Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Owen Duffy wrote:
"Most supply authorities would not allow you to connect a capacitive
load (a leading PF load),-----."

Incorrect. Overexcited synchronous machines are commonly used to correct
the power factor causing reduced line current and lower power loss on
the a-c power transmission line.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

"Roy Lewallen" wrote
No mysterious "reverse r-f energy" is needed to explain this well-known
and well-understood phenomenon.
__________

The quote above reads as though the existence of reverse (reflected)
r-f energy is being denied.

For skeptics, the link below leads to a field report showing a measurement
of the reflection of a narrowband r-f pulse by an analog broadcast TV
antenna, back toward the source. The H.A.D. of the sin² pulse used
represents the shortest transition time that can be accommodated in a ~ 4
MHz transmission channel.

Note that the reflected pulse appears some 6.2 µs after the incident pulse,
which corresponds to a round trip through the ~1,525 feet of 6" OD, 75 ohm,
air pressurized transmission line leading to the antenna in this system.

The return pulse amplitude indicates a far-end match (elbow complex +
antenna) of about 1.05 VSWR (2.3% reflection).

http://i62.photobucket.com/albums/h8...easurement.gif

RF (RCA Broadcast Field Engineer 1965-1980)

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Richard Fry wrote:
"Roy Lewallen" wrote
No mysterious "reverse r-f energy" is needed to explain this
well-known and well-understood phenomenon.
__________

The quote above reads as though the existence of reverse (reflected)
r-f energy is being denied.

Especially strange since the "reverse r-f energy"
is the cause of the impedance responsible for the
mismatch problem. If there were no "reverse r-f energy",
the impedance causing the mismatch wouldn't even exist.
Why is the cause of the offending impedance of no importance?
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

"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

Efficiency and maximum power transfer
 

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*

Efficiency and maximum power transfer
 

Jim Lux wrote:
"in a linear system"

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

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

"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

Efficiency and maximum power transfer
 

"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

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

"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

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Richard Harrison wrote:
Where there is a conjugate match in the transmitter-antenna system, it
exists at every pair of terminals.

Quoting w2du's web page:
“The Conjugate Theorem also shows that in a sequence of
matching networks it is necessary to match at only one
junction *if the networks are non-dissipative*. In actual
practice, since there is usually some dissipation, it
is frequently desirable to adjust at more than one point.”
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

Owen Duffy wrote:
"The Class C amplifier is a switch...
If you say it enough times, will it become true?"

True is true no matter what anyone says. I`ve never seen Terman
misspeak.

On page 255 of his 1955 opus Terman wrote:
"---the Class C amplifier is adjusted so the plate current flows in
pulses that last less than half a cycle."

On page 450 he wrote:
"The high efficiency of the Class C amplifier is a result of the fact
that plate current is not allowed to flow except when the instantaneous
voltage drop across the tube is low; i.e. Eb supplies energy to the
amplifier only when he largest portion of the energy will be absorbed by
the tuned circuit."

Sounds like a switch to me. When switched on, voltage drop across the
tube is low. When switched off, voltage drop across the tube is Eb, but
since current is zero, no power is lost at that instant in the tube.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

Cecil, W5DXP wrote:
"Quoting W2DU`s web page:"

Sure hope Walt finds a publisher soon for his latest edition of
"Reflections". Web TV (a Microsoft company) doesn`t allow me to read
pdf.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

Richard Harrison wrote:
Cecil, W5DXP wrote:
"Quoting W2DU`s web page:"

Sure hope Walt finds a publisher soon for his latest edition of
"Reflections". Web TV (a Microsoft company) doesn`t allow me to read
pdf.
Best regards, Richard Harrison, KB5WZI

Richard, try this: Let Google translate it from PDF to
HTML for you.

Do a Google search for "w2du reflections
chapter 19A".

One of the first entries will be a hit
at http://w2du.com/r3ch19a.pdf .

