wrote in message
...
On Jun 6, 9:15 pm, Walter Maxwell wrote:
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.

Hi Walt,

R is by definition a physical "property of conductors which depends on
dimensions, material, and temperature". So if we multiply both sides
of our "ratio" equation by I^2 to convert to power we get V*I =
I^2*R. Given that V, I, and R are all non-zero, why would you ask us
to believe that I^2*R and V*I could be zero? It's true that V^2/R is
a ratio. And I guess it's probably also true that the equation itself
doesn't dissipate power. But what would you have us believe that that
is supposed to prove?

73, Jim AC6XG

Hello Jim,

I don't understand how my statement in the email above indicates that I^2*R and
V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
resistance that is non-dissipative because a ratio cannot dissipate power.

Walt

Walter Maxwell wrote:
I don't understand how my statement in the email above indicates that I^2*R and
V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
resistance that is non-dissipative because a ratio cannot dissipate power.

"The IEEE Dictionary" is careful to differentiate between
an E/I ratio equaling an impedance vs an "impedor" consisting
of physical components.
--
73, Cecil http://www.w5dxp.com

On Jun 14, 8:46*am, "Walter Maxwell" wrote:

I don't understand how my statement in the email above indicates that I^2*R and
V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
resistance that is non-dissipative because a ratio cannot dissipate power.

Walt

Hi Walt -

If E and I are not zero, then E*I is not zero. But you are correct
that the equations themselves do not dissipate power. :-) Resistors
do, however. If there isn't an actual resistor located where you make
your measurement, then of course there's no power being dissipated
there.

73, ac6xg

Cecil, W5DXP wrote:
"What is the linear source impedance of a class-C amp?"

A conjugate match is necessary for maximum power transfer.

A Class C amplifier is not inherently linear. That is disasterous for an
already modulated AM signal but it is of no importance to an FM signal.
As Richard Fry points out, no tank circuit is required. A low-pass
filter to suppress all harmonics is all that is needed for a clean
signal. No tuning is required of a tank at the operating frequency.

A tank circuit is not selective enough to prevent intermod anyway. When
I worked the morning shift at 790 KHz in Houston, and I fired up the
transmitter, I would hear 740 KHz`audio coming out of the monitor
speaker. They started programming earlier than we did. They were 15
miles away. Our transmitting antenna made a dandy receiving antenna. The
received 740 KHz modulated our final amplifier and we rebroadcast it
although at a level much lower than our own modulation. Any one
listening to 790 KHz who turned up the volume to hear the 740 KHz audio
got their ears knocked off when our modulation started. We had high
level plate modulation of the final amplifier for our own signal. For
the 740 KHz signal the level of modulation was much lower, millivolts
not kilovolts. Either program was cleanly modulated on our carrier. The
only difference was the enormous difference in modulation levels.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"What is the linear source impedance of a class-C amp?"

A conjugate match is necessary for maximum power transfer.

Is the class-C amp conjugately matched during
the 75% of the cycle when it is off? Is there
any such thing as an instantaneous conjugate
match? Don't we have to move downstream from
non-linear sources for our linear math models
to start working?
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
...
Richard Harrison wrote:
Cecil, W5DXP wrote:
"What is the linear source impedance of a class-C amp?"

A conjugate match is necessary for maximum power transfer.

Is the class-C amp conjugately matched during
the 75% of the cycle when it is off? Is there
any such thing as an instantaneous conjugate
match? Don't we have to move downstream from
non-linear sources for our linear math models
to start working?
--
73, Cecil http://www.w5dxp.com

Richard, its a common myth that Class C amps are non-linear, but the truth of
the matter is that although the condition at the input of the pi-network is
decidedly non-linear, the energy storage in the pi-network tank circuit isolates
the input from the output and the result is a totally linear condition at the
output of the pi-network. Evidence proving this is true is that the output of an
unmodulated signal at the output of the network is an almost pure sine wave.
With a Q of at least 12 the difference between a pure sine wave from a signal
generator and that from the pi-network output can not be seen on a dual trace
scope with the traces overlapping.

I don't know about the energy storage in the filters you mention, but I would
assume that if the filter output is a sine wave then the energy storage required
to produce a linear output is sufficient.

Walt, W2DU

Cecil, W5DXP wrote:
"Is the class-C amp conjugately matched during the 75% of the cycle it
is off?"

We must consider the complete cycle.

Working with spark ignition systems (Hettering) you may have encountered
a "dwell meter". It indicates the % of the time ignition points are
closed. When the points are closed, impedance between the meter and the
battery is insignificant. The meter if left continuously connected
through the points would indicate full-scale. When the points open,
their impedance is infinite. Left continuously open, the meter indicates
zero on the dwell scale.

Dwell is measured while the engine is rotating and the meter is being
connected intermittently to the battery through the ignition points.

Intermittent opening and closing of the points causes the same scale
reading that would be caused by replacing the points with some
particular value of fixed resistance (a resistor).
The main difference is that no dissipation occurs in the open ignition
points and precious little energy is lost in the closed points. Voila!
We have produced a dissipationless resistance.
The Class C amplifier is a switch which operates in the same manner. The
Kettering ignition points have a low-resistance ignition coil primary in
series, and the Class-C amplifier has a tuned plate circuit in series,
but both are being switched on and off repeatedly.

Best regards, Richard Harrison, KB5WZI

Walt, W2DU wrote:
"I don`t know about the energy storage in the filter you mention, but I
would assume that if the filter output is a sine wave then the energy
storage required to produce a linear output is sufficient."

As a former FCC official, Walt knows their regulations sharply limit
harmonic content of FM broadcast carrier frequencies. A sine wave must
be pure, otherwise it has harmonic content (not allowed).

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"What is the linear source impedance of a class-C amp?"

A conjugate match is necessary for maximum power transfer.

*in a linear system*

Jim Lux wrote:
"in a linear system"

It produces no significant harmonics, so the system is linear.

Best regards, Richard Harrison, KB5WZI