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).
--
Ian

On 7/6/2015 12:19 AM, 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.


So, it can't be used in a 50 ohm environment? What does that have to do
with anything? The chart has a SWR graph and nowhere does it need source
impedance. If you disagree, please link to one.


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

Please show the EZNEC statement that "assumes the transmission line
matches the transmitter". Look in the help section if you have EZNEC and
can cut and paste or just refer me to the chapter and verse. Also, if
you have EZNEC, you can insert a transmission line with arbitrary
characteristic impedance, put a load on the far end matching the line,
and look at the SWR. It will still be 1:1 because the LOAD matches the
LINE. Not because EZNEC assumes a source impedance. Try it with and
report back here.

There is no way that a source initiates reflections. That is a property
of the line and load only. It may re-reflect a wave reflected from the
load, but that is all.

You can also verify this in LTSPICE if you wish.

What happens if you take any off the shelf commercial amateur radio
transmitter that does not have a built in tuner and:

Attach a 10 Ohm load.

Attach a 200 Ohm load.

Attach a 1,000 Ohm load.

Attach a 1 Ohm load.

Attach a 50 Ohm load.



Please address my questions first before setting up another strawman.

"Jeff Liebermann" wrote in message
...

On Sun, 5 Jul 2015 17:08:56 -0700, "Wayne"
wrote:
I'm off on a different approach.
I have an RF ammeter mounted in a box. The box is in the shack between
the
ATU and the antenna.
I simply adjust the ATU for max current on the ammeter.

Rewind. I just noticed that you're planning to put the RF ammeter
between the ATU and the antenna. That will work, but with
approximately 1200 ohms antenna impedance, you are going to see
50/1200 = 0.04 times the antenna current that you would see on the 50
ohm line between the xmitter and the ATU. If you're running lots of
power, that might work, but offhand, methinks not.

Also, you can't adjust the ATU for maximum current. It adjusts itself
based on it's own internal VSWR sensor. All you can do is watch the
light show and listen to the relays clatter. You might be able to
have some control if it were a motorized antenna tuner. Certainly a
manual antenna tuner would work.

This isn't something planned. I have been using the ammeter between the
tuner and antenna for many years.

With an automatic tuner, there is no feedback from the ammeter to the tuner.
In that case it is simply an indicator that current is present.

Some of my antennas have SWR outside the capabilities of my automatic tuner.
A manual tuner is used in that case.

With a manual tuner, I don't look for a specific reading, just a peak in the
current from the tuner to the antenna.

On 7/6/2015 12:39 AM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,

writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter
side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a
perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place
for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

If the termination matches the line impedance, there is no
reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.

Being essentially a simple soul, that's how I sometimes try to work
out
what happening.

You'll be better off if you killfile the troll. You'll get a lot less
bad information and your life will become much easier.


The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.

I've always understood that the resistive part of a TX output
impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have thought
that
the efficiency of the output stage could never exceed 50% (and aren't
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can vary,
but that is due to normal variances in components, and the difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything, i.e.
high with tubes but low with solid state. But the output impedance can
be converted to 50 ohms or any other reasonable impedance through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of course,
nothing is perfect, so there will be some loss. But the amount of loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not like
having two resistors in a circuit where each will dissipate 1/2 of the
power.

BTW - the resistive part of the impedance is not the same as
resistance.
For a simple case - take a series circuit of a capacitor, an inductor
and a 50 ohm resistor. At the resonant frequency, the impedance will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a
simple
example, but it shows a point.

The resistive part of the impedance is exactly the same as a resistance
as far as the frequency you are using is concerned. And if the
amplifier output impedance *and* the feeder input resistance were *both*
matched to 50 ohms resistive then 50% of the power generated (after
circuit losses due of inefficiency of generation) would be dissipated in
the transmitter. It would, at the working frequency, be *exactly* like
having two equal resistors in the circuit each taking half the generated
power. So the amplifier has a much lower output impedance than 50ohms
and no attempt is made to match it to 50 ohms.


OK, so then please explain how I can have a Class C amplifier with 1KW
DC input and a 50 ohm output, 50 ohm coax and a matching network at the
antenna can show 900 watt (actually about 870 watts due to feedline
loss)? According to your statement, that is impossible. I should not
be able get more than 450W or so at the antenna.

It is true that if the transmitter is thought of as a fixed voltage
generator in series with, say, a 5 ohm resistor then the maximum power
transfer in theory would occur with a 5 ohm load. But to achieve this
the output power would have to be 100 times higher (ten times the
voltage) and half of it would be dissipated in the PA. Not the best way
to run things, it is better to have a voltage generator chosen to give
the right power with the load being much bigger than its generator
resistance.


So why do all fixed-tuning amateur transmitters have a nominal 50 ohm
output instead of 1 or two ohms? And why do commercial radio stations
spend tens of thousands of dollars ensuring impedance is matched
throughout the system?