Under the Google entry for this, carefully look for
the words:
"File Format: PDF/Adobe Acrobat - View as HTML"

The words 'View as HTML' will be linked to Google's
HTML translation of the PDF.

The translation isn't pretty, but the essentials are
there. You will have to fill in some of the blanks
yourself.

Posting this as well for anyone else experiencing this
problem.

Jim, K7JEB

Efficiency and maximum power transfer
 

Richard Harrison wrote:
Cecil, W5DXP wrote:
"Quoting W2DU`s web page:"

Sure hope Walt finds a publisher soon for his latest edition of
"Reflections". Web TV (a Microsoft company) doesn`t allow me to read
pdf.

Do a Google search for NIST definition of "conjugate match"
One of the first entries will be:

[PDF]
Chapter 01
File Format: PDF/Adobe Acrobat - *View as HTML*
Quoting from Robert W. Beatty, NIST, Microwave Mismatch Analysis, (Ref
120):. 1) Conjugate match—The condition for maximum power absorption by
a load, in ...
w2du.com/Appendix09.pdf - Similar pages

Click on View as HTML and the section I quoted should appear
in HTML format.
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

I wrote:
"When switched-off, voltage drop across the tube is Eb"

Owen Duffy wrote:
"That is wrong."

Yes as a general statement, that is wrong. The instantaneous drop across
the tube is the sum of Eb and EL, the signal voltage across the load.

Terman shows this in Fig. 13-1(b) on page 449 of his 1955 opus.

Fig. 13-1(e) shows the plate current pulse which is less than 180
degrees in duration as the tube is biased beyond cut-off.

My switch analogy is imperfect but good enough to exemplify
dissipationless resistance as a part of the output impedance of a Class
C amplifier.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

Richard Harrison wrote:
My switch analogy is imperfect but good enough to exemplify
dissipationless resistance as a part of the output impedance of a Class
C amplifier.

Well, there are mechanical switches, digital switches,
and analog switches. I suspect a class-C amp falls
under the heading of an analog switch.
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Efficiency and maximum power transfer
 

wrote:
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


I always believed that a ratio was a comparative measure between like
units - e.g. forward voltage to reverse voltage, output power to input
power etc. Voltage to current is not a ratio. V/I has dimensions of
resistance - ratios are dimensionless.
Alan

Efficiency and maximum power transfer
 

wrote:
R is by definition a physical "property of conductors which depends on
dimensions, material, and temperature".

That's only one definition. From "The IEEE Dictionary",
the above is definition (A). Definition (B) is simply
"the real part of impedance" with the following Note:
"Definitions (A) and (B) are not equivalent but are
supplementary. In any case where confusion may arise,
specify definition being used."

Definition (B) covers Walt's non-dissipative resistance.
A common example is the characteristic impedance of
transmission line. In an ideal matched system V^2/Z0, I^2*Z0,
or V*I is the power being transferred under non-dissipative
conditions.
--
73, Cecil http://www.w5dxp.com

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

On Jun 13, 10:43*pm, Alan Peake wrote:
wrote:

* 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

I always believed that a ratio was a comparative measure between like
units - e.g. forward voltage to reverse voltage, output power to input
power etc. Voltage to current is not a ratio. V/I has dimensions of
resistance - ratios are dimensionless.
Alan

Good point. You may be right.

73 de jk

Efficiency and maximum power transfer
 

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

Efficiency and maximum power transfer
 

Alan Peake wrote:
"V/I has dimensions of resistance - ratios are dimensionless."

Yes, until we name them. Cycles / seconds is now called Hertz.

My electronics dictionary defines ratio -
"The value obtained by dividing one number by another."

Simple and no qualifications.

From Newton:
Acceleration = force / mass

You must pick the right units or use constants to make the numbers work.

