On Monday, 28 Feb 2005 21:18:18 -500, "Asimov"
wrote:

I'm sorry but it is an erroneous conclusion to think it cools. Don't
you recall mentioning that the "average" voltage and current remains
the same?

Hi OM,

One of those Class A characteristics. It is revealed by the rather
more simple solution to the difficult problem I offered:
What is power in?
What is power out?

This characteristic is part of the lore for those who built their own
equipment as one of several NEVERs
Don't let your amp run without drive;
Don't let your amp run without bias.
Both conditions can lead to melt down.

Unfortunately this kind of Ham memory is fading to the point where
many who witnessed this "internal resistance" are not here to point
out the obvious errors in understanding exhibited by the credit card
generation who build their rigs with plastic.

73's
Richard Clark, KB7QHC

Cecil Moore wrote:
Richard Harrison wrote:

I`ll use Cecil, W5DXP`s argument. Energy must be conserved. Energy in
equals energy out. If some goes to a load it does not stay within the
amplifier to make heat.


From "Electronic Fundamentals and Applications" by
John D. Ryder, regarding Class-A amplifiers:
"As the a-c output increases, the plate loss
decreases and the tube runs cooler."


This is why the load must be matched to the amp. Anyone who has fried a
final will tell you so.
Been there, Done that, Got the bottles to prove it.

Dave WD9BDZ

Richard Clark wrote:

On Mon, 28 Feb 2005 22:53:17 GMT, gwhite wrote:

Obviously people don't have 100 W (or more!) network analyzers looking into the
output and pretending the device is similar to a linear small signal device.

Hi OM,

Well, it is more accurate to say that you don't, that is for sure.
Defining a solution by negative results can fill up a library without
any positive accomplishment. Obviously people don't have a nuclear
reactor, or lunar lander, or bank account to balance the national
debt. The joke of this, of course, is that no one needs a 100 W (or
more!) network analyzer, or nuclear reactor, or lunar lander, or bank
account to balance the national debt to explain a rather more trivial
problem. Which, by the way, has nothing to do with pretending at all.

You entirely missed the point. You don't know the output impedance because you
don't have a way of determining it by swinging the output full-scale. Even for
class A, large signals will/can have rail to rail swing. The device will not be
linear for large swings: sinusoidal input swing will not result in a sinusoidal
output swing. But "impedance" is a sinusoidal (s-domain) concept. So how can
you define an impedance--a sinusoidal concept--when the waveform is not
sinusoidal for an inputted sine wave? The point is that the output impedance is
time dependent ("causes" the non-sinusoid output for sinusoid drive), which
rather makes the concept questionable. As I wrote earlier, one might decide to
consider a time averaged impedance, but I'm not clear on what the utility would
be.

The suggestion that
requires load pull test equipment and that can be expensive
does not negate its existence which commonly proves what you choose to
dismiss as impossible. I have calibrated this gear (called an
artificial or active load), and the gear (called transmitters) it
tests and there are no differences in Physics based upon your
presumption of low-power/high-power demarcations.

There is no "presumption." Linear parameters and theorems totally ignore
practical limitations--this is a fact and you can look it up in just about any
text on circuit analysis. The simple linear model is perfectly okay for small
signal devices. It isn't okay for large signal devices. In any case, load pull
equipment does not make the pretense of defining output impedance of an active
large signal device. It does say what the load needs to be to acquire maximum
power out of the device.

To say
pretending the device is similar to a linear small signal device
is one of those assumptions forced into the argument.

No, it isn't. Thevenins and conjugate matching (for maximum power transfer) are
explicitly linear small signal device models. Their use in RF PA output design
is a misapplication.

There are any
number of ways to do something wrong.

We're talking about one of them. Misapplying small signal linear parameters to
the output of a large signal device.

Trumping none of these straw
men validates another wrong impression passing as theory.

This
returns us to the imposition of impossibilities to answer a rather
mundane concept, eg.
pretending the device is similar to a small nuclear device
pretending the device is similar to a mars rover
pretending the device is similar to the national debt of Lithuania

Who are you quoting and why?

So to return to a common question that seems to defy 2 out of 3
analysis (and many demurred along the way) - A simple test of a
practical situation with a practical Amateur grade transistor model
100W transmitter commonly available for more than 20-30 years now:
1. Presuming CW mode into a "matched load" (any definition will do);

Any definition won't do, and for this discussion the specific "won't do" is
using conjugate matching which is a small signal (linear) model.

