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 26, 6:20*pm, Roy Lewallen wrote:
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

But Roy, consider that the source resistance remains constant at 10
ohms. Then what load resistance will absorb the most power? The answer
is 10 ohms. Any value of load resistance greater or less than 10 ohms
will result in less power delivered. I don't believe it's fair to
change the source resistance when dealing with the Maximum Power
Transfer Theorem.

In your example with a source resistance of 10 ohms and a load
resistance of 50 ohms the power delivered will be 1.39 watts. But when
the load resistance is 10 ohms with the same source resistance the
power delivered is 2.5 watts. As I said above, if the load resistance
is either greater or less than 10 ohms the power delivered will be
less than 2.5 watts. Thus when the source resistance is constant the
maximum power will be delivered when the load is matched to the
source.

Nes pa?

Walt

walt wrote:

But Roy, consider that the source resistance remains constant at 10
ohms. Then what load resistance will absorb the most power? The answer
is 10 ohms. Any value of load resistance greater or less than 10 ohms
will result in less power delivered. I don't believe it's fair to
change the source resistance when dealing with the Maximum Power
Transfer Theorem.

In your example with a source resistance of 10 ohms and a load
resistance of 50 ohms the power delivered will be 1.39 watts. But when
the load resistance is 10 ohms with the same source resistance the
power delivered is 2.5 watts. As I said above, if the load resistance
is either greater or less than 10 ohms the power delivered will be
less than 2.5 watts. Thus when the source resistance is constant the
maximum power will be delivered when the load is matched to the
source.

Nes pa?

Walt

Of course, I know that, and I would hope anyone with even very basic
electrical circuit analysis knowledge does. And anyone with that
knowledge should state as you have,

"FOR A GIVEN SOURCE IMPEDANCE, maximum power transfer occurs when the
source and load impedances are matched (i.e., the load impedance is the
complex conjugate of the source impedance)," which is true.

But the statement which was made was that "Maximum power transfer occurs
when the source and load impedances are matched." This is NOT true, as
the example demonstrates.

It's an important distinction. Instead of declaring what's "fair" and
what isn't with regard to changing source and load impedances, the
maximum power transfer theorem should be stated correctly, in a way
which makes it true.

Roy Lewallen, W7EL

Roy Lewallen wrote:
"Reduce the source imperance to 10 ohms."

FOUL!

In the case of a 10-ohm internal source, the load which extracts maximum
power is 10 ohms.

Best regards, Richard Harrison, KB5WZI

I stand by my position that when the source is an RF power amplifier,
and when all the available power is transferred from the source to the
load, the source impedance is the conjugate of the load impedance. In
a similar instance, if the load is a pure resistance, the source
resistance equals the load resistance.

And referring to a statement Dysart made concerning plate resistance,
Rp, it must be understood that in Class AB, B and C amplifiers, Rp is
NOT the source resistance. In those amplifiers the effect of Rp is a
negative feedback that reduces the effect of plate-current change
resulting from a change in grid voltage, thus reducing the power
output compared to what the output would be if Rp were absent.
Compensation for the power lost to Rp is accomplished by simply
increasing the grid drive.

Consequently, Rp plays no part in achieving a conjugate match at the
junction of the network-output and the load. Although lossless
elements are required to achieve a perfect conjugate match in both
directions, a perfect conjugate match is obtained in the forward
direction with real elements when all available power is being
delivered to the load. This condition is verified in data presented in
Chapter 24 of Reflections 3.

Walt, W2DU

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