Bob,

Sorry for the strong words, but the intent remains the same.

I don't know what happens to the source impedance when the load is
changed, because the system is undefined.

However, read carefully the definition at the beginning of your paper.

The Maximum Power Transfer Theorem:

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 impedance looking into
the two networks at the junction are conjugates of each other
[1]

On Tue, 02 Mar 2004 15:37:25 GMT, Gene Fuller
wrote:

Bob,

Sorry for the strong words, but the intent remains the same.

I don't know what happens to the source impedance when the load is
changed, because the system is undefined.

However, read carefully the definition at the beginning of your paper.

The Maximum Power Transfer Theorem:

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 impedance looking into
the two networks at the junction are conjugates of each other
[1]
.
.
.
[1] W. L. Everitt, "Communication Engineering", McGraw-Hill, 1937

The maximum power transfer theorem describes the impact from change of
the load impedance. It is not the Grand Unified Theory for all the
universe. If someone tries to expand this elegant concept to all sorts
of pathological cases then it is likely that confusion will ensue.

The MPTT analysis is straightforward if the problem is well defined. The
ongoing argument in amateur radio circles is about the source
characteristics of amplifiers and tank circuits. The MPTT does not
address that argument, but rather it is a victim of the silliness.

Including the down-home touch of steam engines adds nothing to the
technical content.
Dear Gene,

No problem, I will try to understand where you're coming from.
Obviously, there have been some experiences in your tour of duty that
have caused you some heartburn - it happens with all of us.

Relative to your points about amplifiers, etc., I hope to have my
article on that topic available before the end of this month.
Actually, it has been "almost ready" for several days now, but there
are some things that need to "cook" awhile before I release it. As
soon as you get into the details of a Class-B linear, you're up to
your butt in alligators, not to mention all of the myths surrounding
them and the matching problem. I'll do my best to keep it "on topic" -
Hi!

73,
Bob

Gene Fuller wrote:
"I don`t know what happens to the source impedance when the load is
changed, because the system is undefined."

Yes. Maximum power transfer is accomplished by making the load impedance
the conjugate of the generator impedance as defined by Thevenin`s
theorem.

The value of the Thevinen impedance is that which might be measured by a
generator`s open-circuit voltage devided by its short-circuit current.
You don`t need to know the generator`s specifics other than, drop in the
output voltage is proportional to the current delivered.

The current which flows in a linear load impedance connected to a
Thevenin generator is the open-circuit voltage divided by the sum of the
generator`s internal impedance and the load impedance. These may be
complex impedances.

At maximum power transfer, internal and load impedances are equal in
resistance and their reactances are conjugate (opposite and equal).

There is no requirement that resistance in either the generator or load
be dissipative, and frequently, lossless resistance is a part of the
generator impedance so that we can get maximum possible power into the
load without losing 50% in the generator.

Best regards, Richard Harrison, KB5WZI

"Lossless resistance?" Would that be zero resistance,
or perhaps a negative resistance, as in the active part of
a tunnel diode's V-I characteristic?

I am a career EE, with a couple of graduate EE degrees;
and this is something entirely new to me. Could you explain
this concept, and/or provide some references?

Thanks, Ed

On Wed, 03 Mar 2004 00:56:57 GMT, "Old Ed"
wrote:

"Lossless resistance?" Would that be zero resistance,
or perhaps a negative resistance, as in the active part of
a tunnel diode's V-I characteristic?

I am a career EE, with a couple of graduate EE degrees;
and this is something entirely new to me. Could you explain
this concept, and/or provide some references?

Thanks, Ed

Hi Ed,

This is the most useless term employed in these threads; what I call a
difference without a distinction.

To put it shortly, it is the resistance observed in an infinite
transmission line (better known as the Characteristic Z) or the
resistance of an antenna (better known as the Radiation Resistance).
You will note that a competent engineer already understands the nature
of this resistance, shrubs may need more words to obtain the same
knowledge.

Some explanations like to force it into the same definition of Z, and
then add more words to denote there is not reactance, and then more
words to add there is no heat generated.

