"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.

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 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. I calculate the loss
to be -0.177 dB or 4%. How much loss would be expected in the feed line
itself if it is a moderate length?

--

Rick

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.

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.

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?

Nothing in this analysis addresses the theoretical maximum possible
efficiency of an arbitrary transmitter and an arbitrary load. In
particular I posted the results of a simulation that showed very clearly
that the loss in the transmitter output impedance can be well below 50%
of the total power drawn from the PSU. Just set the load impedance and
make your output impedance as low as you would like.

It is when you set the output impedance of the transmitter to a fixed
value that a matched load impedance will draw the maximum power from the
transmitter while the loss in the transmitter output will be 50%.

--

Rick

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.

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.

On 7/8/2015 3:38 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.

I was referring to a voltage source.

--

Rick

rickman wrote:
On 7/8/2015 3:38 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.

I was referring to a voltage source.


Instead of arguing about it, one can download QUCS for free which
will simulate the whole thing and one can see what really happens.

Download QUCS for your operating system:

http://qucs.sourceforge.net/

Generate a model consisting of a voltage source with a series resistance
of a few Ohms to simulate a solid state source or a much higher
resistance to simulate a vacuum tube source. Chose a convienient
frequency for the source.

Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html
to calculate an impedance matching network to match the resistance
you've chosen to 50 Ohms.

Put the matching circuit in the model.

Add a transmission line to the model.

Terminate the transmission line with a 50 Ohms resistor.

Add a fixed frequency AC simulation at the desired frequency.

Change various parameters to your heart's content to see what happens.

Change the matching network such that the output of your transmitter
is no longer 50 Ohms and see what happens.

When the QUCS output disagrees with your beliefs, you can argue with
the program.



--
Jim Pennino

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.



No, the SWR bridge is correct. The output of the transmitter is 50 ohms.

You are correct in that if a 75 ohm bridge is used, the indicated SWR
would be 1:1, because everything from that point on is 75 ohms.
However, the mismatch (and reflection) occurs on the transmitter side of
the bridge, not the antenna side. So the bridge will never see it. But
an accurate bridge will show lower power output due to the mismatch.

A mismatch is a mismatch, no matter where in the system it occurs. And
any mismatch will cause less than 100% power to be transferred. The
rest is reflected.

Just look at the specs of any amateur transceiver. They show an
impedance of 50 ohms. So a load of 50 ohms provides for maximum power
transfer; any other impedance causes a mismatch.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

"Jerry Stuckle" wrote in message
...
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

You are correct in that if a 75 ohm bridge is used, the indicated SWR
would be 1:1, because everything from that point on is 75 ohms.
However, the mismatch (and reflection) occurs on the transmitter side of
the bridge, not the antenna side. So the bridge will never see it. But
an accurate bridge will show lower power output due to the mismatch.

A mismatch is a mismatch, no matter where in the system it occurs. And
any mismatch will cause less than 100% power to be transferred. The
rest is reflected.

Just look at the specs of any amateur transceiver. They show an
impedance of 50 ohms. So a load of 50 ohms provides for maximum power
transfer; any other impedance causes a mismatch.

--
The real impedance of the transmitter is not 50 ohms. It is whatever the
device is used in the final stage and the poewr level. For a 100 watt
transmitter it is in the thousand ohm range and for solid state devices it
is very low. The matching circuit is often fixed to be 50 ohms,but could be
made for most any impedance. The older tube circuits were adjustable by the
user for a range of somewhat bleow 50 ohms to around 200 ohms. Could be
more or less depending on the design.

The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm
feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the
transmitter. Whatever power comes out of the transmitter will make it to
the antenna minus the loss of the coax, but not additional loss due to swr.
The power comming out of a 50 ohm transmitter will be less due to mismatch,
but not because of swr of the antenna system which is 1:1.