On Sat, 29 May 2010 16:58:52 -0700 (PDT), walt wrote:

Although I understand that induced
current results from a moving charge, I'm not as comfortable
discussing the Rp issue in terms of charges as I am with currents, so
I'm having a little difficulty in coming to grips with Rp while
reviewing the paper.

Hi All,

For others, there might be wonder at the meaning of induced or
inducted current - or convection current for that matter. It would be
useful to expand here in a new thread.

First. In the interval between cathode and plate, there are only
electrons. In other words, there are no conductors as are found
everywhere else.

Second. When an electron flows between the cathode and the plate,
charge is moving. By convention, this is called current. It goes by
a special label - convection current.

Third. The interval between the cathode and plate is part of a
complete circuit that travels through the power supply, returns to
ground and rises from ground to come to the cathode.

Fourth. For charge movement through the interval between cathode and
plate, there is exactly the same charge movement through ALL wiring
connecting the plate to the power supply, wiring from the power supply
to ground, and wiring from the ground to the cathode.

Fifth. The charge movement throughout the wired circuit, by
convention is called current. it goes by a special label - induced
current.

The distinction between induced current and convection current is
found in their transport media. One is metal, the other is vacuum.
The convection (like convection heat) is electron movement in space
which induces current in the wire. This induction is NOT like
transformer induction. It has absolutely nothing to do with magnetic
linkage. We could call it induction if the entire circuit were
composed of wire. That is, the movement of one electron "induces" the
movement of another electron.

Similarly (and tipping my hand about reflected waves), an extra
electron push in the wiring will induce an electron in the tube to
move. The current is inextricably locked in a dynamic flow. There is
no new physics in this anywhere.

So why two terms? The necessity of this seemingly obscure point is to
exhibit how heat is formed, and why the plate of the tube is a
literal, real resistor in this circuit. Very few correspondents (read
none but myself) to this board dwell on this, and they end up
dismissing the real resistance of this specific and consequential
source as a mathematical fiction suitable for only milliwatt signal
generators.

73's
Richard Clark, KB7QHC

On May 29, 8:25*pm, Richard Clark wrote:
Similarly (and tipping my hand about reflected waves), an extra
electron push in the wiring will induce an electron in the tube to
move. *The current is inextricably locked in a dynamic flow. *There is
no new physics in this anywhere.

The only time photons are not generated is during DC steady-state with
no AC signal applied. If you can see the electron beam inside a vacuum
tube (and I have), the electron beam is necessarily emitting photons.
If the field surrounding an electrical current is detectable, the
electrons are emitting photons.
--
73, Cecil, w5dxp.com

I have shifted the following post to this thread where it fits the
context perfectly:

On Sun, 30 May 2010 06:59:58 -0700 (PDT), K1TTT
wrote:

does it really matter what the mechanism of generating the heat is?
the fact that there is heat generated in the plate means there is
energy lost from the electrons, be it due to impinging on the plate
mechanically or due to the metal resistance it is still a loss of
energy. And since there is more heat when there is more plate current
we can model it as a real resistance. The only question is then, is
it linear or not. that should be a 'relatively' simple measurement,
and one that has likely been done by a researcher somewhere along the
line.

Yes.

This post is sufficiently informing for many. Others may take its
lead and derive EXACTLY how much resistance, heat, power from the data
provided by Walt and confirm the bench experience of real resistance
within the source.

This response quoted above also corrects my mis-observation of my
having the only perception of this universally experienced phenomenon.
By count, we are up to two who acknowledge what is obvious.

73's
Richard Clark, KB7QHC

On May 30, 4:46*pm, Richard Clark wrote:
I have shifted the following post to this thread where it fits the
context perfectly:

On Sun, 30 May 2010 06:59:58 -0700 (PDT), K1TTT
wrote:

does it really matter what the mechanism of generating the heat is?
the fact that there is heat generated in the plate means there is
energy lost from the electrons, be it due to impinging on the plate
mechanically or due to the metal resistance it is still a loss of
energy. *And since there is more heat when there is more plate current
we can model it as a real resistance. *The only question is then, is
it linear or not. *that should be a 'relatively' simple measurement,
and one that has likely been done by a researcher somewhere along the
line.

