Roy Lewallen wrote:

We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the
load. Once you admit that fact, everything else will be moot.

Let's try again. The source is providing 40 watts, 32 watts of which is
delivered to the transmission line. The transmission line is
transferring this 32 watts of power to the load. In the transmission
line, we can calculate that there's 50 watts of "forward power", and 18
watts of "reverse power".

And it is easy to prove that the source has generated 50+18=68
watts that have not been delivered to the load. So I ask
you: Where are those 68 joules/sec located during steady-state
if not in the forward and reflected power waves? Why will
68 joules/sec be dissipated in the system *after* the source
power is turned off?

If those 68 joules/sec that have been generated by the source
but not delivered to the load are not in the forward and reflected
power waves, exactly where are they located? There's really no
sense in continuing this discussion until you answer that question.
Everything else is just a side argument.

The answer to that question will expose the errors in your premises.
You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.
But that exact amount of energy was supplied during the power-on
transient state and will be dissipated during the power-off transient
state. If it's not in the forward and reflected power waves, you
are going to have to store it somewhere else. Where is that
somewhere else?

The source has supplied 68 joules/sec that has not reached the
load. The forward and reflected power waves require 68 joules/sec.
That you don't see the logical connection between those two equal
energy values is amazing.

But I will get you started on an understanding of the component
powers using an S-parameter analysis.

How much of that 18 watts of reverse power is
going through the source resistor to reach the source to "engage in
destructive interference"?

reference the S-parameter equation: b1 = s11*a1 + s12*a2

I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts

The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts

What does it interfere with?

From the S-parameter equation above, it obviously interferes
with s11*a1 .

Please reference HP App Note 95-1, available on the web. It should
answer most of your questions, in particular pages 16 & 17.
|a1|^2 = Power incident on the input of the network
|a2|^2 = Power incident on the output of the network
|b1|^2 = Power reflected from the input port of the network
|b2|^2 = Power reflected from the output port of the network
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil,

You completely ducked the question. How did those waves get there in the
first place? Hint: there are no laws for conservation of waves or
continuity of waves.

It is easy to set up a problem with physically unrealizable inputs. It
is pointless to try to solve such a problem, however.

We've been around this track a couple of times before. Neither of us has
changed.

Bye.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

Cecil,

Nice try.

You first.

Describe how you set up this coherent wave/anti-wave pair that happily
travel together for some indeterminate distance. Then I will describe
what happens when at some arbitrary point and time they decide to
annihilate.


Sure, here's the two coherent reflected waves that cancel at a
Z0-matched impedance discontinuity in a transmission line.

b1 = s11*a1 + s12*a2 = 0

I'm sure you recognize the S-parameter equation for the reflected
voltage flowing toward the source which is the phasor sum of two
other reflected voltages.

They don't travel together for some indeterminate distance. They
are cancelled within the first dl and dt. And they don't annihilate.
They simply cancel in the rearward direction.

Incidentally, if you square both sides of the equation you get

b1^2 = s11^2*a1^2 + s12^2*a2^2 + 2*s11*a1*s12*a2

Pref1 = rho^2*Pfor1 + (1-rho^2)*Pref2 + interference

The forward voltage equation toward the load is b2 = s21*a1 + s22*a2

Richard
Thanks for your explanation, I'm still thinking it over since its been 40
years since I've studied any tube theory. .. I recall many years ago
watching a guy trying to fix an old radio. IIRC it had an 80 rectifier and
right after the radio was turned on and started to warm up a blue cloud
could be seen in the tube as the plates started to turn redder and redder.
He would turn it off, change a part and check it again. I don't think he
fixed it before the 80 went south. Later I figured out that he probably
had a shorted filter cap.
In this case the diode's plate dissipation rating was exceeded and it
melted. The electrons slamming into the plate have a lot of kinetic energy
to transfer. and heat up the plate. This results in lower efficiency as
this power isn't delivered to the load. Now the diodes resistance can be
easily calculated but I'm not sure how to visualize it. Is this where your
term cathode resistance enters the picture?
tnx

--

73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Mon, 06 Jun 2005 20:52:52 GMT, "Henry Kolesnik"
wrote:

But in any event for something to melt we need dissipation and only
resistance can do that!


Hi Hank,

Unfortunately as much as you and I agree on that bedrock principle,
others with Simpson Ohmmeter in hand would glare goggle eyed at us and
say that plate has no resistance to speak of and that no amount of
current through its Ohmic resistance could ever bring about enough
heat to produce the effects so obviously witnessed. One of the most
enraging questions I've asked
"If it is not the value I've offered, what value is it?"

