Revisiting the Power Explanation
 

Keith Dysart wrote:
What drove me to look at alternate explanations for these kinds of
examples
was that the 'reverse power' explanation fails miserably when the
power
gets back to the generator.

The reverse energy wave follows the principles of
conservation of energy and superposition. That we
have no clue what the generator looks like to the
reflected energy wave is not a good reason to
abandon those principles. In fact, allowing
destructive interference to be accompanied by
an energy reflection solves all the problems.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Keith Dysart wrote:
I have yet to question the reflection of EM radiation, just the
existence of "reverse power" in transmission lines.

Then you are certainly engaging in the proverbial Red Herring.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator. If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

Most correspondents agree that what happens depends on the design
of generator; dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

Any level of interference is possible depending upon the
phase angle between the forward E-field and the reflected
E-field. All this is explained in "Optics" by Hecht which
some people have apparently avoided reading/understanding.
Optical physicists solved this problem a century ago. They
don't have the luxury of dealing with voltages and currents
and are forced to deal with power densities. You should
try trodding their paths and enlightening yourself.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Owen Duffy wrote:
If one takes measurements with the instrument, it is true that the power
at a point is "forward power" less "reflected power", and the
manufacturer has scaled the instrument in Watts to facilitate that
calculation, but that does not imply that the value of "forward power" or
"reflected power" has any stand alone value, the ratio of the two is
meaningful, the difference of the two is meaningful, but one alone is
meaningless.

How do you explain the fact that a transmission line
contains exactly the amount of energy needed to support
the actual forward power and reflected power? If it
is not moving at the speed of light (modified by VF)
it is not EM energy.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Owen Duffy wrote:
The voltage at the point may be higher than under matched conditions for
the same load power, and that may cause insulation breakdown.

Caused by the in-phase superposition of forward and
reflected voltages.

The current at the point may be higher than under matched conditions for
the same load power, and that would cause higher loss in conductors and
may result in damage.

Caused by the in-phase superposition of forward and
reflected currents.

None of these explanations require designating "reflected power" at a
point, or implying that it is the energy in "reflected power" that is
totally and solely responsible for the physical damage.

EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Keith Dysart wrote:
And you have done an excellent job of doing so by providing
an explanation that refers only to forward and reflected
voltage and current. Not a mention of "reverse power" in
the explanation..

The forward voltage is in phase with the forward current.
The reflected voltage is in phase with the reflected
current. Vref*Iref*cos(0) = reflected power
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote in
t:

....
EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.

You are not suggesting that the energy contained in a transmission line in
the steady state in the general case is constant, are you?

Owen

Revisiting the Power Explanation
 

Cecil Moore wrote:
Gene Fuller wrote:
It is interesting that you can be so precise at times and so sloppy
at other times. I very carefully limited my discussion to steady state
conditions, which is what everyone is already talking about in this
case. You then conveniently inject modulation into the mix, completely
ignoring what I said.

*Every* real world system has noise modulation
that can be tracked through the system riding on
the forward and reflected traveling waves. Thus
steady-state is never reached in reality and your
argument is therefore just a mind game.

Cecil, every time someone *almost* succeeds in making you stick to the
point, you accuse them of messing with your mind.

Nobody is trying to restrict your freedom of thought or speech. Nobody
wants to, and nobody can. But lots of people are hoping, begging,
pleading that you develop the SELF-discipline to follow an argument all
the way through to its conclusion, without jumping outside of the
boundaries you laid down at the start.

This may indeed trouble your mind; but the process of scientific inquiry
wasn't ever *intended* to feel comfortable. Its only goal is to find
things out and get them right... and that involves staying on track,
even when it takes you outside of the comfort zone. In fact, then most
of all, because it's a sure sign that the track is leading somewhere.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Revisiting the Power Explanation
 

Owen Duffy wrote:
Cecil Moore wrote in
EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.

You are not suggesting that the energy contained in a transmission line in
the steady state in the general case is constant, are you?

Obviously, a leading question. :-) I'm not talking
about instantaneous values here. All my statements
apply only to values averaged over an integer
number of RF cycles in one second.

What I am saying
is that a transmission line obeys the conservation
of energy principle. Whatever energy has not gone
somewhere else is still in the transmission line
and is exactly the sum of the energy required by the
forward wave plus the energy required by the reflected
wave. The steady-state energy stored in a transmission
line with reflections is greater than the steady-
state energy delivered to the load. The extra energy
was sourced during the transient build-up state and
has not yet been delivered to the load. The argument
that 100 watts in and 100 watts out (during steady-
state) doesn't leave any energy left over for the
reflected waves is invalid. Below I give an example
of 100 watts in and 100 watts out with 300 joules
stored in the forward and reflected waves.

I think it would be safe to say that for any one-
wavelength section of line, the average energy
content is constant during steady-state and is
exactly the amount of energy required to support
the forward traveling wave and reflected traveling
wave.

Here's a graphic that I earlier provided that
illustrates the energy buildup to steady-state.

http://www.w5dxp.com/1secsgat.gif
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Ian White GM3SEK wrote:
Nobody is trying to restrict your freedom of thought or speech. Nobody
wants to, and nobody can. But lots of people are hoping, begging,
pleading that you develop the SELF-discipline to follow an argument all
the way through to its conclusion, without jumping outside of the
boundaries you laid down at the start.

Sorry Ian, I don't trust you guys enough to roll
dice in the dark with you and then let you tell
me what value was rolled. :-) You are perfectly
free to play mashed potatoes with RF joules but
please don't ask me to join in. Sometimes I think
you have to be just pulling my leg.

It was Gene who first pointed out the difference
between a traveling wave and a standing wave. Now
he says there is no difference.

