Steve Nosko wrote:
"I think the phrase "available energy" may draw discussion, but I won`t
start it."

I plead guilty. I knew when I wrote it that the choice was poor but
couldn`t think of a better phrase at the moment.

Terman says on page 76 of his 1955 edition:
"Alternatively, a load impedance may be matched to a source of power in
such a way as to make the power delivered to the load a maximum. (see
footnote) The power delivered to the load under these conditions is
termed the "available power" of the power source."

Walter Maxwell straightened me out on my carelessness on this long ago,
and I repeated anyway. Dang me!

Best regards, Richard Harrison, KB5WZI

Steve Nosko wrote:
"With a "conjugate match" the source dissipates 50% of the power and the
load the other 50%."

This is true only if all the source resistance is the type that converts
electrical energy to heat. There is a non-dissipative resistance.
Switched-off time in the Class-C amplifier is part of its internal
resistance.

You can deliver all the available power of the Class-C amplifier to the
load and dissipate less than 50% in the source.

Best regards, Richard Harrison, KB5WZI

Steve Nosko wrote:
"A (tube) amplifier is in conjugate match conditions. It is dissipating
10 watts in its plate. This is the limit of its plate
dissipation----There is also 10 watts in the load.

Now assuming you can, increase the (plate) supply voltage by , say
20%---(may be the fatal flaw)."

Likely so. If everything remains linear, 20% more voltage increases
power by 1.2 squared, or 1.44 times. If the tube was already dissipating
its maximum sustainable power, expect an early failure due to the
overload.

If you were not already in a maximum power transfer condition, and a
condition which might provoke flashover within the tube, a readjustment
of the match might shift more of the available power to the load and
thus relieve the tube of some of the dissipation.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"With a "conjugate match" the source dissipates 50% of the power and the
load the other 50%."

This is true only if all the source resistance is the type that converts
electrical energy to heat. There is a non-dissipative resistance.
Switched-off time in the Class-C amplifier is part of its internal
resistance.

You can deliver all the available power of the Class-C amplifier to the
load and dissipate less than 50% in the source.

Best regards, Richard Harrison, KB5WZI


I think this becomes very academic (or perhaps a better word is awkward)
because originally you framed this as a "conjugate match" situation. Now,
you are in terms of time varying parameters. I think the analysis must stay
in one realm or the other. My mental models have trouble switching back and
forth.
I can't speak to tubes, but I do know that to get the most out of a
transistor power amp (transmitter type) up to about 200 MHz you design the
output match to be for a collector resistance of Vcc^2/(2Po). IF I recall,
this can be derived easily if you assume the transistor pulls all the way to
ground and the output tank swings up to 2 x Vcc. I was shown, and
understood it way back then, but can't recall the path to the solution off
the top of my head.
This, then, always started the discussion of whether this "matched" the
transistors output impedance or some other more esoteric concept. Then talk
of "average resistance" came in and eyes would glaze over....

The designer would then go back to the bench, work to optimize the design to
his requirements, test it over temp, etc. and ship it.

"There comes a time to shoot the Engineer and ship the product." is the
title of a famous editorial from the early 70's (IIR). Got it around here
somewhere...
-- 73
Steve N, K,9;d, c. i My email has no u's.

Steve Nosko wrote:
"Then there`s the solid-state power amplifier standard output resistance
formula.
Rs = Vcc^2 / (2*Po)
The implication should be obvious."

It looks like Ohm`s law to me, P=Vsq / R.

The implication of (2*Po) is that 50% of the power is in the source and
50% of the power is in the load. If so, it`s a Class-A amplifier
formula, but the semiconductors could be biased to cut-off (Class-B) to
reduce dissipation in the transistors when they are idle. The best
collector load resistance is often not that which produces maximum
output, but that which produces maximum "undistorted output".

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"I think the phrase "available energy" may draw discussion, but I won`t
start it."

I plead guilty. I knew when I wrote it that the choice was poor but
couldn`t think of a better phrase at the moment.........
Best regards, Richard Harrison, KB5WZI

I think this is what causes much of the discussion here. Terms are not
always as clear cut as things like: voltage or resistance. We know those,
but some of the others are more vague...or just what concept they point to
is not obvious. It is also sometines difficult to determine what the KEY
point or word is of a question or statement.
When I read yours, I made the assumption that "available power" was not
where your question pointed. I went along the; "Can we get a non 50-50
division? and not waste so much in the stage" path.
It is clear that different responders "Hear" different question.

Cryminy crum! I think my brain has forgotten how to hit the keys "on" in
the correct order. I keep getting things like "questino" for
"question"....or is it a keyboard timing issue??? As my fingers fly
along...
--
Steve N, K,9;d, c. i My email has no u's.

I wrote a long boring post on "effective aperture". I apologize. I
regret not becoming familiar with Kraus`s "Antennas" long ago. Kraus
makes many ideas clear with few words.

On page 43 of his 1950 edition of "Antennas" Kraus writes:
"The ratio of the power W in the terminating impedance to the power
density of the incident wave will be defined as the effective aperture
Ae.
Thus Ae = W/P
If W is in watts and P is in watts per square meter, the Ae is in watts
per square meters."

