The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9

Kewl, then I'll just run a tap directly to the 5000 ohm plates and start
a long chat up ... what the heck is all those pi matching components in
the way of the rf? Probably some loss there! ROFLOL!!!

JS

It appears that you agree with that part of my post but you are
drawing
an invalid conculsion from it. I never suggested that the passive
matching
network usually found in a transmitter output is unnecessary.

Bob

"Bob" wrote in message
ups.com...
On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.
The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9


In that case...

Half power is only lost when terminated to a 50-ohm load;
i.e. no standing waves. What happens when there's a
mismatch and reflected energy



I'll go stand in a corner...

"Uncle Peter" wrote in message
...

"Bob" wrote in message
ups.com...
On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.
The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9


In that case...

Half power is only lost when terminated to a 50-ohm load;
i.e. no standing waves. What happens when there's a
mismatch and reflected energy



I'll go stand in a corner...


A mismatch where, between the feedline and the antenna, feedline and source,
source and impedance seen at the input to the feedline.

Jimmie

In .com, billcalley said:
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

Two problems:

1) The transmitter may well have output impedance matching the
characteristic impedance of the transmission line. RF power reflected
back in this case gets converted to heat in the output stage of the
transmitter, in addition to whatever heat the output stage already has to
dissipate.

1a) The reflection may increase requirement of the output
tubes/transistors to both drop voltage and dissipate power. This can be a
problem for many transistors, especially a lot of bipolar ones. It is not
necessarily sufficient to stay within power, current, voltage and thermal
ratings. Many bipolar transistors have reduced capability to safely
dissipate power at voltages that are higher but within their ratings -
sometimes even at voltages as low as 35-50 volts. This problem tends to
be worse with bipolar transistors that are faster and/or better for use
with higher frequencies. The keyphrase here is "forward bias second
breakdown", a problem of uneven current distribution within the die at
higher voltage drop.

2) It appears to me that transmitters can have output stage output
impedance differing from the intended load impedance.
An analog is common practice with audio amplifiers - output impedance is
often ideally as close to zero as possible, as opposed to matching the
load impedance.

If zero output impedance is achieved in an RF output stage, I see a
possible benefit - reflections do not increase output stage heating but
get reflected back towards the antenna. Then again, the impedance of the
input end of the transmission line could be low or significantly reactive
depending on how the load is mismatched and how many wavelengths long the
transmission line is, and that can increase heating of the output stage.
In a few cases transmitted power can also increase.

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )

*Sigh*

The same misconceptions keep coming up, as they have countless times on
this newsgroup and I'm sure they will for decades or perhaps centuries
to come. After one of the many previous discussions, I wrote a little
tutorial on the topic. Originally in the form of plain text files, I've
combined it into a pdf file for easier viewing. You can find it at
http://eznec.com/misc/Food_for_thought.pdf.

On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED
IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several
examples which clearly show that there's no relationship between the
"reverse power" and the source dissipation. The remainder of the
tutorial explains why.

Any theory about "forward" and "reverse" power, what they do, and their
interaction with the source, will have to explain the values in the
example chart on page 8. Does yours?

Roy Lewallen, W7EL

Don Klipstein wrote:
In .com, billcalley said:
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

Two problems:

1) The transmitter may well have output impedance matching the
characteristic impedance of the transmission line. RF power reflected
back in this case gets converted to heat in the output stage of the
transmitter, in addition to whatever heat the output stage already has to
dissipate.

1a) The reflection may increase requirement of the output
tubes/transistors to both drop voltage and dissipate power. This can be a
problem for many transistors, especially a lot of bipolar ones. It is not
necessarily sufficient to stay within power, current, voltage and thermal
ratings. Many bipolar transistors have reduced capability to safely
dissipate power at voltages that are higher but within their ratings -
sometimes even at voltages as low as 35-50 volts. This problem tends to
be worse with bipolar transistors that are faster and/or better for use
with higher frequencies. The keyphrase here is "forward bias second
breakdown", a problem of uneven current distribution within the die at
higher voltage drop.

2) It appears to me that transmitters can have output stage output
impedance differing from the intended load impedance.
An analog is common practice with audio amplifiers - output impedance is
often ideally as close to zero as possible, as opposed to matching the
load impedance.