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

One can send two coherent light beams in opposite
directions almost collinear to each other and observe
the standing waves. Hecht has a graphic of such in
"Optics". The two light beams, forward and reverse,
emerge past the standing wave space undisturbed and
unaffected by the superposition.

How does your theory hold up when measurement of voltage
and current is impossible and everything occurs in free
space for anyone to observe with his/her own eyes? If it
doesn't work for EM light waves then it also doesn't
work for EM RF waves except as a shortcut.

So please take the above example from optics and show
me how the reverse traveling wave is different on
either side of the standing wave space.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

On Wed, 21 Mar 2007 08:18:14 -0500, "Richard Fry" wrote:

"Walter Maxwell" wrote
(RF): And if so, would that also mean that such a tx would not be prone
to producing r-f intermodulation components when external signals
are fed back into the tx from co-sited r-f systems?

This issue is irrelevant, because the signals arriving from a co-sited
system would not be coherent with the local source signals, while load-
reflected signals are coherent. The destructive and constructive
interference that occurs at the output of a correctly loaded and tuned
PA requires coherence of the source and reflected waves to achieve
the total re-reflection of the reflected waves back into the direction
toward the load.

But even for coherent reflections, if the PA tank circuit has very low loss
for incident power (which it does), why does it not have ~ equally low loss
for load reflections of that power? Such would mean that load reflections
would pass through the tank to appear at the output element of the PA, where
they can add to its normal power dissipation.

Also, does not the result of combining the incident and reflected waves in
the tx depend in large part on the r-f phase of the reflection there
relative to the r-f phase of the incident wave? And the r-f phase of the
reflection is governed mostly by the number of electrical wavelengths of
transmission line between the load reflection and the plane of
interest/concern -- which is independent of how the tx has been
tuned/loaded.

If the ham transmitter designs that your paper applies to produce a total
re-reflection of reverse power seen at their output tank circuits, then
there would be no particular need for "VSWR foldback" circuits to protect
them. Yet I believe these circuits are fairly common in ham transmitters,
aren't they? They certainly are universal in modern AM/FM/TV broadcast
transmitters, and are the result of early field experience where PA tubes,
tx output networks, and the transmission line between the tx and the antenna
could arc over and/or melt when reflected power was sufficiently high.

RF
Richard, your statement above begs the question, "Are you aware of the phase relationships between forward and
reflected voltages and between forward and reflected currrents that accomplish the impedance-matching effect
at matching points such as with stub matching and also with antenna tuners?

When the matching is accomplished the phase relationship between the foward and reflected voltages can become
either 0° or 180°, resulting in a total re-reflection of the voltage. If the resultant voltage is 0°, then the
resultant current is 180°, thus voltage sees a virtual open circuit and the current sees a virtual short
circuit. The result is that the reflected voltage and current are totally re-reflected IN PHASE with the
source voltage and current. This is the reason the forward power in the line is greater than the source power
when the line is mismatched at the load, but where the matching device has re-reflected the reflected waves.

This phenomenon occurs in all tube transmitters in the ham world when the tank circuit is adjusted for
delivering all available power at a given drive level. When this condition occurs the adjustment of the
pi-network has caused the relationship between the forward and reflected voltages to be either 0° or 180° and
vice versa for currents, as explained above. When this condition occurs, destructive interference between the
forward and reflected voltages, as well as between the forward and reflected currents, causes the reflected
voltage and current to cancel. However, due to the conservation of energy, the reflected voltage and current
cannot just disappear, so the resulting constructive interference following immediately, causes the reflected
voltage and current to be reversed in direction, now going in the foward direction along with and in phase
with the forward voltage and current.

In transmitters with tubes and a pi-network output coupling circuit there is no 'fold back' circuitry to
protect the amp, because none is needed, due to the total re-reflection of the reflected power. It is only in
solid-state transmitters that have no circuitry to achieve destructive and constructive interference that
requires fold back to protect the output transistors.

This has been a quick and dirty explanation of the phase relations that accomplish impedance matching.
However, I have explained it in much more detail in my book "Reflections--Transmission Lines and Antennas."
Yes, I know the book has been sold out and now unavailable, but I have put several chapters on my web page
avaliable for downloading. The pertinent chapters covering this issue are Chapters 3, 4, and 23, available at
www.w2du.com. I hope that reviewing these chapters will be helpful in clearing up some of the
misunderstandings that are clearly evident in some of the postings on this thread.

Walt, W2DU

Revisiting the Power Explanation
 

Walter Maxwell wrote:
In transmitters with tubes and a pi-network output coupling circuit there is no 'fold back' circuitry to
protect the amp, because none is needed, due to the total re-reflection of the reflected power. It is only in
solid-state transmitters that have no circuitry to achieve destructive and constructive interference that
requires fold back to protect the output transistors.

One can illustrate the destructive and constructive
interference with a solid-state transmitter and
no tuner. Consider the following example using
S-parameter terms.

100W--50 ohm line--+--1/2WL 300 ohm line--50 ohms
a1-- --a2
--b1 b2--

Since there is zero reflected power on the 50 ohm
line, we know that "total destructive interference"
(as described by Hecht in "Optics", 4th edition, page
388) exists toward the source at point '+'.
s11 = (300-50)/(300+50) = 0.7143 = -s12
b1 = (s11)(a1) + (s12)(a2) = 0

Note that given a1, s11, and s12, we can calculate the
magnitude and phase of a2 needed to make b1=0. That
is the Z0-match condition.

The conservation of energy principle says that, (in
a transmission line with only two directions) "total
constructive interference" must exist in the opposite
direction to the "total destructive interference" and
that they must be of equal magnitudes. That tells us
what *must* happen to the energy associated with the
a2 reflected wave.