In other words (too many of them) the issue is driven from the physics
of heat which if anyone peeled back the onion layers, then they would
find they have not actually escaped from it all, and more words are
forced into the definition to argue what dissipation means.

Principally, a new instrument has been added to the Ohm Meter, the
thermometer, to prove you have in fact measured the value of a carbon
composition resistor. No one actually does this; no one actually
could offer a suitable caloric answer if their life depended on it;
and certainly no one could tell you what the resistance is from a
thermometer reading. But they would demand it is necessary none the
less. ;-)

73's
Richard Clark, KB7QHC

Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

Suppose we adjust a variable d-c supply to full-scale indication on an
external meter. Next, install a chopper (lossless on-off circuit) driven
at a high frequency to produce a square wave interruption of the d-c
with a 50% duty cycle, and insert the chopper contacts in series with
the external meter.

The chopper connects the external meter 50% of the time and disconnercts
it 50% of the time. The meter reads 50% of full scale.

Another way to reduce the scale reading to 50% is to insert a resistor
in series with the meter. If it is an 0-1 ma meter and if it has an
internal resistance of 1000 ohms, insertion of a 1000-ohm resistor in
series with the meter will reduce the meter scale indication to 50%.

The chopper as part of the meter source eliminates current to the meter
50% of the time.

The resistor which has the same effect and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

The power in the load, a meter in our example, is the same using either
the resistor or the chopper. The resistor is analogous to a Class-A
amplifier. The chopper is analogous to a Class-C amplifier.

The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

Best regards, Richard Harrison, KB5WZI

Ummmm... sorry, Richard.

I am very familiar with choppers, having designed and built quite
a few in my time. But no engineer I have ever met would refer to
a chopper as a "lossless resistor."

An (ideal) resistor is in the class of passive, linear, time-invariant
devices. A "lossless resistor" would be a 0 Ohm resistor.

A chopper is a time-variant device, and depending on the details,
could be modelled as linear or non-linear on top of that.

The two devices (resistor and chopper) have very little in common,
besides being usable in electronic circuits. One of many differences
(a BIG one for us radio guys) is that the resistor would generate
no radio noise (other than thermal); but the chopper would generate
horrific radio hash, from the chopper frequency to daylight.

As for analogies to power amplifiers, an ideal Class A amplifier
would normally be modelled as an active, linear, time-invariant
device; so it would have some commonality with a resistor.
But the difference between passive and active is huge, of course.

A Class C amplifier would normally be modelled as an active,
non-linear, time-invariant device. There is very little similarity
left to a resistor, except for the time-invariant part. But a sidewalk
is normally time-invariant, too. So we could just as well say a
sidewalk is like a "lossless resistor." Or maybe that would be a
sidewalk with ice on it. ;-)

But to each his own...

Regards, Ed

"Richard Harrison" wrote in message
...
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

Suppose we adjust a variable d-c supply to full-scale indication on an
external meter. Next, install a chopper (lossless on-off circuit) driven
at a high frequency to produce a square wave interruption of the d-c
with a 50% duty cycle, and insert the chopper contacts in series with
the external meter.

The chopper connects the external meter 50% of the time and disconnercts
it 50% of the time. The meter reads 50% of full scale.

Another way to reduce the scale reading to 50% is to insert a resistor
in series with the meter. If it is an 0-1 ma meter and if it has an
internal resistance of 1000 ohms, insertion of a 1000-ohm resistor in
series with the meter will reduce the meter scale indication to 50%.

The chopper as part of the meter source eliminates current to the meter
50% of the time.

The resistor which has the same effect and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

The power in the load, a meter in our example, is the same using either
the resistor or the chopper. The resistor is analogous to a Class-A
amplifier. The chopper is analogous to a Class-C amplifier.

The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

Best regards, Richard Harrison, KB5WZI

OOPS!

Richard,

Now I understand the source of your "lossless resistance" remark in the
class C thread. I respectfully disagree. There is nothing of the sort. I
will admit, however that in a Switched case, the concept of resistance *may*
be thought of this way, but it is not a good way to look at it. I believe
you are taking two things which are not similar and calling them the same.
I will point out the error below. It can be verified by experiment rather
simply if you wish.



