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.

"John Smith" wrote in message
...
Actually, not.
1. the line is mismatched at both ends,

John, You missed something. The problem was stated;

"Isn't a 50 ohm transmitter conjugately matched to a 50 ohm load
when fed through 1/2 wavelength of 450 ohm ladder-line?"


The TRANSMITTER "sees" 50 ohms therefore it is. Some discussion can be made
for the exact phrasing which says, in effect; "...50 ohm transmitter ...to a
50 ohm load". By this I mean using the phrase "to a 50 ohm load" is a bit
misdirecting. In the stated case, the Tx is NOT actually connected *to* the
load. It is connected to a 1/2 wave of line. At the Tx output, there is a
conjugate match.
I believe the intended situation is that a 1/2 wave of ANY line repeats the
load Z at its input. From your comment #2, it appears that you do not know
this.


2. If I have a 50 ohm load at each end, and a line of impedance Z, 1/2
wavelength long, what does Z have to be for max power transfer?

Actually, it IS anything! Yea, Yea there are other *practical*
considerations...balanced...unbalanced....


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

Nothing
constrains Z to be a value. Z at 50 is great, Z at 500 is not bad (have
to
do some matching)
This problem the way it is stated 50 1/2 line 50 is implying that any
type
of line will do as long as it is 1/2 long, then Z can be 0.01 and Z=20,000

So as long as the matching transformers are in,(assuming the right ones)
it
is matched max power too.
The Tx out 50 ohm unbalanced is matched to the 450 balanced line 1/2
wavelength long and another transformer back down to 50.


/////////////\\\\\\\\\\\\\//////////////\////////\\\\\\\\\\\\\\\\\\\\\\\\\\\
\\\\\\\\/////////////////////\/\/\/\\/\/\/\\/\\\/\\/\//\/\/\/\/\/\/\


"JDer8745" wrote in message
...
"Isn't a 50 ohm transmitter conjugately matched to a 50 ohm load
when fed through 1/2 wavelength of 450 ohm ladder-line?"

==============

Heck yes, I'm assuming u mean 50 + j 0 Ohms.

73 de Jack, K9CUN

and *I* got suckered in... Oh well...

-- 73
Steve N, K,9;d, c. i My email has no u's.

"Robert Lay W9DMK" wrote in message
...
On Wed, 3 Mar 2004 22:08:54 -0600, "John Smith"
wrote:

Actually, not.
1. the line is mismatched at both ends, 450 balanced to 50 ohm
singelended.
If you put in or assume transformers for the balanced to unbalanced 450
to
50 ohm conversion at both ends, then it is matched (still assuming that
Tx
and Ant are 50 ohm resistive.) I have an equation that calculates the
loss
from a text book. (if the section is mismatched it is not conjugally
matched)

I think we have some fertile ground here, ready for plowing.

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

OK, OK, OK. How come voltage dB and power dB are different, huh? Yhy is it
that power dB carry twice the weight of voltage dB?

dB= 10 LOG(power ratio) = 20 LOG(voltage ratio)

Knock yourself out.

--
Steve N, K,9;d, c. i My email has no u's.
he he he
"J. McLaughlin" wrote in message
...
When a candidate for an EE faculty position visits, someone, usually at
lunch, will bring the conversation around to the MPTT. If he or she
does anything other than giggle they do not get my vote.

Let us leave this tar-baby out in the field. 73 Mac N8TT

--
J. Mc Laughlin - Michigan USA

Cecil --

I think your right, 1:1 should work fine, with 1/2 wavelength, low loss
line.

I was matching to a long line with charestic impedance of 450. But, I think
the length of the line must be long (or have loss) before the charestic
impedance of the line shows up to really matter. I used to have a program
that showed this effect. I installed a 450MHz 500 foot line in Saudi, and
measurements had to be careful, because we had a lot of line loss. (we were
heating up the Heliax, more than radiating power out the antenna)

But on a short run, lower frequency, there is not enough of the line to make
a large effect. (I think it has to be more than about 2-3 dB line loss...),
so the Tx output is seeing primarily the load(antenna). And then the 1/2
wavelength will causing a "null" at the Tx output. When I run across the
program/formula I post it. It is probably in the same book, same section,
too.
'''''''''''''''''''''''''''''''''''''''''''''''''' ''''''''''''''''''''''''''
''''''''''''''''''

"Cecil Moore" wrote in message
...
John Smith wrote:
Agreed that adding a 9:1 balun (impedance match) will have little
mismatch
loss at x and all the Tx power will be "presented" to the antenna port.

