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, 9:52*am, Richard Clark wrote:
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.


Yup. Measure 800V at 0.26A and the power is 208 W. No mention of
resistance. This is good.

From Walt's data indicating 100W and the schematic showing a 5 ohm
cathode resistor, one
might infer that each tube is dissipating (208-100-(5*.26))/2-53.35
W. Feel free to include
the heater and grid dissipations for a more accurate evaluation.

From the 8146B spec sheet, depending on the operating point, Rp (i.e.
slope of plate E-I curve)
is between 10k and 30k so for the two tubes in the circuit it would be
5k to 15k ohms. But this
number is not much use for computing dissipations.

....Keith

On Wed, 2 Jun 2010 15:43:37 -0700 (PDT), Keith Dysart
wrote:

On Jun 2, 9:52*am, Richard Clark wrote:
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.


Yup. Measure 800V at 0.26A and the power is 208 W. No mention of
resistance. This is good.

From Walt's data indicating 100W and the schematic showing a 5 ohm
cathode resistor, one
might infer that each tube is dissipating (208-100-(5*.26))/2-53.35W.

Hi Keith,

Each tube is dissipating 53.35W? What a fire-eater! Each tube in an
RF application is rated for between 18W to 35 Watts maximum. However,
those maximum values are continous service, not tune up. Further, Ham
usage with long duty cycles would average the plate dissipation over
the long haul to well below 50+ W.

When I review the RCA specifications for a 6146B, published Feb 1964,
I find that both Walt's Plate voltage and Plate current exceed
absolute maximum ratings of (depending upon service) 750V at 250mA.
However, this would seem to be tolerably within generous meter
inaccuracy (if I were to peg it as low as 5%). Also, the 250mA meter
mark is an ad-hoc adjustment set at a condition of maximum power, not
by independantly confirming the actual current. That is of little
consequence to me and Walt's numbers arrived at straight from the
Kenwood service manual are suitable.

However, when we return to the "dissipation," through both RF and
Heat; then we are very close to the classic 50% efficiency of a
matched (by Conjugate Z basis) source.

The heat from the plates satisfy every physical interpretation of a
classic resistor, and an ordinary Ham transmitter exhibits what is
classically called source resistance.

*********

So far, all of Walt's data and extrapolations of R are right on. I
have seen similar reports through alternative methods seeking the same
determination that agree. However, I don't see how he then jumps the
tracks to hedge R not being real and being borne by the plate of the
tube.

73's
Richard Clark, KB7QHC

On Jun 2, 9:36*pm, Richard Clark wrote:
On Wed, 2 Jun 2010 15:43:37 -0700 (PDT), Keith Dysart

wrote:
On Jun 2, 9:52 am, Richard Clark wrote:
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.

Yup. Measure 800V at 0.26A and the power is 208 W. No mention of
resistance. This is good.

From Walt's data indicating 100W and the schematic showing a 5 ohm
cathode resistor, one
might infer that each tube is dissipating (208-100-(5*.26))/2-53.35W.

Hi Keith,

Each tube is dissipating 53.35W? * What a fire-eater! *

Those 108 W did have to go somewhere.

Each tube in an
RF application is rated for between 18W to 35 Watts maximum. *However,
those maximum values are continous service, not tune up. *Further, Ham
usage with long duty cycles would average the plate dissipation over
the long haul to well below 50+ W.

When I review the RCA specifications for a 6146B, published Feb 1964,
I find that both Walt's Plate voltage and Plate current exceed
absolute maximum ratings of (depending upon service) 750V at 250mA.
However, this would seem to be tolerably within generous meter
inaccuracy (if I were to peg it as low as 5%). *Also, the 250mA meter
mark is an ad-hoc adjustment set at a condition of maximum power, not
by independantly confirming the actual current. *That is of little
consequence to me and Walt's numbers arrived at straight from the
Kenwood service manual are suitable.

However, when we return to the "dissipation," through both RF and
Heat; then we are very close to the classic 50% efficiency of a
matched (by Conjugate Z basis) source.
The heat from the plates satisfy every physical interpretation of a
classic resistor,

Every? A classis resistor satisfies the equations
R = V/I = deltaV/Delta/I
Given the V/I curves published for the 6146B, I see no reason to
expect
the plate resistance to satisfy these two equations.

and an ordinary Ham transmitter exhibits what is
classically called source resistance.

*********

So far, all of Walt's data and extrapolations of R are right on. *I
have seen similar reports through alternative methods seeking the same
determination that agree. *However, I don't see how he then jumps the
tracks to hedge R not being real and being borne by the plate of the
tube.

But what exactly do you mean by 'real'?
R=V/I?
R=deltaV/deltaI?
Heat is dissipated?

....Keith

On Jun 2, 6:43*pm, Keith Dysart wrote:
On Jun 2, 9:52*am, Richard Clark wrote:

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.

