On Mon, 24 May 2010 07:06:44 -0700 (PDT), Keith Dysart
wrote:

I have often wondered if the manufacturer's
tuning
procedures had anything to do with maximizing output power transfer,
or
were they, in fact, optimizing some other aspect.

This resolves quickly in measurement - no need to wonder unless it
offers some secondary benefit of not measuring things.

An alternative is to simply examine conventional design
considerations. One can add to Plate current by throwing a lot of
power into the grid. More plate current yields more output power
results, but grid lifetime plumments.

One can do innumerable things to force an artificial outcome that
strains to prove a distorted logic. Examining a suite of sources, in
initial conditions that are average for their application quickly
reveals a common design paradigm.

******

The fundamental answer to your question is the manufacturer ultimately
designs for market domination, or maximum investment return (the two
don't necessarily converge). Thus the marketplace gives us a spectrum
of choice and the norm of the distribution reveals cautious design
that has its eye on a value exchange expressed in money. THAT is the
only optimization you can expect = in an honest barter, you get what
you pay for.

73's
Richard Clark, KB7QHC

On May 24, 10:55*am, walt wrote:
Keith, would you please elaborate on why you believe my analysis of
transmitter output impedance is flawed? And what is the basis for your
belief that my explanations in Reflections require large chunks of
linear circuit theory to be discarded. Could it be because you
consider the source resistance in the transmitter to be dissipative,
as in the classical generator? If so, you must be made to realize that
the source resistance of the transmitter is non-dissipative, which is
the reason that its efficiency can exceed 50%.

No problems there. There has been much confusion in this area and
anything
that reduces this confusion is beneficial.

Or are you considering the output characteristic of the transmitter to
be non-linear? This is not the case, because the effect of energy
storage in the tank circuit isolates the non-linear input from the
output circuit, which is linear as evidenced by the almost perfect
sine wave appearing at the output of the tank.

This may be the root of my disagreement. Certainly the output can be
an
arbitrarily perfect sine wave, but this simply depends on the
characteristics of the filter and not on whether the system is linear.

But the way the filter transforms the impedances is the crux of the
issue.

It is my understanding that the input impedance to a filter can be
computed by starting with the load impedance applied to the filter and
then, using the rules for series and parallel connected components,
compute the way through the filter until reaching the input and the
result is the input impedance to the filter.

Similarly, the output impedance of the filter can be computed by
starting with source impedance driving the filter, series and
paralleling
the components until reaching the output and the result is the output
impedance of the filter.

The desired impedance for the input to the filter is that impedance
which
produces the desired load on the tube. And the component values are
computed to produce this load on the tube when the correct load is
attached to the output.

For the output impedance of the filter, the question then becomes:
What
is the source impedance driving the filter? If the source is
constructed
as a Class A amplifier, then it depends on the controlling device,
and
for the simplest of circuits would be Rp of the tube. (Just for
clarity,
in this discussion Rp is the slope of the plate E/I curve with
constant
grid voltage. In an ideal tube, these lines are equidistant apart and
the
slopes are the same. Real tubes, of course, are not so well behaved,
but
this should not affect the basic discussion.)

Since the component values for the filter were chosen to provide the
optimum load to the tube, and the optimum load value has no relation
to
Rp, there is no reason to expect the filter will transform Rp to be
the conjugate of the load impedance.

For amplifiers where conduction is not for 360 degrees (Class AB, B,
C),
the controlling device is no longer time-invariant so the rules for
linear circuit analysis no longer apply. None-the-less, for example,
consider a Class AB amplifier where the tube is only cut off for 1
degree. This short cut-off would not have much affect so the analysis
for Class A would apply. As the cut-off period increases the behaviour
will diverge more and more from that of the Class A amplifier.

Simulations produce some interesting results:

Another way of measuring the source impedance is to observe the effect
on a reflected wave entering the amplifier from the load. With a
Class
C amplifier, simulation reveals that the effect on the reflected wave
depends on the phase of that wave with respect to the drive signal
applied to the tube. As the phase of the reflected wave is changed,
the reflection co-efficient experienced by the wave changes. Truly a
non-linear behaviour. Intriguingly, when the conduction angle is
exactly
180 degrees, this effect largely disappears, and the result is much as
if the source impedance of the tube was 2 times Rp, which seems to
make some sense since the tube is only conducting one-half of the
time.

One last question: Are you basing your dissatisfaction of Reflections
from reviewing the 2nd or 3rd edition? Chapter 19 has been expanded in
the 3rd edition, in which I presented additional proof of my position
on the subject that you should be aware of. If you haven't yet seen
the addition that appears in the 3rd ed, please let me know so that I
can send you a copy of the addition.

I have been reading the .pdfs at w2du.com along with correspondence
and
other writings in QST, QEX and newsgroups.

The expanded Chapter 19 at w2du.com offers more experimental evidence
that seems to support the hypothesis that the transmitter is conjugate
matched to the load after tuning,

But given, from circuit analysis, that the output impedance can not be
well defined for any but a Class A amplifier, the fascinating question
is why is there experimental evidence that agrees with the premise
that
the output impedance of a tuned transmitter is the conjugate match of
the load?

One simple example to consider which has similar behaviour is a bench
power supply that also has a constant current limiter. Set such a
power
supply to produce a voltage of 100V (more precisely a maximum voltage)
and a current limit of 2A. Apply a variable load. Maximum power will
be drawn when the load resistance is 50 ohms. Varying the resistance
on either side of 50 ohms will reduce the power which might be
misconstrued to suggest that the power supply has an output impedance
of 50 ohms, when, in fact, it has a infinite output impedance when
the load is below 50 ohms and a zero output impedance when the load
is above.

I have looked for such a simple explanation in the circuits of the
transmitters used in the experiments but was not able to find one.
So I am still puzzled by the observations.

Also include your email address so I can send it.
Keith.dot.dysart.at.gmail.com .dot. = . .at. = @

…Keith

On May 24, 6:58*pm, Richard Clark wrote:
On Mon, 24 May 2010 07:06:44 -0700 (PDT), Keith Dysart

wrote:
I have often wondered if the manufacturer's
tuning
procedures had anything to do with maximizing output power transfer,
or
were they, in fact, optimizing some other aspect.

This resolves quickly in measurement - no need to wonder unless it
offers some secondary benefit of not measuring things. *

An alternative is to simply examine conventional design
considerations. *One can add to Plate current by throwing a lot of
power into the grid. *More plate current yields more output power
results, but grid lifetime plumments.

One can do innumerable things to force an artificial outcome that
strains to prove a distorted logic. *Examining a suite of sources, in
initial conditions that are average for their application quickly
reveals a common design paradigm.

******

The fundamental answer to your question is the manufacturer ultimately
designs for market domination, or maximum investment return (the two
don't necessarily converge). *Thus the marketplace gives us a spectrum
of choice and the norm of the distribution reveals cautious design
that has its eye on a value exchange expressed in money. *THAT is the
only optimization you can expect = in an honest barter, you get what
you pay for.

73's
Richard Clark, KB7QHC

You have gone to a bit higher level than I intended with my question
and
I agree with you conclusions at that level. But my question was more
basic.

When designing the filter for a PA, among other things, one uses the
desired load to be applied to the tube and the disired load impedance
to be supported and selects filter components to perform the desired
transformation.

When operating the radio, the operator has meters that measure some
values, some knobs that control some component values and a procedure
for adjusting these knobs.

It is not at all obvious what exactly the result of performing the
procedure is. Does it result in the same load being applied to the
tube that was computed by the designer? There are some hints that
the procedure will result in the load applied to the tube being
real, but beyond that, what exactly are the circuit conditions
that result?

....Keith

On May 24, 6:15*pm, Keith Dysart wrote:
This may be the root of my disagreement. Certainly the output can be
an arbitrarily perfect sine wave, but this simply depends on the
characteristics of the filter and not on whether the system is linear.

Since anything except a class-A amplifier is non-linear and since we
are talking about linear analysis, it seems we need to locate a point
in the system where V is a sine wave, I is a sine wave, and V/I is the
constant impedance at that point. IMO, that is the first point at
which we can use a linear math analysis and maybe that point is what
Walt is talking about. It's certainly not going to be the plate of a
class-C amplifier and it may not even be the load-line of the class-C
amplifier. There is probably some point in an otherwise non-linear
system where a linear analysis becomes possible. I think that point is
what Walt considers to be the linear source point, wherever that point
might be located.

In fact, here is my personal take on the subject. Given an antenna
system that presents 50+j0 ohms looking into 50 ohm coax, the internal
impedance of the source doesn't matter. For any voltage source,
irrespective of the source impedance, if reflected energy doesn't
reach the source, the source impedance doesn't matter (except for
efficiency). Seems to me, the highest efficiency would be achieved by
a source with zero ohms of source impedance.
--
73, Cecil, w5dxp.com

On Mon, 24 May 2010 16:23:26 -0700 (PDT), Keith Dysart
wrote:

It is not at all obvious what exactly the result of performing the
procedure is. Does it result in the same load being applied to the
tube that was computed by the designer?

Hi Keith,

By and large, Yes.

There are some hints that
the procedure will result in the load applied to the tube being
real, but beyond that, what exactly are the circuit conditions
that result?

I am a little lost on that. The load applied is the load applied
(sorry for the Zen). If you mean that the load is transformed by
tuning to a real R for the Plate to see, then, yes, that is operative.

However, that is not the end of it. That R is seen as the loss of a
now-poorer Q for the Plate tank. This is the distinction between
loaded and unloaded Q. The Plate tank Q expressed in terms of loaded
Q, to be effective, is quite low in comparison to its unloaded value.
This value of loaded Q is roughly between 10 and 20 where the
components in isolation (unloaded) could easily achieve 10 to 30 times
that.

The term "loaded" includes BOTH the plate and the applied load
(whatever is presented to the antenna connection). The only time the
unloaded Q of the Plate tank is at peak value is when it is sitting in
isolation from the chassis, circuitry, and even mounts - which means
it is not very useful in that configuration, except as a trophy. Many
silver plate their tanks as trophies (because this rarely results in
better operation).