All of the energy incident upon point '+' from both
directions, |a1|^2 + |a2|^2, is directed toward the
load by the interference patterns at the Z0-match
point '+'. We hams commonly refer to that condition
as being 100% re-reflected.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote:
All of the energy incident upon point '+' from both
directions, |a1|^2 + |a2|^2, is directed toward the
load by the interference patterns at the Z0-match
point '+'. We hams commonly refer to that condition
as being 100% re-reflected.

The above is true in the special case of a Z0-match.
In general, |a1|^2 + |a2|^2 = |b1|^2 + |b2|^2
and since |b1|^2 = 0, the above expression is
correct.

*Quoting from HP Ap Note 95-1*:

|a1|^2 = Power incident on the input of the network
(i.e. Forward power on the 50 ohm line)

|a2|^2 = Power reflected from the load
(i.e. Reflected power on the 300 ohm line)

|b1|^2 = Power reflected from the input port of the network
(i.e. Reflected power on the 50 ohm line)

|b2|^2 = Power incident on the load
(i.e. Forward power on the 300 ohm line)

end quote from HP Ap Note 95-1
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote in
:

....
s11 = (300-50)/(300+50) = 0.7143 = -s12
b1 = (s11)(a1) + (s12)(a2) = 0

Cecil,

I see you are back to using S parameters to disguise the fact you are using
about Vf and Vr in trying to support your "power wave" explanation of what
happens on the transmission line.

S parameters are ratios of Vf and Vr.

Owen

Revisiting the Power Explanation
 

Cecil Moore wrote in
et:

Owen Duffy wrote:
Cecil Moore wrote in
EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.

You are not suggesting that the energy contained in a transmission
line in the steady state in the general case is constant, are you?

Obviously, a leading question. :-) I'm not talking
about instantaneous values here. All my statements
apply only to values averaged over an integer
number of RF cycles in one second.

This gets confusing. You are talking about "the amount of EM wave energy
contained in a transmission line" and now you qualify it with "values
averaged over an integer number of RF cycles in one second". Average
energy over time is POWER... are you talking about power and foxing us by
calling it energy. I am confused.

Owen

Revisiting the Power Explanation
 

Cecil Moore wrote:


It was Gene who first pointed out the difference
between a traveling wave and a standing wave. Now
he says there is no difference.


Cecil,

Utter nonsense. I have never said any such thing.

What I *did* say, and it is still true today, is that there is no
difference between a standing wave and its *constituent* traveling wave
components. It is purely a matter of mathematical convenience. There is
no underlying hidden physics available from trying to look at only one
of the traveling wave components. If you sort out an individual
component then the standing wave is no longer there.

ad hominem Your debating style is a bit rough, but it is clear from
this thread that others have observed your tricks. /ad hominem

73,
Gene
W4SZ

Revisiting the Power Explanation
 

Owen Duffy wrote:
I see you are back to using S parameters to disguise the fact you are using
about Vf and Vr in trying to support your "power wave" explanation of what
happens on the transmission line.

Others use the term "power wave", Owen, but *I DO NOT*
so please stop accusing me of something of which I am
not guilty. I use the term "EM RF energy wave" for
the traveling waves under discussion.

When anyone can prove that RF energy waves don't exist
or are not associated with EM energy or don't move at
the speed of light, I will retire from the argument.
Good luck on that one.

S parameters are ratios of Vf and Vr.

Exactly! No disguise intended - it's just additional
support from the well respected field of S-parameter
analysis for the distributed network wave reflection
model. The only difference is that the S-parameter
Vf and Vr values are normalized to Z0 so when they
are squared they indeed do yield watts.

Your tone seems to reject the S-Parameter analysis
as a valid model of reality. Any model that has to
resort to rejecting the S-Parameter analysis as well
as the distributed network wave reflection model is
certainly suspect. Did you ever see the movie, "One
Bridge Too Far"? This "reflected wave energy doesn't
exist" argument reminds me of that movie.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Owen Duffy wrote:
This gets confusing. You are talking about "the amount of EM wave energy
contained in a transmission line" and now you qualify it with "values
averaged over an integer number of RF cycles in one second". Average
energy over time is POWER... are you talking about power and foxing us by
calling it energy. I am confused.

I have been convinced by Jim, AC6XG, to abandon the word
"power" because of the difference in definitions between
the field of physics and the field of RF engineering.

Jim would argue with you and say that average energy over
time is NOT necessarily POWER and is only power if actual
work is done which, of course, is not done by a reflected
wave.

So you need to go off and argue with Jim over the
definition of "power". Instead of talking about power,
Jim has convinced me to talk about watts or joules/sec
which he says are not necessarily power. The confusion
comes from the field of physics, not from me. While you
are talking to Jim, get him to explain the definition
of "transfer".
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote in
:

Your tone seems to reject the S-Parameter analysis
as a valid model of reality. Any model that has to

Not at all.

S parameters are Vf and Vr based and when properly applied will produce
exactly the same analysis outcome.

It is the application of S parameters in the "power flow analysis" that
is a reach, it might be convenient, but it does not legitmise the
argument that forward and reflected "power waves" exist separately.

A quote from HP (which you seem to respect):

===quote
Notice that the square of the magnitude of these
new variables has the dimension of power. |a1|^2
can then be thought of as the incident power on
port one; |b1|^2 as power reflected from port one.
These new waves can be called traveling power
waves rather than traveling voltage waves.
Throughout this seminar, we will simply refer to
these waves as traveling waves.
===equote

There is a difference between "can then be thought of as..." and
"are...".

Owen

Revisiting the Power Explanation
 

Gene Fuller wrote:
Utter nonsense. I have never said any such thing.

Yes, you did, in the part that you deleted. Here it
is again:

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

What I *did* say, and it is still true today, is that there is no
difference between a standing wave and its *constituent* traveling wave
components.

