Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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.

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

On Wed, 2 Jun 2010 20:17:41 -0700 (PDT), walt wrote:

And even though 'real',
Rp is non-dissipative, and therefore does not dissipate any power

Hi Walt,

The physical plate is in series with the load, the physical plate
conducts the current in the load, the physical plate and the load both
dissipate equal amounts of heat. I would presume the load is a
physical resistor just as the physical plate is a resistor.

Is that load resistor dissipating power?

For your logic to be consistent, then the dummy load does not
dissipate power, because it dissipates heat identical in amount and by
the same physical, atomic process as steel.

If you are saying the physical plate is not a "carbon" resistor, then
that is exceedingly specific and wholly original requirement that I
doubt you could cite any authority demanding. It would exact that only
the metal carbon qualifies and the metal steel does not when it comes
to source resistance.

As I already know that there are no references available for you to
cite, and this unique qualification is original to you, as your
hypothesis you have to defend it with data.

Where to begin?

73's
Richard Clark, KB7QHC

Plate Resistance
 

This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.

What's the resistance of the diode?

Is the diode's resistance V/I = 0.7/.01 = 70 ohms? Well, yes. The diode
is dissipating power, equal to V*I = 7 mW. If we removed the diode and
replaced it with a 70 ohm resistor, the voltage and current would be
exactly the same as before, and the resistor would dissipate exactly as
much as the diode. If the resistor's resistance is 70 ohms, then the
diode's resistance is surely 70 ohms also.

Is the diode's resistance "non-dissipative"? No, it dissipates exactly
as much as the equivalent resistor.

Now connect another current source across the biased diode, but make it
an AC source and connect it through a capacitor. Make the AC source
current 0.1 mA RMS, which is much less than the bias current of 10 mA.
What AC voltage should we expect? Well, it's not 0.1 mA * 70 ohms,
because 70 ohms is the DC resistance V/I, and the AC current is
encountering a resistance of dV/dI, where dV/dI is the ratio of the
*change* in voltage to the *change* in current. This is the dynamic, or
AC resistance, and it equals the slope of the V/I curve of the diode.
The AC and DC resistance are the same for a simple resistor, but not the
nonlinear diode. The dynamic resistance actually changes for each
instantaneous value of the AC waveform, but as long as the AC current
waveform is much less than the DC bias current, the change won't be
much. It turns out that the AC resistance for a diode is about 0.026/Idc
where Idc is the bias current. So in our case, the AC resistance is
about 0.026/0.01, or about 2.6 ohms. Therefore we'd see an AC voltage
across the diode of 0.1 mA * 2.6 = 0.26 volt.

The AC voltage is in phase with the AC current through the capacitor, so
the AC current source is supplying power to the diode, an amount equal
to Iac * Vac = 0.1 mA * 0.26 V = 0.026 mW. What happens to this power?
The answer is that the diode turns it into heat. Is the AC resistance
"dissipative"? Of course it is, it's dissipating the AC power supplied
by the AC current source. As before, we can replace the resistance, this
time the AC resistance, with a resistor -- we'll just have to connect it
through a capacitor so it doesn't see any of the DC bias. Now the
equivalent diode is a 70 ohm resistor with another 2.6 ohm resistor in
parallel through a capacitor. This will behave just the same as the
diode, with respect to voltage, current, and power dissipation, both AC
and DC, at the specified operating point.

The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Roy Lewallen, W7EL

Plate Resistance
 

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

Plate Resistance
 

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

Plate Resistance
 

On Thu, 3 Jun 2010 04:59:24 -0700 (PDT), Keith Dysart
wrote:

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

Hi Keith,

As demonstrated by Walt's data.

73's
Richard Clark, KB7QHC

Plate Resistance
 

Roy Lewallen wrote:
This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.

What's the resistance of the diode?

The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.



Another fine example is a "negative resistance" device like a tunnel
diode, gas discharge tube (neon bulb), or a free burning arc.

Obviously, the arc dissipates power, but the V/I slope is negative.

Plate Resistance
 

On Wed, 02 Jun 2010 22:01:13 -0700, Roy Lewallen
wrote:

The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Hi Roy,

That is a good analogy of power, non-linearity and dynamic resistance.

