On 7/2/2015 8:14 AM, Jerry Stuckle wrote:
On 7/2/2015 6:31 AM, John S wrote:
On 7/1/2015 4:23 PM, Jerry Stuckle wrote:
On 7/1/2015 12:26 PM, Jeff Liebermann wrote:
On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle
wrote:

Yes, it's most effective to match the feedline to the antenna at the
antenna connection. But it's also important to match the
transmitter to
the feedline.

This latter piece is often ignored because people will use a feedline
who's characteristic impedance matches the transmitter already (i.e. 50
ohm line for a 50 ohm transmitter).

However, there are exceptions. For instance, if you're feeding a 75
ohm
antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna
will provide a good match (ideally, 1:1). But there will be a 1.5:1
mismatch to a 50 ohm transmitter. In this case it would be better to
have the matching network at the transmitter.

We may have had this discussion before. Matching a 75 ohm load to a
50 ohm source might be academically interesting, but the actual loss
is almost negligible. for a VSWR of 1.5, the return loss is 14dB and
the load mismatch attenuation is 0.177dB. That's about what I would
expect to lose in two coax connector pairs.

You could also feed the antenna with 50 ohm feedline and place the
matching network at the antenna. The effect would still be a 1:1 SWR,
but the lower impedance of the coax would create higher i^2R losses;
not
important if you're talking a short line, but a longer one would lower
output at the antenna.

True, but for roughly equivalent sizes of coax cables, the 75 ohm
cable has less loss and the equivalent 50 ohm cable. If you want to
handle high power, use 50 ohms. If you want low loss, use 75 ohms:
http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm

Note that these are for air dielectric cables.

Things are not so neat if we consider the dielectric. See the bottom
paragraph and graphs:
http://www.microwaves101.com/encyclopedias/why-fifty-ohms
Dielectric Dielectric const Minimum loss impedance
solid PTFE 2.2 50 ohms
foam PTFE 1.43 60
air 1.0 75
RG-6/u CATV 75 ohm foam coax still has slightly less loss than the
equivalent 50 ohm cable, but not as much as I've previously claimed.

This is cute:
http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/

"A good rule of thumb is that if the device being connected
via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal."



Jeff, do you always miss the forest for the trees? That was an EXAMPLE.
The same would be true if you were feeding a 300 ohm yagi with 300 ohm
twinlead and a transmitter with a 10 ohm output impedance.

Transmitter output impedance does not determine SWR.


Transmitter output impedance vs. feedline impedance does determine SWR
at one end of the system. If you have a mismatch, you will have a
non-1:1 SWR.

And BTW - when calculating, you forgot about the transmitters which cut
back power to protect the finals. Many will do so even with a 1.5:1 SWR.




Not true. Post some links to support your position, please.

On 7/2/2015 11:07 AM, John S wrote:
On 7/2/2015 8:14 AM, Jerry Stuckle wrote:
On 7/2/2015 6:31 AM, John S wrote:
On 7/1/2015 4:23 PM, Jerry Stuckle wrote:
On 7/1/2015 12:26 PM, Jeff Liebermann wrote:
On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle
wrote:

Yes, it's most effective to match the feedline to the antenna at the
antenna connection. But it's also important to match the
transmitter to
the feedline.

This latter piece is often ignored because people will use a feedline
who's characteristic impedance matches the transmitter already
(i.e. 50
ohm line for a 50 ohm transmitter).

However, there are exceptions. For instance, if you're feeding a 75
ohm
antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna
will provide a good match (ideally, 1:1). But there will be a 1.5:1
mismatch to a 50 ohm transmitter. In this case it would be better to
have the matching network at the transmitter.

We may have had this discussion before. Matching a 75 ohm load to a
50 ohm source might be academically interesting, but the actual loss
is almost negligible. for a VSWR of 1.5, the return loss is 14dB and
the load mismatch attenuation is 0.177dB. That's about what I would
expect to lose in two coax connector pairs.

You could also feed the antenna with 50 ohm feedline and place the
matching network at the antenna. The effect would still be a 1:1
SWR,
but the lower impedance of the coax would create higher i^2R losses;
not
important if you're talking a short line, but a longer one would
lower
output at the antenna.

True, but for roughly equivalent sizes of coax cables, the 75 ohm
cable has less loss and the equivalent 50 ohm cable. If you want to
handle high power, use 50 ohms. If you want low loss, use 75 ohms:
http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm


Note that these are for air dielectric cables.

Things are not so neat if we consider the dielectric. See the bottom
paragraph and graphs:
http://www.microwaves101.com/encyclopedias/why-fifty-ohms
Dielectric Dielectric const Minimum loss impedance
solid PTFE 2.2 50 ohms
foam PTFE 1.43 60
air 1.0 75
RG-6/u CATV 75 ohm foam coax still has slightly less loss than the
equivalent 50 ohm cable, but not as much as I've previously claimed.

This is cute:
http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/


"A good rule of thumb is that if the device being connected
via coaxial cable is a receiver of some kind, 75 Ohm Coax is
ideal."



Jeff, do you always miss the forest for the trees? That was an
EXAMPLE.
The same would be true if you were feeding a 300 ohm yagi with
300 ohm
twinlead and a transmitter with a 10 ohm output impedance.

Transmitter output impedance does not determine SWR.


Transmitter output impedance vs. feedline impedance does determine SWR
at one end of the system. If you have a mismatch, you will have a
non-1:1 SWR.

And BTW - when calculating, you forgot about the transmitters which cut
back power to protect the finals. Many will do so even with a 1.5:1
SWR.




Not true. Post some links to support your position, please.

Basic physics that anyone with even an inkling of AC theory should
understand. Any time you have an impedance mismatch in a system - in
this case, whether at the antenna or the transmitter end of a feedline -
you will not have perfect power transfer. What is not transferred will
be reflected. This causes an SWR greater than 1:1.

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.

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

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

"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 7/2/2015 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.

On 7/2/2015 3:24 PM, Ralph Mowery wrote:
"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.


Incorrect. The real impedance of the transmitter is 50 ohms. The
impedance of the final stage may be above or below that, and is matched
to the 50 ohm standard. What happens before the match is unimportant.
The only important part of the discussion is the 50 ohm output.

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.



Incorrect. The connection between the 50 ohm transmitter and the 75 ohm
coax is also part of the antenna system. The system starts at the
transmitter output (actually the output of the final stage - but since
this is converted to the 50 ohm standard, you can effectively consider
the output of the matching network to be the start of the antenna
system), not the coax.

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

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