Reg, G4FGQ wrote:
"What allows a class-C amplifier to exceed 50% efficiency is a small
operating angle."

Exactly, and during the majority of the degrees it`s switched completely
off. It draws no current and suffers no "IsquaredR loss" during the
amplifier off-time. Impedance is approximately E/I, but I is the average
I, which is much less than the bursts of I during the conduction angle.

The switched-off time makes the I in the denominator of E/I very small
indeed and the solution to Ohm`s law is a high impedance without the
dissipation of a resistance that remains in place continuously while
agitating the atoms of a poor conductor to limit current.

Instead, we have a low-resistanc in high conducton for short spurts.
On-time is limited, instead of conduction, to produce a certain
effective resistance.

An automobile Kettering ignition system may use a dwell-meter to
indicate how much of the time the points are closed. An ohmmeter
indicates the resistance between its test prods. The two test circuits
are almost the same although limitation of the deflection of the
dwell-meter is different from limitation of the deflection of the
ohmmeter due to the difference between limited conduction angle ignition
points, and the continuous conduction through a current-limiting
resistor. There`s an analogy between Class C and Class A amplifiers in
there somewhere.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Reg, G4FGQ wrote:
"What allows a class-C amplifier to exceed 50% efficiency is a small
operating angle."

While this is too vague, Richard tries to add detail, but mis-steps just
a bit... and Steve goes into an extended "You ain't quite correct blurb..."


Exactly, and during the majority of the degrees it`s switched completely
off. It draws no current and suffers no "IsquaredR loss" during the
amplifier off-time. Impedance is approximately E/I, but I is the average
I, which is much less than the bursts of I during the conduction angle.

We must be careful with the word "average" here.

First, my "class C" model is a follows:
I liken it to digital or "switched modes". While I have never scoped the
plate to observe this... When the tube is cut off for a large part of the
cycle, there is a high voltage on the tube (I believe it swings higher than
the supply dou to the "ringing" of the plate tuned circuit), but no current.
Hence, ExI=0.
When the tube is on, it is slammed hard on by the "high" grid signal and
there is a high plate current, but the plate voltage is very low (anybody
know how low and if I am all wet? ... tubes aren't quite like transistors
in the digital mode)--therefore ExI=somthing, but since the E is low, it is
lower than in class A during that part of a cycle. There may also be some
effect due to the fact that the plate tank is swinging low allowing the
plate voltage to be even lower.

Did you know that in class A, the plate power dissipated goes DOWN by
the amount that is delivered to the load??? Cool! huh? Isn't physics neat!

Second, it is the RMS current through the tube which will waste power,
so it is what we must be concerned with. Yes, if the tube is off the
current is zero at that time, but the RMS must be considered and it does not
go down as fast you might think. As an example, for the same current
pulses, but spaced out to half the duty cycle, the average drops to half,
but the RMS only drops to .707. There is a square root in there.

[[Anybody see the "AC Watt meter article in QST]] It is an OOPS! Most
power supplies don't draw sine wave current. It is pulses. I have been in
contact with both Bob Shrader (the author) and Stu Cohen (Tech editor) and I
just finished an analysis and am going to make more measurements to verify,
but the numbers Bob published can be as much as 1/3 the true power values
(depending upon the DVM he used and the current waveform of the supplies he
measured.

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

wa-da ya blokes think.



The switched-off time makes the I in the denominator of E/I very small
indeed and the solution to Ohm`s law is a high impedance without the
dissipation of a resistance that remains in place continuously while
agitating the atoms of a poor conductor to limit current.

Instead, we have a low-resistanc in high conducton for short spurts.
On-time is limited, instead of conduction, to produce a certain
effective resistance.

Another way of saying just wjat I did above, but "effective resistance"
is one way of thinking about it and this resistance must be calculated using
the RMS values.

An automobile Kettering ignition system may use a dwell-meter to
indicate how much of the time the points are closed. An ohmmeter
indicates the resistance between its test prods.

I'd be willing to place a bet (knowing how an analog ohm meter works,
that the *diflection* of the two meter pointers is the same (see below).
Both meters respond to the averacge current through them and both will show
full scale when the points are open (I think thta is the correct polarity).
Here's the "below":
There is, however, the confusion added by the coil/cap waveform for which
the ohm meter is not equiped to limit - whereas, I believe the dwell meter,
if well designed, will have something to limit so as to remofe it as a
complicatin.

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

The two test circuits
are almost the same although limitation of the deflection of the
dwell-meter is different from limitation of the deflection of the
ohmmeter due to the difference between limited conduction angle ignition
points, and the continuous conduction through a current-limiting
resistor. There`s an analogy between Class C and Class A amplifiers in
there somewhere.

Best regards, Richard Harrison, KB5WZI

Steve Nosko wrote:
"First, my "class C" model is as follows: I liken it to digital or
"switched modes"."

I do too.

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

RMS is the effective value, not the average value, of an a-c ampere. It
is defined as 0.707X the peak value of the waveform. It is derived from
the average of the squared current over a half cycle, as the heating
value of an ampere is proportional to the current squared.

Speaking inversely, the ratio of maximum to effective value for a sine
wave is 1.414, which is the square root of 2.

Ordinarily, with nonsinusoidal currents, the ratio of maximum to
effective value is not the square root of 2.

Best regards, Richard Harrison, KB5WZI

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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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 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

"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

Steve Nosko wrote:
"I wonder if a "true RMS" DVM can handle this?"

So do I. The true rms meter seems a wonderful development to me.

Steve also wrote:

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

My speculation is that the effective value of a nonsinusoidal waveform
could be found by summation of its sinusoidal constituents.

But, it`s not difficult to find an effective value for not only
sinusoidal periodic waveforms but for
nonsinusoidal periodic waveforms as well. One can graphically take a
large number of equally spaced ordinates of the form, using at least one
complete alternation, Richard Clark. Both alternations are not needed
but could be used as a minus times a minus is a plus and each of the
ordinate values must be squared because power is a function of the
current squared, so both alternations when their ordinates are squared
produce positive values. Next, we sum the squared ordinate values and
divide by the number of ordinates. You get the average value of the
squared curve which is what we are looking for.

Or, you can construct the squared curve and integrate the area under the
squared curve by using a planimeter. Dividing the area by the baseline
length gives the same average value of the squared curve as iusing the
graphic ordinates above. Voila: rms.

Unless I`ve opened a new can of worms with this posting, I don`t know of
any difference of opinion I have with Steve.

Best regards, Richard Harrison, KB5WZI