As for the DC operating point being muddled into the discussion, I
don't think I have seen anyone raise that separately - which is
perhaps what you mean.

For Walt, and all,

So, to take that part, and give this RF final stage a better view of
the dynamic resistance's part in source resistance, let's consider how
much power is dissipated in the plate that is strictly due to the bias
and a zero input condition.

There are few typical characteristics that match Walt's data, and I
have to abstract both from the TS520S schematics, service manual, and
the published data for the 6146B for this. As I had to juggle these
considerations to give Walt some perspective of the plate resistance
being far lower than his anticipated 5 kOhm to 14 kOhm, so I must
similarly reckon for the DC baseline.

By all things considered, we have a grid one bias of -60V. This
exceeds the RCA published data for class AB, but not for C (which is
in excess of this at -100V or more). My readings of all these
variables puts the plate current for a zero input at 20-25-30mA which
yields from something under 20W to something over 20W for a plate heat
load. Subtract this from the 53+W Keith figured and we have a ball
park 30W heat load attributable to RF. This is a back of the napkin
kind of balance sheet, mind you. I say this because the drive has a
cooling factor in that it drives down that zero input plate current
during part of a cycle. Having said that, it is enough to simply note
that there is significant heat dissipated in the plate that is
attributable to the RF current that passes through it and the load.

73's
Richard Clark, KB7QHC

Plate Resistance
 

On Jun 3, 10:39*am, Richard Clark wrote:
On Thu, 3 Jun 2010 04:59:24 -0700 (PDT), Keith Dysart

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

Hi Keith,

As demonstrated by Walt's data.

73's
Richard Clark, KB7QHC

Richard, you ask me where to begin. Well, I begin with Terman, where I
quote him in Chapter 19, Sec 19.3, in Reflections 2. Since attachments
are prohibited in this forum I'm going to try to copy Sec 19.3 here,
which explains in detail why plate resistance Rpd is dissipative,
while Rp is non-dissipative--it explains the difference between the
two. In addition, I don't believe the diode is a correct model for
describing the operation of a Class B or C amplifier. Now the quote:

Sec 19.3 Analysis of the Class C Amplifier

The following discussion of the Class C amplifier, which reveals
why the portion of the source resistance related to the
characteristics of the load line is non-dissipative, is based on
statements appearing in Terman’s Radio Engineers Handbook, 1943 ed.,
Page 445, and on Terman’s example of Class C amplifier design data
appearing on Page 449. Because the arguments presented in Terman’s
statements are vital to understanding the concept under discussion, I
quote them here for convenience: (Parentheses and emphasis mine)

1. The average of the pulses of current flowing to an
electrode represents the direct current drawn by that electrode.
2. The power input to the plate electrode of the tube at any
instant is the product of plate-supply voltage and instantaneous plate
current.
3. The corresponding power (Pd) lost at the plate is the product
of instantaneous plate-cathode voltage and instantaneous plate
current.
4. The difference between the two quantities obtained from items
2 and 3 represents the useful output at the moment.
5. The average input, output, and loss are obtained by averaging
the instantaneous powers.
6. The efficiency is the ratio of average output to average
input and is commonly of the order of 60 to 80 percent.
7. The efficiency is high in a Class C amplifier because
current is permitted to flow only when most of the plate-supply
voltage is used as voltage drop across the tuned load circuit RL, and
only a small fraction is wasted as voltage drop (across Rpd) at the
plate electrode of the tube.