The constituent traveling wave components possess changing
phase. The standing wave doesn't possess changing phase.
You are contradicting yourself.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Gene, W4SZ wrote:
"No need to account for any mythical power in the reflected waves."

My dictionary defines reflected power as:
"The power flowing back to the generator from the load."

Maximum power theorem is defined as:
"The maximum power will be absorbed by one network from another joined
to it at two terminals, when the impedance of the receiving network is
varied, if the impedances (looking into the two networks at the
junction) are conjugates of each other."

Clearly a generator (transmitter) connected to a load through a lossless
line sees Zo of the line as its load until the instant that reflected
power returns to the generator from the load.
Suppose the round-trip delay of the line makes the reflected voltage
exactly in phase with the transmitter output. further suppose the
reflection was total so that the reflected voltage exactly equals the
transmitter output. In this special case, we might as well be connecting
identical battery cells in parallel. No current is going to flow. The
generator is seeing infinite impedance.

What is the generator load that extracts maximum power from a
transmitter? A conjugately matched load, of course. To determine the
impedance of a transmitter, one only needs to find the load which
extracts maximum power. The transmitter impedance is its conjugate.

Best regards, Richard Harrison, KB5WZI

Revisiting the Power Explanation
 

Owen Duffy wrote:
It is the application of S parameters in the "power flow analysis" that
is a reach, it might be convenient, but it does not legitmise the
argument that forward and reflected "power waves" exist separately.

Nth reminder to you: Please stop implying something
I didn't say. I have said that forward and reflected RF
traveling energy waves exist separately. If you can find
an example of me using the term "power wave" in the 21st
century, I will send you a $100 bill.

A quote from HP (which you seem to respect):
Throughout this seminar, we will simply refer to
these waves as traveling waves.

There is a difference between "can then be thought of as..." and
"are...".

EXACTLY! You and I are generally in agreement except
when you accuse me of nonsense like "power waves".
Please cease and desist! I simply refer to these
waves as traveling energy waves, NOT POWER WAVES!
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote in
:

Owen Duffy wrote:
This gets confusing. You are talking about "the amount of EM wave
energy contained in a transmission line" and now you qualify it with
"values averaged over an integer number of RF cycles in one second".
Average energy over time is POWER... are you talking about power and
foxing us by calling it energy. I am confused.

I have been convinced by Jim, AC6XG, to abandon the word
"power" because of the difference in definitions between
the field of physics and the field of RF engineering.

Jim would argue with you and say that average energy over
time is NOT necessarily POWER and is only power if actual
work is done which, of course, is not done by a reflected
wave.

So you need to go off and argue with Jim over the
definition of "power". Instead of talking about power,
Jim has convinced me to talk about watts or joules/sec
which he says are not necessarily power. The confusion
comes from the field of physics, not from me. While you
are talking to Jim, get him to explain the definition
of "transfer".

A neat diversion from the issue re the "amount of energy" qualified later
as a average over time which is a different quantity, Joules vs Watts to
many of us.

The fact is that the energy stored in a transmission line in the steady
state is in the general case, a time variable, and you cannot state the
energy (in joules) at a point in time knowing only forward and reflected
power and the one way propagation time.

So Cecil,

Is it all about semantics? Is the lack of a shared language the cause of
difficulty understanding your concepts. You wouldn't be alone, Art
experiences the same difficulties with convention.

Owen

Revisiting the Power Explanation
 

On Thu, 22 Mar 2007 15:55:40 GMT, Walter Maxwell
wrote:

On Wed, 21 Mar 2007 08:18:14 -0500, "Richard Fry" wrote:

"Walter Maxwell" wrote
(RF): And if so, would that also mean that such a tx would not be prone
to producing r-f intermodulation components when external signals
are fed back into the tx from co-sited r-f systems?

This issue is irrelevant, because the signals arriving from a co-sited
system would not be coherent with the local source signals, while load-
reflected signals are coherent. The destructive and constructive
interference that occurs at the output of a correctly loaded and tuned
PA requires coherence of the source and reflected waves to achieve
the total re-reflection of the reflected waves back into the direction
toward the load.

Hi Walt,

It is not irrelevant, merely illustrative of the concept of reflection
that is consistent with a coherent source.

Your points of phase are the sine non quo to the discussion, but all
too often arguers only take the half of the 360 degrees available to
argue a total solution. Even more often, they take only one or two
degrees of the 360.

But even for coherent reflections, if the PA tank circuit has very low loss
for incident power (which it does), why does it not have ~ equally low loss
for load reflections of that power? Such would mean that load reflections
would pass through the tank to appear at the output element of the PA, where
they can add to its normal power dissipation.

This is the symmetry of the illustration of external signals. You
used external signals yourself as part of your case study; hence the
relevance has been made by you.

Also, does not the result of combining the incident and reflected waves in
the tx depend in large part on the r-f phase of the reflection there
relative to the r-f phase of the incident wave? And the r-f phase of the
reflection is governed mostly by the number of electrical wavelengths of
transmission line between the load reflection and the plane of
interest/concern -- which is independent of how the tx has been
tuned/loaded.

And we return to the sine non quo for the discussion: phase.

If the ham transmitter designs that your paper applies to produce a total
re-reflection of reverse power seen at their output tank circuits, then
there would be no particular need for "VSWR foldback" circuits to protect
them. Yet I believe these circuits are fairly common in ham transmitters,
aren't they? They certainly are universal in modern AM/FM/TV broadcast
transmitters, and are the result of early field experience where PA tubes,
tx output networks, and the transmission line between the tx and the antenna
could arc over and/or melt when reflected power was sufficiently high.

RF

Richard, your statement above begs the question, "Are you aware of the phase relationships between forward and
reflected voltages and between forward and reflected currrents that accomplish the impedance-matching effect
at matching points such as with stub matching and also with antenna tuners?

