Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load resistance
must be equal to the generator resistance. Under such conditions efficiency is
50% (half power dissipated in the generator, half delivered to the load).

To achieve a higher efficiency, the load resistance should be made higher than
the generator resistance, although this would turn into a lower power delivered
to the load (the maximum power transfer condition is now no longer met). This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output power.

That said, now the question.

Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically 60% for class-B, 70% for class-C).
This would seem to contradict the above cited fact that, under maximum power
transfer condition, efficiency is 50%.

Pertinent comments are welcome.

73

Tony I0JX - Rome, Italy

On Jun 6, 2:12 pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load resistance
must be equal to the generator resistance. Under such conditions efficiency is
50% (half power dissipated in the generator, half delivered to the load).

To achieve a higher efficiency, the load resistance should be made higher than
the generator resistance, although this would turn into a lower power delivered
to the load (the maximum power transfer condition is now no longer met). This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output power.

That said, now the question.

Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically 60% for class-B, 70% for class-C).
This would seem to contradict the above cited fact that, under maximum power
transfer condition, efficiency is 50%.

Pertinent comments are welcome.

73

Tony I0JX - Rome, Italy

Simple: a transmitter is not an ideal voltage source with a resistor
in series.

I'm playing with a switching power supply design that delivers about a
kilowatt at 100 volts. The output is designed specifically to have a
negative resistance, so the output voltage increases as the current
drawn increases. The output dynamic impedance is about -1 ohms
(adjustable, actually). The linear model is a 100 volt battery in
series with -1 ohms. With an 11 ohm load, I get 10 amps load current,
with the battery thus delivering 1000 watts, the load dissipating 1100
watts, and the -1 ohm resistance dissipating -100 watts. Which shows
the absurdity of thinking of a dynamic output resistance being
anything like a real resistance. In my switching supply, I can adjust
the dynamic output resistance between a small negative value and a
rather larger positive value, with very little change in efficiency.

Although transmitters MAY have dynamic output resistances similar to
the recommended load resistance, that's not a necessary condition, and
has little to do directly with efficiency.

Cheers,
Tom

On Jun 6, 4:12�pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load resistance
must be equal to the generator resistance. Under such conditions efficiency is
50% (half power dissipated in the generator, half delivered to the load).

To achieve a higher efficiency, the load resistance should be made higher than
the generator resistance, although this would turn into a lower power delivered
to the load (the maximum power transfer condition is now no longer met). This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output power.

That said, now the question.

Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically 60% for class-B, 70% for class-C).
This would seem to contradict the above cited fact that, under maximum power
transfer condition, efficiency is 50%.

Pertinent comments are welcome.

73

Tony I0JX - Rome, Italy

In addition to Tom's comments, an RF Power Amplifier's efficiency is
defined as (Pout/Pin)X100%. Pout is RF and Pin is usually DC. So if
you pump 1000 watts DC in to a class B RF amp and get 600 watts rms
out you are 60% efficient. Your model of a Voltage Source in series
with an internal resistance does not apply here.

The maximum power theorem gives conditions where power in the load, is
equal to internal power in the generator. Not always a good idea. 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.

Good question.
Gary N4AST

Antonio Vernucci wrote:
Usually, when a transmitter is tuned for maximum output power,
efficiency results to be higher than 50% (typically 60% for class-B,
70% for class-C). This would seem to contradict the above cited fact
that, under maximum power transfer condition, efficiency is 50%.

The maximum power transfer theorem only applies to
linear sources. What is the linear source impedance
of a class-C amp?
--
73, Cecil http://www.w5dxp.com

Dear Tom K7ITM:

Thank you. Thank you. Thank you.

Let us put away forever the idea that a normal transmitter fits the
passive, linear model of an ideal voltage source in series with an actual
resistor.

The active device(s) in an amplifier (tube, BJT, FET, ...) needs to see
a certain impedance (at a given frequency) in order to have desirable
performance characteristics. Too many have extrapolated from that
information (found in data sheets for the active devices) the conclusion
that the active device has an internal Z that is the complex-conjugate of
the load Z.

