Brian Reay wrote:
wrote:
Brian Reay wrote:
On 06/07/15 01:21, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

That is true but is doesn't address the point. There should still be
somewhere to represent the source impedance, albeit normalised.

The purpose of a Smith chart it to match a SOURCE to a LOAD.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.

Likewise, that is a sweeping statement which evades the point.

The main purpose of EZNEC is to design an antenna for amateur radio use
and all commercial amateur radio transmitters have an output impedance
of 50 Ohms.


Neither of those responses address the points.

It is clear you cannot support your assertions.

EZNEC can set the source impedance to any value one desires, but the
default is 50 Ohms as most people are interested in 50 Ohm systems
as the majority of transmitters are designed for 50 Ohm loads.

EZNEC calculates the SWR for an antenna presented to the SOURCE
in the model.

For most simulations, the SOURCE is placed at the antenna terminals,
which represents what an attached transmission line will see.

The transmission line most commonly used is 50 Ohm coax.

EZNEC will also allow you to put a SOURCE at one end of a transmission
line with the other end of the line at the antenna terminals.

In this simulation, EZNEC simulates what the transmitter would see if
it were connected to the transmission line/antenna system.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

SWR = (1 + |r|)/(1 - |r|)

Where r = reflection coefficient and

r = (Zl - Zo)/(Zl + Zo)

Where Zl = complex load impedance, Zo = complex source impedance.


--
Jim Pennino

John S wrote:
On 7/7/2015 1:58 PM, wrote:
John S wrote:
On 7/5/2015 7:21 PM, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.


The EZNEC help file is very comprehensive. Please find any reference to
your assertion that there is an assumption of source impedance there and
provide information for us to verify your assertion.

Why don't you email the author and get his take on your assumptions?



Why don't YOU? You are the one in need of knowledge. If I do it and
report back here you will just doubt it or find something else to argue
about. Better you should do it first-hand.


EZNEC calculates the SWR presented to the SOURCE which is usually
placed at the antenna terminals.

EZNEC also calculates the SWR presented to the SOURCE which can be
modeled as a SOURCE at one end of a transmission line and the antenna
at the other end.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

I am tired of typing in the same equations over and over again.



--
Jim Pennino

On 7/8/2015 6:09 AM, John S wrote:
On 7/7/2015 10:05 AM, Jerry Stuckle wrote:
On 7/7/2015 3:05 AM, John S wrote:
On 7/6/2015 1:03 PM, wrote:
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some
numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into
the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power
into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not
the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out
of a
power (energy?) source, the source having an internal resistance
is a
BAD THING. You don't design the source to have an internal
resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available
from the
source when the load resistance equals the source resistance, you
can
only do so provided that the heat you generate in the source does
not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?

Yes, quite a lot, you get a whole new set of laws.

If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you
apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of
current. How does the AC change the law?


You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps
flowing (open circuit).
You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing,
with the current leading the voltage by 90 degrees.

You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps
flowing (short circuit).
You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing
with the voltage leading the current by 90 degrees.

You place the capacitor and inductor in series.
Fed with DC, you get 0 amps flowing (open circuit).
Fed with 1MHz AC, you get infinite current flowing (short circuit).

You place the capacitor and inductor in parallel.
Fed with DC, you get infinite current flowing (short circuit).
Fed with 1MHz AC you get 0 amps flowing (open circuit).

There is a huge difference between ac and dc!


Yes, but the LAWS have not changed. The components have changed. So,
changing the components changes the laws of physics?

Suppose you apply .01Hz AC RMS to the components you specified. What then?

The point is - the rules for AC are different than the rules for DC.

I'm not going to waste my time figuring out the calculations - you can
do that. But the bottom line will be there will be some impedance in
every case. It will be neither zero nor infinity, as it would be in a
DC circuit.

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

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

In message , John S
writes
On 7/8/2015 6:32 AM, Jeff wrote:


So, at 1Hz the law has changed, eh? What new law do I need to use?


The laws are exactly the same for dc as ac, just at dc the frequency
dependant parts tend to the point that they are of no consequence.

Jeff

Thanks, Jeff. I agree, of course.

The thinking that the laws of physics suddenly change when AC is so slow
that it suddenly becomes DC is probably about as sound as my
tongue-in-cheek suggestion that a SWR suddenly ceases to exist if there
is no reflection. ;o)
--
Ian

On 08/07/15 19:36, wrote:
Brian Reay wrote:
wrote:
Brian Reay wrote:
On 06/07/15 01:21, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

That is true but is doesn't address the point. There should still be
somewhere to represent the source impedance, albeit normalised.

The purpose of a Smith chart it to match a SOURCE to a LOAD.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.

Likewise, that is a sweeping statement which evades the point.

The main purpose of EZNEC is to design an antenna for amateur radio use
and all commercial amateur radio transmitters have an output impedance
of 50 Ohms.


Neither of those responses address the points.

It is clear you cannot support your assertions.

EZNEC can set the source impedance to any value one desires, but the
default is 50 Ohms as most people are interested in 50 Ohm systems
as the majority of transmitters are designed for 50 Ohm loads.

EZNEC calculates the SWR for an antenna presented to the SOURCE
in the model.

