Ken Smith wrote:

In article , gwhite wrote:
Ken Smith wrote:


The strongest argument for dropping the impedance matching concept is PA
efficiency, and therefore maximum signal swing. Obtaining maximum swing is a
load line issue.

What do you mean by "maximum signal swing" in this context. I can get a
bigger swing by leaving the output completely unloaded and hence causing
the actual efficiency to be zero.

LOL. Sure, the purpose of a power amp is to actually extract power. This is a
good start.

No, the purpose of the power amp is to deliver power, not extract it.

Well really, once the device and supply have been determined, we can indeed view
it as extraction. We'll load it to extract the most. If you want to mince
words and call it "deliver," that is fine.

Perhaps a simplistic (and of course idealized) class A example would help. And
I want to remind that this is a simplification of the first order design cut.

Don't bother with the over simplified Class A case. RF power
amplification is rarely done class and and it is a digression from the
actual topic.

Well, class A is certainly done. Two cases are where the extra little bit of
linearity is desired and at high frequencies, were PAE starts to take a bite as
the gain drops below 10 dB.

Our circuit loaded with 10 ohms delivers twice as much power as with the lesser
5 ohms or greater 20 ohms. That is, extracted output power is peaking at some
finite non-zero value. This is also easily seen to be most efficient point for
this simplistic example.

At some point as you decrease the resistance, the output will drop to zero
as the amplifier fails or it will start to decrease in some more
controlled manner as the protection circuits take control. If we assume
the latter case, it is easy to see that the power reaches a maximum value
and then decreases as the resistance is lowered. The point at which the
power is at the maximum is the point at which the load is matched. If you
make a small change in the load and observe the voltage and current when
that small change is made, you will see that that is indeed the output
impedance of the amplifier. I think this is the part you are not
grasping.

No, this is exactly where I'm saying you are incorrect. You are not getting the
practical limitations and are mistakenly applying linear concepts. It doesn't
work if you want to extract maximum power from the DC supply through a real
device, converting the DC power into RF power.

Ken Smith wrote:

In article , gwhite wrote:
[...]
Here's the original quote [Ken]:

"When the correct matching is done, the antenna works as an impedance mathcing
network that matches the output stages impedance to the radiation resistance."

Yes, I stand by and have just in another part of the thread once again
explained that indeed the impedance is matched. ie: If you make a small
change in the impedance in any direction the power decreases.

Driven to max swing, this is true. But it is because of asymmetrical clipping,
not because of conjugate mismatch. For lower drives, what you say won't
necessarily be true *unless* you've mis-designed according to conjugate match
ideals. Your argument is circular.

If you design for conjugate match, you're right. I'm saying: don't do that. If
I design for load line match and you design for conjugate max (both pf us using
the same device and supply), I will get a higher peak power than you will.
However, you'll get to be right about how your amp acts regarding diverging from
conjugate load. But it is irrelevent: you made a fundamental mistake.

Increasing
the resistance is the obvious one. The other three are because the
protection circuits act. The OP had a completed transmitter he was
connecting to a length of wire.

G. White wrote:
"This example is intended to be illustrative rather than exact."

Nobody contradicts that maximum power from an amplifier requires an
impedance match, so far.

Nobody contradicts that efficiency in an ideal Class A amplifier can
never exceed 50%, so far.

Some texts show waveform distortion typical of a single-ended Class A
amplifier and note that "maximum undistorted power" requires a load
impedance 2 to 3 times that of the amplifying device.

G. White gives a peak to peak voltage of 20 and a power out of 5 watts.
RMS volts are 7.07 in this case. The load then is the square of the
volts divided by the power. This is 50/5 or 10 ohms for the load. If
this matches his amplifier, 5 watts is the maximum output.

G. White said that he did not consider the output-Z of the amplifier in
its loading. Ignoring an amplifier`s impedance does not make it go away.
Ignoring an amplifier`s impedance does not revoke the maximum power
transfer theorem, either.

Best regards, Richard Harrison, KB5WZI

Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current)

At present I have no facilities to make such measurements.

Perhaps a few curious readers, who do have facilities, might make crude
measurements on their HF rigs and report approximate impedance values on one
or two bands on this newsgroup. Some sort of average could be obtained.
----
Reg, G4FGQ
===================================

"Richard Harrison" wrote in message
...
Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A
amplifier)."