The wording found in another string of messages communicates the right
idea. A normal transmitter "wants to see" a certain Z. That Z is most
often 50 ohms.

RIP 73, Mac N8TT

P.S. An IEEE paper explored this issue in the last year.
--
J. McLaughlin; Michigan, USA
Home:
"K7ITM" wrote in message
...
On Jun 6, 2:12 pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load
resistance
must be equal to the generator resistance. Under such conditions
efficiency is
50% (half power dissipated in the generator, half delivered to the load).

To achieve a higher efficiency, the load resistance should be made higher
than
the generator resistance, although this would turn into a lower power
delivered
to the load (the maximum power transfer condition is now no longer met).
This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output
power.

That said, now the question.

Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically 60% for class-B, 70% for
class-C).
This would seem to contradict the above cited fact that, under maximum
power
transfer condition, efficiency is 50%.

Pertinent comments are welcome.

73

Tony I0JX - Rome, Italy

Simple: a transmitter is not an ideal voltage source with a resistor
in series.

I'm playing with a switching power supply design that delivers about a
kilowatt at 100 volts. The output is designed specifically to have a
negative resistance, so the output voltage increases as the current
drawn increases. The output dynamic impedance is about -1 ohms
(adjustable, actually). The linear model is a 100 volt battery in
series with -1 ohms. With an 11 ohm load, I get 10 amps load current,
with the battery thus delivering 1000 watts, the load dissipating 1100
watts, and the -1 ohm resistance dissipating -100 watts. Which shows
the absurdity of thinking of a dynamic output resistance being
anything like a real resistance. In my switching supply, I can adjust
the dynamic output resistance between a small negative value and a
rather larger positive value, with very little change in efficiency.

Although transmitters MAY have dynamic output resistances similar to
the recommended load resistance, that's not a necessary condition, and
has little to do directly with efficiency.

Cheers,
Tom

This has been a good example of a common pitfall in modeling. The error
made in this case was to attempt to apply an unsuitable model (a voltage
source in series with a resistance) to a system to be modeled (a
transmitter). As the OP showed, the attempt leads to an impossible
result. The classic example of this is the "proof" that a bumblebee
can't fly, based on a flawed model and immediately shown to be false by
simply observing that they do, indeed, fly. Yet we see people falling
into this trap daily, not only in modeling electrical circuits, but also
in modeling such diverse processes as human behavior, economic systems,
and roulette wheel numbers.

Unfortunately, the bad results of applying unsuitable models aren't
always so obvious as they were here. So it's always wise to check to see
if the model fits before putting faith in the results.

Roy Lewallen, W7EL

On Fri, 06 Jun 2008 20:58:13 -0500, Cecil Moore wrote:

Antonio Vernucci wrote:
Usually, when a transmitter is tuned for maximum output power,
efficiency results to be higher than 50% (typically 60% for class-B,
70% for class-C). This would seem to contradict the above cited fact
that, under maximum power transfer condition, efficiency is 50%.

The maximum power transfer theorem only applies to
linear sources. What is the linear source impedance
of a class-C amp?

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. The input at the pi-network in the xmtr is non-linear, but
the fly-wheel effect of the network tank isolates the input from the output, resulting in a linear condition
appearing at the output. Except for a very slight deviation from a sine wave due to a small amount of harmonic
content, the voltage E and current I at the output are essentially a sine wave, which one can easily prove
with a good oscilloscope, proving the output to be linear. I'm speaking for tube rigs with pi-network tanks,
not for solid-state rigs.