"Richard Harrison" wrote in message
...
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

...d-c supply to full-scale indication on an external meter.

Therefore we have the situation whe
V out = Ifs x Rm
V out set to the IR drop of the meter at fullscale. All ok so far.


Next, install a chopper (lossless on-off circuit) ... with a 50% duty
cycle...
... The meter reads 50% of full scale.

Still ok so far.


Another way to reduce the scale reading to 50% is to insert a resistor

Yep, still ok.



The resistor which has the same effect ...

Start of the error. I disagree with "same effect". One is a steady
state current (resistor), the other a square wave of current (chopper),
right? The time average current IS the same. ...HOWEVER...



and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The word "effective" here is not standard as it is the same word we
use to denote "producing the same power". Again, it certainly is that the
AVERAGE current is obviously the same for both cases.



The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

This is indeed correct. The chopper DOES change produce the same
AVERAGE current as the resistor. The added resistor looses some power, the
chopper does not. Now the zinger.


The power in the load, a meter in our example, is the same using either
the resistor or the chopper.

Here is where the error is. The POWER in the load (meter) in the
chopper case is not the same as it is in the resistor case! It is...Hold on
to your hat... TWICE as much as in the resistor case!

In the meter only case load power is 1 miliwatt.
In the resistor case the load power is 0.25 miliwatt (half voltage x half
current = 25%)
Or P=I^2 * R which is also 0.25 miliwatt.
In the chopper case the load power is 0.5 Miliwatt!

In the chopper case, the RMS current is = Imax x square-root(duty-cycle).
This is 1ma * 0.707 or 0.5 miliwatt.

See
http://www.irf.com/technical-info/an949/append.htm
For verification of the square root of the duty cycle formula.

You can easily verify this since a power difference should be easily
observed by finger-and -clock method. Set up the experiment with a resistor
and power supply which gives some power which can be felt as a temperature
rise within, say 10 seconds. I'm guessing somewhere around 0.2 to 1.0 watt.
This is with the two resistors in the circuit. Then subsitute the chopper
(a relay). Twice the power should be evident in a faster rise in
temperature. It should heat up in half the time. I maintain that if it
doesn't, then you did it incorrectly.



The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

This is what I believe would be called a self conflicting term.
Resistance removes power from the circuit in question in all cases, period.
And YES on the proverbial infinite T-line, the power is gone from the
source, never to return. It looks just like a resistor (Steve says with
great confidence, knowing full well that others may disagree, but
nonetheless firm in his beliefe) It *is* LOST.

Perhaps a concept called "Average resistance" might be a subject to discuss.
In matching a class C power amplifier, there has been myriad of words spent
on just what, if anything, is being matched by the output matching network.
Is there an "output" impedance there or not and if so, what is
it....(rhetorical, of course)

Sorry Richard, but I believe you have it wrong this time. Steady state and
pulsed must be understood with slightly different principles.
--
73, Steve N, K,9;d, c. i My email has no u's.

Steve Nosko wrote:
"The power in the load (meter) in the chopper case is not the same as it
is in the resistor case!"

A square wave has the same heating value as d-c because it has identical
amplitude and both alternations are equally effective in producing heat.
Power is indifferent to polarity.

The sine wave has an amplitude 1.414X its effective value. The heat
produced over a period is the average of the sum of the instantaneous
powers of increments within a cycle.

Someone argued that resistance is an agency which removes power from the
scene. Thus, Zo is a resistance.

I would rather define resistance as the ratio of the applied emf to the
resulting current in the circuit. The current must be in-phase with the
applied voltage..

With my definition, the Zo of the transmission line as defined by the
square root of L/C makes sense to me. It is a lossless resistance.

Best regards, Richard Harrison, KB5WZI

Working very hard to resist his long-winded gene, and apparently failing,
Steve responds:



Hi Richard,



"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"The power in the load (meter) in the chopper case is not the same as it
is in the resistor case!"

Sorry Richard, but I show how this is true below. Stay tuned...

