Efficiency and maximum power transfer
 

Richard Fry wrote:
"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?

the output network might be non-dissipative, but that doesn't mean that
the voltages and/or currents won't exceed the limitations of the components.

Consider a gedanken.. you have an active device acting as a shunt DC
regulator that puts out 1 amp at 1000 volts into a matching network, the
other side of which is a 50 ohm load. The power supply is a 5000 volt
power supply with a 2000 ohm dropping resistor, and 1 amp is also
flowing through the active device. The 2 amps total flowing through the
dropping resistor results in the 1000 volt output into the network. The
implication is that the matching network, with the 50 ohm output, is
presenting a 1000 ohm load.

Now, say you change the load impedance on the network so that the
impedance on the input side is zero ohms. Now, the voltage across the
active device is zero volts, the voltage across that dropping resistor
is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and
possibly fail.

Or, say you change the load impedance so that the input of the network
presents an infinite impedance. The same 1 amp is flowing through the
active device (say it's a controlled current source), but, since there's
no load current, the voltage drop across the resistor is now 2000 volts
instead of 4000 volts, and the voltage across the active device is now
3000 volts instead of 1000. If the device can only take 2000 volts,
it's cooked.

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

"Jim Lux" wrote in message
...
Richard Fry wrote:
"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?

the output network might be non-dissipative, but that doesn't mean that
the voltages and/or currents won't exceed the limitations of the components.

I fail to understand how the limitations of the components have anything to do
with the output resistanceof the network. Can you please explain how this is
relevant?

Consider a gedanken.. you have an active device acting as a shunt DC
regulator that puts out 1 amp at 1000 volts into a matching network, the
other side of which is a 50 ohm load. The power supply is a 5000 volt
power supply with a 2000 ohm dropping resistor, and 1 amp is also
flowing through the active device. The 2 amps total flowing through the
dropping resistor results in the 1000 volt output into the network. The
implication is that the matching network, with the 50 ohm output, is
presenting a 1000 ohm load.

Now, say you change the load impedance on the network so that the
impedance on the input side is zero ohms. Now, the voltage across the
active device is zero volts, the voltage across that dropping resistor
is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and
possibly fail.

Or, say you change the load impedance so that the input of the network
presents an infinite impedance. The same 1 amp is flowing through the
active device (say it's a controlled current source), but, since there's
no load current, the voltage drop across the resistor is now 2000 volts
instead of 4000 volts, and the voltage across the active device is now
3000 volts instead of 1000. If the device can only take 2000 volts,
it's cooked.

Sorry Jim, I'm not familiar with a 'gedanken'. Is it distant cousin of a Class B
or C amplifier? Your discussion above doesn't appear to have any relation
whatsoever to those types of RF amplifiers. Are you using your discussion in an
attempt to prove that the output resistance of those types of amps is not
non-dissipative? Or what are you trying to prove? I don't get it.

Walt, W2DU

Efficiency and maximum power transfer
 

Owen Duffy wrote:
"Have you hopped on another tram?"

Maybe. I agree. Radio power amplifiers have clean sinusoidal outputs,
otherwise they would generate unacceptable harmonics. All except Class A
amplifiers which have continuous plate current flows and a maximum
efficiency of 50%, have plate currents which flow in pulses.

In Class B, the plate current in individual tubes flows in pulses of
approximately a half cycle. Actual efficiency is about 60%.

In Class C, the plate current pulses last less than a half cycle.
Practical efficiencies are in the range of 60 to 80 per cent.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

Walter Maxwell wrote:
"Jim Lux" wrote in message
...
Richard Fry wrote:
"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?
the output network might be non-dissipative, but that doesn't mean that
the voltages and/or currents won't exceed the limitations of the components.

I fail to understand how the limitations of the components have anything to do
with the output resistanceof the network. Can you please explain how this is
relevant?

Consider a gedanken.. you have an active device acting as a shunt DC
regulator that puts out 1 amp at 1000 volts into a matching network, the
other side of which is a 50 ohm load. The power supply is a 5000 volt
power supply with a 2000 ohm dropping resistor, and 1 amp is also
flowing through the active device. The 2 amps total flowing through the
dropping resistor results in the 1000 volt output into the network. The
implication is that the matching network, with the 50 ohm output, is
presenting a 1000 ohm load.