Some people are persuaded that resistance = loss. Not so at all.
Resistance is just a name given to the ratio of voltage to current. A
perfect reactance produces a voltage drop but no power is lost. A
transmission line can be lossless enough to qualify and have a Zo = sq
rt of L/C. Reg Edwards used to say that if your perfect line were long
enough you could measure its Zo with an ohmmeter. Reg was right because
no reflection would ever return to change the current supplied by the
ohmmeter.

Free-space has a lossless Zo of 120 pi (or 377 ohms) according to page
326 of Saveskie`s "Radio Propagation Handbook". This is a ratio which is
related to volts and amps but is actually the ratio of the electric
field strength to the magnetic field strength in an EM wave. The volts
and amps are in phase so it has the units of a pure resistance.

Best regards, Richard Harrison, KB5WZI

Efficiency and maximum power transfer
 

"Owen Duffy" wrote in message
...
(Richard Harrison) wrote in news:23000-
:

Jim Lux wrote:
"in a linear system"

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

That is a new / unconventional definition of 'linear'.

The term is usually used in this context to mean a linear transfer
characteristic, ie PowerOut vs PowerIn is linear.

Considering a typical valve Class C RF amplifier with a resonant load:

Conduction angle will typically be around 120°, and to achieve that, the
grid bias would be around twice the cutoff voltage.

If you attempted to pass a signal such as SSB though a Class C amplifier
that was biased to twice the cutoff value, there would be no output
signal when the peak input was less than about 50% max drive voltage, or
about 25% power, and for greater drive voltage there would be output. How
could such a transfer characteristic be argued to be linear?

Owen

Owen, 'linear transfer characteristic' isn't the only context for the use of the
word 'linear'. Even though the input circuit of a Class C amplifier is
non-linear, the output is linear due to the energy storage of the tank circuit
that isolates the input from the output, therefore, the output is linear. Proof
of this is that the output signal is a sine wave. In addition, the voltage and
current at the output terminals of the pi-network are in phase. Furthermore, the
ratio E/I = R appearing at the network output indicates that the output source
resistance R is non-dissipative, because a ratio cannot dissipate power. This
resistance R is not a resistor.

Walt

Efficiency and maximum power transfer
 

wrote in message
...
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

Jim, have you reviewed the new section Chapter 19A that appears on Cecil's
website that he uploaded on June 7? It appears there posted as 'Chapter 19A from
Reflections 3'. If you haven't reviewed it I urge you to do so, especially the
last portion where I report the measurements I made with a complex impedance
loading the amplifier. These measurements prove two things: 1) that the output
resistance of the amp is non-dissipative, and 2) that no reflected energy
reaches the amp tube, and in fact the measurements show that the tube doesn't
even see the reflected power. When you see the numbers and understand the
procedure I used in obtaining them you will be hard pressed to disagree with the
results.

Walt, W2DU

Efficiency and maximum power transfer
 

"Alan Peake" wrote in message
...


wrote:
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


I always believed that a ratio was a comparative measure between like
units - e.g. forward voltage to reverse voltage, output power to input
power etc. Voltage to current is not a ratio. V/I has dimensions of
resistance - ratios are dimensionless.
Alan

Alan, I disagree with you when you say that 'voltage to current' is not a ratio.
IMHO, your are definine 'ratio' to narrowly. Below is a quote from Google:

Ratio
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This article is about the mathematical concept. For the Swedish institute, see
Ratio Institute. For the academic journal, see Ratio (journal).
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The ratio of width to height of typical computer displays
A ratio is a quantity that denotes the proportional[citation needed] amount or
magnitude of one quantity relative to another.

Ratios are unitless when they relate quantities of the same dimension. When the
two quantities being compared are of different types, the units are the first
quantity "per" unit of the second - for example, a speed or velocity can be
expressed in "miles per hour". If the second unit is a measure of time, we call
this type of ratio a rate.

Fractions and percentages are both specific applications of ratios. Fractions
relate the part (the numerator) to the whole (the denominator) while percentages
indicate parts per 100.

Note, Alan, the expression "When the two quantities being compared are
different types......

Walt, W2DU