I nearly forgot. The only dissipation in the amp tube(s) is due to the filament-to-plate current as the
electrons bombard the plate. The efficiency is determined by the ratio of the DC input power to the RF output
power. The maximum power is delivered when the load resistance equals the output resistance R of the source.
But since resistance R is non-dissipative it is not a factor in determining efficiency. The only factors in
determining efficiency are the RF output power and the dissipation in the tube caused by the electrons
striking the plate. The non-dissipative output resistance is the reason Class B and C amps can have an
efficiency greater than 50 percent. If the output resistance were dissipative it would be the determining
factor in efficiency, which could never be greater than 50 percent if the load resistance was equal to the the
output resistance.

A report of my measurements will soon be available in Reflections 3, from measurements taken since those
reported in Chapter 19 of Reflections 2.

Cecil, if I send you a copy of the new chapter that has the report of my newer measurements do you have any
way to make it available to the guys on this thread?

Walt, W2DU

wrote in
:

....
The maximum power theorem gives conditions where power in the load, is
equal to internal power in the generator. Not always a good idea. 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

"Antonio Vernucci" wrote in
:

Someone may regard the following question a bit OT, but as it deals
with impedances I have considered that the antenna newsgroup could be
the most appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance
in series. It is well known that, for maximum power transfer, the load
resistance must be equal to the generator resistance. Under such
conditions efficiency is 50% (half power dissipated in the generator,
half delivered to the load).

To achieve a higher efficiency, the load resistance should be made
higher than the generator resistance, although this would turn into a
lower power delivered to the load (the maximum power transfer
condition is now no longer met). This can be verified in practice: by
decreasing the antenna coupling in a transmitter, one obtains a higher
efficiency though with a lower output power.

That said, now the question.

But your statements are not true. The model you propose for a transmitter
does not apply in general. Whilst it would be possible to build a
transmitter like that, most transmitters that hams use are not built like
that.

So... it is a loaded question of a type, a question premised on a
falsehood.


Usually, when a transmitter is tuned for maximum output power,
efficiency results to be higher than 50% (typically 60% for class-B,
70% for class-C). This would seem to contradict the above cited fact
that, under maximum power transfer condition, efficiency is 50%.


If the equivalent source impedance is not important, ie it does not need
to be fixed by the design, there here is an analysis.

If you take the case of a grounded cathode triode in class C with a
steady signal, the conduction angle is usually somewhere around 120°. The
anode current waveform is a little like a truncated sine wave, but even
for the range of grid voltages where anode current is greater than zero,
the transfer characteristic is not exactly linear, and the wave will be
further distorted.

If the nature of the anode load is that it is some equivalent R at the
fundamental and zero impedance at all other frequencies, the power output
can be determined by finding the fundamental component of the anode
current waveform, squaring it, and multiplying it by R. The input power
is the average anode current multiplied by the DC supply voltage.
Efficiency is OutputPower/InputPower. By varying the grid bias, drive
voltage, load impedance and supply voltage for a given triode, different
efficiencies will be found, and the maximum could be well over 80%.

Nothing in this approach to design attempts to fix the equivalent source
impedance, the design is performed without regard to that characteristic.

Nevertheless, some argue that the output network performs magic and
achieves source matching naturally without designer intervention, and
does this irrespective of parameters like the dynamic anode resistance,
and the effects of feedback (such as cathode degeneration in grounded
grid amplifiers which in turns depends on the source impedance of the
exciter).

Owen

Owen Duffy wrote in
:

....
voltage. Efficiency is OutputPower/InputPower. By varying the grid
bias, drive voltage, load impedance and supply voltage for a given
triode, different efficiencies will be found, and the maximum could be
well over 80%.

If you want to explore this approach to design, I have implemented a
spreadsheet which in turn implements the methods described in CPI/Eimac's
"Care and Feeding of Power Tubes" it is described and can be downloaded at
http://www.vk1od.net/RFPATPC/index.htm . The spreadsheet is populated with
the example 4CX20000 in Class C from the above publication.

The spreadsheet calculates the anode current wave form from an
interpolation based on a number of points from the published
characteristics, and calculates the fundamental component of anode current
using an FFT. It then calculates anode efficiency and overall efficiency
given operating parameters.

Owen