Based on these statements the discussion and the data in
Terman’s example that follow explain why the amplifier can deliver
power with efficiencies greater than 50 percent while conjugately
matched to its load, a condition that is widely disputed because of
the incorrect assumptions concerning Class B and C amplifier operation
as noted above. The terminology and data in the example are Terman’s,
but I have added one calculation to Terman’s data to emphasize a
parameter that is vital to understanding how a conjugate match can
exist when the efficiency is greater than 50 percent. That parameter
is dissipative plate resistance Rpd. (As stated earlier, dissipative
resistance Rpd should not be confused with non-dissipative plate
resistance Rp of amplifiers operating in Class A, derived from the
expression Rp = delta Ep/delta Ip.)
It is evident from Terman that the power supplied to the
amplifier by the DC power supply goes to only two places, the RF power
delivered to load resistance RL at the input of the pi-network, and
the power dissipated as heat in dissipative plate resistance Rpd
(again, not plate resistance Rp, which is totally irrelevant to
obtaining a conjugate match at the output of Class B and C
amplifiers). In other words, the output power equals the DC input
power minus the power dissipated in resistance Rpd. We will now show
why this two-way division of power occurs. First we calculate the
value of Rpd from Terman’s data, as seen in line (9) in the example
below. It is evident that when the DC input power minus the power
dissipated in Rpd equals the power delivered to resistance RL at the
input of the pi-network, there can be no significant dissipative
resistance in the amplifier other than Rpd. The antenna effect from
the tank circuit is so insignificant that dissipation due to radiation
can be disregarded. If there were any significant dissipative
resistance in addition to Rpd, the power delivered to the load plus
the power dissipated in Rpd would be less than the DC input power, due
to the power that would be dissipated in the additional resistance.
This is an impossibility, confirmed by the data in Terman’s example,
which is in accordance with the Law of Conservation of Energy.
Therefore, we shall observe that the example confirms the total power
taken from the power supply goes only to 1) the RF power delivered to
the load RL, and 2) to the power dissipated as heat in Rpd, thus,
proving there is no significant dissipative resistance in the Class C
amplifier other than Rpd.

Data from Terman’s example on Page 449 of Radio Engineers Handbook:

(1) Eb = DC Source Voltage = 1000 v.
(2) Emin = Eb - EL = 1000 - 850 = 150 v. [See Terman, Figs 76(a) &
76(b)]
(3) Idc = DC Plate Current = 75.l ma. 0.0751a.
(4) EL = Eb - Emin = 1000 -150 = 850 v. = Peak Fundamental AC Plate
Voltage
(5) I1 = Peak Fundamental AC Plate Current = 132.7 ma. 0.1327 a.
(6) Pin = Eb x Idc = DC lnput Power = l000 x O.0751 = 75.l w.
(7) Pout (Eb - Emin)/2 = ELI1/2 = Output Power Delivered to RL =
[(1000 -150) x 0.1327]/2 = 56.4 w.
(8) Pd = Pin - Pout = Power Dissipated in Dissipative Plate Resistance
Rpd = 18.7 w,
(9) Rpd = 18.7W/0.0751^2 = Dissipative Plate Resistance Rpd = 3315.6
ohms
(10) RL = (Eb - Emin)/I1 = EL/I1 = Load Resistance = 850/0.137 = 6405
Ohms (6400 in Terman)
(11) Plate Efficiency = Pout x 100/Pin = 56.4 x 100/75.1 =
75.1%

Note that Terman doesn't even mention non-dissipative plate
resistance Rp, and therefore it cannot be considered the source
resistance.
Note also in line (10) that RL is determined simply by the ratio
of the fundamental RF AC voltage EL divided by the fundamental RF AC
current I1, and therefore does not involve dissipation of any power.
Thus RL is a non-dissipative resistance. (For more on non-dissipative
resistance see Appendix 10.)
Referring to the data in the example, observe again from line
(10) that load resistance RL at the input of the pi-network tank
circuit is determined by the ratio EL/I1. This is the Terman equation
which, prior to the more-precise Chaffee Fourier Analysis, was used
universally to determine the approximate value of the optimum load
resistance RL. (When the Chaffee Analysis is used to determine RL.
from a selected load line the value of plate current I1 is more
precise than that obtained when using Terman’s equation, consequently
requiring fewer empirical adjustments of the amplifier’s parameters to
obtain the optimum value of RL.) Load resistance RL is proportional to
the slope of the operating load line that allows all of the available
integrated energy contained in the plate-current pulses to be
transferred into the pi-network tank circuit. (For additional
information concerning the load line see Sec 19.3a below.) Therefore,
the pi-network must be designed to provide the equivalent optimum
resistance RL looking into the input for whatever load terminates the
output. The current pulses flowing into the network deliver bursts of
electrical energy to the network periodically, in the same manner as
the spring-loaded escapement mechanism in the pendulum clock delivers
mechanical energy periodically to the swing of the pendulum. In a
similar manner, after each plate current pulse enters the pi-network
tank curcuit, the flywheel effect of the resonant tank circuit stores
the electromagnetic energy delivered by the current pulse, and thus
maintains a continuous sinusoidal flow of current throughout the tank,
in the same manner as the pendulum swings continuously and
periodically after each thrust from the escapement mechanism. The
continuous swing of the pendulum results from the inertia of the
weight at the end of the pendulum, due to the energy stored in the
weight. The path inscribed by the motion of the pendulum is a sine
wave, the same as at the output of the amplifier. We will continue the
discussion of the flywheel effect in the tank circuit with a more in-
depth examination later.
Let us now consider the dissipative plate resistance Rpd, which
provides the evidence that the DC input power to the Class C amplifier
goes only to the load RL and to dissipation as heat in Rpd (Again, not
Rp.) With this evidence we will show how a conjugate match can exist
at the output of the pi-network with efficiencies greater than 50
percent. In accordance with the Conjugate Matching Theorem and the
Maximum Power-transfer Theorem, it is well understood that a conjugate
match exists whenever all available power from a linear source is
being delivered to the load. Further, by definition, RL is the load
resistance at the tank input determined by the characteristics of the
load line that permits delivery of all the available power from the
source into the tank. This is why RL is called the optimum load
resistance. Thus, from the data in Terman’s example, which shows that
after accounting for the power dissipated in Rpd, all the power
remaining is the available power, which is delivered to RL and thence
to the load at the output of the pi-network. Therefore, because all
available deliverable power is being delivered to the load, we have a
conjugate match by definition. In the following Sec 4 we will show
how efficiencies greater than 50 percent are achieved in Class C
amplifiers operating into the conjugate match.

