .5
> in each line together and you have one half of the power
add the two .5's and you have the same loss.
> lost now as compared to a single line with all of the
> current flowing through it to the load.
>
look at a simpler case:
1. you have 1 coax that exhibits 3db power loss from input to
output at a given frequency. so you get out 1/2 the power you
put into it.
2. put another piece in parallel with it. divide the power so
that 1/2 of it goes into each line. each line will still loose
3db of the power put into it (coax is a linear element at a
specific frequency after all). so at the end of each piece of
coax you get 1/2 minus 3db or 1/4 of the original power. combine
the two lines and you get 2*1/4 or the same 1/2 power back out.
3. repeat above for 4 lines, and you still get the same 3db
of loss from input to output.
the only advantages of using two coax lines in parallel is if
you already have a balanced load and don't want to put a balun
at the load, and you have twice as much coax laying around(at
twice the price of course, you can't get away with cheaper
coax because you still have to deal with the full loss). it
also does have an advantage over bare twin lead in that it is
not affected by nearby conductors(assuming of course that
proper care is taken to keep currents inside the coax)
--
David Robbins K1TTT (ex KY1H)
k1ttt@berkshire.net or robbins@berkshire.net
http://www.berkshire.net/~robbins/k1ttt.html
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