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

Matched across what boundary? from output pi network to load?
From active device to pi network?
I'd venture that between active device and pi network, there isn't a
conjugate match, at any given instant, and perhaps not even considered
over the entire cycle (without resorting to some things like "apparent
impedance" which doesn't have a real clean definition).

In order to provide a true "conjugate match" to a device that is
changing state, the load must also be changing state: i.e. the match is
single valued, for any Zload, there is a single Zmatch that maximizes
power transfer; except perhaps for some trivial cases, like Z=zero or
infinity, but in such cases, the load doesn't dissipate ANY power, so
what is there to maximize.

Furthermore, if one looks at situations where you have, for instance, a
very low source impedance (a stiff voltage bus) or a very high source
impedance (a constant current source), power transfer is maximized to a
given load impedance when the reactive components are conjugate. In such
a case, the source and load resistances are not equal.



One might look at
http://p1k.arrl.org/~ehare/temp/conj...ch_theorum.pdf
http://mysite.orange.co.uk/g3uur/index.html





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.

I would say "apparent resistance".. the "conjugate match" and any other
linear circuit analysis can't necessarily be "averaged". Something like
Kirchoff's current law or voltage law (or Ohm's law, for that matter)
has to be true at any instant.

The challenge faced by folks faced with analyzing "real" circuits is
that you have to be careful about how you turn a circuit that is likely
time-varying AND nonlinear into a linearized approximation. For
instance programs like SPICE's transient analysis uses linear circuit
theory (via matrix analysis) in combination with an iterative
differential equation solver, and tries to treat the circuit as linear
at a given instant. (granted, newer versions of SPICE and its ilk are a
bit more sophisticated, since they can handle nonlinear terms in the
matrix).


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.

Complicated substantially by the fact that the active device in a RF
amplifier generally doesn't act as an ideal switch. So the piecewise
linearization you describe isn't totally applicable. For instance, a BJT
acts like a constant current source if base drive is fixed, but in RF
circuits, the base drive isn't fixed. In FET circuits you worry about
the gate capacitance.


This is why there are all sorts of variants of SPICE modified for
switching power supplies.



Best regards, Richard Harrison, KB5WZI

(Richard Harrison) wrote in news:26007-4851914B-
:

....
The Class C amplifier is a switch ...

If you say it enough times, will it become true?

Owen

Richard Harrison wrote:
Jim Lux wrote:
"in a linear system"

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

I'm not trying to be difficult, but I just resurrected the
Fourier equation for the output of a class-B amplifier. It is:

i = 0.318Im + 0.500Im*sin(A) - 0.212Im*cos(2A) - 0.0424Im*cos(4A) - ...

The second harmonic is 42% of the amplitude of the
fundamental frequency at the plate of the tube amp.
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
...
Richard Harrison wrote:
Jim Lux wrote:
"in a linear system"

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

I'm not trying to be difficult, but I just resurrected the
Fourier equation for the output of a class-B amplifier. It is:

i = 0.318Im + 0.500Im*sin(A) - 0.212Im*cos(2A) - 0.0424Im*cos(4A) - ...

The second harmonic is 42% of the amplitude of the
fundamental frequency at the plate of the tube amp.
--
73, Cecil http://www.w5dxp.com

OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
plate of the tube, can you determine the amplitude of the 2nd harmonic at the
output of a pi-network having a Q of 12?

Walt, W2DU

"Cecil Moore" wrote in message
...
Richard Harrison wrote:
Jim Lux wrote:
"in a linear system"

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

I'm not trying to be difficult, but I just resurrected the
Fourier equation for the output of a class-B amplifier. It is:

i = 0.318Im + 0.500Im*sin(A) - 0.212Im*cos(2A) - 0.0424Im*cos(4A) - ...

The second harmonic is 42% of the amplitude of the
fundamental frequency at the plate of the tube amp.
--
73, Cecil http://www.w5dxp.com

OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
plate of the tube, can you determine the amplitude of the 2nd harmonic at the
output of a pi-network having a Q of 12?

Walt, W2DU

Walter Maxwell wrote:
OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
plate of the tube, can you determine the amplitude of the 2nd harmonic at the
output of a pi-network having a Q of 12?

I think I understand ideal class-A operation pretty well.

What I am trying to understand is: In class-B operation,
if the frequency of interest is made up of less than 1/2
of the current through the amplifier tube, is more than
1/2 of the current not conjugately matched?
--
73, Cecil http://www.w5dxp.com

"Richard Fry" wrote
Note that the reflected pulse appears some 6.2 µs after the incident
pulse, ...
_________

Correction: the reflected pulse appears 3.1 µs after the incident pulse (the
time base was 0.5 µs/cm).

The rest of the statements there remain valid.

RF

Jim Lux wrote:
"Matched across what boundary?"

Where there is a conjugate match in the transmitter-antenna system, it
exists at every pair of terminals. That is, the impedances looking in
opposite directions are conjugates. The resistive parts of the impedance
are equals and the reactances looking in opposite directions are
opposites of each other.

Best regards, Richard Harrison, KB5WZI