Steve Nosko wrote:

"Cecil Moore" wrote:
Even SQRT(L/C)???? The Z0 of transmission line is a resistance,

I don't consider that term (resistance) suitable for this situation.
"real part of Z" is better.

Steve, the "real part of Z" is the IEEE definition of "resistance",
i.e. they are exactly the same thing.

Z = R + jX Guess what the R stands for? Methinks the problem is that
people confuse "resistor" and "resistance". A resistance is not
necessarily a resistor. More often than not, it is simply a V/I ratio.
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
"Remember, the Zo is properly called "Characteristic impedance and Surge
impedance."

No doubt impedance is used because some are uncomfortable calling a
resistance a resistance when it wastes no energy. I think this is
unfortunate because it weakens the important idea that resistance is
first and foremost a ratio of voltage to current in which the voltage
and current are in-phase. Those who insist that every resistance wastes
electrical energy by converting it to heat are still firmly stuck in the
d-c era.

Terman writes on page 88 of his 1955 edition:
"The characteristic impedance Zo is the ratio of voltage to current in
an individual wave--- it is also the impedance of a line that is
infinitely long or the impedance of a finite length of line when ZL=Zo.
It will be noted that at radio frequencies the characteristic impedance
is a resistance that is independant of frequency."

Best regards, Richard Harrison, KB5WZI

"Cecil Moore" wrote in message
...
Steve Nosko wrote:
"Cecil Moore" wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

I don't know from "momentum" of power. If you asking what is going on
within the section of ladder line... The easy explanation is that the
waves
act such that the two ends look line 50 ohms

It's a pretty simple question. The energy wave reflected from the load
possesses energy and momentum, both of which must be conserved. There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax. So what changes the direction
and momentum of the energy wave reflected from the load?

oh wow, all the way to momentum calculations now. is it time to point you
all at a design for a pure electromagnetic rocket engine that uses the
momentum of traveling waves to push itself along??

Nick, WA5BDU wrote:
"3) Since (1) and (2) are both good, they must be equivalent to each
other."

(1) is the maximum power transfer theorem.

(2) is an efficiency statement.

(1) and (2) are dissimilar.

(1) says that the only way to get all the power you can from a source is
to adjust the load resistance to equal that of the source and to
eliminate any reactance which opposes power transfer. This is
demonstrable; no need to argue.

(2) Resistance of the type which converts electrical energy to heat and
that is in the path of energy flow extracts energy and reduces
efficiency. It is indubitably minimized to maximize efficiency.

The argument in this thread is on the existence of lossless resistance.
Proof of the existence of lossless resistance is demonstrable. We have
power sources delivering maximum power without dissipating as much power
within as is being delivered to the load. That is conclusive.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
[...]
The argument in this thread is on the existence of lossless resistance.
Proof of the existence of lossless resistance is demonstrable. We have
power sources delivering maximum power without dissipating as much power
within as is being delivered to the load. That is conclusive.

Best regards, Richard Harrison, KB5WZI


Ahhh! Thanks, Richard. Now I see the essence of the discussion. I believe
there is an inherent conflict in the terms you have, namely; "delivering
maximum power" and "without dissipating as much power...as ...the load".

This is an interesting tack, but I don't see the demonstration you refer
to. I am missing just what it is that gets to the "Loss-less resistance"
conclusion.

We know that the MPT Theorem says the source and load powers must be equal
for THAT situation...so. I think (I hope) you agree that in this case, the
limiting factor for the delivered power is the source's "Internal"
resistance. Now lets look at a case where the source Z is LESS than the
load Z.

I will take for a first example (at the risk of being accused of being
"stuck" in DC) a standard series regulated power supply. (this is the same
example I described with my "change the load Z to a higher value" example
earlier.) The supply can be designed to supply many watts, right! Yet they
have very low output resistances, right!

Lets look he

1- Is it putting out its "maximum power" when loaded to its rating?

2- Is it dissipating an amount equal to the load?

3- What is the limiting factor for its output current (what is it that
allows us to only draw its rated current?)?

SO, are you saying that we have some "Loss-Less Resistance", in this case,
which equals the load resistance?

If we were to put a load on it which equaled the 0.05 ohm output resistance,
then what situation do we have? Maximum power transfer?



In the above case, the series pass transistor power dissipation ability is
the limiting factor on the output power of the supply, not the internal
resistance. I can make the same case for a switching supply and my "raise
the supply voltage" example is the same thing.

SO...?

--
Steve N, K,9;d, c. i My email has no u's.

On 9 Mar 2004 10:01:04 -0800, (Nick Kennedy) wrote:

I see this paper as a variation on a line of reasoning that goes like
this:

1) A conjugate match results in maximum power delivered to a load, so
it is good.