Yup. Measure 800V at 0.26A and the power is 208 W. No mention of
resistance. This is good.

From Walt's data indicating 100W and the schematic showing a 5 ohm
cathode resistor, one
might infer that each tube is dissipating (208-100-(5*.26))/2-53.35
W. Feel free to include
the heater and grid dissipations for a more accurate evaluation.

From the 8146B spec sheet, depending on the operating point, Rp (i.e.
slope of plate E-I curve)
is between 10k and 30k so for the two tubes in the circuit it would be
5k to 15k ohms. But this
number is not much use for computing dissipations.

...Keith

Keith, I measured 1400 ohms at the input of the pi-network. The output
power was 100 w at 50 ohms, meaning an output voltage of 50 v and
current 1.414 a. The ratio of output to input resistances of the pi-
network is 28, meaning the voltage at the network input is 374.17 v.
Thus, neglecting the small loss in the network, the input current to
the network is 0.267 a. and the power entering the network is 100 w.

I prefer to consider the source resistance of the power amp to be at
the output of the network, which, when adjusted to deliver 100 w into
the 50 + j0 load, the source resistance is 50 ohms. On the other hand,
if you wish to consider the source resistance at the input of the
network, then it's 1400 ohms, not somewhere between 5k and 15k ohms.

Consequently, if Rp is between 5k and 15k, Rp cannot be considered the
source resistance of the amp. From this info I still believe that Rp
is not the source resistance of an RF power amp.

I understand that the efficiency of the amp is determined by the
difference between plate power input output power, which is the power
lost to dissipation in the plate that is 108 w in the example I
presented. Efficiency does not include filament and grid powers.

Walt, W2DU

On Jun 2, 10:01*pm, walt wrote:
On Jun 2, 6:43*pm, Keith Dysart wrote:



On Jun 2, 9:52*am, Richard Clark wrote:

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.

Yup. Measure 800V at 0.26A and the power is 208 W. No mention of
resistance. This is good.

From Walt's data indicating 100W and the schematic showing a 5 ohm
cathode resistor, one
might infer that each tube is dissipating (208-100-(5*.26))/2-53.35
W. Feel free to include
the heater and grid dissipations for a more accurate evaluation.

From the 8146B spec sheet, depending on the operating point, Rp (i.e.
slope of plate E-I curve)
is between 10k and 30k so for the two tubes in the circuit it would be
5k to 15k ohms. But this
number is not much use for computing dissipations.

...Keith

Keith, I measured 1400 ohms at the input of the pi-network. The output
power was 100 w at 50 ohms, meaning an output voltage of 50 v and
current 1.414 a. The ratio of output to input resistances of the pi-
network is 28, meaning the voltage at the network input is 374.17 v.
Thus, neglecting the small loss in the network, the input current to
the network is 0.267 a. and the power entering the network is 100 w.

I prefer to consider the source resistance of the power amp to be at
the output of the network, which, when adjusted to deliver 100 w into
the 50 + j0 load, the source resistance is 50 ohms. On the other hand,
if you wish to consider the source resistance at the input of the
network, then it's 1400 ohms, not somewhere between 5k and 15k ohms.

Consequently, if Rp is between 5k and 15k, Rp cannot be considered the
source resistance of the amp. From this info I still believe that Rp
is not the source resistance of an RF power amp.

I understand that the efficiency of the amp is determined by the
difference between plate power input output power, which is the power
lost to dissipation in the plate that is 108 w in the example I
presented. Efficiency does not include filament and grid powers.

Walt, W2DU

Richard, perhaps our disagreement on Rp is only in a misunderstanding
concerning 'real'. Contrary to what you believe I said, I consider
resistance Rp as real, but not as a resisTOR. And even though 'real',
Rp is non-dissipative, and therefore does not dissipate any power--it
only represents power that is not developed in the first place by
causing a reduction of the plate voltage developed across the load
current increases--so how could it be 'source' of the power delivered
to a load? In the absence of Rp the output power would be greater by
the amount lost by the presence of Rp.

Walt

On Wed, 2 Jun 2010 19:01:17 -0700 (PDT), walt wrote:

On the other hand,
if you wish to consider the source resistance at the input of the
network, then it's 1400 ohms, not somewhere between 5k and 15k ohms.

Hi Walt,

If I simply look at the spec sheet for the 6146, and note that in
push-pull it has a plate-to-plate Load Resistance of 3600 Ohms to 6050
Ohms (depending upon many variables); and if I simple reconfigure the
two tubes from series to parallel; then the Load Resistance falls
somewhere in the range of 900 Ohms to 1500 Ohms.

I am having a hard time trying to tease out how you would explain that
the plates do not dissipate their half of the power when your numbers
demonstrate every conventional expectation of literal and physical
source resistance.

Do you have data for the peak-to-peak voltage swings at the grid and
plate for the final?

73's
Richard Clark, KB7QHC