"Roger Hayter" wrote in message
...
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!)

For a transmitter I wold think you start with the tube or transistor and
decide on the power level. From there you start designing the matching
network to go either from the thousand ohm range for a tube or the below
say 10 ohm range for a solid state device depending on the supply voltage
range.

Seems that people are mixing in Norton and Thevenin circuits to explain what
is going on , which is not this case.

As I stated before you are not burning up half the power getting a match in
that equvilent circuit with the series resistor. If this were the case, it
would take a lot more than a DC input of 2000 watts to get 1200 watts out of
the amplifier.

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.





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.



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.




--
Roger Hayter

wrote:

Roger Hayter wrote:
wrote:

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.

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.



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.

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.


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!

--
Roger Hayter

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

snip

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'.

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 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.

You are mixing circuit theory with transmission line theory.

Designed to operate with a low load SWR means the output impedance is
designed to be about 50 Ohms, i.e. commonaly available coax.


--
Jim Pennino

If there is incident energy upon a correctly matched transmitter's output,
it need not absorb 100% of the energy.

If the transmitter is an ideal, linear, Thevenin/Norton source, it must, as
is consistent with linearity, and the power transfer theorem, and
transmission line theory, and all that.

But, it is quite common for transmitters to reflect incident power. This
characteristic is captured by the scattering parameters, namely s22, the
output reflectance. (Well, that would be Gamma_22, but they're equivalent
parameters in the end.)

This corresponds, in turn, with the dynamic output impedance, which you may
recall from audio amplifiers, need not equal the "best load" impedance. For
instance, an amplifier for 8 ohm loads might have a dynamic impedance of
0.03 ohms, a very good voltage source in comparison -- acting like a short
circuit to incident energy (and therefore reflecting energy, out of phase,
back towards the loudspeaker, thus giving it a high "damping factor").

The same is true of RF amplifiers, except rather than constant voltage
characteristics due to the use of voltage negative feedback, the
characteristic is usually constant current (high impedance), due to the lack
of negative feedback (or the use of current feedback) and the high intrinsic
impedance of the devices (i.e., the collector / drain / plate resistance).

As the energy "piles up on" and reflects off the transmitter, internal
voltages and currents may get into dangerous ranges, causing excessive
dissipation or breakdown; this is partly why transmitters shouldn't be
operated with high SWR (the other part being, efficiency and power capacity
suck).

The only amplifiers that have a relatively matched intrinsic output
imepdance are vacuum tube triode amplifiers. Though even these tend to have
a poor match, either being operated in class AB with R_L Rp, or class B/C
with R_L Rp (and lots of grid current to push plate current up there).
Because, again, nonlinear devices don't need to obey the linearity theorems,
and can have impedances different from the "best load" value.

Tim

--
Seven Transistor Labs
Electrical Engineering Consultation
Website: http://seventransistorlabs.com

"rickman" wrote in message
...
On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
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."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various qualifications
for a simplification apply. But to do that, the qualifiers have to be
fully understood and no one here is showing what the qualifiers are much
less that they are met. So until we get a real explanation I will stick
with what I recall. In the end, to settle this we may have to use the
math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe the
description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

--

Rick

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

Roger Hayter wrote:
wrote:

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.

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.



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.

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.


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!


No, only YOU agree that only half the RF generated reaches the load. A
transmitter with an output impedance of 25 ohms feeding a 50 ohm antenna
system would exhibit a 2:1 SWR, with associated power loss.

If what you say were true, the transmitter could have a 1 ohm output
impedance and feed virtually any antenna equally well. This is
demonstrably false. An impedance mismatch in an antenna system is an
impedance mismatch, no matter where it occurs in the system.

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

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

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?



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.

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

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

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.

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.

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.


--
Jim Pennino

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.






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.)

--
Roger Hayter

wrote:

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

snip

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'.

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.)

The transmitter actually applies a mis-match to signals coming from the
antenna, but this does not affect the SWR as seen from the transmitter
end.




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.

You are mixing circuit theory with transmission line theory.

Designed to operate with a low load SWR means the output impedance is
designed to be about 50 Ohms, i.e. commonaly available coax.


--
Roger Hayter