It seems he is on the face of it, doesn't it? Afterall, he is quite
explicit to this in the statement you are challenging.

When the matching is accomplished the phase relationship between the foward and reflected voltages can become
either 0° or 180°, resulting in a total re-reflection of the voltage. If the resultant voltage is 0°, then the
resultant current is 180°, thus voltage sees a virtual open circuit and the current sees a virtual short
circuit. The result is that the reflected voltage and current are totally re-reflected IN PHASE with the
source voltage and current. This is the reason the forward power in the line is greater than the source power
when the line is mismatched at the load, but where the matching device has re-reflected the reflected waves.

Nothing here contradicts anything either of you have to say.

This phenomenon occurs in all tube transmitters in the ham world when the tank circuit is adjusted for
delivering all available power at a given drive level.

This introduces the two concepts of the "need for match" and the
"match obtained." They are related only through an action that spans
from one condition to the other. They do not describe the same
condition, otherwise no one would ever need to perform the match:

When this condition occurs the adjustment of the
pi-network has caused the relationship between the forward and reflected voltages to be either 0° or 180° and
vice versa for currents, as explained above. When this condition occurs, destructive interference between the
forward and reflected voltages, as well as between the forward and reflected currents, causes the reflected
voltage and current to cancel. However, due to the conservation of energy, the reflected voltage and current
cannot just disappear, so the resulting constructive interference following immediately, causes the reflected
voltage and current to be reversed in direction, now going in the foward direction along with and in phase
with the forward voltage and current.

If a tree were to fall onto the antenna, a new mismatch would occur.
Would the transmitter faithfully meet the expectations of the Ham
unaware of the accident? No, reflected (0-179 degrees) energy would
undoubtedly offer a 50% chance of excitement in the shack. The
consequences of dissipation would be quite evident on that occasion.
For the other 180 (180-359) degrees of benign combination; then
perhaps not.

In transmitters with tubes and a pi-network output coupling circuit there is no 'fold back' circuitry to
protect the amp, because none is needed, due to the total re-reflection of the reflected power.

That would more probably be due to cost averse buying habits of the
Amateur community, and the explicit assumption of risk by them to
react appropriately in the face of mismatch. Tubes were far more
resilient to these incidents than transistors of yore.

It is only in
solid-state transmitters that have no circuitry to achieve destructive and constructive interference that
requires fold back to protect the output transistors.

They too have access to the services of a transmatch that is probably
more flexible than the tubes' final. If they didn't use a tuner, then
the foldback would render many opportunistic antennas as useless.
Again, as a cost item, this solution (fold-back) is dirt cheap and was
driven by the market economies of a more onerous and costly repair
through a lengthy bench time to replace the transistor (which has an
exceedingly high probability of a quicker failure for a poor job).

73's
Richard Clark, KB7QHC

Revisiting the Power Explanation
 

Owen Duffy wrote:
The fact is that the energy stored in a transmission line in the steady
state is in the general case, a time variable, and you cannot state the
energy (in joules) at a point in time knowing only forward and reflected
power and the one way propagation time.

The time-averaged energy stored in a transmission line
doesn't vary with time (by definition) unless something
changes.

If the forward and reflected powers are based on fixed
RMS values of voltage and current with fixed sources and
loads, then certainly, time-averaged energy values can be
calculated. That's the nature of *irradiance* in optics
where instantaneous values are impossible to measure. The
Power Flow Vector is analogous to irradiance.

Is it all about semantics? Is the lack of a shared language the cause of
difficulty understanding your concepts.

No, the basic problem is that you keep accusing me of
saying something I never said. I have never used the
term "power wave" except in postings like this one.
I have no idea what is the definition of "power wave".

How it feels:

It feels to me like you are one of a number of people
willing to reject the basic principles of physics in
order to mount an argument with me or anyone else
who have been convinced by those laws of physics that
reflected energy waves exist in reality - you know,
the ones you actually see when looking in a mirror?
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote:
Gene Fuller wrote:
Utter nonsense. I have never said any such thing.

Yes, you did, in the part that you deleted. Here it
is again:

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

What I *did* say, and it is still true today, is that there is no
difference between a standing wave and its *constituent* traveling
wave components.

The constituent traveling wave components possess changing
phase. The standing wave doesn't possess changing phase.
You are contradicting yourself.

Cecil,

Sorry, I thought that English was your native tongue. Would you like me
to try to translate into another preferred language?

Hint: I stand by everything I have said in the quotes above. No
contradictions at all.

73,
Gene
W4SZ

Revisiting the Power Explanation
 

Gene Fuller wrote:
Sorry, I thought that English was your native tongue. Would you like me
to try to translate into another preferred language?

Translation: I've been caught red-handed by
my own American English words. Maybe converting
to an obscure foreign language would give me more
wiggle room.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

"Owen Duffy" wrote
So Cecil,
Is it all about semantics? Is the lack of a shared language the cause of
difficulty understanding your concepts. You wouldn't be alone, Art
experiences the same difficulties with convention.
Owen

Great deduction, Owen! Perhaps Art would benefit from a copy
of "Gaussian Optics, with Definitions". Thinkiing about that makes
me wonder....if Cecil were to feed a one-second long transmission
line, 50 Zo, terminated by Art's Gaussian Array, how long before
the (power)(energy)(whatever) starts sloshing ??

Mike W5CHR

Revisiting the Power Explanation
 

Mike Lucas wrote:
"Owen Duffy" wrote
So Cecil,
Is it all about semantics? Is the lack of a shared language the cause of
difficulty understanding your concepts. You wouldn't be alone, Art
experiences the same difficulties with convention.

Great deduction, Owen!

