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

Bill,

an exellent treatment on this question has been published in QEX December 94
under the title "Where does the power go?"

73 de Ulrich, DF6JB

"billcalley" schrieb im Newsbeitrag
oups.com...
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?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

Thanks!

-Bill

Richard Clark wrote:
I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

It really doesn't matter except for overall
efficiency. A 10 ohm source outputting 100 volts
into a local load of R +/- jX sources the same
amount of power as a 100 ohm source outputting
100 volts into the same local load.
--
73, Cecil http://www.w5dxp.com

Don Klipstein wrote:
If zero output impedance is achieved in an RF output stage, I see a
possible benefit -

Yep, 100% efficiency would be quite a benefit.
--
73, Cecil http://www.w5dxp.com

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

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

Ulrich Bangert wrote:
an exellent treatment on this question has been published in QEX December 94
under the title "Where does the power go?"

Unfortunately, that article doesn't explain where the
power does go. A much better treatment of the subject
is in "Optics", by Hecht. To understand where the
power does go, one must understand destructive and
constructive interference. Please see my transmission
line example in another posting.

The energy content of a transmission line during
steady-state is always exactly enough to support
the forward traveling wave and the reverse traveling
wave without which there would be no standing wave.
--
73, Cecil http://www.w5dxp.com

In , Cecil Moore wrote:
Don Klipstein wrote:
If zero output impedance is achieved in an RF output stage, I see a
possible benefit -

Yep, 100% efficiency would be quite a benefit.

There are audio amplifiers with output impedance around .1 ohm, driving
8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical
limit for efficiency of a class B amp driving a resistive load with a
sinewave is 78.54%.

- Don Klipstein )

Don Klipstein wrote:
Cecil Moore wrote:
Yep, 100% efficiency would be quite a benefit.

There are audio amplifiers with output impedance around .1 ohm, driving
8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical
limit for efficiency of a class B amp driving a resistive load with a
sinewave is 78.54%.

Of course, that was a tongue-in-cheek posting.
But if you could design a Thevenin equivalent
source with a 0.1 ohm source impedance, wouldn't
the efficiency calculate out to be pretty high?
--
73, Cecil http://www.w5dxp.com