The maximum power transfer at equal impedance theorem only applies if
you started with a *fixed* output voltage generator. We don't; we
start with a load impedance (50 ohm resistive), then we decide what
power output we want, and we choose the voltage to be generated
accordingly. (Thank you for giving me the opportunity to think about
this!)


Actually, it doesn't matter if it's a fixed or a variable output voltage
- maximum power transfer always occurs when there is an impedance match.

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.


Your simplified response shows you never took an EE course in your life.

I suggest you take some EE courses and learn how things work. It's very
obvious you don't have any more knowledge than you get with ohm's law.

But if what you say is correct, then I should be able to get a lot of
power out of my 100 watt transmitter feeding a 1 ohm antenna. Never
mind the 50:1 SWR.

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

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

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.

If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.

That is demonstrably false.

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

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

On 7/6/2015 11:59 AM, Jerry Stuckle wrote:
On 7/6/2015 12:39 AM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

It is true that if the transmitter is thought of as a fixed voltage
generator in series with, say, a 5 ohm resistor then the maximum power
transfer in theory would occur with a 5 ohm load. But to achieve this
the output power would have to be 100 times higher (ten times the
voltage) and half of it would be dissipated in the PA. Not the best way
to run things, it is better to have a voltage generator chosen to give
the right power with the load being much bigger than its generator
resistance.


So why do all fixed-tuning amateur transmitters have a nominal 50 ohm
output instead of 1 or two ohms? And why do commercial radio stations
spend tens of thousands of dollars ensuring impedance is matched
throughout the system?

The maximum power transfer at equal impedance theorem only applies if
you started with a *fixed* output voltage generator. We don't; we
start with a load impedance (50 ohm resistive), then we decide what
power output we want, and we choose the voltage to be generated
accordingly. (Thank you for giving me the opportunity to think about
this!)


Actually, it doesn't matter if it's a fixed or a variable output voltage
- maximum power transfer always occurs when there is an impedance match.

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.


Your simplified response shows you never took an EE course in your life.

I suggest you take some EE courses and learn how things work. It's very
obvious you don't have any more knowledge than you get with ohm's law.

But if what you say is correct, then I should be able to get a lot of
power out of my 100 watt transmitter feeding a 1 ohm antenna. Never
mind the 50:1 SWR.

Would anyone else like to explain to Jerry the fallacy of his argument?
I get tired of explaining the obvious sometimes.

--

Rick

On 7/5/2015 7:08 PM, Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote:

As for a simpler way, I'd recommend a remote auto-matcher like an SGC at
the antenna base. It will minimise coax losses and should give you a
good match, at least for most bands. I've used a similar set up (with
radials) and achieved a good match even on 80m.

If your radio has a built in tuner, then it can be used to 'tweak' the
match in the event the radio isn't 'seeing' 1.5:1. Turn it off
initially. Let the SGC find a match. If it isn't ideal, use the local
ATU for a final tweak. I never found this was required but YMMV.

Not everyone is a true believer in antenna tuners:
http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html


Interesting.
I'm off on a different approach.

I have an RF ammeter mounted in a box. The box is in the shack between
the ATU and the antenna.

I simply adjust the ATU for max current on the ammeter.

Hey, Wayne -

As a matter of curiosity on my part, can you find a way to measure the
ammeter's resistance and let me know the full-scale value?

Many thanks,
John

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?


If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.


There is no VSWR nor ISWR if the load matches the line.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.

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?

Because RF transmitters deliver high frequency AC to a transmission line.


--
Jim Pennino

John S wrote:
On 7/6/2015 12:19 AM, 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.


So, it can't be used in a 50 ohm environment? What does that have to do
with anything? The chart has a SWR graph and nowhere does it need source
impedance. If you disagree, please link to one.


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

Please show the EZNEC statement that "assumes the transmission line
matches the transmitter". Look in the help section if you have EZNEC and
can cut and paste or just refer me to the chapter and verse. Also, if
you have EZNEC, you can insert a transmission line with arbitrary
characteristic impedance, put a load on the far end matching the line,
and look at the SWR. It will still be 1:1 because the LOAD matches the
LINE. Not because EZNEC assumes a source impedance. Try it with and
report back here.

There is no way that a source initiates reflections. That is a property
of the line and load only. It may re-reflect a wave reflected from the
load, but that is all.

You can also verify this in LTSPICE if you wish.

What happens if you take any off the shelf commercial amateur radio
transmitter that does not have a built in tuner and:

Attach a 10 Ohm load.

Attach a 200 Ohm load.

Attach a 1,000 Ohm load.

Attach a 1 Ohm load.

Attach a 50 Ohm load.



Please address my questions first before setting up another strawman.

Start with Electromagnetics by Kraus and Carver, Chapter 13.


--
Jim Pennino