Unfortunately, Owen may be the one with the semantic problem.
He has quoted me as uttering the phase, "power wave", which
I have never, to the best of my memory, ever uttered. A
$100 bill is awaiting proof of his assertions.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

On Mar 22, 12:31 pm, (Richard Harrison)
wrote:

What is the generator load that extracts maximum power from a
transmitter? A conjugately matched load, of course. To determine the
impedance of a transmitter, one only needs to find the load which
extracts maximum power. The transmitter impedance is its conjugate.

Best regards, Richard Harrison, KB5WZI

We have a new signal generator here at the Labs, but it came without
any documentation. We've been hoping to find out more about it, and
your suggestion gave us a clue for a test to try. We have a limited
set of loads, only from 40 to 60 ohms, but we can add some known
reactance in series with that. When we put an accurate 50 ohm load on
the generator, we measured exactly 1 watt with our accurate RF power
meter. When we changed the load to 40 ohms, the power went up to
1.238 watts, and with a 60 ohm load, it was .839 watts. When we add
reactance, either capacitive or inductive, the power goes down, so
we're pretty sure the generator output doesn't look reactive. Beaker
is having a little trouble with the math on this one. Could you help
him out? What sort of output impedance does this generator have?

Oh, and we thought to do another experiment. While the generator was
operating, we sent a short burst of RF down a 50 ohm transmission line
to the generator's output. Because the power measurements seem to say
the generator's output resistance is very low, we thought we would see
a return pulse, delayed by the round-trip time down the 15 meters of
high-quality cable. But we didn't see any echo. What could be going
on? We double-checked everything, and even looked into the line next
to the generator with a high impedance probe. We can see the burst
going in there, but still nothing comes back.

From the labs,
Dr. Honeydew

Revisiting the Power Explanation
 

"Dr. Honeydew" wrote in
oups.com:

On Mar 22, 12:31 pm, (Richard Harrison)
wrote:

What is the generator load that extracts maximum power from a
transmitter? A conjugately matched load, of course. To determine the
impedance of a transmitter, one only needs to find the load which
extracts maximum power. The transmitter impedance is its conjugate.

Best regards, Richard Harrison, KB5WZI

We have a new signal generator here at the Labs, but it came without
any documentation. We've been hoping to find out more about it, and
your suggestion gave us a clue for a test to try. We have a limited

If you have a suspicion that the generator might be 50 ohms, and you have
a 50 ohm load, then the output voltage should fall by 6.02dB when the
nominal load is connected (compared to o/c).

One of the simple tests that is often used to verify that a generator is
loaded properly is that adding a second nominal load in parallel reduces
terminal voltage by 3.52dB. This test is often easy to perform, use a
power meter for the first termination, watch the reduction in indicated
power when a double termination is applied. Be mindful that transmission
line lengths need to be short wrt operating frequency.

Some numbers:

1st termination reduces output by 6.02dB;
2nd termination reduces output by 3.52dB;
3rd termination reduces output by 2.50dB;

Owen
....

Revisiting the Power Explanation
 

On 22 Mar 2007 20:08:59 -0700, "Dr. Honeydew"
wrote:

What sort of output impedance does this generator have?

With the "accurate RF power meter" determination of 1W, it appears to
be a precision source (of 1W). The output impedance IS the 50 Ohm
resistance you applied to it. This loading at the output terminals is
a conventional usage of a precision source.

73's
Richard Clark, KB7QHC

Revisiting the Power Explanation
 

Dr. Honeydew wrote:
"We have a new signal generator here at the Labs, but it came without
documentation,"

Terman elaborates on Thevenin`s theorem on page 74 of his 1955 opus.
From Fig. 3-19 (b) on page 75, I would suggest loading the generator
with resistance until its open-circuit voltage is cut in half. That
resistance should equal the internal impedance of the generator which
you have already determined is resistive. Provided you don`t maintain
that sort of overload too long on your 1-watt source, you should do no
damage to it.

Best regards, Richard Harrison, KB5WZI

Revisiting the Power Explanation
 

On Mar 21, 4:25 pm, Dan Bloomquist wrote:
Keith Dysart wrote:
A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

I explained this last week, albeit for a different reason. I'll paste:
***
Hi Richard,
He says it in the last sentence. But here is an example. Take a 50 ohm
thevinin source. Power off, it looks like 50 ohms back into it. Take a
second thevinin source to represent a reflection and drive 5 volts into
the first source. Now set your first source 180 degrees to the
reflection and drive forward 5 volts.

(s)-----/\/\/\--------(c)-----/\/\/\--------(r)

(s)source (c)connection (r)reflection. With (s) 180 degrees out of phase
from (r), (r) will see a short at (c). It is because of the power
generated at the source that the impedance into it can look purely
reactive. And, you can use 5 ohms with 1 volt at the source, (c) will
still look like a short to (r). The source resistance doesn't matter as
long as a 'match' is made.

And for the same reason, why the 50 ohm line doesn't look like 50 ohms
is because of reflected power. Drive an open quarter wave line and it
looks like a short because the reflected voltage is 180 degrees out from
the source.
***

So you don't like my example? Well I can use yours then.

While you use the word 'power', the real analysis in your example is
all
done with volts. This is excellent and helps demonstrate my point that
'reverse power' is not needed as an explanation. We can carry on from
the analysis you have done and compute some powers. The real power at
(c)
is 0 Watts (the voltage is 0 at all times so using P=VI, the power
must
be zero). Assuming that when you say "drive 5 volts", you mean that
the
voltage source in the Thevenin equivalent generator is set to 10 V,
the 'forward power' at (c) is 0.5 W and the 'reverse power' is 0.5 W.
When subtracted, these produce the expected result of 0 W which agrees
with the actual computed power. All is well. (And yes, the Bird works
for determining this result).

