On 7/8/2015 2:38 PM, John S wrote:
On 7/8/2015 10:48 AM, rickman wrote:
On 7/8/2015 10:09 AM, John S wrote:
On 7/2/2015 1:38 PM, rickman wrote:
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

Try this - connect the output of an HF transmitter to an SWR bridge.
Now connect a piece of 75 ohm coax such as RG-59 to the output of the
SWR meter, and connect that to a 75 ohm resistive load. Do you think
the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1.



What you have described is a case of using the wrong swr bridge. You
are
trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm
bridge is
used it will show a 1:1 SWR.

The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no
reflected power.

My knowledge of antenna systems is limited, but I do know that this is
correct, there will be no reflection from the antenna.

If there is no reflections from the antenna, how can there be a loss in
the source end? There is NO power returned according to your own
statement.

I don't see any contradiction. The power comes from the source through
the source impedance. The source impedance will create a loss, no?


If the
transmitter output is 50 ohms there will be a loss in this matching
that
will result in less power being delivered to the feed line, but that
will not result in reflections in the feed line.

Why? What causes the loss? The transmitter output resistance? So that
would mean that one can never achieve more that 50% efficiency at the
transmitter's OUTPUT! And that would mean that a 1000W transmitter is
dissipating 500 watts under the BEST circumstances. Good luck on getting
that to work to your satisfaction.

Maybe "loss" isn't the right term then. The output of a 50 ohm source
driving a 75 ohm load will deliver 4% less power into the load than when
driving a 50 ohm load. That comes to -0.177 dB. Is there any part of
that you disagree with?

All of it. Let's say you have a 1A source and it has a 50 ohm impedance
in series with its output. With a 50 ohm load it will provide 50W to the
load. With a 75 ohm load it will provide 75W to the load. The only
difference is that the 50 ohm load will cause the source voltage (before
the series impedance) to be 100V while the 75 ohm load will require 112V
(before the series impedance). If the series impedance is 0 +/- j75
ohms, it will have no power loss. If the series impedance is 50 + j0 it
will have a 50W loss.

Oops! Source voltage will be 70.7V for 50 ohms and 90V for 75 ohms and
dissipation-less output impedance.