2. Report the DC power consumed before hitting the key;
3. Report the DC power consumed while holding the key.

Hey, at least you're recognizing that DC power is important. Where in conjugate
matching ideas or Thevenins theorem do you see any concern of DC power? That's
right, you don't because they a simple small signal models where DC power and
voltage have no bearing because the signals are so small, relatively speaking.

Concurrently note:
A. Report Heat Sink Temperature for a previously idle/rcv condition;
B. Report Heat Sink Temperature after 10 minute key-down.

For a hypothetical "100W" model (again, a contemporary, common example
for Amateur use) available through standard commercial venues:
2. About 20W - 30W
3. About 200W - 250W
A. About 20 degrees C (or room temperature)
B. Well above 37 degrees C (or skin temperature)

Now, if we are to be any judge of efficiency (Thevenin does not have
to be invoked, condemned, or venerated); then it runs close to 50%
(±10%). Others can invoke their favorite deity to explain.

*You* brought up Thevenins and armchair philosophy regarding it, not me. I said
Thevenins was irrelevent, and now you appear to agree with me. Ken effectively
brought up conjugate matching, not me. The original comment I was challenging
was:

"...the antenna works as an impedance mathcing network that matches the output
stages impedance to the radiation resistance."

I simply wanted to make it clear that the "matching" done was not an issue of
"output impedance" per se. It is an issue of how the transistor is to be loaded
to extract maximum ouput power.

Now, if we are to be any judge of dissipation (no requirement for
advanced degree); then heat as a loss by virtue of less than 100%
efficiency is quite evident. Others can invoke photons to describe
why.

To forestall any armchair engineers, yes, this efficiency is System
efficiency. However, I would be surprised if a practical common
Amateur grade transistor model transmitter commonly available for more
than 20-30 years now has any configuration that does not apply supply
voltage directly to the final transistors; and instead adds a
significant current path outside of this load (citations to available
schematics would be compelling, but any argument without this would be
speculation). It takes very little effort to subtract out the power
drain of the receive mode (being very representative of the similar
power demand of supporting circuitry for transmit up to the driver
stage). Barring such amazing evidence of a significant power drain
not found in the finals, it follows that a simple computation of
efficiency has its merit and has been met.

Exactly. It is about DC to RF efficiency, as I've been pointing out since my
first post, and which you initially commented was "nonsense" but now seem to
agree with. "Impedance matching" meant in the normal sense of conjugate
matching for maximum transfer of power is a misapplied small signal
concept/model. I think that is all I've really been saying.

The most simple way to describe it is -

With a class-A amplifier the power taken from the DC supply remains constant
regardless of signal power output.

The device(s) internal dissipation decreases by the same amount as the
signal power output (drive level) increases.

If there's any change in power taken from the DC supply as the power output
changes then the device is not operating under class-A distortion-less
conditions.

KISS. KISS. KISS. KISS.

To slightly change the subject, anybody who mentions a conjugate match
doesn't know what he's waffling about.

And if I remember correctly, a 6N7, a class-B dual-triode, metal tube was
specially designed to have a very high Mu and could be operated under very
simple zero-bias conditions. Each of the two triodes handled one half of the
complete audio sinewave.

With grids in push-pull and the anodes in parallel, you could get 10 watts
out of it as an HF frequency doubler.

But it would still take the skin off your fingers.

Ahhh! Happy days!
----
Punchinello.

"Cecil Moore" bravely wrote to "All" (01 Mar 05 09:21:18)
--- on the heady topic of " Say what you mean."

CM From: Cecil Moore
CM Xref: aeinews rec.radio.amateur.antenna:26191

CM From "Electronic Fundamentals and Applications" by
CM John D. Ryder, regarding Class-A amplifiers:
CM "As the a-c output increases, the plate loss
CM decreases and the tube runs cooler."

harmonic generation...

A*s*i*m*o*v

.... Be nice to your kids. They'll choose your nursing home.

Reg Edwards wrote:
With a class-A amplifier the power taken from the DC supply remains constant
regardless of signal power output.

I wish I had said that, Reg. thanks ...
--
73, Cecil http://www.qsl.net/w5dxp

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

On Tue, 01 Mar 2005 18:06:18 GMT, gwhite wrote:

It is about DC to RF efficiency,

Put a number to it.

as I've been pointing out since my
first post, and which you initially commented was "nonsense"

Hi OM,

And so it remains with additional elaborations not quoted here.

but now seem to agree with.

