On May 24, 7:30*am, Keith Dysart wrote:
On May 23, 4:31*am, Owen Duffy wrote:



Owen Duffy wrote in news:Xns9D81BC11E3183nonenowhere@
61.9.191.5:

"Sven Lundbech" wrote in
. dk:

...
As mentioned earlier, most of the stuff is old hat to me - but I
really look forward to dig into the chapters concerning tx output
impedance. A highly controversial subject for decades.

Here is a simple little test for the hypothesis that Zs=50+j0 that uses
equipment found in many if not most HF ham shacks.

Oh, the URL:http://vk1od.net/blog/?p=1028.
Owen

While the analysis of transmitter output impedance in Reflections is
flawed,
experiments (claimed to be repeatable) described in Reflections appear
to
support the conclusions of the flawed analysis.

It would be highly valuable if the results of these experiments could
be
explained in a manner that aligns with established understandings.

Such an explanation might start by describing the circuit conditions
that
result from following the manufacturer’s tuning procedures. After
all,
these usually depend on measuring currents and voltages so are only
indirectly related to power. Perhaps the resulting conditions are not
as they are usually assumed to be.

Try as I might, I have not been able to derive a mechanism to explain
the observations in Reflections. But the explanations offered in
Reflections require large chunks of linear circuit theory to be
discarded,
so this does not seem to be an appropriate path.

...Keith

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%.

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.

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. Also include your email address
so I can send it.

Keith, you are the only person I know of who appears to have found
flaws in my presentation on this subject. Which is why I'm anxious to
know exactly why you believe my presentation is flawed.

Walt Maxwell, W2DU

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, 6:15*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

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 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

Keith Dysart wrote:
"For the most part, "maximum power transfer is just an interesting
ideosyncracy of linear circuit theory."

In the world of 50 and 60 Hz, we don`t want all the power plant can
supply when we flip on a light switch.

The RF world is usually different.

Maximum power transfer only occurs when source and load match
conjugately, and the match proves the load and source impedances are
equals. It is well known and easily shown that a match results in
maximum power transfer.

If one has a 100-watt transmitter he probably wants 100 watts out of it
sometimes and may only be able to do so when his antenna is matched to
his transmitter,

Maxumum power treansfer is more than an "interesting ideosyncracy".

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Keith Dysart wrote:
"For the most part, "maximum power transfer is just an interesting
ideosyncracy of linear circuit theory."

In the world of 50 and 60 Hz, we don`t want all the power plant can
supply when we flip on a light switch.

The RF world is usually different.

Maximum power transfer only occurs when source and load match
conjugately, and the match proves the load and source impedances are
equals. It is well known and easily shown that a match results in
maximum power transfer.
. . .

It's also easily shown that it doesn't.

Consider a 10 volt voltage source having a 50 ohm source resistance,
feeding a 50 ohm resistive load. Power at the load is 0.5 watt, is it not?

Reduce the source impedance to 10 ohms.

Now what is the power dissipated in the load?
Is it greater or less than it was when the source and load impedances
were matched?

Roy Lewallen, W7EL

On May 24, 11:32*pm, (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.

A 60 Hz Power Generation Plant operates at high efficiency, not at the
maximum power transfer point. If they were 50% efficient, they would
go out of business. (That's what Edison expected.) Why is maximum
power transfer desirable in ham transmitters? Is such a design the
highest power/cost ratio? Is it possible to build an output amp with a
10 ohm source impedance designed to be 80% efficient? 1 ohm source
impedance designed to be 98% efficient? Is co$t the driving parameter?
--
73, Cecil, w5dxp.com

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, 5:29*am, Keith Dysart wrote:
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.

Well, consider the following two systems. Z01 is 50 ohms and Z02 is
300 ohms. The two systems are identical except for the circuits hidden
inside the two identical source black boxes. Both sources are
supplying a 100v sine wave to the system.

Source1----Z01----+----1/4WL Z02----1800 ohm

Source2----Z01----+----1/4WL Z02----1800 ohm

Every passive voltage, current, power, and impedance measurement is
identical in both systems. As far as we can passively measure, both
systems are identical and linear. The only thing we don't know is what
is inside the two source boxes..

Inside the Source1 box is a linear ideal 50 ohm Thevenin equivalent
source delivering an ideal 100v sine wave. Inside the Source2 box is a
non-linear class-C amplifier filtered to provide an ideal 100v sine
wave.

Without changing the system conditions, can one make a passive
measurement to determine which system is conjugately matched and which
one is not conjugately matched? If one cannot tell the difference, are
they both conjugately matched, or both not conjugately matched, or
what?

Here's my take. A 50 ohm Z0-match exists in both systems and all
conditions are identical on the load side of that Z0-match. In
particular, at any point in the system on the load side of the Z0-
match, the impedance looking in one direction is the conjugate of the
impedance looking in the other direction. That is a characteristic of
a conjugate match. So are both systems conjugately matched between the
Z0-match and the load? If it walks and quacks like a duck ...
--
73, Cecil, w5dxp.com