For most simulations, the SOURCE is placed at the antenna terminals,
which represents what an attached transmission line will see.

The transmission line most commonly used is 50 Ohm coax.

EZNEC will also allow you to put a SOURCE at one end of a transmission
line with the other end of the line at the antenna terminals.

In this simulation, EZNEC simulates what the transmitter would see if
it were connected to the transmission line/antenna system.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

SWR = (1 + |r|)/(1 - |r|)

Where r = reflection coefficient and

r = (Zl - Zo)/(Zl + Zo)

Where Zl = complex load impedance, Zo = complex source impedance.



The matter at hand isn't SWR it is the output impedance of PAs.

That seems to be you sticking point. You are assuming the PA is a
transmission line, rather than an active source.

You confusion isn't helped by the habit of some manufacturers including
SWR in the PA spec.

Modern PAs are designed to drive a load of 50 ohms, they don't have a
source impedance of 50 ohms. If they are driven into the wrong load,
they can operated outside there safe area of operation. If the power
isn't reduced, they can be damaged.

On 7/8/2015 1:39 PM, wrote:
John S wrote:
On 7/7/2015 1:58 PM, wrote:
John S wrote:
On 7/5/2015 7:21 PM, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.


The EZNEC help file is very comprehensive. Please find any reference to
your assertion that there is an assumption of source impedance there and
provide information for us to verify your assertion.

Why don't you email the author and get his take on your assumptions?



Why don't YOU? You are the one in need of knowledge. If I do it and
report back here you will just doubt it or find something else to argue
about. Better you should do it first-hand.


EZNEC calculates the SWR presented to the SOURCE which is usually
placed at the antenna terminals.

EZNEC also calculates the SWR presented to the SOURCE which can be
modeled as a SOURCE at one end of a transmission line and the antenna
at the other end.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

No. It is defined as Vmax/Vmin on the line. Show an equation that
defines SWR as the matching of the source to the line.

I am tired of typing in the same equations over and over again.

Then you are excused from participating.

On 7/8/2015 1:18 PM, wrote:
John S wrote:
On 7/7/2015 1:52 PM, wrote:
Brian Reay wrote:

Do the experiment.

Did it decades ago in electromagnetics lab with calibrated test equipmemnt,
not with amateur radio equipment.

Post the original lab notes, please. That way we cannot challenge the
accuracy of your memory.

Sorry, that was decades ago.

If you are so convinced, do the experiments yourself and post the results.

Or you could read an electromagnetics text on transmission lines and
show me the errors of my statements.

I did, decades ago. The results are that you are wrong. You surely trust
my memory as well as I trust yours, yes?

Brian Reay wrote:
On 08/07/15 19:36, wrote:
Brian Reay wrote:
wrote:
Brian Reay wrote:
On 06/07/15 01:21, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

That is true but is doesn't address the point. There should still be
somewhere to represent the source impedance, albeit normalised.

The purpose of a Smith chart it to match a SOURCE to a LOAD.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.

Likewise, that is a sweeping statement which evades the point.

The main purpose of EZNEC is to design an antenna for amateur radio use
and all commercial amateur radio transmitters have an output impedance
of 50 Ohms.


Neither of those responses address the points.

It is clear you cannot support your assertions.

EZNEC can set the source impedance to any value one desires, but the
default is 50 Ohms as most people are interested in 50 Ohm systems
as the majority of transmitters are designed for 50 Ohm loads.

EZNEC calculates the SWR for an antenna presented to the SOURCE
in the model.

For most simulations, the SOURCE is placed at the antenna terminals,
which represents what an attached transmission line will see.

The transmission line most commonly used is 50 Ohm coax.

EZNEC will also allow you to put a SOURCE at one end of a transmission
line with the other end of the line at the antenna terminals.

In this simulation, EZNEC simulates what the transmitter would see if
it were connected to the transmission line/antenna system.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

SWR = (1 + |r|)/(1 - |r|)

Where r = reflection coefficient and

r = (Zl - Zo)/(Zl + Zo)

Where Zl = complex load impedance, Zo = complex source impedance.



The matter at hand isn't SWR it is the output impedance of PAs.

Nope, the matter at hand is the definition of SWR.


That seems to be you sticking point. You are assuming the PA is a
transmission line, rather than an active source.

Nope, the matter at hand is the definition of SWR, which is defined
in terms of SOURCE impedance and LOAD impedance.

It does not matter in the slightest if the SOURCE impedance is the
output of a transmitter or the end of a transmission line.

You confusion isn't helped by the habit of some manufacturers including
SWR in the PA spec.

Manufacturers specify the LOAD impedance for the transmitter, and
sometimes the SWR range that the transmitter will handle.

Modern PAs are designed to drive a load of 50 ohms, they don't have a
source impedance of 50 ohms. If they are driven into the wrong load,
they can operated outside there safe area of operation. If the power
isn't reduced, they can be damaged.

Once again, SWR is defined in terms of SOURCE impedance and LOAD
impedance. The normal LOAD for a transmitter is one end of a piece
of coax with an antenna on the other end.

The SWR at the near end of a piece of coax may or may not be the
same as the SWR at the far end of the coax.


--
Jim Pennino

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 12:47 PM, wrote:
John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.

So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and
DC end?