Yes.

This post is sufficiently informing for many. *Others may take its
lead and derive EXACTLY how much resistance, heat, power from the data
provided by Walt and confirm the bench experience of real resistance
within the source.

This response quoted above also corrects my mis-observation of my
having the only perception of this universally experienced phenomenon.
By count, we are up to two who acknowledge what is obvious.

73's
Richard Clark, KB7QHC

occam's razor is usually the sharpest knife.

On May 30, 12:46*pm, Richard Clark wrote:
I have shifted the following post to this thread where it fits the
context perfectly:

On Sun, 30 May 2010 06:59:58 -0700 (PDT), K1TTT
wrote:

does it really matter what the mechanism of generating the heat is?
the fact that there is heat generated in the plate means there is
energy lost from the electrons, be it due to impinging on the plate
mechanically or due to the metal resistance it is still a loss of
energy. *And since there is more heat when there is more plate current
we can model it as a real resistance. *The only question is then, is
it linear or not. *that should be a 'relatively' simple measurement,
and one that has likely been done by a researcher somewhere along the
line.

Yes.

This post is sufficiently informing for many. *Others may take its
lead and derive EXACTLY how much resistance, heat, power from the data
provided by Walt and confirm the bench experience of real resistance
within the source.

This response quoted above also corrects my mis-observation of my
having the only perception of this universally experienced phenomenon.
By count, we are up to two who acknowledge what is obvious.

73's
Richard Clark, KB7QHC

Hi Richard,

On reviewing the paper you sent me I’ve given a lot of thought
concerning the source of the induced current flowing in the external
resistance, R, the load of the plate of the tube. You say the source
of the I^2R power dissipated in R is plate resistance, Rp. However, as
I read Par 7.2 on Pages 140 and 141, the reality of Rp is described in
there, but only indirectly, thus requiring a deep inquiring mind to
ferret it out. For me, it takes a leap and a stretch to understand
that Rp is the source of the I^2R power delivered to R.

In Par 7.2 the author says this power is supplied by the kinetic
energy of the electrons in the beam. It takes no great amount of
mental concentration to understand that that statement is true. But
where I see a tangential approach to describing Rp is in the statement
that “Since each electron in the beam faces a decelerating electric
field due to I^2R, it loses an amount of energy eIR during its flight
from the grid to electrode P.” This energy is lost, of course, because
the plate voltage is decreasing due to the increase in voltage drop IR
across R, caused by resistance Rp. We know, of course, that although
Rp is real, it is non-dissipative because it is not a physical
resistor, but is only the result of the ratio deltaE/deltaI.

Again quoting from Par 7.2, “It is evident that IR must be smaller
than the voltage to which the electrons have been accelerated in order
that they reach P.” I understand this to be caused by Rp. However,
this still doesn’t make clear that Rp is the source of the I^2R power.
Only until we see the quote “…the total power lost by the electrons in
the interelectrode region is (equation 7.12), which is equal to the
heat power dissipated in the resistance (R) by the induced current.”
does the reduced power seem to have any relevance to Rp.

Now continuing the quote from Par 7.2, “The remaining part of the
electron’s kinetic energy is transformed into heat energy when they
strike electrode P.” It’s my understanding that this is the energy
that’s dissipated in the plate, causing it to heat.

However, the relationship between energy lost in the interelectrode
region and the energy dissipated in R seems to say that Rp is the
source of the energy dissipated in R, but it is difficult for me to
accept that relationship. I find it difficult to accept that a loss in
energy due to decelerating field in the interelectrode region could be
the source of the energy dissipated in R.

Richard, can you help me out here in understanding this concept, if
it’s really true? It seems to me that the loss of energy due to Rp is
simply energy that was never developed in the first place due to the
deceleration of the electric field in the interelectrode region.

