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
From Wikipedia, the free encyclopedia
.. Learn more about using Wikipedia for research .Jump to: navigation, search
This article is about the mathematical concept. For the Swedish institute, see
Ratio Institute. For the academic journal, see Ratio (journal).
This article or section is in need of attention from an expert on the
subject.
Please help recruit one or improve this article yourself. See the talk
page for details.
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WikiProject

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