However, first I'll address this digression. You say:
A square wave has the same heating value as d-c because it has identical
amplitude and both alternations are equally effective in producing heat.
Power is indifferent to polarity.

I believe you have changed your premise. The chopper example you
gave is not this type of square wave. Your example has what might be called
a "ONE sided alternation". Again, more below.

The sine wave has an amplitude 1.414X---
I want to stay on the chopper example you posed.

Someone argued that resistance is an agency which removes power from the
scene. Thus, Zo is a resistance.

I did. I maintain that *to the sourse* - that is, from the source's point
of view, an R or a Zo (all real) are indistinguishable. The current is in
phase with the voltage and power is removed from the source. Yes the power
isn't lost from the system, yet, so I do understand your resistance (sorry)
to calling it a loss. It is traveling down the line, but it IS gone from the
source.

I would rather define resistance as the ratio of the applied emf to the
resulting current in the circuit. The current must be in-phase with the
applied voltage.
With my definition, the Zo of the transmission line as defined by the
square root of L/C makes sense to me. It is a lossless resistance.
Best regards, Richard Harrison, KB5WZI
So, you hold true the two means to the definitions. That's ok, but I
maintain that the chopper example is correctly analyzed by my method. I see
that you are trying to build mental models of these situations, but I think
your model does not do justice to the physics... IT seems a reasonable
conclusion that - because the value of Zo is the ratio of two lossless
parameters that therefore the Zo itself is "lossless". In the sence that
the power is not lost from the system ans delivered to some type of load,
this is true. So I understand this analogy, though I believe it does not
allow a good understanding of electronics.

Now, I return to the Meter and resistor/chopper premise you posed earlier:



Rather than succumbing to jokes, I submit that while you
base your understanding on the facts, your approach is from the wrong
direction. I suggest that your "loss-less resistance" concept is the wrong
way to look at it. When we analyze a situation we must be careful to have
all the correct basics - and not violate any basic "laws". I believe there
is only one kind of resistance (as far as the circuit is concerned) and
therefore all resistance follows the same rules -- all the time. I don't
need one understanding for loss-full resistance and another for loss-less
resistance.

I actually have pondered trying to use a model called
"average resistance", but I think it is not the right way to go. Here's how
I came to this conclusion.... Short answer: it does not work.



Long answer:

I DO know that one not-to-be-violated principle is that ALL
resistance removes or looses power - that there is no such thing as
loss-less resistance. If we place ourselves 'inside' a circuit, any time
some of the I is in phase with V (across an element of the circuit), power
exits out that element. A real Zo does this, a resistor does this and a
motor does this. The power exits the source and becomes energy traveling
down a T-line, heat or mechanical energy, respectively. (I hope I am using
all the correct terminology here) This MUST be an overriding principle and
my solution must not violate it. If it does, then I went in the wrong
direction. I am not going to say that the switched situation is more
complex because it is actually quite simple. You just went the wrong way in
a desire to understand it.



You give three situations which can still be used to get to the correct
conclusion.



1) A simple 1V. source and a 1K load (a meter) [ IL=1 ma. PL= 1 mW ]

2) The above with an added 1k resistor [ IL=0.5 ma. VL=0.5 V. PL=0.25
mW. ]

3) #1 with a 50% chopper.[ ILavg.=0.5 ma. ... PL=0.5mW. ]



You tried to understand #3 (chopper) by comparing it to #2
(added resistor) both having the same AVERAGE current. You say they are the
same. This sounds reasonable, at first,, but switched or pulsed circuits
can be misleading.



I expected a response from someone to the effect that I
threw in a square root out of nowhere and was confusing the matter by
throwing math at it. Bear with me, and I will show that the added resistor
and the chopper situations are definitely NOT the "same". While I knew the
root is correct, I actually puzzled over this. Why isn't #2 (added
resistor) and #3 (chopper) the correct comparison? I do know that the
chopper situation does have a square root in the power equation. In fact,
when doing RMS calculations there will always be a root in the equation
which is not in the 'average' equation. Here's a way to look at it which
gives the correct conclusion.



