RadioBanter - max power transfer theorem

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.