It is also a cliche unless another outcome was expected. As there are
other types of match, "good" becomes a speculation of value
judgements.

2) A connection where the load has much higher resistance than the
Thevenin equivalent source resistance results in high efficiency, so
it is also good.

At this point it is well to point out that there is no expectation of
"efficiency" implied or expressed in conjugation. Good follows from
application, application does not drive good.

3) Since (1) and (2) are both good, they must be equivalent to each
other. Therefore a conjugate match is what it is not. This is an
apparent contradiction.

"is what it is not" I am the Walrus? Coo-coo ca choo!

Two goods have no inherent relationship. Scratching an itch, and
solving world hunger also have no equivalency. Contradictions and
correlations can be forced across innumerable topics by this standard.

4) The contradiction is resolved by postulating a special kind of
resistance that adds to the source resistance. However, it has no
physical effect and exists only to resolve the contradiction in (3).

This has descended into the conventional mixing of theories to argue
one example. There is no reason to call on any resistance to perform
matching for efficiency's sake. There is no reason to enforce
efficiency upon a conjugate match.

A Conjugate Match is simply the observation of the inverse Load
Impedance presented to it by the source. The quality of that is that
the combination of inverse reactances yields a purely resistive
solution. Conjugation is an artifice to protect the source from
circulating currents and elevated potentials, it has no other purpose.

The Zo Match is not a Conjugate Match although it may be in special
circumstances. A Conjugate Match is not a Zo Match although it may be
in special circumstances. The Zo Match closely mimics the Thevenin
topology.

The Thevenin Theorem never contained a resistance to later argue its
necessity. Thevenin offered a source with an Impedance. An Impedance
"may" contain resistance, but it is not a necessity. The argument for
dissipationless resistances within the source automatically reject
themselves from the Thevenin category through being non-linear.
Attempts to re-embrace this linearity are convincing (the flywheel
argument of the finals' plate tank); but it does not deny the physical
resistance rendering heat that inhabits the source interior.

73's
Richard Clark, KB7QHC

Dave wrote:
oh wow, all the way to momentum calculations now.

I don't think we should expect to see any momentum "calculations" from
Cecil. He can't even explain how the energy magically reverses
direction and heads back toward the load. That's why he always phrases
that as a question. The fact is, in a matched system, energy obviously
isn't reflected from the load.

73, Jim AC6XG

Yikes! Long wind alert!

"Cecil Moore" wrote in message
...
Steve Nosko wrote:
"Cecil Moore" wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

I don't know from "momentum" of power. If you asking what is going on
within the section of ladder line... The easy explanation is that the
waves
act such that the two ends look line 50 ohms

It's a pretty simple question. The energy wave reflected from the load

First I must understand where we are looking. I think is is best to
take small steps in the descriptions. (or is this a troll, Cecil) You want
to look at a point on the 450 ohm line, is that correct?

Assuming yes... I think you are saying that we have a 450 ohm line and a 50
ohm load and therefore there is a reflection. Is that where you are?

....pause...



I would rather take one step at a time and stick to the above as the main
discussion, but will comment as follows anyway, so the following may be a
digression.

possesses energy and momentum, both of which must be conserved.

Again, I do not know what momentum is for an EM wave on coax and don't
believe it is necessary for this discussion. Waves move along the line and
we are trying to come to a common understanding of what happens at the
junctions, I think... As far as conservation is concerned, I assume a
loss-less line.


There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax.

I think you contradicted yourself here. Are you saying that there IS or
IS NOT reflected energy/waves on the 50 ohm section (to the left of the
"--x--"??

I say there is none. It is loaded by the 1/2 wave of 450 ohm line which
presents a 50 ohm appearance. To be honist, I don't think I can remember
what happens at this junction. I think there is a jump in voltage since 450
is greater than 50 - for the forward wave. In any case, the sum result of
the forward wave and the reflected wave is such that E/I=50 at this
junction. This is just to the right of the "--x--".



So what changes the direction
and momentum of the energy wave reflected from the load?

You lost me here. Perhaps you are asking; "If there is a wave
reflected at the end with the 50 ohm, back toward the left, how does this
power then get absorbed IN that load?" Is this the question?

The wave, in the 450 section, reflected from the load is NOT the power being
delivered TO the load, so it does not have to be "changed' to be delivered
there. The wave traveling to the right on the 450 ohm section is
*significantly* diffeernt that what is absorbed in the load. I'd be hard
pressed to give a value since school was some odd...mumble mummble years
ago... Perhaps this is a stumbling block. The wave traveling to the right
on the 450 section can be much larger than the power delivered to the load.
Weird, but true. It has to be if there is some bounced back to the left and
still deliver 100W to the load. I'm sure there is some math to show
this...reflection coef and all...

73,

--
Steve N, K,9;d, c. i My email has no u's.