Now consider someone who believes in the reality of 'forward' and
'reflected power'. There is 0.5 W of 'forward power' which reaches
the generator at the right. Since the impedance of this generator is
the
same as the characteristic impedance of the line (or the left
generator
in this example because there is no line), there is no reflection
so the 'power' must go into the generator. Similarly with no
reflection,
the 'reverse power' goes into the generator. Where does this power go?
If 'reverse power' is real, then it needs to be accounted. Since there
are many examples for which this accounting can not be done
it leads to the inescapable conclusion that 'reverse power' does not
really exist. (Please though, this does not mean that forward and
reverse voltages and currents do not exist. The fundamental error
being committed is that it is not valid to multiply the voltages and
currents computed using superposition to produce powers that actually
represent flowing energy).

Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

....Keith

Revisiting the Power Explanation
 

On Mar 22, 12:20 am, Cecil Moore wrote:

This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

I suggest that you have quite mixed up your impedances here
thus rendering all further analysis invalid.

At any point in the system there are 4 impedances.

There is the characteristic impedance looking left and the
characteristic impedance looking right. For systems in
sinusoidal steady-state, there is also the effective impedance
looking left and the effective impedance looking right.

The characteristic impedance is dependant only the elements of
the system. It does not depend on the length of the line or
the frequency of excitation. At changes in the characteristic
impedance, reflections occur. This impedance can be used for
transient as well as steady-state analysis of a system.

The effective impedance is dependant on the characteristic
impedances of the components of the system, as well as line
length and excitation frequency. It can only be used for
steady-state analysis. It does not cause reflections.

Until these impedances are kept straight in discussions, there
is no hope for correctness.

....Keith

Revisiting the Power Explanation
 

"Dr. Honeydew" wrote
Oh, and we thought to do another experiment. While the generator was
operating, we sent a short burst of RF down a 50 ohm transmission line
to the generator's output. Because the power measurements seem to say
the generator's output resistance is very low, we thought we would see
a return pulse, delayed by the round-trip time down the 15 meters of
high-quality cable. But we didn't see any echo. What could be going
on? We double-checked everything, and even looked into the line next
to the generator with a high impedance probe. We can see the burst
going in there, but still nothing comes back.
____________

Below is a link leading to a documented, real-world RF pulse measurement to
think about.

The reflection results from an antenna mismatch at the far end of ~1,600
feet of air-dielectric transmission line. A directional coupler and
calibrated, step attenuator allow for accurate measurement of the weak
return pulse amplitude in the presence of the much greater incident pulse.

Note that the reflection is (barely) seen even in the incident display,
time-aligned with its location in the reflected display.

Reflections DO exist in such systems, but measuring them requires
appropriate instrumentation and methodology.

http://i62.photobucket.com/albums/h8...easurement.gif

RF

Revisiting the Power Explanation
 

Keith Dysart wrote:
Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

Destructive interference at the generator output
terminal is all the accounting one needs.

Here's a mental aid to help see what happens to
the reflected energy. Assume the output impedance of
the generator and the characteristic impedance of the
line are Z1. Assume THE GENERATOR SEES Z2 AS A LOAD.
Z2 is a general case impedance and could have any
value depending upon the load ZLoad which is not
equal to Z1. The load is mismatched and therefore
Pref1 is not zero.

Gen---Z1 t-line----ZloadZ1
Pfor1--
--Pref1

You say reflected energy flows into the generator.
I'll show you why it does NOT and that total
destructive interference is responsible for
eliminating reflections at the generator.

Add any length of lossless transmission line
with a characteristic impedance of Z2 at the output
of the generator (nothing changes) so it looks like:

Gen---Z2 t-line---+---Z1 t-line----loadZ1
Pfor2-- Pfor1--
--Pref2=0 --Pref1

Pfor1 and Pref1 are the same values as above.

The Z2 t-line has NO reflected waves because it
is terminated in its characteristic impedance Z2.
The Generator continues to see Z2 as a load. One
has to admit that nothing at the Generator output
has changed. THE GENERATOR SEES EXACTLY THE SAME
LOAD IMPEDANCE IN BOTH CASES. NOTHING HAS CHANGED
IN THE SYSTEM EXCEPT OUR ABILITY TO ANALYZE IT.

Now a simple energy analysis will indicate what
happens to the reflected waves on the Z1 t-line. Since
Pref2=0, the system is Z0-matched to Z2 at point '+'.
Total destructive interference is what happens to the
reflected waves on the Z2 t-line. All of the reflected
energy is re-reflected back toward the load at the Z2-
match point '+'.

NONE OF THE REFLECTED ENERGY REACHES THE GENERATOR SO
NONE OF THE REFLECTED ENERGY CAN BE ABSORBED BY THE
GENERATOR IMPEDANCE.

So there you have it - a detailed mathematically based
reason why the reflected energy does not make it into
the generator in your example. To fully understand why
zero reflected energy flows in the generator, one needs
to understand superposition along with destructive and
constructive interference. This is covered in more detail
in my WorldRadio Energy Analysis article at:

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

I would recommend that everyone interested in this subject
read the "Optics", by Hecht chapters on superposition and
interference of EM waves. It is the best treatment that
I have seen in writing.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Keith Dysart wrote:
On Mar 22, 12:20 am, Cecil Moore wrote:
This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

I suggest that you have quite mixed up your impedances here
thus rendering all further analysis invalid.

No, I may have confused you but I did not mix up the
impedances. What the generator and reflected wave "sees"
above refers to the effective or virtual impedance, the
ratio of V/I designated by Z1, Z2, etc. The characteristic
impedances are designated by Z01, Z02, etc.

At any point in the system there are 4 impedances.

