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.


--
Jim Pennino

On 05/07/2015 21:17, 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.

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.

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

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

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 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

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

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

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

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

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