Walt

On Mon, 31 May 2010 12:54:30 -0700 (PDT), walt wrote:

You say the source
of the I^2R power dissipated in R is plate resistance, Rp. However, as
I read Par 7.2 on Pages 140 and 141, the reality of Rp is described in
there, but only indirectly,

Hi Walt,

The point I have been emphasizing is that the plate resistance is
real, where real is the conventional meaning of heat following
current. The authors expresses this in all the conventional usages
and mathematical expressions of heat, velocity, kinetics, mass,
collision, current, voltage - everything you would find in an ordinary
resistor.

As for indirectness, no single section encapsulates the entire topic.

However, the remainder of the chapter does. Don't stop early to
examine the sidewalk to ignore the destination.

As so often happens in our newsgroup, the discussion of many topics
stalls at the freshman level and is regarded as complete in every
detail and nuance. "Physical Electronics," is far in advance of those
thoughts.

73's
Richard Clark, KB7QHC

On Jun 1, 12:31*pm, Richard Clark wrote:
On Mon, 31 May 2010 12:54:30 -0700 (PDT), walt wrote:
You say the source
of the I^2R power dissipated in R is plate resistance, Rp. However, as
I read Par 7.2 on Pages 140 and 141, the reality of Rp is described in
there, but only indirectly,

Hi Walt,

The point I have been emphasizing is that the plate resistance is
real, where real is the conventional meaning of heat following
current. *

There seems to be some confusion about the meaning of resistance in
these
posts.

At the start of my electronics education (and I suspect most of
yours), I
was taught that resistance was:
R = V/I

This works for resistors, which have a linear V-I curve which passes
through the origin. This R has the property that it can be used in
R*I^2 to compute the power dissipated in the resistor.

Later in my training, a more sophisticated definition of resistance
was
introduced:
R = deltaV/deltaI
where the V-I curve is not a straight line or does not pass through
the
origin (and some devices even have a negative resistance).
With this, more sophisticated definition of resistance, it is often
not
correct to use R in R*I^2 to compute the dissipation.

It is always correct to compute the dissipation by multiplying the
voltage
by the corresponding current at a particular point on the V-I curve
but
V/I at this point is not the resistance unless the V-I curve is
straight
and passes throught the origin.

The plate V-I characteristic curves for a tube are quite non-linear
and
the use of plate resistance (i.e. deltaV/deltaI) for the computation
of
power is an invalid operation.

V*I of V and I selected at any point on the curve will yield the
power,
but V/I will not be the resistance at this point. You need to dV/dI
for
that.

....Keith

On Wed, 2 Jun 2010 04:21:13 -0700 (PDT), Keith Dysart
wrote:

V*I of V and I selected at any point on the curve will yield the
power,
but V/I will not be the resistance at this point. You need to dV/dI
for
that.

Hi Keith,

You want to try that again with actual data?

Start with Walt's:
2 Finals 6146B

In this condition the DC plate voltage is 800 v and plate current is 260 ma. DC input power is therefore 800·V·0.26·A = 208·W.

73's
Richard Clark, KB7QHC

On Jun 2, 6:21*am, Keith Dysart wrote:
Later in my training, a more sophisticated definition of resistance
was introduced:

The "IEEE Dictionary" gives two definitions for "resistance". (A) a
dissipative resistance, and (B) a non-dissipative resistance. Then it
says: "Note: Definitions (A) and (B) are not equivalent but are
supplementary. In any case where confusion may arise, specify
definition being used." I believe that Walt has specified his
definition.
--
73, Cecil, w5dxp.com

On Jun 2, 10:07*am, Cecil Moore wrote:
On Jun 2, 6:21*am, Keith Dysart wrote:

Later in my training, a more sophisticated definition of resistance
was introduced:

The "IEEE Dictionary" gives two definitions for "resistance". (A) a
dissipative resistance, and (B) a non-dissipative resistance. Then it
says: "Note: Definitions (A) and (B) are not equivalent but are
supplementary. In any case where confusion may arise, specify
definition being used."

Somewhat orthogonal to the previous post, but reasonable definitions.

I believe that Walt has specified his definition.

Agreed.

....Keith