Roy Lewallen wrote:
But what happens if we reduce the source resistance to 50 ohms?
Now the source delivers 66.7 watts, of which 22.2 is dissipated
in the source resistance and 44.4 in the load resistance.

The well known and often misapplied rule about maximizing power transfer
by matching the source and load impedances applies only when you're
stuck with a fixed source resistance and can only modify the load.

What if we assume that we are stuck with the 50 ohm source
impedance above? How do we get greater than 50% power transfer?
--
73, Cecil http://www.w5dxp.com

"Roy Lewallen" wrote in message
news:JrCdnR01Yp20tNHVnZ2dnUVZ_sednZ2d@easystreeton line...
There's a common misconception that, for a linear circuit, the maximum
efficiency and/or power available from a voltage source occurs when the
source resistance equals the load resistance (or, more generally, when
they're complex conjugates). But this isn't universally true, as I'll
show with a simple example.

Suppose we have a 100 volt perfect voltage source in series with a
variable source resistance, and a fixed load resistance of 100 ohms. If
we make the source resistance 100 ohms, the source delivers 50 watts, 25
of which are dissipated in the source resistance and 25 watts in the
load. The efficiency, if you consider the source resistance dissipation
as wasted, is 50%. But what happens if we reduce the source resistance
to 50 ohms? Now the source delivers 66.7 watts, of which 22.2 is
dissipated in the source resistance and 44.4 in the load resistance. The
power to the load has increased, and the efficiency has increased from
50 to 66.7%. The efficiency and load power continue to increase as the
source resistance is made smaller and smaller, reaching a maximum when
the source resistance is zero. At that point, the source will deliver
100 watts, all of which is dissipated in the load, for an efficiency of
100%.

The well known and often misapplied rule about maximizing power transfer
by matching the source and load impedances applies only when you're
stuck with a fixed source resistance and can only modify the load.

Roy Lewallen, W7EL

What seems to be overlooked here is that the source resistance at the output
terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which
is the reason they can be loaded for delivering all available power for a given
grid drive, and still have efficiencies greater than 50 percent. One of the
myths circulated for years, and still prevelant, is that the reason for Class B
and C amps to have efficiencies greater than 50 percent is that the load
resistance must be greater than the source resistance. Tain't so.

I've proved the above to be true with extensive measurements using laboratory
grade instruments. Reports on those measurements are reported in Chapter 19 in
Reflections 2, and additional measurements taken after Reflections 2 was
published are reported in Chapter 19A, to be published soon in Reflections 3.
This additional chapter is listed here in the rraa for your information. If
anyone is interested in reading Chapter 19 in Reflections 2 it appears in my
website at www.w2du.com.

Walt, W2DU

On Jun 7, 7:14Â*pm, Owen Duffy wrote:
wrote :





On Jun 7, 12:43�am, Owen Duffy wrote:
wrote
innews:73353273-a079-499c-89df-c11975b37c78@z66g200
0hsc.googlegroups.com:

...

The maximum power theorem gives conditions where power in the load,
is equal to internal power in the generator. �Not always a good
ide
a. �A
50HZ generator capable of Megawatts of power would dissiapate 1/2
in the generator and 1/2 in our houses if they designed them to
conform to the MPT. �The 50HZ generators would melt. �Utilities
design their
Generators to have nearly 0.0 ohms internal impedance.

Actually, the AC power distribution system from alternator down has a
manged substantial equivalent source impedance.

The source impedance serves to limit fault currents, which reduces
the demands on protection devices.

Sure, the network is not operated under Jacobi MPT conditions, but
neither

does it have near zero source impedance.

Owen

Not really sure I agree. Â*A multi-megawatt 60HZ generator by necessity
has near zero source impedance. Â*The ones I am familar with require
forced air cooling on their output buses. Â*If you are pumping out
Mega- watts, then any non -zero source impedance results in serious
heat. I^2R.
Gary N4AST

Gary, you use the terms impedance and resistant as if they were
equivalent.