On Tue, 9 Mar 2004 18:01:12 -0600, "Steve Nosko" Ahhh! Thanks,
Richard. Now I see the essence of the discussion. I believe
there is an inherent conflict in the terms you have, namely; "delivering
maximum power" and "without dissipating as much power...as ...the load".

This is an efficiency statement.

This is an interesting tack, but I don't see the demonstration you refer
to. I am missing just what it is that gets to the "Loss-less resistance"
conclusion.

We know that the MPT Theorem says the source and load powers must be equal

It says nothing of the sort. The comparisons are Z based.

for THAT situation...so. I think (I hope) you agree that in this case, the
limiting factor for the delivered power is the source's "Internal"
resistance. Now lets look at a case where the source Z is LESS than the
load Z.

I will take for a first example (at the risk of being accused of being
"stuck" in DC) a standard series regulated power supply. (this is the same
example I described with my "change the load Z to a higher value" example
earlier.) The supply can be designed to supply many watts, right! Yet they
have very low output resistances, right!

You started with a Z and ended with an R. As you are confined to a DC
argument that is allowable, but the logic does not extend to an RF
source.

Lets look he

1- Is it putting out its "maximum power" when loaded to its rating?

Rating is a marketing issue.

2- Is it dissipating an amount equal to the load?

Only if the load R equals the internal Ohmic loss.

3- What is the limiting factor for its output current (what is it that
allows us to only draw its rated current?)?

Thevenin explicitly describes the source Z as the ratio of open
circuit voltage divided by short circuit current. Substitute R for Z
in this special case.

The internal Ohmic loss defines the total current available. This is
not necessarily the rated current as the system components may not
tolerate the heat even though the current is "available." This is the
distinction of the "marketing."

SO, are you saying that we have some "Loss-Less Resistance", in this case,
which equals the load resistance?

You have crossed the line from DC to other less restricted examples.
Your logic will fail at some point. The "Loss-Less Resistance"
cannot avoid Ohmic losses; but through clever engineering, DC supplies
(switchers) can reduce the amount of current through them. If you put
as large of a conventional heat sinking on switchers, their capacity
would rise (that marketing term of "rated"), but then it intrudes on
the quality of switchers for being small (and is a facile response
that belies more complex issues).

If a switcher is 99% efficient, it still dissipates 1% power in heat
from Ohmic loss.

If we were to put a load on it which equaled the 0.05 ohm output resistance,
then what situation do we have? Maximum power transfer?

You've switched back (no pun) to the DC argument? I presume so. Can
you envision any other outcome? If we are to allow an infinitely
adaptable source, then this becomes a tediously futile exercise.
However, that is often the argument offered (hence the violation of
Thevenin's requirement for linearity).

In the above case, the series pass transistor power dissipation ability is
the limiting factor on the output power of the supply, not the internal
resistance.

It would be difficult to separate the two wouldn't it? This delves
back into the marketing term of "rated." You can certainly design a
finals deck without a heat sink to power limit it. Anyone can design
for failure to prove that outcome.

The actual, real answer is that the power out is current density
limited at the emitter-collector junction. Nearly all failures of
solid state power designs devolve to melt-down, a catastrophic thermal
failure borne of resistance. The current density is an abstract
expression of the inability of the device to shed heat through another
resistance: thermal resistance which is expressed in degrees per Watt.
If the junction temperature rises to melting through the concurrent
availability of so many Watts - BINGO! Failure. In the parlance of
the wire trade, this is called the fusing current.

I can make the same case for a switching supply and my "raise
the supply voltage" example is the same thing.

Only if you neglect heat, sinking, thermal resistance, melt-down and
the host of real concerns.

SO...?

Hi Steve,

If you care to look at any power transistor operating curves,
absolutely NONE run to "rated" capacity. Across the board, they are
"de-rated" at the upper limit (observe the dashed lines that clip the
maximum region). Like the specification of GainBandWidth, power
ratings are mutable against a myriad of considerations, most of which
do not allow full blown power.

73's
Richard Clark, KB7QHC

On Tue, 9 Mar 2004 18:34:43 -0600, "Steve Nosko"
wrote:

Yikes! Long wind alert!
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load
I don't know from "momentum" of power. If you asking what is going on
within the section of ladder line...
....
take small steps in the descriptions. (or is this a troll, Cecil)
....

Is that where you are?
...pause...
....
Again, I do not know what momentum is for an EM wave on coax and don't
believe it is necessary for this discussion.
....
I think you contradicted yourself here. Are you saying that there IS or
IS NOT
....
You lost me here. Perhaps you are asking
....
Is this the question?

This is going to take a loooooooooong time, Steve. Follow your
instincts and accept it only for its entertainment value.

73's
Richard Clark, KB7QHC