Best regards, Richard Harrison, KB5WZI

Steve, have we now moved to an antenna that has a preamplifier on it for
listening ? If so I need not continue to struggle to follow the thread day
by day to determine its implications to the subject at hand.
The half power thingy I presume is understood by all so IS something very
exciting to be revealed that shows that the dipole is really an efficient
radiator after all, but only if you put a class C amplifier on it ? There
are stacks of information in books and lots of words in a dictionary, but
sooner or later one has to give a reason as to why they are reading out loud
if it is meant to be instructive or explanaatory with respect to the thread,
i.e. dipole and its impedance. You can start a different thread which would
help out for when one checks out the archives unless the intent
is meant to be destructive
Regards
Art


.. "Steve Nosko" wrote in message
...
Hi Richard...

"Richard Harrison" wrote in message
...
[...]
"----Second, it is the RMS current through the tube which will waste
power, so that is what we must be concerned with."

I don`t believe current through a Class C amplifier consists of an
ordinary sine wave.

And I didn't say that it does nor do I believe it does. I'm inclined
to
take my 100MHz storage scope to to the 6146's of my TS830s and see for
myself.
Your words imply (at least I infer) you are thinking that only a sine
wave has an RMS value. Every wave of any shape has an effective or RMS
value - its heating or "power causing" value.


[...] I think it consists of short unidirectional pulses.
The tuned "tank circuit" is the source of sine waves.

This certainly has to be correct. The tank will most likely cause some
sine-like VOLTAGE waveform, but the tube current has to be pulses of some
shape. This is a very timely discussion in view of the AC power meter QST
article and the extensive investigation I just completed on several pulse
shapes..


RMS is the effective value, not the average value, of an a-c ampere.

I will differ here. The RMS value is more appropriately described as
the power producing value of ANY wave form. Pulses can produce heat just
as
well as sine wave AC. We all know this from a practical view since tubes
can only conduct in one direction and the plates DO get hot.



...as the heating
value of an ampere is proportional to the current squared.

This is actually a simplification. P=ExI Power is the product of
voltage and current *only*. Because this is a second order effect, in a
resistance it can be related to either voltage squared or current
squared...
because that captures the second order character. Maybe there's a better
way
to say it mathematically, but I don't know it.
When we get to non sine shapes, then we have to fall back on the
actual
definition. root [avg of square] ...with the integral and all.
http://www.ultracad.com/rms.pdf

[...snip...]

Ordinarily, with nonsinusoidal currents, the ratio of maximum to
effective value is not the square root of 2.
Best regards, Richard Harrison, KB5WZI

Doing the math for pulses with the shape of sine, triangle (a single
slope with sudden end) and trapezoid (a sudden start to one level then a
slope to a peak and a sudden end), I decided to look at the RMS to AVERAGE
ratio since average is what a common meter will measure in Bob Shrader's
article (AC watt meter Jan 04 QST).
I was particularly interested in the sine-shaped pulses of various
duty
cycle because the current of common power supplies occurs in short pulses
with a sine-like shape that are near the peak of the voltage waveform.
It was interesting that for all these shapes, this ratio was very
similar. One relatively simple thing to understand which came out of the
analysis was that the average value is directly proportional to the duty
cycle as you might reasonably postulate. Where duty cycle is the ratio of
"on" time to off time. Where "on" time is the time that ANY current
flows.
Whereas the RMS is proportional to the Square root of the duty cycle.
e.g.
drop the duty cycle to half and the RMS drops to .707.

I have to do some verification, but it sure looks as though Bob's
numbers can be as much as three times what he quoted, depending on the
waveshape and some measurements I made.

http://www.irf.com/technical-info/an949/append.htm
Trapezoid=rectangular. Also for the phase controlled sine, the things
that
look like tau and a small n are both pi i.e. sin [pi x (1-D)] cos [pi x
(1-D)] and denominator of 2 x pi


Some average & RMS values here.
http://home.san.rr.com/nessengr/techdata/rms/rms.html

More (better) average formulas:
http://www.st.com/stonline/books/pdf/docs/3715.pdf
NOW I know where the average value of a sine wave comes from = (2/pi)
The Greek delta = d.

A calculator for RMS:
http://www.geocities.com/CapeCanaveral/Lab/9643/rms.htm

Richard Harrison wrote:
Thus Ae = W/P
If W is in watts and P is in watts per square meter, the Ae is in watts
per square meters."

Seems to me that Ae would be in square meters. :-)
--
73, Cecil, W5DXP

"Steve Nosko" wrote
With a "conjugate match" the source dissipates 50% of the power and the
load
the other 50%. This can't be changed.
=============================
Without disagreeing with what you say -

A conjugate match is not relevant in the present discussion because there is
seldom, if ever, a conjugate match between a PA and its antenna system.

The tuning-up process is NOT intended to produce such a match.

Tuning up is just the simple process of adjusting the transmitter load
resistance to be equal to its designed-for load resistance, usually an
arbitrary 50 ohms.

The internal resistance of a transmitter is NOT 50 ohms. It is not a design
feature. It is whatever happens to appear after the designer has met a
series of other requirements. The designer himself does not know what the
internal resistance is unless, out of curiosity, he bothers to measure or
calculate it.
----
Reg, G4FGQ