Alternators have a designed value of leakage reactance, and they also
have resistance. The combination make the equivalent source impedance,
and it is sufficient to limit fault current to something typically in the
range of 20 to 50 times the rated output current.

Transmission lines and transformers in the transmission and distribution
networks are usually designed in the same way.

It is not zero, and it is not purely resistive. Most supply authorities
would not allow you to connect a capacitive load (a leading PF load), so
another concept, conjugate matching (in the Jacobit MPT sense) is also
not practiced.

Understanding the electricity network does not really give an insight
into a typical ham radio transmitter, they do not share the same design
objectives.

Owen- Hide quoted text -

- Show quoted text -

We are obviously talking about two different things, that as you say
have little to do with a Ham Radio transmitter.
Gary N4AST

"Walter Maxwell" wrote
What seems to be overlooked here is that the source resistance at the
output
terminals of the pi-nework in Class B and C amplifiers is non-dissipative,
which
is the reason they can be loaded for delivering all available power for a
given
grid drive, and still have efficiencies greater than 50 percent. One of
the
myths circulated for years, and still prevelant, is that the reason for
Class B
and C amps to have efficiencies greater than 50 percent is that the load
resistance must be greater than the source resistance. Tain't so.
____________

Walt - what is your thinking on the point that untuned, solid-state
amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or
more at the device level?

In fact the solid-state, analog FM broadcast transmitters supplied by Harris
Corporation and others need no tuning to produce their rated output power
into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz.
Even the harmonic filter needs no changes, and maintains harmonics at -80
dBc or better. They are frequency agile, and can be reset from one carrier
frequency to another, anywhere in the FM band with a transition time of a
few seconds

The overall AC input to r-f output efficiency of these transmitters exceeds
60% (includes the exciter, control system, IPA, and cabinet fans).

RF

Richard Fry wrote:
"Walter Maxwell" wrote
What seems to be overlooked here is that the source resistance at the
output
terminals of the pi-nework in Class B and C amplifiers is
non-dissipative, which
is the reason they can be loaded for delivering all available power
for a given
grid drive, and still have efficiencies greater than 50 percent. One
of the
myths circulated for years, and still prevelant, is that the reason
for Class B
and C amps to have efficiencies greater than 50 percent is that the load
resistance must be greater than the source resistance. Tain't so.
____________

Walt - what is your thinking on the point that untuned, solid-state
amplifiers also can have PA DC-to-RF power conversion efficiencies of
70% or more at the device level?

In fact the solid-state, analog FM broadcast transmitters supplied by
Harris Corporation and others need no tuning to produce their rated
output power into a 1.3:1 SWR or less, anywhere in the FM broadcast band
88-108 MHz. Even the harmonic filter needs no changes, and maintains
harmonics at -80 dBc or better. They are frequency agile, and can be
reset from one carrier frequency to another, anywhere in the FM band
with a transition time of a few seconds

The overall AC input to r-f output efficiency of these transmitters
exceeds 60% (includes the exciter, control system, IPA, and cabinet fans).

RF

I know this question was directed to Walt, but I'd like to mention that
I've designed and built solid state class C amplifiers at the 5 - 10
watt level which have measured efficiencies of greater than 85%.

Roy Lewallen, W7EL

"Richard Fry" wrote in message ...
"Walter Maxwell" wrote
What seems to be overlooked here is that the source resistance at the
output
terminals of the pi-nework in Class B and C amplifiers is non-dissipative,
which
is the reason they can be loaded for delivering all available power for a
given
grid drive, and still have efficiencies greater than 50 percent. One of
the
myths circulated for years, and still prevelant, is that the reason for
Class B
and C amps to have efficiencies greater than 50 percent is that the load
resistance must be greater than the source resistance. Tain't so.
____________

Walt - what is your thinking on the point that untuned, solid-state
amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or
more at the device level?

