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.

As we all agree, under equal output impedance and load impedance
conditions, onty half the RF generated reaches the load. This is sim;ly
not acceptable or likely for any real-world transmitter. Do 50kW radio
station output valves dissipate 50kW? I hope not!

You are attempting to mix circuit theory and transmission line theory.


You rather have to if you are going to connect a practical circuit to a
practical transmission line!



The "valves" in a transmitter are not connected to the transmission
line. The "valves" in a transmitter are a voltage source connected
to an impedance matching network which then connects to a transmission
line.

Fair enough. Do you want to dissipate 50kW in the matching circuit




A 50kW radio station does not generate 50kW of power, it generates
a voltage that results in 50kW being dissipted into a 50 Ohm load.

There is a difference.

Not much, seeing it also has to supply the in-phase current to maintain
that voltage across the resistive 50 ohm load.







--
Roger Hayter

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


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

Roger Hayter wrote:
wrote:

Brian Reay wrote:
On 05/07/2015 21:17, wrote:
Roger Hayter wrote:

snip

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'.

Nope. If they didn't have a 50 Ohm source impedance, the SWR with
50 Ohm coax and a 50 Ohm antenna would be high. It is not.


The SWR looking into the cable from the transmitter is unaffected by the
source impedance. Indeed, it is exactly the same if the transmitter is
not connected (though you have to connect some kind of generator in
order to measure it, it matters little what kind it is.)

If the input end is not connected, SWR is meaningless.

SWR bridges are calibrated to the source impedance, not the load.


--
Jim Pennino

On Sun, 5 Jul 2015 20:58:58 +0100, (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.

I previously posted how to measure the output impedance of an RF power
amplifier. Methinks this would be a good time to repeat it...

It might be interesting to measure the output impedance of your HF
xmitter. All you need is a dummy load, and an RF voltmeter, RF probe
and voltmeter, or oscilloscope.
1. Turn down the xmitter RF output to some level where you won't blow
up your test equipment and so that it doesn't go into high VSWR
protect mode. My guess is about 10 watts is about right.
2. Measure the RF voltage across the output connector both with a
load (Vload) and without a load (Vno_load).
3. If measuring peak voltage, convert RMS by multiplying by 0.707. If
measuring peak-to-peak, divide by 2 and then multiply by 0.707.

Output_Impedance = 50 ohms (Vno_load - Vload) / Vload

Easy enough. Just don't let the RF output level or VSWR get high
enough to trigger the ALC or VSWR protection. I haven't done this in
a long time, and don't recall exactly what this simple measurement
produced on radios that I've designed. I do know that the output of
my (marine radio HF) PA's always went to a low pass filter, which was
designed to be terminated by 50 ohm in/out. If the PA didn't present
50 ohms to the filter input, the filter response would not be flat,
and there would be additional losses. Since that was not the case, I
can deduce that the amplifier output was at least close to 50 ohms.

Anyone wanna make the measurement? My bench is a currently a disaster
area so I can't do it without a major cleanup.




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

Roger Hayter wrote:
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.

As we all agree, under equal output impedance and load impedance
conditions, onty half the RF generated reaches the load. This is sim;ly
not acceptable or likely for any real-world transmitter. Do 50kW radio
station output valves dissipate 50kW? I hope not!

You are attempting to mix circuit theory and transmission line theory.


You rather have to if you are going to connect a practical circuit to a
practical transmission line!

Nope.

In the real world you pick a system impedance to match the transmission
line you want to use.

You design the transmitter with circuit theory to match the line.

You design the antenna with electromagetic theory to also match the line.

The "valves" in a transmitter are not connected to the transmission
line. The "valves" in a transmitter are a voltage source connected
to an impedance matching network which then connects to a transmission
line.

Fair enough. Do you want to dissipate 50kW in the matching circuit

What anyone wants is irrelevant to physics.

A 50kW radio station does not generate 50kW of power, it generates
a voltage that results in 50kW being dissipted into a 50 Ohm load.

There is a difference.

Not much, seeing it also has to supply the in-phase current to maintain
that voltage across the resistive 50 ohm load.