Seeming is a rather insubstantial thing to hang your theories on.

"Impedance matching" meant in the normal sense of conjugate
matching for maximum transfer of power

And this reveals the error of "Seeming" because the so-called meaning
you ascribe is this same nonsense. Pay more attention to reading
instead of writing. It has been pointed out more than once, and by
several, that Matching comes under many headings. The most frequent
violation is the mixing of concepts and specifications (your text is
littered with such clashes).

is a misapplied small signal
concept/model. I think that is all I've really been saying.

And I preserved this clash quoted above as an example. If there is
any misapplication, you brought it to the table with this forced
presumption. The misapplication of S parameters to a large signal
amplifier is one thing, to project this error backwards into the
fictive theory that there is some difference between large and small
signal BEHAVIOR (not modeling) is tailoring the argument to suit a
poorly framed thesis.

None of your dissertation reveals any practical substantiation, hence
it falls into the realm of armchair theory. We get plenty of that
embroidered with photonic wave theory that is far more amusing.

73's
Richard Clark, KB7QHC

In article , gwhite wrote:
[...]
You entirely missed the point. You don't know the output impedance because you
don't have a way of determining it by swinging the output full-scale.

You don't have to swing the output full-scale to measure the impedance.
Any change in the load, no matter how small, will cause a change in the
output voltage and the output current. From these you can calculate the
output impedance at the current operating point.

When a transistor is operating under large signal conditions into a tuned
load, there is still an output impedance and this impedance still
discribes what will happen for small changes in the load.


--
--
forging knowledge

"Richard Harrison" bravely wrote to "All" (01 Mar 05 08:29:27)
--- on the heady topic of " Say what you mean."

RH From: (Richard Harrison)
RH Xref: aeinews rec.radio.amateur.antenna:26189

RH The Class A amplifier gets all its power from the d-c supply and it is
RH constant, signal or no signal. With signal power output, some of the
RH power in exits to the load.

Let's look at it from the dynamic point of view. When there is no
signal the circumstances are as you describe. When the signal swing
turns the device nearly OFF there is a maximum voltage across the
device but the current is a minimum or close to zero.

Conversely when the signal swings in the opposite polarity the device
turns ON hard but the voltage is at a minimum or near zero too. In
either case the "ideal" device dissipates no extra power because the
two products are always zero. In fact the power difference is
always zero, so there is no reason for the device to cool.

However, no device is ideal, it has losses, and generates harmonics.
If the rectification is such that the average current decreases then
the device will cool. There is a gotcha however, because the harmonics
rob power from the desired output.


RH I`ll use Cecil, W5DXP`s argument. Energy must be conserved. Energy in
RH equals energy out. If some goes to a load it does not stay within the
RH amplifier to make feat.

As you said, if the amplifier dissipates 100W when idle and it can
drive 100W into a load then the power source will still supply 100W
when so doing, of course being an ideal device and neglecting all
other losses. It is easy to test by paralleling a substantial
capacitance and measuring the supply current before the cap. The
current should remain the same or rise only very slightly.


RH Asimov also wrote:
RH "---linear source not operating Class A?"

RH I`ll give an example. The Class B amplifier is biased near current
RH cut-off. Current is near zero when the signal is. Yet, output can
RH favorably vie with that from a Class A amplifier for purity. I learned
RH that nearly 60 years ago when I built my first 6N7 phonograph
RH amplifier.

Class B is hardly linear if it only amplifies 1/2 of the sinewave.
Time for our periodic 10 year review. g

A*s*i*m*o*v

.... That's a cute trick.

I read in sci.electronics.design that Ken Smith
wrote (in )
about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005:
In article , gwhite wrote:
[...]
You entirely missed the point. You don't know the output impedance because you
don't have a way of determining it by swinging the output full-scale.

You don't have to swing the output full-scale to measure the impedance.
Any change in the load, no matter how small, will cause a change in the
output voltage and the output current. From these you can calculate the
output impedance at the current operating point.

When a transistor is operating under large signal conditions into a tuned
load, there is still an output impedance and this impedance still
discribes what will happen for small changes in the load.


This incremental impedance is one of several different impedances that
can be defined for a non-linear source. No one is more valid
conceptually than another, but some are of more practical significance
than others.

The point is that if you want to talk/write about one of these
impedances, you need, to prevent misunderstanding, use a precise term,
such as 'incremental output source impedance' and define it.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk