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

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

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 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

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

(Richard Harrison) wrote in
:

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.

Lets plug some real world numbers in...

Take a DC supply of 1000V, and a valve that saturates at 200V, the RF
approximately sinusoidal voltage swing on the anode is from 200V to
1800V. (For avoidance of doubt, whilst the RF voltage on the anode is
approximately sinusoidal, the anode current waveform is not.)

If the conduction angle is 120 degrees (typical for Class C amplifiers),
the valve starts conducting at about 1000-800*sin((180-120)/2) or 600V
instantaneous anode voltage... and continues conducting as the
instantaneous anode voltage passes through the minimum and rises again,
cutting off when then instantaneous anode voltage again reaches 600V. In
this case, the anode voltage during conduction varies between 400 and
600V, 40% to 60% of the supply voltage.

The switch analogy is not a good one.

When switched off, voltage drop across the tube is Eb, ...

That is wrong. The anode voltage is approximately a sinusoidal voltage
swing of almost (70% to 90%) Eb zero to peak superimposed on the DC
supply voltage (Eb).

You have taken a quote from Terman and weaved your own flawed extensions
(being the switch analogy and the statement about instantaneous anode
voltage).

Owen

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

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

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

(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