Reg Edwards wrote:
The internal resistance of a generator is independent of its internal
voltage. Just reduce the drive level to some small, no particular value,
and stop making excuses.

( Somebody will say it IS dependent. But we are interested only in
ball-park accuracy. And in any case the operating point will be within the
normal range of operation between no-drive and full drive, or between no
modulation and full modulation.)

That might work better for Class-AB than Class-C. If one
reduces drive level on a Class-C, one is reducing the 'on'
time of the device, thus changing the internal impedance
probably somewhat inversely proportional to the 'on' time.

If I remember correctly, my IC-706 would not fold back
under any load condition when run at minimum power.
--
73, Cecil http://www.qsl.net/w5dxp

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"Cecil Moore" bravely wrote to "All" (02 Mar 05 20:05:49)
--- on the heady topic of " Say what you mean."

CM From: Cecil Moore
CM Xref: aeinews rec.radio.amateur.antenna:26289

CM Asimov wrote:
Watts to plasma.

CM Sorry, it's solid state - no screen. Let's face it - unloaded
CM amplifiers tend to burn up no matter what class they are running.

Why don't you repeat that to the folk who think their amp cools when
it outputs watts?

A*s*i*m*o*v

.... New computer? But I like my vacuum tubes... They keep me warm.

Asimov wrote:
"I was confusing it with Class B which does warm up the heatsink"

True. Class B amplifiers can be single-ended for R-F if used with the
right tank circuit to supply the missing 1/2-cycle. More often they are
push-pull to reduce harmonics.

SSB final amplifiers are biased to near current cut-off. During
modulation gaps, current falls to nearly zero. When current is zero, so
is the dissipation.

Reg mentioned the blistering heat of the 6N7. Thai`s my recollection too
when getting nearly 10 watts of audio from the small twin-triode
octal-based metal tube. Reduce the audio volume and the heat is reduced
simultaneously.

Best regards, Richard Harrison, KB5WZI

Richard Clark wrote:


As I've noted in the past, you can fill a library with negative
assertions...

The troublesome assertion is not the negative one. It is that RF PA's are
conjugate matched. Neither you nor Ken has provided a single example of such a
design that also extracts the maximum amount of "linear" power from a device and
essentially its power supply (after all, that is what it is: a _power_ amp).
Your example said nothing about output-Z, which suggests you have no clue, since
you didn't even remotely address the issue.

For Ken's part, he recently obfuscated by dismissing an example that was
primarily intended to be illustrative, but yet holding the salient points. He
completely ignored (or didn't understand) the clipping issue. Further
obfuscation was provided by talking about "protection circuitry," which may or
may not exist in a circuit, but adds zero to a discussion regarding how the PA
is to be loaded. "Protection" is a non-stater because the PA is either off or
impaired.

Ken's argument is circular. He say's that if a design is done for conjugate
match,
then it will behave as if it is conjugately matched. Well of course (or at
least sort of under specific test conditions and circuits)! It is
self-fullfilling prophecy but it unfortunately makes no statement regarding
obtaining the maximum power out of the circuit in the sense of turning DC power
into RF power (yes, *extracting* power from the DC supply and transformed to
RF). This is paramount to PA design. To use the device to maximum efficacy, as
Cripps puts it, a load-line match is needed. Ken's "conjugate match" design
won't do that, and that's why PA's aren't designed that way.

The bottom line is that if I design an amp via load line techniques using the
same device and power supply as Ken (him using conj-match), my amp will deliver
higher unclipped PEP than his. That is the factual result you resist. Now if
you want to pay for extra power and big devices, that's your business--go ahead
and attempt to conj-match your amp--but engineers who design PA's don't do that.

Another idealized and hypothetical example to elucidate the load-line principle
is offered.

Let's say we have a 10 W FET we'll build into a class A circuit. An RF choke is
used to supply drain current. We DC bias it to Vd = 10 V and Id = 1 A. Just
for argument sake, let's say it has a constant internal resistance of 110 ohms
and the device will break down at 25 V. According to the most idealized and
standard load-line theory, we should load it to rL = Vd/Id = 10 Ohms. This
idealization includes the definition of positive and negative clipping --
whichever comes "first" -- of being the operational limit for output voltage
swing. Clipping is associated with severe distortion.

Since we need rL to be 10 ohms, and Ri = 110 ohms, we need to make the actual
load resistor equal to: RL = 11 Ohms. Let's check that result and see if it
meets the clipping constraint for maximum available power.

positive swing = Id*rL = 1*10 = 10 V
negative swing = Vd = 10 V
Power delivered to RL: Pload = 10^2/(2*11) = 4.55 W
The efficiency is a little under 50% because of the internal resistance. Note
the Load resistance is decidely not the conjugate of the internal resistance.

Let's spot check the load to see if it at least appears to be the peak available
power, by testing two loads "immediately" on either side of our optimum 11 ohms.

