Richard Harrison wrote:
I don`t believe current through a Class C amplifier consists of an
ordinary sine wave. I think it consists of short unidirectional pulses.
The tuned "tank circuit" is the source of sine waves.

Yep, the Class C amp is like the energy pulse from a pendulum
clock spring. The tank/filter circuitry is like the pendulum.
--
73, Cecil http://www.qsl.net/w5dxp



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Dan, K6MHE wrote:
"So according to you, as an example, a resonate 14 or 1/2 monopole will
extract more energy than a non-resonate 5/8 monopole."

A 5/8-wavelength monopole has a directive gain over a 1/4-wavelength
self-resonant monopole.

However, the 5/8-wavelength monopole must be resonated externally to
remove its reactance to take advantage of its extended length so that
the volts per meter it is exposed to can make a large current flow in
the antenna.

Best regards, Richard Harrison, KB5WZI

On Tue, 24 Feb 2004 06:17:01 -0800, Dan Richardson
wrote:

On Mon, 23 Feb 2004 17:47:15 -0600 (CST),
(Richard Harrison) wrote:

Not in the "grand scheme of things", perhaps, but a resonant antenna
extracts more energy when swept by a traveling wave than an
off-resonance conductor does at the same distance from the transmitter.

So according to you, as an example, a resonate 1/4 or 1/2 monopole
will extract more energy than a non-resonate 5/8-wave monopole.

Enjoy your dreams, I'm QRT

73
Danny, K6MHE

Hi All,

As a test of modeling and actual, real data, I put this to the test:

1.) I established two full quarter wave antennas;
2.) each with a set of 48 radials;
3.) tuned to 1.7 MHz;
4.) separated them by 20KM (100 wavelengths);
5.) set one to have a source of 1000W;
6.) loaded the other at the base with a 50Ohm resistor;
all in EZNEC with enough segments to hit the 500 limit.

When I called for the load data, EZNEC calculated it to be:
Power = 0.0008166 watts
(already confirmed accurate to FCC Field trials)

THEN, I increased the same antenna's length to 5/8th Wavelength
and observed its load's response to the identical field:
Power = 0.0001185 watts

Through the simple observation of power ratios, the untuned 5/8th
Wavelength antenna suffers to the tune of:
-8.38dB

Now, if this discussion revolves on the myopic consideration of
resonant antennas, devoid of other issues like tuners, matching, lines
that connect and so forth; then it is painfully obvious that NO
ANTENNA absorbs ANY POWER (by virtue of the load having been defined
out of the problem). This is further confirmed in the daily
experience of a million yagi owners whose passive elements offer just
exactly that lossless coupling. When interfaced to a real load, any
SWLer will confirm through simple tuning (and through modeling) that
the received power peaks through adjustment. The 5/8th antenna is no
different.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
However, the 5/8-wavelength monopole must be resonated externally to
remove its reactance to take advantage of its extended length so that
the volts per meter it is exposed to can make a large current flow in
the antenna.

In other words, the linear system transfers the most energy when the
source (the antenna) is matched to the load (the receiver). It cannot
transfer more than 1/2 the energy, but it can transfer less.

It is conceivable that a receiver's input impedance could be optimized
to the feedpoint impedance of a 5/8WL monopole but I don't know of any
receiver like that.
--
73, Cecil http://www.qsl.net/w5dxp



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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

I wrote:
"RMS is the effective value, not the average value of an a-c ampere.

Steve Nosko responded:
"I will differ here. The RMS value is more appropriately described as
the power producing value of ANY wave form.."

A periodic function is not necessarily a sine wave but when I look up
alternating current in my dictionary, I find the diagram of a sine wave
and it is tagged: "alternating current".

By fourier series, any complex waveform may be resolved into a
fundamental plus a finite number of terms involving its harmonics.

My electronics dictionary says:
"The rms value of a periodic quantity is equal to the square root of the
average of the squares of the instantaneous values of the quantity taken
throughout one period. If that quantity is a sinewave, its rms amplitude
is 0.707 of its peak value."

