Jerry Stuckle wrote:
On 7/5/2015 6:48 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle
wrote:

So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually
any EE degree, and anyone claiming an EE degree should be able to tell why.

Actually, there are low output impedance RF power amps. They're quite
common in NMR power amps to reduce coupling between stages:

Ultra-low output impedance RF power amplifier for parallel excitation
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/

Ultra-low Output Impedance RF Power Amplifier Array
http://cds.ismrm.org/ismrm-2007/files/00172.pdf



We're not talking amateur transmitters, troll. Try to stay on topic -
if that's at all possible.

What kind of transmitters should be talk about in an amateur group?

Are amateur transmitters different than any other kind of transmitter?


--
Jim Pennino

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.

--
Jim Pennino

On Sun, 05 Jul 2015 20:05:17 -0400, Jerry Stuckle
wrote:

On 7/5/2015 6:48 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle
wrote:

So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually
any EE degree, and anyone claiming an EE degree should be able to tell why.

Actually, there are low output impedance RF power amps. They're quite
common in NMR power amps to reduce coupling between stages:

Ultra-low output impedance RF power amplifier for parallel excitation
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/

Ultra-low Output Impedance RF Power Amplifier Array
http://cds.ismrm.org/ismrm-2007/files/00172.pdf

We're not talking amateur transmitters, troll. Try to stay on topic -
if that's at all possible.

NMR is Nuclear Magnetic Resonance. To the best of my knowledge,
medical imaging in not performed by amateurs. I realize that the
translation from Chinese to English is somewhat lacking in the above
articles. I think if you make the effort, you might learn something
about low output impedance power amplifiers, which seems to be a topic
drift that *YOU* started with the above "1 ohm" question. Since the
original question of the purpose of the matching transformer on the 43
ft vertical has been correctly answered at least 3 times, I would
guess that some topic drift might be acceptable. Do continue.

Since the original question is addressed to Rick, I won't ruin your
fun by providing an answer. No need to thank me.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

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.

On 7/5/2015 8:23 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 20:05:17 -0400, Jerry Stuckle
wrote:

On 7/5/2015 6:48 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle
wrote:

So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually
any EE degree, and anyone claiming an EE degree should be able to tell why.

Actually, there are low output impedance RF power amps. They're quite
common in NMR power amps to reduce coupling between stages:

Ultra-low output impedance RF power amplifier for parallel excitation
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/

Ultra-low Output Impedance RF Power Amplifier Array
http://cds.ismrm.org/ismrm-2007/files/00172.pdf

We're not talking amateur transmitters, troll. Try to stay on topic -
if that's at all possible.

NMR is Nuclear Magnetic Resonance. To the best of my knowledge,
medical imaging in not performed by amateurs. I realize that the
translation from Chinese to English is somewhat lacking in the above
articles. I think if you make the effort, you might learn something
about low output impedance power amplifiers, which seems to be a topic
drift that *YOU* started with the above "1 ohm" question. Since the
original question of the purpose of the matching transformer on the 43
ft vertical has been correctly answered at least 3 times, I would
guess that some topic drift might be acceptable. Do continue.

Since the original question is addressed to Rick, I won't ruin your
fun by providing an answer. No need to thank me.


We're not talking about NMR. We're discussing amateur ratio.

Try to stay on target, troll.

And no, I didn't start ANY topic on output impedance drift. But you're
too stoopid to understand the topic at hand, so I can see how you can
come to that conclusion.

Fortunately, intelligent people understand the thread and the fact I
didn't start the topic on impedance drift. The fact you think I did
proves your stoopidity.

As well as the fact that amateurs have anything to do with NMR.

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

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

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.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

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.

--

Rick

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

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.

In case anyone wants to duplicate the simulation...

Version 4
SHEET 1 880 680
WIRE 48 48 0 48
WIRE 144 48 48 48
WIRE 304 48 224 48
WIRE 336 48 304 48
WIRE 0 80 0 48
WIRE 336 80 336 48
WIRE 0 192 0 160
WIRE 336 192 336 160
WIRE 48 304 0 304
WIRE 144 304 48 304
WIRE 304 304 224 304
WIRE 336 304 304 304
WIRE 0 336 0 304
WIRE 336 336 336 304
WIRE 0 448 0 416
WIRE 336 448 336 416
FLAG 0 192 0
FLAG 336 192 0
FLAG 48 48 V1
FLAG 304 48 Vload1
FLAG 0 448 0
FLAG 336 448 0
FLAG 48 304 V2
FLAG 304 304 Vload2
SYMBOL voltage 0 64 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 200 100K)
SYMBOL res 320 64 R0
SYMATTR InstName RL1
SYMATTR Value 50
SYMBOL res 128 64 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName RS1
SYMATTR Value 50
SYMBOL voltage 0 320 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 200 100K)
SYMBOL res 320 320 R0
SYMATTR InstName RL2
SYMATTR Value 50
SYMBOL res 128 320 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName RS2
SYMATTR Value 1
TEXT -34 472 Left 2 !.tran 10us


--

Rick

On 7/5/2015 5:19 PM, Brian Reay wrote:
On 05/07/2015 21:17, wrote:
Roger Hayter wrote:

The impedance of a transmitter output will be nothing like 50 ohms
resistive, as this would result in an efficiency well below 50%, with
all the normal amplfier losses plus the actual RF power produced being
50% dissipated in the PA. This is why matching in the forward direction
coexists with a mjaor mismatch in the reverse direction. This is good
because if there is any reflected wave we don't want it to add yet more
to the PA dissipation. But it does explain what is happening, and why
there are increased losses in the feeder as well as the matching
networks.

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.

They are designed to drive into a 50 ohm load, that doesn't mean they
have a 50 ohm source impedance. Otherwise efficiency would be rather
'disappointing'.

The PA stages are designed to operate safely with a load equivalent to a
SWR of (typically) 1.5:1 . Any higher, and it means the load is out of
spec, and the PA leaves its safe area of operation (assuming there is no
mechanism to reduce the power). This is were the myth of RF 'entering'
the PA came from - people thinking that a high SWR meant the reflected
RF was getting into the PA and causing damage. In fact, it 'sees' a
mismatch and therefore can't enter the PA.

I designed a line driver circuit once that used "synthetic impedance" to
simulate a 50 ohm output impedance with just a 12.5 ohm resistor.
Basically it uses positive feedback to sense the load current through
the output resistor and manage the drive to appear to be a higher
voltage source with a higher output impedance while only dissipating the
power of the smaller resistor as well as utilizing a lower power supply
voltage.

As to your mismatch theory, a mismatch doesn't mean *no* power enters
the amp from the feed line. It just means not all the power will be
transferred into the amp.

--

Rick

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.


--
Jim Pennino