Now, say you change the load impedance on the network so that the
impedance on the input side is zero ohms. Now, the voltage across the
active device is zero volts, the voltage across that dropping resistor
is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and
possibly fail.

Or, say you change the load impedance so that the input of the network
presents an infinite impedance. The same 1 amp is flowing through the
active device (say it's a controlled current source), but, since there's
no load current, the voltage drop across the resistor is now 2000 volts
instead of 4000 volts, and the voltage across the active device is now
3000 volts instead of 1000. If the device can only take 2000 volts,
it's cooked.

Sorry Jim, I'm not familiar with a 'gedanken'. Is it distant cousin of a Class B
or C amplifier? Your discussion above doesn't appear to have any relation
whatsoever to those types of RF amplifiers. Are you using your discussion in an
attempt to prove that the output resistance of those types of amps is not
non-dissipative? Or what are you trying to prove? I don't get it.

Walt, W2DU



I was attempting to discuss the comment above:
"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?"



"gedanken" refers to an intellectual exercise describing an experiment
with (usually) contrived or idealized circumstances so that the
underlying concepts can be understood. Classics are Schroedinger's cat,
Maxwell's demon, etc.



http://en.wikipedia.org/wiki/Thought_experiment for more details and
history than anyone could want.

Efficiency and maximum power transfer
 

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 catastrophic failures to the output
circuit components often seen when high power transmitters operate
without suitable SWR protection into highly mismatched loads?

This question has been asked many times (several times by you, as I
recall) and answered (several times at least by me) on this forum. The
answer is that when a transmitter is terminated in an impedance too far
from the one it's designed to see, voltages and/or currents in the
transmitter and tank circuit rise to unacceptably high values. This
causes the damage or destruction.

I don't think this is difficult to understand. So the reason for the
need to keep asking is that it's apparently not being believed,
presumably because of the deeply compelling need to see waves of
"reflected power" entering the transmitter to wreak havoc. But the
simple answer is true. This is very much like trying to convince an
astrology believer that he stubbed his toe because he didn't watch where
he was going rather than because Mars was lined up with Jupiter.

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?

Power from one antenna is coupled to the other antenna by a process
known as "mutual coupling". This power enters the other transmitter via
normal transmission along the transmission line, where it reaches the
final stage for mixing. No mysterious "reverse r-f energy" is needed to
explain this well-known and well-understood phenomenon.

Roy Lewallen, W7EL

Efficiency and maximum power transfer
 

Roy Lewallen wrote:
This question has been asked many times (several times by you, as I
recall) and answered (several times at least by me) on this forum. The
answer is that when a transmitter is terminated in an impedance too far
from the one it's designed to see, voltages and/or currents in the
transmitter and tank circuit rise to unacceptably high values. This
causes the damage or destruction.

But the offending impedance equals (Vfor+Vref)/(Ifor+Iref), a
virtual impedance caused by the superposition of the reflected
wave and the forward wave.

I don't think this is difficult to understand.

What is difficult to understand is the denial of the role that
the energy content of the reflected wave plays in causing the
offending impedance. If it were not for the reflected waves,
that offending impedance would never happen.
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

I wrote:
"In Class C, the plate current pulses last less than a half cycle.
Practical efficiencies are in the range of 60 to 80 per cent."

Owen Duffy wrote:
"Again, not rectangular pulses, not nearly."

A pulse does not need to be rectangular. According to my electronics
dictionary:
"Pulse - 1. The variation of a quantity having a normally constant
value. This variation is characterized by a rise and decay of finite
duration. 2. An abrupt change in voltage, either positive or negative,
which conveys information to a circuit. (See also Impulse.)"

A rectified sine wave could properly be called a string of pulses being
constantly off between pulses then rising and falling between pulses for
a short finite duration.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

This illustrates, I hope, only one, small point:

Owen says:

For most practical valves in continous mode, they cannot develop their
rated maximum power at very small conduction angles without exceeding
rated cathode current, so there is often little benefit in operating at
conduction angle much below 120 degrees.


Recall that valve/tube diodes in rectifier service were not used with
capacitive filters because the rated peak to average current ratio was
something like 1.5:1. Modern, solid-state diodes can have ratios of 40:1
and can be happy with capacitive filters in rectifier service.

Peak current limits of high-voltage, high-power tubes is a real physical
limit. The cathode can only emit so much.

73, Mac N8TT
--
J. McLaughlin; Michigan, USA
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