My dictionary also says:
"Effective current---That value of alternating current which will give
the same heating effect as the corresponding value of direct current.
For sine-wave alternating currents, the effective value is 0.707 times
the peak value."

The average value for one alternation of a sine wave is 2/pi or 0.637
times the maximum value.

I make lots of mistakes, but fortunately, I don`t think I made any of
consequence in this thread. I probably didn`t go into enough detail, but
I was just an engineer, not a professor.

Best regards, Richard Harrison, KB5WZI

On Tue, 24 Feb 2004 14:30:02 -0600 (CST),
(Richard Harrison) wrote:

I make lots of mistakes, but fortunately, I don`t think I made any of
consequence in this thread. I probably didn`t go into enough detail, but
I was just an engineer, not a professor.

Best regards, Richard Harrison, KB5WZI

Hi Richard,

The one I noted (mistake) was in your reference:
It is derived from the average of the squared current over a half cycle
which necessarily forces both a doubling, and a symmetry that is not
demanded of native RMS determinations. It then follows that the
commonplace illustration of the mains Sine wave completes the
illusion. Few EE students migrated beyond this simplicity because the
world is nasty place to measure power.

RMS by and of its mathematical nature through the squaring operation
negates any requirement for "half cycle" determinations (no issue of
negatives). It also preserves the natural order (of two forced by the
half). If you think about it, any biased sine wave impinging upon a
load imparts the power loss of the bias at the 180° portion of the
Sine cycle. RMS copes with this, the notion of half cycles does not.

The simple determination of RMS is the graphical integration of the
area under the curve. There are as many "correction factors" for RMS
as there are shapes, and they all derive from this simple concept.

When the computational horsepower requirement becomes enormous (there
are many here that give up too easily with complexity); it is the
provence of the "Old School" to suggest that since RMS is all based on
the notion of power, you simply measure the caloric result and ignore
shape altogether. This may be done with thermo-electric piles or
other measurable property transformers that perform the complexity of
integration through physics*. I can anticipate those who dearly
embrace the complexity that they shudder to face (such contradictions
of their love-hate relationships) when I hear Crest, or pulse/power
factor (or duty cycle) uttered. Clearly the problem will have
migrated from Power to some other consideration, but is dressed as an
RMS debate.

73's
Richard Clark, KB7QHC

* The Chaberlain and Hookum meter; the Sangamo meter, the Wright
Demand meter; the GE Demand meter; the Edison chemical meter (where
the weight of a cathodic reduction revealed 1.224 grams = one
ampere-hour)...

Richard Harrison wrote:
If we have a Class C amplifier feeding power to the same antenna and
enjoying a conjugate match, we can have a source that takes less than
50% of the available energy. So, the transmitting antenna system can be
more efficient than the receiving antenna system, it seems to me.

Comparing a Class C amplifier to a linear receive antenna source is comparing
apples and oranges. If one makes the transmitter Class-A linear (like the
receive antenna is linear) then the antenna is reciprocal for transmit and
receive. (A Class C or Class B or even Class AB *single stage* is not linear).
Putting two of them in push-pull doesn't make the individual single stages linear.
--
73, Cecil http://www.qsl.net/w5dxp



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"Richard Harrison" wrote in message
...
I wrote:
"RMS is the effective value, not the average value of an a-c ampere.

Steve Nosko responded:
"I will differ here. The RMS value is more appropriately described as
the power producing value of ANY wave form.."

Richard,

I'm not sure what your point of contention is here for most of these
comments. Perhaps I was not clear in my lone objection. All I was
correcting was your reference to "a-c ampere". My intent was to point out
that pulses to one side of zero also have an RMS value. If you do the
"square root ...average...squares..instantaneous
values..quantity..throughout one period." thing, you will get it. So
therefore, tube's current pulses have some RMS value; and that's what'll
heat the tube and cause the efficiency stuff. That's where I was going. If
we can define or approximate the shape, then we could calculate an RMS. I
wonder if a "true RMS" DVM can handel this?