If zero output impedance is achieved in an RF output stage, I see a
possible benefit - reflections do not increase output stage heating but
get reflected back towards the antenna. Then again, the impedance of the
input end of the transmission line could be low or significantly reactive
depending on how the load is mismatched and how many wavelengths long the
transmission line is, and that can increase heating of the output stage.
In a few cases transmitted power can also increase.

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )

Roy Lewallen wrote:
*Sigh*

The same misconceptions keep coming up, as they have countless times on
this newsgroup and I'm sure they will for decades or perhaps centuries
to come. After one of the many previous discussions, I wrote a little
tutorial on the topic. Originally in the form of plain text files, I've
combined it into a pdf file for easier viewing. You can find it at
http://eznec.com/misc/Food_for_thought.pdf.

On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED
IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several
examples which clearly show that there's no relationship between the
"reverse power" and the source dissipation. The remainder of the
tutorial explains why.

Any theory about "forward" and "reverse" power, what they do, and their
interaction with the source, will have to explain the values in the
example chart on page 8. Does yours?

Mine does. All of your values can be understood by looking
at the destructive and constructive interference and applying
the irradiance (power density) equations from the field of
optics. You see, optical engineers and physicists don't have
the luxury of measuring voltage and current in their EM waves.
All they can measure is power density and interference and
thus their entire body of knowledge of EM waves rests upon
measurements of those quantities. Those power density and
interference theories and equations are directly applicable
and 100% compatible with RF theories and equations. Any
analysis based on power density and interference will yield
identical results to the ones you reported in your "food for
thought" article which includes the following false statement:

"While the nature of the voltage and current waves when
encountering an impedance discontinuity is well understood,
we're lacking a model of what happens to this "reverse power"
we've calculated."

We are not lacking a model of what happens to this
"reverse power" we've calculated. The model is explained
fully in "Optics", by Hecht. When one has standing waves of
light in free space, it is hard to hide the details under
the transmission line rug.

In general, it is just as easy, and sometimes easier, to deal
with the energy values and then calculate voltage and current
as it is to start with voltage and current and then calculate
the power.

All this is explained in my WorldRadio article at:

http://www.w5dxp.com/energy.htm

The great majority of amateur antenna systems are Z0-matched.
For such systems, an energy analysis is definitely easier to
perform than a voltage analysis. Here's an example:

100W------50 ohm---+---Z050 ohms-----load
Pfor1=100W-- Pfor2--
--Pref1=0W --Pref2

The power reflection coefficient is 0.5 at point '+'.
The power reflection coefficient is 0.5 at the load.

What are the values of Pfor2 and Pref2? What is the physics
equation governing what happens to Pref2 at point '+'?
--
73, Cecil http://www.w5dxp.com

On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened
(Don Klipstein) wrote in
:

snipped good stuff

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )

All true.
Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

On Mon, 12 Mar 2007 11:00:04 GMT, Jan Panteltje
wrote:

On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened
(Don Klipstein) wrote in
:

snipped good stuff

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )

All true.
Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

Jan and Don,

Both of you gentlemen really need to read Walter's book Reflections
(any edition) and put that myth to rest once and for all.

Danny, K6MHE

Danny Richardson wrote in
:

(any edition) and put that myth to rest once and for all.

Wishful thinking Danny, myths are the stuff of ham radio, aren't they?

Owen

Jan Panteltje wrote:
Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

Some gurus will say that it's the voltage and/or current
that is destroying the final, not the reflected energy.
They have yet to explain how those dangerous voltages
and/or currents can exist without assistance from the
ExH joules/second in the reflected energy wave. Depending
upon phase, the E in the ExH reflected wave is what causes
the overvoltage due to SWR. The H in the ExH reflected
wave is what causes the overcurrent due to SWR.

The impedance seen by the source is

Z = (Vfor+Vref)/(Ifor+Iref)

Where '+' indicates phasor (vector) addition.

The above equation also gives the impedance anywhere
along the transmission line and anywhere along a
standing-wave antenna.
--
73, Cecil http://www.w5dxp.com