The phase of a transmitter is what it is and this is a red herring.


--
Jim Pennino

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

I've done some admittedly crude testing of various matching
contrivances by measuring the resultant field strength for a given RF
power level (measured at the antenna connector). Although not
conclusive or spectacular, the early model automagic antenna tuner was
rather lossy.

Incidentally, I contrived a rather crude but effective way to measure
relative overall efficiency. I measured the power consumption from
the AC line with a Kill-a-Watt meter (in watts, not VA) and adjusted
the CW RF output for some reference level on the field strength meter.
While this would not give me a real number for the efficiency, it does
produce relative numbers for comparing antenna matching devices.
Unfortunately, I can't find my results, but I do recall that the
winner was a simple 4:1 torroidal matching xformer.

Remember, you really want a low SWR for two reasons, one because modern
radios demand it but also to reduce coax (or feeder) loss. With an
matcher at the antenna feed point, coax losses are minimised. An ATU at
the TX end does nothing to reduce coax losses in real terms.

Coax losses below 500 MHz are mostly in the I^2R losses of the copper
(as limited by skin effect). Above 500MHz, the dielectric gets
involved. See Fig 4:
http://www.maximintegrated.com/en/app-notes/index.mvp/id/4303
Higher RF currents, caused by low impedance terminations, will cause
higher I^2R losses. These higher losses are why very low impedances
are not popular for RF power devices. This has NOTHING to do with
matching. For a given RF current through the coax, the contribution
of the coax to overall losses will be independent of the VSWR.



--
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 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.

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.

--
==================
Remove the "x" from my email address
Jerry Stuckle

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

On 7/5/2015 6:33 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/5/2015 5:37 PM, Roger Hayter wrote:
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.


Because it doesn't have a "50 ohm output" it has an output designed for
a 50 ohm load.


No, it has a 50 ohm output - as did the transmitter I designed back in
my EE class days. But if what you say is correct, how is it designed
for 50 ohms? If you claim it as a low output impedance, then it should
work equally well on a 75 ohm antenna, or even a 1,000 ohm antenna, as
long as the feedline matches. I can assure you that is NOT the case.


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?

See comment above. If you look at the spec. it probably will not
specify the output impedance, just the load impedance.


See comment above. What makes the load impedance special? Why
shouldn't it work with any sufficiently high load impedance - in fact,
the higher, the better?

What makes the load resistance special is that the voltage output of the
transmitter will drive the correct load resistance with just the right
amount of current to provide the design output power without dissipating
too much heat. Too high a load resistance may simply not take enough
power, but also may upset the operating conditions of the PA in
exact-ciruit dependent ways. Too low a load resistance will draw too
much current and overheat the amplifier. The design has absolutely
nothing to do with making the output impedance equal to the load
resistance.


So, what happens when I cut my 100W transmitter down to 10W? The
voltage output is different - but the impedance doesn't change.

You really need to learn transmitter design and impedance matching.






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.

Maximum power transfer for a given voltage generator, not maximum power
transfer for a given dissipation available.

You know electronics. Just do two simple circuits on the back of an
envolope, or cigarette pack if available. A voltage generator in
series with a 5 ohm internal resistance and a 50 ohm load. And another
one with 50 ohm internal resistance and a 50ohm load. Make the output
power 100 W, that is an RMS voltage of 70 volts across the load. Now
calculate the voltage of the generator, and the total power produced,
for both cases.


So if what you say is true, I should be able to feed a 300 ohm antenna
through 300 ohm feedline and a 1:1 balun with no ill effects.

You may think you know electronics - but you do not understand
transmission theory. I don't need to draw circuits on a cigarette pack
- all I need to do is hook up my wattmeter to my transmitter to prove
you wrong.

The output impedance of a practical RF power amplifier has exactly zero
to do with transmission line theory. (The *effect* of a practical PA
output impedance on the transmission line is where we came in, but that
only arises *after* we've sorted out the PA output impedance.)


It has everything to do with matching the output of the transmitter to
the transmission line and load. But you've already shown you don't
understand that part.

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

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

"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.