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

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

Good to see that everyone agrees that a generator with a resistor in series is
an unsuitable model for an RF transmitter.The easy part of the work is done!

Now the more difficult part. As, by the Thevenin theorem, any complex circuit
comprising resistors, voltage generators and current generators is equivalent to
a generator with a resistor in series, evidently the transmitter model must
comprise elements other than just resistors, voltage generators and current
generators.

Can one suggest how such a model looks like? (even a plain one, that does not
take into account second- or superior-order effects).

73

Tomy I0JX - Rome, Italy

On Jun 8, 6:52 am, "Antonio Vernucci" wrote:
Good to see that everyone agrees that a generator with a resistor in series is
an unsuitable model for an RF transmitter.The easy part of the work is done!

Now the more difficult part. As, by the Thevenin theorem, any complex circuit
comprising resistors, voltage generators and current generators is equivalent to
a generator with a resistor in series, evidently the transmitter model must
comprise elements other than just resistors, voltage generators and current
generators.

Can one suggest how such a model looks like? (even a plain one, that does not
take into account second- or superior-order effects).

73

Tomy I0JX - Rome, Italy

I believe you have mis-stated the Thevenin theorem. First, it applies
only to linear circuits. Fine -- over some narrow range at least, a
transmitter does indeed look like a linear circuit. But more
importantly, it describes ONLY what you observe at an external pair of
terminals, with no other connections, NOT what goes on inside the
"black box" containing those elements you mentioned.

A very simple example is a voltage source (a perfect battery) and two
resistors in series across the battery; the external terminals for
this example will be at opposite ends of one of the resistors. Let's
say the battery is 2 volts and each resistor is 2 ohms. That will
look like a Thevenin equivalent 1V in series with 1 ohm. Note that it
also looks like a one amp source in parallel with a one ohm resistor.
But it does not behave INTERNALLY like either of those. Consider also
the same internal circuit, except drop the voltage source to 1V and
add a 1/2A current source across the output terminals, polarity so
that there's no drop across the resistor between the voltage and the
current source (with no external load). Now figure the internal
dissipation for each of those two cases, with no load, with a 1 ohm
load, and with a short-circuit load.

Cheers,
Tom