In fact the solid-state, analog FM broadcast transmitters supplied by Harris
Corporation and others need no tuning to produce their rated output power
into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz.
Even the harmonic filter needs no changes, and maintains harmonics at -80
dBc or better. They are frequency agile, and can be reset from one carrier
frequency to another, anywhere in the FM band with a transition time of a
few seconds

The overall AC input to r-f output efficiency of these transmitters exceeds
60% (includes the exciter, control system, IPA, and cabinet fans).

RF

Hello Richard,

Sorry, Richard, I have no knowledge of solid-state untuned amps, so my thinking
on them is zero, nada. As you'll note, all of my discussion on the subject
concerns only tube amps with a pi-network output, and I've specifically stated
these conditions.

If you've read Chapter 19 and its addition as Chapter 19A, do you agree with my
position that the output resistance at the output terminals of the pi-network is
non-dissipative?

Walt, W2DU

"Walter Maxwell" wrote
The source resistance appearing at the output of either a Class B or C
amplifier is R = E/I, where E is the peak voltage at the output terminals
and I is the peak current at the output. Or RMS values can also be
used.Since E/I is simply a ratio, R is also a ratio. And we know that a
ratio cannot dissipate power, or turn electrical energy into heat, thus the
output resistance R is non-dissipative. I have made many measurements that
prove this. It is also the reason why reflected power does not dissipate in
the tubes, because it never reaches the tubes. The reflected power simply
causes a mismatch to the source, causing the source to deliver less power
than it would if there were no mismatch.
__________

If the source resistance of a tuned r-f PA stage was truly non-dissipative,
and the tx simply supplied less power into poor matches, how would that
explain the catastrophic failures to the output circuit components often
seen when high power transmitters operate without suitable SWR protection
into highly mismatched loads?

Another reality is that r-f power from two co-sited, tuned transmitters on
two frequencies in the same band can be present in each others output stage
due to antenna coupling, which causes r-f intermodulation between them. The
non-linear (mixing) process occurs at the active PA stage. If reflected
(reverse) r-f energy never reaches the PA stage as you assert, then how
could this IM generation occur?

RF

"Richard Fry" wrote in message ...
"Walter Maxwell" wrote
The source resistance appearing at the output of either a Class B or C
amplifier is R = E/I, where E is the peak voltage at the output terminals
and I is the peak current at the output. Or RMS values can also be
used.Since E/I is simply a ratio, R is also a ratio. And we know that a
ratio cannot dissipate power, or turn electrical energy into heat, thus the
output resistance R is non-dissipative. I have made many measurements that
prove this. It is also the reason why reflected power does not dissipate in
the tubes, because it never reaches the tubes. The reflected power simply
causes a mismatch to the source, causing the source to deliver less power
than it would if there were no mismatch.
__________

If the source resistance of a tuned r-f PA stage was truly non-dissipative,
and the tx simply supplied less power into poor matches, how would that
explain the catastrophic failures to the output circuit components often
seen when high power transmitters operate without suitable SWR protection
into highly mismatched loads?

You must be talking about solid-state tx when you mention SWR protection. I
don't know of any tube tx that have such protection. Any catastrophic failures
in tube tx with pi-network output circuits due to poor impedance match at the
output without retuning to match the the network to the load simply leaves the
tx detuned away from resonance. The result is excessive plate current that would
be reduced to normal by resonating the tank circuit. No operator in his right
mind would allow the tx to be operated with the tank not tuned for the resonant
dip in plate current.

Another reality is that r-f power from two co-sited, tuned transmitters on
two frequencies in the same band can be present in each others output stage
due to antenna coupling, which causes r-f intermodulation between them. The
non-linear (mixing) process occurs at the active PA stage. If reflected
(reverse) r-f energy never reaches the PA stage as you assert, then how
could this IM generation occur?