Oh Boy, I really screwed up. Replace that 9:1 balun with a 1:1 balun.
I was very stupid to have said 9:1. A 9:1 balun would reduce that 50
ohm Z0-match to 5.56 ohms thus creating a mismatch.
--
73, Cecil, W5DXP

Steve Nosko wrote:
OK, OK, OK. How come voltage dB and power dB are different, huh? Yhy is it
that power dB carry twice the weight of voltage dB?

dB= 10 LOG(power ratio) = 20 LOG(voltage ratio)

Because log(v^2) = 2*log(v)? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Seems to me this thread has gone off track, as many do after the first three or
four responses.

This thread was started with Bob's fine treatise on proof of the existence of
non-dissipative resistance, a resistANCE that cannot be duplicated with a
resisTOR. But the thread went astray.

Non-dissipative resistance is not well accepted or understood by many otherwise
well informed engineers, because it has had little or no (or even incorrect)
treatment in EE courses. This is because the profs by habit generally used only
the classical generator as the source in treating network theory. The classical
generator is always considered to have a dissipative internal resistive source.
Consequently, the profs never considered treatment of a source having other than
a 50 percent maximum efficiency. All the equations I'm familiar with show
clearly that 50 percent is all yer gonna get.generally

Unfortunately, the profs of my ken never tried to understand the reason the
efficiencies of Class B and C amps exceed 50 percent. Fortunately, Bob Lay comes
along with proof that there really is resistance established only by the RATIO
of voltage to current with no dissipation whatever to heat or radiation.

Bob's paper furnishes further proof of my own proof that dissipationless
resistance exists. I presented my proof in Chapter 19, and Appendices 9, 10, and
11 in Reflections 2. For those who are interested in reviewing my writings
there, but who don't have a copy of Reflections 2, those references are on my
web page at http://home.iag.net/~w2du.

For a short preview, take note that in Class B and C amps the DC source power
goes to only TWO places, 1) the power that is dissipated in the cathode/plate
resistance, which transitions to heat, and 2) the power dissipated in the load.
Note also, the source resistance at the output of the tank circuit is linear,
time invariant, and determined solely by the voltage/current ratio at the output
of the tank. The reason is that that the energy storage of the tank isolates the
non-linear input portion of the amp from the linear output portion. With
sufficient Q the output wave form is a near-perfect sine wave, thus verifying
linearity.

Also, data obtained from my measurements of the source resistance of Class B and
C power amps shows that when the plate tuning and loading conrtrols are adjusted
to deliver all the available power at a nominal drive level, the source and load
resistance are equal. Consequently, if the power dissipated in the cathode/plate
resistance plus the power dissipated in the load equals the DC input power,
there can be no power dissipated in the source resistance.

And finally, with the source resistance non-dissipative, it cannot absorb any
power reaching it from a mismatched termination down stream, which is why no
reflected power is absorbed in the amp.

Eric Nichols, KL7AJ, operates multi megawatt ionospheric sounding rigs in
Alaska. The water cooling the water-cooled amp tubes is carefully measured
calorimetrically. He has observed over a period of many years that the water
temperature remains the same, no matter what the SWR at the input of load line
is, proving there is no reflected power absorbed in the amplifier. Of course the
tank circuit elements are re-adjusted to deliver all the available power, what
ever the input impedance of the line may be.

Howabout if we give Bob's paper a somewhat more open-minded review. Crap? That
criticism of Bob's paper is where the crap is!

Walt, W2DU

Steve Nosko wrote:
"How come voltage dB and power dB are different?"

They aren`t.

10log(power ratio) = 20log(voltage ratio)

They are equal because power is proportional to voltage squared. To
square with a logarithm, you multiply by (2).

Best regards, Richard Harrison, KB5WZI

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