Now, let's return to my statement about what Q is "effective" AND that
it measures out at roughly 10 to 20. This is straight out of Terman
if you need a citation. As for explanation (also found in Terman),
you have to consider that the Plate tank is the gate-keeper (as well
as transformer of Z) of power. If you have too high a Q, the power is
not getting THROUGH the tank as it must, and necessarily it remains in
the tank (as energy, albeit).

Consider further that ALL resonant circuits can be cast from series
circuits to parallel circuits or parallel to series (a fact lost on
some inventors of antennas). To describe the Plate tank in series
terms as I do, then the plate resistance and load resistance combine
in series through a simple circular path through ground. There are
parallel tank designs where the resistances combine in parallel. The
net result is the same insofar as Q is concerned.

Consult Terman if that is confusing. No doubt others will either more
clearly cite him, or add to the confusion.

73's
Richard Clark, KB7QHC

On Mon, 24 May 2010 16:15:52 -0700 (PDT), Keith Dysart
wrote:

I have looked for such a simple explanation in the circuits of the
transmitters used in the experiments but was not able to find one.
So I am still puzzled by the observations.

Hi Keith,

Consult Terman. That should be sufficient for both the instructor and
the student to use as a benchmark design for tubes, at least. The
design of the transistorized finals' deck of Ham grade HF rigs has
been stable for nearly 4 decades.

Any departure from HF Ham grade equipment capable of 100W is going to
lead to equivocal statements.

73's
Richard Clark, KB7QHC

Cecil Moore, W5DXP wrote:
"Seems to me, the highest efficiency would be achieved by a source with
zero source impedance."

Me too, but zero source impedance does not match the load as required
for maximum power transfer. The best combination is then a source
impedance matching the load and which is also pracrically lossless. The
Class C amplifier does this by acting as a switch which is infinite in
impedance when open during a large part of the RF cycle and a near short
circuit to a low impedance (near zero Z) D.C. power source for the short
part of the RF cycle it is switched on. It is the time averaged
impedance which counts. Is this linear? No way, but the tank circuit is
able to remove the harmonics and turn current pulses into a low
distortion sine wave. Efficiency? Terman says on page 450 of his 1955
opus that Class C eddiciency is typically 60% to 80%. Compare that with
50% efficiency in a Class A amplifier.

Best regards, Richard Harrison, KB5WZI

On 25/05/2010 01:55, walt wrote:
....
Or are you considering the output characteristic of the transmitter to
be non-linear? This is not the case, because the effect of energy
storage in the tank circuit isolates the non-linear input from the
output circuit, which is linear as evidenced by the almost perfect
sine wave appearing at the output of the tank.

Well, for the purposes of application of linear circuit theory,
linearity means that V is linearly related to I, or at least dV/dI over
the operating range is substantially constant.

If the circuit is not linear in those terms, then you cannot form a
valid Thevenin equivalent circuit, and discussion of the Thevenin
equivalent series source impedance is a nonsense... it cannot be used.

Owen

On May 24, 7:49*pm, Cecil Moore wrote:
On May 24, 6:15*pm, Keith Dysart wrote:

This may be the root of my disagreement. Certainly the output can be
an arbitrarily perfect sine wave, but this simply depends on the
characteristics of the filter and not on whether the system is linear.

Since anything except a class-A amplifier is non-linear and since we
are talking about linear analysis, it seems we need to locate a point
in the system where V is a sine wave, I is a sine wave, and V/I is the
constant impedance at that point. IMO, that is the first point at
which we can use a linear math analysis and maybe that point is what
Walt is talking about. It's certainly not going to be the plate of a
class-C amplifier and it may not even be the load-line of the class-C
amplifier. There is probably some point in an otherwise non-linear
system where a linear analysis becomes possible. I think that point is
what Walt considers to be the linear source point, wherever that point
might be located.

Recalling that if a conjugate match is achieved at one ponit in a
system
it is achieved at all points....

It does not seem possible for a system to be non-linear at one end and
turn in to a linear system at some other point.

In fact, here is my personal take on the subject. Given an antenna
system that presents 50+j0 ohms looking into 50 ohm coax, the internal
impedance of the source doesn't matter. For any voltage source,
irrespective of the source impedance, if reflected energy doesn't
reach the source, the source impedance doesn't matter (except for
efficiency). Seems to me, the highest efficiency would be achieved by
a source with zero ohms of source impedance.

True, if the source impedance originates in dissipative components and
it is a voltage source. For a current source, infinite impedance
offers
the best efficiency.

....Keith

On May 25, 12:32*am, (Richard Harrison)
wrote:
Cecil Moore, W5DXP wrote:

"Seems to me, the highest efficiency would be achieved by a source with
zero source impedance."

Me too, but zero source impedance does not match the load as required
for maximum power transfer.

It seems to me that much too much is made of 'maximum power transfer'
in
the RF world. In the world of 50 and 60 Hz, where significantly more
energy is moved, 'maximum power transfer' is never mentioned.
Efficiency
is much more of interest.

For the most part, 'maximum power transfer' is just an interesting
ideosyncracy of linear circuit theory.

....Keith