There is the characteristic impedance looking left and the
characteristic impedance looking right. For systems in
sinusoidal steady-state, there is also the effective impedance
looking left and the effective impedance looking right.

Yes, nothing I said disagrees with that. I will try to
draw an ASCII block diagram from now on.

The characteristic impedance is dependant only the elements of
the system. It does not depend on the length of the line or
the frequency of excitation. At changes in the characteristic
impedance, reflections occur. This impedance can be used for
transient as well as steady-state analysis of a system.

Yes, nothing I said disagrees with that.

The effective impedance is dependant on the characteristic
impedances of the components of the system, as well as line
length and excitation frequency. It can only be used for
steady-state analysis. It does not cause reflections.

Yes, nothing I said disagrees with that.

Until these impedances are kept straight in discussions, there
is no hope for correctness.

OK, I will insert the impedances for you in what you quoted
above.

This is a lot like the 1/2WL W7EL example in his food for
thought articles.

50wGen---1/2WL open-circuit stub---Z=infinity
Pfor=50w--
--Pref=50w

The generator is *NOT* matched to the load as it sees an
open circuit (effective Z=infinity) and cannot continue
to stuff 50 watts into the open circuit. The generator
is as mismatched as it can possibly be. The reflected wave
also sees that open circuit (Z=infinity) and is 100% reflected.
The reflected wave encounters the source wave which puts the
voltages in phase and the currents 180 degrees out of phase.
That's what happens at an open circuit.

Since the generator is not delivering any power and there is
a forward power and a reflected power, the reflected power is
supplying the forward power, i.e. being 100% re-reflected. The
re-reflection is associated with total destructive interference
toward the generator and total constructive interference
toward the load. Anything else violates the conservation of
energy principle.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Keith Dysart wrote:

On Mar 21, 4:25 pm, Dan Bloomquist wrote:

Keith Dysart wrote:

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

So you don't like my example?

Your example assumes that the reflected power will see the 50 ohms of
the generator. And I had shown you a condition where it will see a short
no mater what the 'output' impedance of the generator.

While you use the word 'power', the real analysis in your example is
all
done with volts. This is excellent and helps demonstrate my point that
'reverse power' is not needed as an explanation. We can carry on from
the analysis you have done and compute some powers. The real power at
(c)
is 0 Watts (the voltage is 0 at all times so using P=VI, the power
must
be zero).

It is a short. It is just that simple. If you have any kind of lab, take
a long piece of coax and drive it with a pulse and watch with a scope.
Find out for yourself that energy will reflect off a short.

Assuming that when you say "drive 5 volts", you mean that
the
voltage source in the Thevenin equivalent generator is set to 10 V,
the 'forward power' at (c) is 0.5 W and the 'reverse power' is 0.5 W.
When subtracted, these produce the expected result of 0 W which agrees
with the actual computed power. All is well. (And yes, the Bird works
for determining this result).

Now consider someone who believes in the reality of 'forward' and
'reflected power'. There is 0.5 W of 'forward power' which reaches
the generator at the right. Since the impedance of this generator is
the
same as the characteristic impedance of the line (or the left
generator
in this example because there is no line), there is no reflection
so the 'power' must go into the generator.

No. (c) is a short. It is just that simple. The power does not flow past
it 'into the other generator'. If you have any doubt that power is
reflected from a short you have not made the observation. And if you
have not made an observation that contradicts this, you can not make the
claim of a contradiction.

snip assumptions based on a false premise

Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

If you stick to the premise that the reflected power sees the generator
impedance it can't be 'proven wrong'. I can say that gravity doesn't
exist all day long but that won't make it so. The observation will
'prove' me wrong.

Revisiting the Power Explanation
 

On Mar 22, 10:43 pm, Richard Clark wrote:
On 22 Mar 2007 20:08:59 -0700, "Dr. Honeydew"
wrote:

What sort of output impedance does this generator have?

With the "accurate RF power meter" determination of 1W, it appears to
be a precision source (of 1W). The output impedance IS the 50 Ohm
resistance you applied to it. This loading at the output terminals is
a conventional usage of a precision source.

73's
Richard Clark, KB7QHC


Thank you, Mr. Clark. But I'm having trouble reconciling what you
wrote with what Mr. Harrison posted about the output impedance being
equal to the conjugate of the load which dissipates the most power.
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so
we don't see how your answer and Mr. Harrison's posting can both be
correct.

From the labs,
Dr. Honeydew

Revisiting the Power Explanation
 

On Mar 20, 3:37 pm, Walter Maxwell wrote:
One of the issues discussed in this thread that Owen originated concerned whether or not reflected power
enters the power amp and dissipates as heat in the plates of the amp. Some of the posters apparently are
unable to appreciate that the reflected power does not cause heating of the amp, unless the reflected power
detunes the amp and the amp is left detuned from resonance, which of course is not the correct manner of
operating the amp.

In the last post of the original thread I presented the details of an experiment I performed (one of many
using the same procedure) on a Kenwood TS-830S transceiver that proves how and why reflected power in no way
causes heating of the amp when the amp is properly adjusted in the presence of the reflected power.

Usually, such a presentation as in the last post in that thread evokes a great deal of response, as for
example, Art Unwin's. So I'm somewhat surprised, and a little disappointed that my post has resulted in total
silence. Have my efforts in helping to solve the problem gone for naught?

Walt, W2DU

I have followed this thread with interest There are some who seem to
dispute the existance of reflected power and its ability to do damage.
My answear is that reflected power can certainly do so any one want to
run a completly detuned antenna to prove me wrong may do so at their
expence and risk. as to it causing heat in the amp of course not the
extra work that the amp has to do to "push" the signal to the antenna
does. if a semiconductive device is over worked then it will produce
heat.
The old thermonic valves are far more tolerant than semiconductors.
mike M0DMD