Let RL = 10 ohms
positive swing = Id*rL = 1*9.17 = 9.17 V
negative swing = Vd = 10 V
Since we positive clip at 9.17 V, we are limited by our design clipping
constraint to only driving the PA such that 2*9.17 V is the maximum available
voltage swing.
Power delivered to RL: Pload = 9.17^2/(2*10) = 4.20 W

Let RL = 12 ohms
positive swing = Id*rL = 1*10.82 = 10.82 V
negative swing = Vd = 10 V
Since we negative clip at 10 V, we are limited by our design clipping constraint
to driving the PA such that 2*10 V is the maximum available voltage swing.
Power delivered to RL: Pload = 10^2/(2*12) = 4.17 W

Sure enough, the power peaked at a load of 11 ohms, just like load-line theory
says it will. Now let's see what the available power hit of conjugate matching
is.

By definition, conj-match insists RL = Ri = 110 ohms. Again we are limited in
our clipping constraint by static drain current, and supply voltage,
specifically 10 V.

Our negative swing limit is, as ever, 10 V (the drain voltage).

positive swing = Id*rL = 1*55 = 55 V

This would breakdown the device, but the lower negative swing will force us to
back down the drive to meet the design defined clipping constraint.

Pload = 10^2/(2*110) = 0.455 W

Conjugate matching resulted in a 10*log(0.455/4.55) = 10 dB available power
hit. Power amplifiers are not designed with conjugate matching in mind. You
don't need to re-invent the wheel. Just follow well established principles when
doing cookie cutter PA design.

The list could go on,...

LOL. Given your pattern, I am sure it will.

You sighed with content at being offered a "relevent
question/statement" Your re-iterative response contains the same (how
could it be otherwise?) slack of precision that started this. Want to
try again?

Not really. The problem isn't precision, it is you can't, or refuse, to
comprehend what is being said, which I presume is why you instead write with the
most bizarre terms and phrasology that has nothing of import to the topic at
hand.

You could have as easily expressed what sense they ARE matched,

For what seems like the billionth time now: they are load-line matched.

...but instead this time offer what Basis
of Matching you are attempting to
describe.

I've given a didactic example (actually a couple), you just don't--or more
likely won't--get it. If you don't like my example, you can refer to Cripps,
who is considered one of the preeminant RF PA experts in the world.

Even more simplistic is Malvino's discussion on pp177-185 of the first edition
((c) 1968) of "Transistor Circuit Approximations." It is basically a technician
level description, so perhaps it is well-suited to you. In academics, load-line
theory is presented down to tech level courses and up across to engineering.
That some engineers and techs aren't clear on the load-line concept for PA's (or
*any* circuit needing a wide symmetrical swing) is notwithstanding.

This is the more rigorous approach that eliminates vague
descriptions and uses standard terms. If you have to query about what
"Basis" means (used by professionals - namely metrologists who can
quantify Output Z of all sources) - then we can skip it as a topic out
of the reach of amateur discussion.

I see you still don't know what impedance is. In any case, it doesn't mean that
looking into a properly designed PA output with a network analyzer confirms the
conj-match precept, it doesn't.

Impedance is a *linear* conception, a portion of linear theory, and again by
definition:

Z = V/I

V and I are sinusoids (phasors). But with power amps, substantial non-linearity
exists (destroying the linearity assumption of impedance), thus applying a
linearly defined concept to a non-linear milieu is a misapplication. You are
attempting, as is Ken, to stuff a square peg down a round hole. Why?

The concept is even questionable for the most linear of the power amps: class
A. In any case, given real devices with real supplies, the conj-match ideal is
next to worthless. While I could agree that the borderline may be fuzzy
regarding where and when to drop the impedance notion, it still stands that the
concept is not useful in determining how to optimally load an RF PA.

At this point you own the conj-match assertion as much as Ken. Prove it! You
can't because it is fundamentally incorrect.

Note:
Again, RF PA's should be load-line matched.
Does not qualify as a Basis.

Load-line matching is such a basic electronic concept it is unbelievable how
oblivious you are to the concept. Read a basic book. Don't rely on me: look it
up and do your own design!

It is suggestive of one, but because you indiscriminately
mix several Basis within your discussions, it is your
responsibility to be precise.

You just like to hear yourself talk. I've been explicit and precise. You just
don't know anything about the elementary electronics principle of load line
matching. I presume this is why your comments have zero substantive
responsiveness.

If you can accomplish this, then we can
proceed to review how little it all matters.

If you keep ignoring what I've written, and that which is written in elementary
electronics texts, you can remain happily ignorant of understanding the
simple-basic-fundamental concept presented. Your choice.

Barring resolving any of these issues of precise language,...]

The guy ignorant of the definition of impedance and that s-domain theory *is*
linear circuit theory (and more goodies) is talking about "precise language."
Amusing.

I notice
that you rather enjoy...

No, I don't enjoy it at all. Your lack of electronic understanding is dismal,
especially given your tone. It would have been a lot easier for me if Ken
hadn't made the erroneous
statement in the first place and made a correct one instead. That would have
been my preferance.

..fruitless jousting with them than challenging my
support of Ken's (supposed) statement that you say is your focus:
However, responding to the bald statement, I find nothing
objectionable about it.

That's because you don't understand the difference between impedance matching
and ac load line matching.
We will leave that as another dead-end.

I suspect you will. I already understand it -- you're the one who doesn't.