Well, I've never been given a quantitative answer, however I've seen
enough carefully crafted mathematical proofs in this group to replace
substantive results so easily seen. There is some irrefutable logic
in circulation that clearly reveals that what we've experienced just
couldn't be.

Glasses will be need to be readjusted for such extreme myopic
aberrations. There are two principles involved in what is called
Plate Resistance, and the first and foremost is not even related to
the plate at all. It is called the work function of the cathode
emissivity. So, in fact it is more proper to refer to this usual loss
as Cathode Resistance, not Plate Resistance. The cathode is the
fundamental limit on power generated.

What Plate Resistance is, is the ill termed substitution for Plate
dissipation. If folks want to work their Simpson, they would blow an
aneurysm trying to measure the resistance from cathode (filaments have
the same work function issue too) to plate. In fact, the hobby horse
argument of it is not resistance at all, but some figurative charting
artifice called a "load line" usually appears in the last gasp.

Plate Dissipation is resistance clear and simple in spite of the
failure of conventional tools to measure a common physical property.
Newton would have recognized it, it is called inertia.

Once the work function is overcome (the job of the grid), then Plate
voltage dominates through the acceleration of charge beyond the grid,
toward the plate. That stream of electrons (and there is no doubt
about actual current flow in easily counted, significant populations
of electrons) is elevated to 90% the speed of light. This current
flow is entirely different from what current flows in the remainder of
the Plate load. That is also known as displacement current and
electrons are shuffling along at a placid meter per second rate.
Plate current and displacement current are equal in amplitude and
phase, but not in motion nor kinetics.

NOW. When that same stream encounters the Plate - WHAM! If anyone
here has walked into the wall, and NOT encountered resistance, then we
will call you Casper.

Inertia reveals that to slow a mass in a distance results in
acceleration (negative in this instance) and that property is called
Force. Force over time expends calories and is expressed in any
number of systems and units - Watts is one, Degrees is another. We
could abstract to Horsepower and Candelas (the plate glows too).

We know the speed, not many here would give it much though, but none
would know the length interval of going at that speed to going zero
(0). It is roughly two atoms distance into the metal of the plate. I
will leave those calculations of Force to the student to compute or I
can provide it from notes of correspondence with Walt Maxwell and
Richard Harrison from a round robin discussion several years ago.

Hank, does this fulfill your earlier question as to "what" is
happening? I first gave you many examples, I hope this segue into
real physics fills in their actuality. Too many correspondents demand
that I open the source and point at a 50 Ohm carbon composition
resistor that is the "source resistance."

73's
Richard Clark, KB7QHC

Gene Fuller wrote:
You completely ducked the question. How did those waves get there in the
first place? Hint: there are no laws for conservation of waves or
continuity of waves.

I answered the question in another posting. The waves got there
during the power-on transient state. Conservation of energy is
assumed. Hope you don't disagree with that principle.

If we make Roy's lossless 50 ohm feedline one second long (an
integer number of wavelengths), during steady-state, the source
will have supplied 68 joules of energy that has not reached the
load. That will continue throughout steady state. The 68 joules
of energy will be dissipated by the system during the power-off
transient state.

What you guys are trying to do is hide 68 joules of energy that
cannot be destroyed. Where can you hide it in a transmission line
to prove that it is not there in the forward and reflected waves?
What is your agenda in trying to deny/hide/disguise/ignore that
68 joules of energy?

In the one second example, the forward and reflected waves require
68 joules of energy for their existence in the feedline. The source
has supplied 68 joules of energy that has not yet reached the load
so it must necessarily still be in the feedline. Wonder where
the energy in the forward and reflected waves came from? Shirley,
you jest!

Incidentally, QEX wants to publish my article.
--
73, Cecil http://www.qsl.net/w5dxp


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Please reference HP App Note 95-1, available on the web.

There's only one in every half a million of radio amateurs who have a
copy of this document, or have ever heard of its existence, let alone
having any chance of finding a readable copy of it within the next
dozen years. By which time they will have changed their hobby to
keeping tropical fish or making Newtonian telescopes. Or just died.

In all likelihood they won't be able to make any sense out of it
anyway.

Cec, why do you bother to mention it? (smiley)
----
Reg.

Reg Edwards wrote:

W5DXP wrote:
Please reference HP App Note 95-1, available on the web.

There's only one in every half a million of radio amateurs who have a
copy of this document, or have ever heard of its existence, let alone
having any chance of finding a readable copy of it within the next
dozen years.

Give me a break, Reg. Download it for free and enjoy from:

http://www.sss-mag.com/pdf/hpan95-1.pdf

In all likelihood they won't be able to make any sense out of it
anyway.

My dog isn't able to make sense of it either - poor dog.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Roy Lewallen wrote:

[No, I didn't. Cecil wrote the following paragraph.]

We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant. It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor or a DC current through an inductor (which, in fact, is
exactly what the feedline stored energy consists of) -- it doesn't have
any effect on an AC analysis.

[I did write this one.]

Let's try again. The source is providing 40 watts, 32 watts of
which is delivered to the transmission line. The transmission line
is transferring this 32 watts of power to the load. In the
transmission line, we can calculate that there's 50 watts of
"forward power", and 18 watts of "reverse power".


And it is easy to prove that the source has generated 50+18=68 watts
that have not been delivered to the load.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, not 60. 32 of those are
delivered to the load and 8 to the source resistor. I guess you mean 68
joules -- but as I said, it's irrelevant.

So I ask you: Where are
those 68 joules/sec located during steady-state if not in the forward
and reflected power waves? Why will 68 joules/sec be dissipated in
the system *after* the source power is turned off?

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.


If those 68 joules/sec that have been generated by the source but not
delivered to the load are not in the forward and reflected power
waves, exactly where are they located? There's really no sense in
continuing this discussion until you answer that question. Everything
else is just a side argument.

You tell me -- they can be anywhere you'd like. Just answer the simple
questions about the power "waves".


The answer to that question will expose the errors in your premises.

What exactly is my premise, please? All I've done is to give the
currents and powers at significant points in the circuit. Are any of the
values incorrect? It's you who has the premise, not me.

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption. I'm questioning the existence of
traveling waves of average power, and so far you've failed to give any
evidence to convince me otherwise.

But that exact amount of energy was supplied during the power-on
transient state and will be dissipated during the power-off transient
state. If it's not in the forward and reflected power waves, you are
going to have to store it somewhere else. Where is that somewhere
else?

You tell me. My analysis doesn't need to consider the stored energy at
all. Apparently yours does, so have at it.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

The forward and reflected power waves require 68 joules/sec.

The "forward power" is 50 watts. The "reverse power" is 18 watts. It
requires 32 watts to sustain this. That's the amount of power flowing
through the transmission line, from source to load. That's 32
joules/second, not 68. If the line were open circuited, the forward and
reverse powers would be equal, and it would take no power to sustain them.

That you
don't see the logical connection between those two equal energy
values is amazing.

But I will get you started on an understanding of the component
powers using an S-parameter analysis.


Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

How much of that 18 watts of reverse power is going through the
source resistor to reach the source to "engage in destructive
interference"?


reference the S-parameter equation: b1 = s11*a1 + s12*a2

I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts


Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts

What's (s22*a2)^2? The forward power wave? The reverse power wave? If
it's something else, does it have a name? Where does it go? Is it
getting dissipated in the source resistor, reflected at the transmission
line/resistor interface, reflected at the source/resistor interface, get
radiated, or what?

What does it interfere with?


From the S-parameter equation above, it obviously interferes with
s11*a1 .

What's s11*a1? It must be something inside the source. The source is
just that, a source. It has (AC) voltage and current, 100 volts of
voltage and 0.4 amps of current. Does s11*a1 reside inside every source,
or only some special ones? Apparently 11.52 watts of this s11*a1 gets
cancelled by the reverse power wave. How much of it is left over?


Please reference HP App Note 95-1, available on the web. It should
answer most of your questions, in particular pages 16 & 17. |a1|^2 =
Power incident on the input of the network |a2|^2 = Power incident on
the output of the network |b1|^2 = Power reflected from the input
port of the network |b2|^2 = Power reflected from the output port of
the network

Nope.

Enough hand waving and evasion, we've been here before. [Of all the
questions, the sole quantitative answer was "(s12*a2)^2 = 11.52 watts".]
I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence. Have fun -- I've got actual work to do while you
take care of the visionary leadership part.

Roy Lewallen, W7EL

On Tue, 07 Jun 2005 15:01:26 GMT, "Henry Kolesnik"
wrote:

Now the diodes resistance can be
easily calculated but I'm not sure how to visualize it. Is this where your
term cathode resistance enters the picture?

Hi Hank,

Getting electrons to "boil" off the cathode (or filament, same thing)
is not a simple task otherwise there would be no filaments needed.
Even with filaments, Edison current is not very considerable unless
you add a monomolecular layer of metal to the surface of either the
filament, or the cathode.

You may note that some tubes are described as having "Thoriated"
filaments. This is that monomolecular addition. Its purpose is to
lower the W, the work function of the interface.

It is far from odd how physics demonstrates at every stage and in
every discipline that interfaces where there is mismatch, there is
difficulty in transfering power. When an electron from the interior
of the metal crystal approaches the surface, it is repelled by that
interface due to the potential of the work function, and is attracted
by the bulk material behind it. The Thoriated surface offers a
matching mechanism between the bulk metal and the free space beyond
the surface. In classic Optics this is known as Index Matching.