To understand #3 (chopper), it is better to compare it to _#1_. Like
this:

When the switch is closed, Power in the load is PL= 1.0 mW. When open,
PL= 0.0. Therefore based on an intuitive understanding of how things work,
it surely seems that the average Power in the load must be half or 0.5 mW.
Well, this IS correct.

Now look closely at case #2 (added resistor) . The Power in the load is
0.5 V times 0.5 ma., or 0.25 mW. So while the average current is the same
as #3 (chopper), the average power is not (it's half that of the chopper
case). Therefore, the chopper case and the added resistor case are not
really the same. Something is definitely different. It is the power.

Applying the RMS formula to the chopper case gives the same result. The
chopper case is not your regular square wave with a positive and negative
"alteration". You are correct that the standard square wave is = to DC.
The chopper case IS NOT.

It turns out that when we move into the RMS realm, a square root always
appears in the formula. In case you can't see that Int' Rect paper, the
chopped case RMS is:

RMS = Imax * sqr-root(duty-cycle ratio).

So for a 50-50 chopper it is Irms = 0.707ma. This gets the 0.5 mw number.



Intl Rect paper URL:

http://www.irf.com/technical-info/an949/append.htm

Formulas roughly in mid document.



OK, so what about the "resistance" of the chopper? Can we assign it a
numerical value? I suggest that while this is a very tempting path , it
should be discarded and you only think in terms as I describe. That is, for
the steady state cases (#1 & 2) we use ohms law and for #3 (chopper) we
shift gears slightly and apply principles (still correct ones) which
properly describe that situation.



Another stumbling point is the idea of current being "In-Phase" for the
chopper case. Phase (when we start getting into power factor
considerations) is a property of sine signals and applying it to the chopper
case, I believe, is a confusing use. I can imagine a concept where - the
wave shape of the current being the same shape as the voltage wave indicates
a resistive circuit - as being valid.





Now, we may want to ask; "OK. Then what IS the "Equivalent resistance"
of this 50-50 chopper. Since it is limiting the current to this average
value of 50%, it MUST have some "resistance" associated with it--shouldn't
it? Well, let us look two ways. If there is a equivalent resistance, then
we should get the same value any way we figure it.



Looking at the AVERAGE current, we'd be led to think it is 1k since the
"current" is cut in half. However, looking at the power, it appears to be
414.43 ohms. (this is what it would take to get 0.5 mW. in the 1k load)

Now this looks strange. That is, a 50-50 duty cycle chopper and two
different ways to arrive at this "equivalent resistance". Pick other
chopper duty cycles and I think we may find that a square root crops up (I
haven't derived it, yet). There is no square-root in the average current
formula. It is simply the duty cycle ratio. This suggests that the concept
of an "equivalent resistance" has some problem. If there is a truly
equivalent resistance, then any method of analysis should arrive at the same
value. I didn't need any "loss-less resistance" to get the numbers.



I submit that it is a bad mental model and should be discarded.



We all must form what I call "mental models" of electronics. If our models
are good, then we can use them and come to the correct conclusion as we look
at new or more complex situations. The formulas always work when properly
applied. I guess the catch is: which formulas are the correct ones. I
think some may think that arrogance tells me which is which. If we pick the
wrong view, we can be led astray. This is the case here.



I believe the conclusion should be:

In the simpler steady state DC (or AC) situations, we use ohms law and think
in terms of our normal understanding.

In the switch-mode situation, we still use ohms law. However, we base our
understanding on different concepts which do not overlap completely with the
steady state case, but are nonetheless correct basics (and equivalent or
loss-less resistance are not among them). Also that using two applicable,
but different methods, we arrive at only one answer.



For a VERY light look at the chopper situation, the "equivalent resistance"
idea may be ok for some simplistic explanation to a non technical person,
but for someone trying to understand electronics, it falls apart.



Are you swayed, or am I whistling up a drain pipe? If we agree to disagree,
so be it, but how do you justify the difference in *power* between case 2
and 3 when you maintain they are the "same"? I would like to focus on that.
--
Steve N, K,9;d, c. i My email has no u's.