"One of the principal differences between linear RF amplifier design and PA
design is that, for optimum power, the output of the device is not presented
with the impedance required for a linear conjugate match. That causes much
consternation and has been the subject of extensive controversy about the
meaning and nature of conjugate matching. It is necessary, therefore, to swallow
that apparently unpalatable result as early as possible (Section 1.5), before
going on to give it more extended interpretation and analysis (Chapter 2)." --
Cripps, p1


The quote is on Page 1. Swallow it now. Learn something for a change.

I read in sci.electronics.design that gwhite wrote
(in ) about '1/4 vs 1/2 wavelength
antenna', on Thu, 3 Mar 2005:
By definition, conj-match insists RL = Ri = 110 ohms. Again we are
limited in our clipping constraint by static drain current, and supply
voltage, specifically 10 V.

Our negative swing limit is, as ever, 10 V (the drain voltage).

positive swing = Id*rL = 1*55 = 55 V

This would breakdown the device, but the lower negative swing will force
us to back down the drive to meet the design defined clipping
constraint.

Pload = 10^2/(2*110) = 0.455 W

And the power dissipated in the device is also 0.445 W. Matching
according to the 'maximum power theorem' or conjugate matching, results
in equal power in the PA and load. That's why it isn't useful for power
amplifiers.

Doesn't everyone know that an audio amplifier that id designed to feed
an 8 ohm load MUST have an output source impedance of 0.0000001 ohms or
less. An output source impedance of 8 ohms would dramatically decrease
the electromagnetic damping on the loudspeaker voice-coil - by the huge
factor of .... two!(;-)
--
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

On Thu, 03 Mar 2005 20:25:16 GMT, gwhite wrote:

You sighed with content at being offered a "relevent
question/statement" Your re-iterative response contains the same (how
could it be otherwise?) slack of precision that started this. Want to
try again?

Not really.
....
I notice that you rather enjoy fruitless jousting with them than challenging my
support of Ken's (supposed) statement that you say is your focus:
However, responding to the bald statement, I find nothing
objectionable about it.

That's because you don't understand the difference between impedance matching
and ac load line matching.
We will leave that as another dead-end.

I suspect you will.

Hi OM,

224 line postings to produce this little qualitative information? :-)

73's
Richard Clark, KB7QHC

On Thu, 3 Mar 2005 20:53:48 +0000, John Woodgate
wrote:

Doesn't everyone know that an audio amplifier that id designed to feed
an 8 ohm load MUST have an output source impedance of 0.0000001 ohms or
less.

Hi John,

I hope that was a joke.

73's
Richard Clark, KB7QHC

Richard Clark wrote:


224 line postings to produce this little qualitative information? :-)

I see one line here with no content. You don't have any argument because you
have zero understanding. No one can cure that but you. But can you?

"Stupid is as stupid does." -- Forrest Gump

John Woodgate wrote:

I read in sci.electronics.design that gwhite wrote
(in ) about '1/4 vs 1/2 wavelength
antenna', on Thu, 3 Mar 2005:
By definition, conj-match insists RL = Ri = 110 ohms. Again we are
limited in our clipping constraint by static drain current, and supply
voltage, specifically 10 V.

Our negative swing limit is, as ever, 10 V (the drain voltage).

positive swing = Id*rL = 1*55 = 55 V

This would breakdown the device, but the lower negative swing will force
us to back down the drive to meet the design defined clipping
constraint.

Pload = 10^2/(2*110) = 0.455 W

And the power dissipated in the device is also 0.445 W.

I think it is 1A*10V - 0.455 W = 9.545 W
^^^^^^ ^^^^^^^
DC input Power
Power delivered
to RL


The resistance dissipated in the "internal AC resistance" is equal to RL in the
conj-match condition. Of course, we're ignoring input power here, which is
"small" when the gain is +20 dB.

Matching
according to the 'maximum power theorem' or conjugate matching, results
in equal power in the PA and load. That's why it isn't useful for power
amplifiers.

Amusingly for my hypothetical class A conj-match example, the "equal power
dissipation" isn't such a big deal, since it is class A and the fractional power
dissipated in either the internal AC resistance or the external load resistance
is rather small compared to DC dissipation (less than 10%).

Doesn't everyone know that an audio amplifier that id designed to feed
an 8 ohm load MUST have an output source impedance of 0.0000001 ohms or
less. An output source impedance of 8 ohms would dramatically decrease
the electromagnetic damping on the loudspeaker voice-coil - by the huge
factor of .... two!(;-)

Nice one.

On Thu, 03 Mar 2005 20:53:48 +0000, John Woodgate wrote:
I read in sci.electronics.design that gwhite wrote
....
Doesn't everyone know that an audio amplifier that id designed to feed
an 8 ohm load MUST have an output source impedance of 0.0000001 ohms or
less. An output source impedance of 8 ohms would dramatically decrease
the electromagnetic damping on the loudspeaker voice-coil - by the huge
factor of .... two!(;-)

Yeah - isn't that why the TOOB amps had those taps? So you could get that
rich, full-bodied TOOB sound? ;-)

Cheers!
Rich