A periodic function is not necessarily a sine wave but when I look up
alternating current in my dictionary, I find the diagram of a sine wave
and it is tagged: "alternating current".

Sure. I'm ok with that. A sine is indeed one example of a periodic
function.

By fourier series, any complex waveform may be resolved into a
fundamental plus a finite number of terms involving its harmonics.

Also ok, but not sure how it plays into the RMS discussion. See below.

My electronics dictionary says:
"The rms value of a periodic quantity is equal to the square root of the
average of the squares of the instantaneous values of the quantity taken
throughout one period. If that quantity is a sinewave, its rms amplitude
is 0.707 of its peak value."

No issue here. (from me, that is) The "average" turns out to be an
integration and I had to drag out the son's calc book to remember how to do
it to see some of the answers myself.

My dictionary also says:
"Effective current---That value of alternating current which will give
the same heating effect as the corresponding value of direct current.
For sine-wave alternating currents, the effective value is 0.707 times
the peak value."

Still none.


The average value for one alternation of a sine wave is 2/pi or 0.637
times the maximum value.

Yea. untill I went through the trouble of doing the calculation, I had
never known the source of the 0.637 number. Now I know. Pretty cool to
actually do it! The integration for this is pretty simple. Took me a few
looks to figure out where the pi came from. It's the old indefinite to
definite integral thingy.


I make lots of mistakes, but fortunately, I don`t think I made any of
consequence in this thread. I probably didn`t go into enough detail, but
I was just an engineer, not a professor.

I'm not a "professor", just an Enginer and "adjunct faculty" for two
community colleges for some years. (that's a fancy name for part time
teacher)
73, Steve

"Richard Clark" wrote in message
...

The one I noted (mistake) was in your reference:
It is derived from the average of the squared current over a half cycle
which necessarily forces both a doubling, and a symmetry that is not
demanded of native RMS determinations. It then follows that the
commonplace illustration of the mains Sine wave completes the
illusion. Few EE students migrated beyond this simplicity because the
world is nasty place to measure power.

RMS by and of its mathematical nature through the squaring operation
negates any requirement for "half cycle" determinations (no issue of
negatives). It also preserves the natural order (of two forced by the
half). If you think about it, any biased sine wave impinging upon a
load imparts the power loss of the bias at the 180° portion of the
Sine cycle. RMS copes with this, the notion of half cycles does not.

I assumed Richard's intent here is that you only have to do the
calculation for one full period of the periodic component to derive the RMS
value - and if it is symmetrical, then only one half period will suffice.
This all assumes a symmetrical AC shape. .. I see no reason why this would
not be true.

However a DC biased periodic shape requires another squaring and root
operation if you capture all the components. It gets a bit more harry



The simple determination of RMS is the graphical integration of the
area under the curve. There are as many "correction factors" for RMS
as there are shapes, and they all derive from this simple concept.

Here I'll take issue with the ONE WORD "graphical". You can integrate
if you can describe the function of the wave shape mathematically.


When the computational horsepower requirement becomes enormous (there
are many here that give up too easily with complexity); it is the
provence of the "Old School" to suggest that since RMS is all based on
the notion of power, you simply measure the caloric result and ignore
shape altogether. This may be done with thermo-electric piles or
other measurable property transformers that perform the complexity of
integration through physics*. I can anticipate those who dearly
embrace the complexity that they shudder to face (such contradictions
of their love-hate relationships) when I hear Crest, or pulse/power
factor (or duty cycle) uttered. Clearly the problem will have
migrated from Power to some other consideration, but is dressed as an
RMS debate.

73's
Richard Clark, KB7QHC

Yikes! Not sure where you went on that last bit, Richard C...
Now, I ask. Do the power meters on the outside of our houses take all those
factors into consideration and REALLY show TRUE watt hours? I have one in
the basement and I think I figured out why I was seeing twice the reading I
should have (letting a light bulb sit on for awhile) ... I counted the
teeth to get the ratio of the gear train, just to find that it is printed
(somewhat cryptically) on the face) (I made a two wire / three wire
connection error)

73, Steve K9DCI