RF

:You said two tx operating on two different frequencies, which means the RF
signals from the two tx are not phase coherent. In this condition the signal
from each tx does enter the other. On the other hand, the wave reflected from a
mismatched termination is phase coherent with the source wave, resulting in the
addition of the reflected wave to the source wave when either the antenna tuner
or the pi-network is adjusted to deliver all available power at the desired grid
drive.

W2DU

"Richard Fry" wrote in message
...
"Walter Maxwell" wrote
The source resistance appearing at the output of either a Class B or C
amplifier is R = E/I, where E is the peak voltage at the output terminals
and I is the peak current at the output. Or RMS values can also be
used.Since E/I is simply a ratio, R is also a ratio. And we know that a
ratio cannot dissipate power, or turn electrical energy into heat, thus
the output resistance R is non-dissipative. I have made many measurements
that prove this. It is also the reason why reflected power does not
dissipate in the tubes, because it never reaches the tubes. The reflected
power simply causes a mismatch to the source, causing the source to
deliver less power than it would if there were no mismatch.
__________

If the source resistance of a tuned r-f PA stage was truly
non-dissipative, and the tx simply supplied less power into poor matches,
how would that explain the catastrophic failures to the output circuit
components often seen when high power transmitters operate without
suitable SWR protection into highly mismatched loads?

if the load impedance seen at the transmitter terminals is outside the range
that it was designed for you end up with either arcing from excessive
voltage or meltdown from high currents. remember, the pa output as long as
it is connected to a linear load and you consider only the sinusoidal steady
state condition can be completely replaced by a lumped load impedance, hence
no reflections necessary to figure it out... same load, same result, hence
the reflections have no effect on the internals of the pa.


Another reality is that r-f power from two co-sited, tuned transmitters on
two frequencies in the same band can be present in each others output
stage due to antenna coupling, which causes r-f intermodulation between
them. The non-linear (mixing) process occurs at the active PA stage. If
reflected (reverse) r-f energy never reaches the PA stage as you assert,
then how could this IM generation occur?

non-coherent input would of course pass through the matching network going
into a pa, there is nothing that precludes that. that current would cause
mixing in the non-linear tube or transistor and therefore im generation.
but that incoming rf is not reflected, it is the incident wave, with nothing
coherent to interfer with why would it not pass into the pa?

Now, have fun with this one... two transmitters on exactly the same
frequency feeding a common load through equal lengths of coax... what
happens if you change the phase relationship between them? make it
simpler, remove the coax and connect both pa outputs togther with a single
load, now no reflections to worry about... does it matter? do the two pa's
feed all their power into each other? where do the reflections go??

Richard Fry wrote:
"If the source resistance of a tuned r-f PA stage was truly
non-dissipative, and the tx simply supplied less power into poor
matches, how would that explain the catastropic failures to the output
circuit components often seen when high power transmitters operate
without suitable SWR protection into highly mismatched loads?"

The PA is a switch. Almost no voltage across it when it is closed and no
current through it when it is open. Some of its impedance is dissipative
and some is non-dissipative.

A conjugate match to its total impedance is the way to deliver maximum
power from the transmitter to its load. Alexander H. Wing wrote on page
43 of "Thansmission Lines, Antennas, and Wave Guides":
"If a dissipationless network is insrted between a constant voltage
generator of internal impedance Zg1 and a load ZR such that maximum
power is delivered to the load, at every pair of terminals the
impedances looking in opposite directions are conjugates of each other."

An operating transmitter is normally adjusted for conjugate match with
its load.

Normal plate dissipation occurs when electrons strike the anode and
there is little damage to the tube when the current and cooling are
within limits.

Let an arc strike across the transmission line and it may effectively
become a short circuit which may impose an enormous mismatch in an
instant to the transmitter. That`s why a d-c supply is often connected
in series with a relay coil across the transmission line. The arc
completes the d-c circuit energizing the relay which breaks the
interlock circuit. The transmitter instantly is shut down until it is
manually restarted.

Tubes are often destroyed by internal arcs if overloads don`t act in
time.

Best regards, Richard Harrison, KB5WZI