The no-signal loss of a Class A amplifier is 100%. It equals volts x
amps and appears in the amplifier. Now feed a signal to the ideal
amplifier set just below the clipping level. Average d-c power is
unchanged from the unloaded and no-signal conditions.

Connect a matched load resistor to the amplifier output. If physically
small, the resistor may become warm with heat that were it not for the
load would be otherwise dissipated in the amplifier. Input power to the
Class A amplifier is unchanging.

Finding the internal resistance theoretically is simple. It is simply
the open-circuit output voltage divided by the short-circuit current.

Open-circuit voltage at full output and short-circuit current may be
severe. Internal resistance can be found under less stressful
conditions. Internal resistance will drop the voltage to any load
reasistance. Use the voltage-divider formula to calculate the internal
resistance.

With pure resistances, half the open-circuit volts are dropped by the
internal resistance when the load is a match.

A power amplifier`s internal impedance can be determined.

Power output from a Class A amplifier cools it.

Best regards, Richard Harrison, KB5WZI

Asimov wrote:

"Cecil Moore" bravely wrote:
CM Why do the instructions on my stereo amp warn against
CM running the amp with no speakers attached?

Because then the screen tries to carry the plate signal, the reactance
in the output transformer is not damped, and because a tube is
sensitive to voltage, this quickly leads to a molten hole in the side?
Watts to plasma.

Sorry, it's solid state - no screen. Let's face it - unloaded
amplifiers tend to burn up no matter what class they are running.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current)

Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Reg Edwards wrote:

Richard, your method of measuring internal impedance of an amplifier
sounds
interesting. (o/c volts divided by s/c current)


Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.

Some of Motorola's devices might actually be able to stand
Reg's little test. The MRF150, for instance is advertised as
being able to withstand a 30:1 VSWR at all phase angles. I
think Reg should offer to pay for any damage sustained as the
result of this test.
73,
Tom Donaly, KA6RUH

"Richard Harrison" bravely wrote to "All" (02 Mar 05 10:35:29)
--- on the heady topic of " Say what you mean."

RH From: (Richard Harrison)
RH Xref: aeinews rec.radio.amateur.antenna:26249

RH Asimov wrote:
RH "Let`s look at it from the dynamic point of view (loss in a Class A
RH amplifier)."

RH The no-signal loss of a Class A amplifier is 100%. It equals volts x
RH amps and appears in the amplifier. Now feed a signal to the ideal
RH amplifier set just below the clipping level. Average d-c power is
RH unchanged from the unloaded and no-signal conditions.

RH Connect a matched load resistor to the amplifier output. If physically
RH small, the resistor may become warm with heat that were it not for the
RH load would be otherwise dissipated in the amplifier. Input power to
RH the Class A amplifier is unchanging.
[,,,]
RH Power output from a Class A amplifier cools it.

Okay, it cools. Enough said. I was confusing it with Class B which
does warm up the heatsink. ;-)

A*s*i*m*o*v

.... May you find the light and walk the mountain tops.

"Rich Grise" bravely wrote to "All" (03 Mar 05 23:00:40)
--- on the heady topic of " 1/4 vs 1/2 wavelength antenna"

RG From: Rich Grise
RG Xref: aeinews rec.radio.amateur.antenna:26329
RG sci.electronics.design:1386
RG 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!(;-)

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

No, the taps are there to maximize the power output. If one uses the
8 ohm tap with a 4 ohm speaker the power is reduced and the same if
one uses a 16 ohm speaker. Audio output matching has little to do with
plate resistance in a beam tube or pentode. It has everything to do
with plate voltage swing and maximum plate current. The less plate
resistance the more power can get to the load. Beam tube curves look a
lot like a transistor's collector saturation curves.

However, some tube amplifiers had a current feedback control which
would increase the output impedance to equal the speaker's. Needless
to say it severely reduced the speaker damping resulting in exagerated
frequency response artifacts.

A*s*i*m*o*v

.... Without ignorance, knowledge is powerless.

Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier
sounds
interesting. (o/c volts divided by s/c current)

Reg, I tried that one time in college. I got as far as torching
two ARC-5 1625's driving an open circuit so I don't know what
would have happened with a short circuit.
--
73, Cecil

==============================
Cecil,
Presumably you hadn't heard of Ohms Law.

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.)
----
Reg, G4FGQ