Hans,
The key qualifier is that the **matched** loss of the combined pair is
equal to the **matched** loss of the individual coaxes! The trick is
that Zo has changed between the individual case and the combined case.
The short explanation is that each individual coax is operating under
matched conditions, each flows half the total load current, each handles
half the total power and contributes half the total loss. But there are
two of them, so the total loss is double the individual loss!
Say we combine two 50 Ohm coaxes in parallel to feed a 25 Ohm matched
load. We've essentially doubled the cross section of copper, but for a
given power the total current flowing into the 25 Ohm load is higher
than it would have been into a 50 Ohm load. The net result is that the
total % power loss is the same.
Steve G3TXQ
On 04/08/2016 18:50, Hans Hammarquist via TowerTalk wrote:
Jim,
That's something I don't really understand. Why do you get the same loss in coax if you
run them parallel? If you run "normal" wires in parallel you reduce the losses
due to the lower current in each from the conductor RI^2 loss and the V^2 loss in the
dielectric, even if that's not a big deal at, say, 60 Hz. Why not the same for coax? Yes,
if I parallel two coax I need to use some matching to get the 50 ohm (or whatever I have).
Hans
_______________________________________________
_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk
|