To give examples as to how well a monomolecular addition performs:

A Tungsten filament (no treatment) offers a current density of ½A/cM²
This makes for a baseline.

A Thoriated Tungsten filament offers a current density of 4A/cM²

Then we step back to a cylindrical cathode employing Barium.
Such a cathode offers a current density of ½A/cM²

That seems rather regressive to use a cathode, but temperatures are
telling. The simple Tungsten filament is operating at 2500° K and the
cathode needs only to simmer along at 1000° K. This gives
considerably longer life and more efficiency (most of the power for
heating is lost through radiation). Needless to say, cathodes find
more application in low power circuits, or their surfaces are treated
with other low work function metals for greater emission.

Now, when you add a potential gradient, you also lower the work
function of the surface (but it is always an advantage to have it
lowered going into this game). This is called the Schottky effect.
One might be tempted to simply ask, why don't we up the voltage and
discard the filament? This device would be called a Cold Cathode but
the potential gradient then rises to the level where you run the risk
of secondary emission.

When that electron stream strikes the plate and raises the
temperature, it is just short enough power to present this secondary
emission. But if we were to run at 17KV or so, then the electron
stream would be so aggressive as to produce high energy effects such
as X-Ray emission.

So, to return to the resistance of this all, we have physical
impositions of cathode surface area (probably offering the prospect of
being greater than filament surface area), current density, and
potential difference. Discarding all the extraneous surface units
(cM²) and employing the proper division (E/I) we have a resistor that
glows in the dark under extremes of operation. How this fails to be
source resistance is strictly handwaving and the schematic symbolic
mysticism of demanding a carbon composition resistor.

73's
Richard Clark, KB7QHC

On Tue, 07 Jun 2005 11:44:29 -0700, Richard Clark
wrote:

When that electron stream strikes the plate and raises the
temperature, it is just short enough power to present this secondary
emission. But if we were to run at 17KV or so, then the electron
stream would be so aggressive as to produce high energy effects such
as X-Ray emission.

Something felt wrong here. I should have said:
...just short enough energy to present this secondary emission.

Roy Lewallen wrote:
Cecil Moore wrote:
We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant.

Do you think it is random coincidence that the amount of energy
stored in the feedline is *EXACTLY* the amount of energy required
by the forward and reflected waves???? The fact that you think
it is irrelevant is simply a flight into a wet dream fantasy.

It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor ...

What a coincidence! It's *EXACTLY* the amount of energy required
by the forward and reflected waves that you say don't exist - and
it's RF photons, not DC, so it must travel at the speed of light!
You cannot store photons in a capacitor.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, ...

Of course, but during the power-on transient period, 68 watts is NOT
delivered to the load. Please don't make me waste my time calculating
the forward and reflected power during the power-on transient period.
When you actually do those calculations, you will agree with me that,
during the power-on transient phase, 68 watts of power has been stored
in the feedline and remains there until the power-off transient phase.
It is *EXACTLY* the amount of power required by the forward and
reflected waves.

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.

Which must necessarily include the energy stored in the feedline during
the power-on transient condition because it is *still there during
steady-state*. Ignoring the energy stored in the feedline during the
power-on transient phase is both irrational and illogical. It could
even be bad for your mental health.

What exactly is my premise, please?

from my earlier posting:

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption.

I'm sorry, Roy, but that is just BS! Either you admit there is enough
energy to support the steady-state forward and reflected waves or you
don't.

My analysis doesn't need to consider the stored energy at all.

And that is exactly why your analysis is wrong. You have, once again,
been seduced by the steady-state model and are spreading old wives'
tales as a result. Hopefully, you don't really want to do that.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

68 joules/sec in your original example. 68 joules in my one-second-
long feedline example.

Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

The 68 joules were stored in the feedline during the power-on
transient phase. They are still there during steady-state. An
S-parameter analysis yields the correct results because it
includes the reflected power, |a2|^2, as an energy source,
something you deny. Wonder what the S-parameter analysis folk
know that you don't choose to admit?

Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details. Stand by. With
an unprejudiced open mind, you might actually learn something.

What's (s22*a2)^2?

Same as (1-rho^2)*Pref2 in ham terms. Reflected power from the
load that is re-reflected back toward the load by an impedance
discontinuity.

What's s11*a1?

Same as Vfor1*rho in ham terms. Forward voltage that is reflected
back toward the source by an impedance discontinuity.

I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence.

Do you realize what intellectual shape you would be in if you had
adopted that attitude when you were one year old? :-)
--
73, Cecil http://www.qsl.net/w5dxp


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