Walt, W2DU

Plate Resistance
 

On Thu, 3 Jun 2010 12:08:42 -0700 (PDT), walt wrote:

Data from Terman’s example on Page 449 of Radio Engineers Handbook:

(1) Eb = DC Source Voltage = 1000 v.
From your own data
Eb = DC Source Voltage = 800 v.
(2) Emin = Eb - EL = 1000 - 850 = 150 v. [See Terman, Figs 76(a) &
76(b)]
You do not supply context for me to apply to your data.
(3) Idc = DC Plate Current = 75.l ma. 0.0751a.
Idc = 260mA
(4) EL = Eb - Emin = 1000 -150 = 850 v. = Peak Fundamental AC Plate
Voltage
You do not supply your minimum nor maximum plate voltage swing.
(5) I1 = Peak Fundamental AC Plate Current = 132.7 ma. 0.1327 a.
You do not supply your peak AC Plate current.
(6) Pin = Eb x Idc = DC lnput Power = l000 x O.0751 = 75.l w.
Pin = 208W
(7) Pout (Eb - Emin)/2 = ELI1/2 = Output Power Delivered to RL =
[(1000 -150) x 0.1327]/2 = 56.4 w.
You report Pout, but we cannot use this formula for lack of data.
Pout = 100W
(8) Pd = Pin - Pout = Power Dissipated in Dissipative Plate Resistance
Rpd = 18.7 w,
From your report of Pin and Pout:
Pd = 108W
(9) Rpd = 18.7W/0.0751^2 = Dissipative Plate Resistance Rpd = 3315.6
ohms
From what is reported by you:
Rpd = 108W/(.260mA)²
Rpd = 1597 Ohms
(10) RL = (Eb - Emin)/I1 = EL/I1 = Load Resistance = 850/0.137 = 6405
Ohms (6400 in Terman)
You do not report Emin but you report Load Resistance
RL = 50 Ohms
(11) Plate Efficiency = Pout x 100/Pin = 56.4 x 100/75.1 =
75.1%
Plate Efficiency = 48%

By all reckoning according to your reference from Terman, using what
you report, it appears that you have exhibited a Conjugate basis Z
match as you claim, and that the plate Rpd by the same reckoning is
the same as we formerly arrived at Rp. It would appear that over the
course of some dozen years between publications that Terman simplified
the term Rpd to Rp which, according to you, is not found in your
volume, and as such this migration of terms seems logical by the
numbers agreeing in both volumes for different labels. Inasmuch as
Rpd does not appear in Terman's later work, nor in any of the Tube
specifications since that era, Rpd appears to be an orphan.

You report your own Rp (now Rpd) or its equivalent by resistor
substitution (a valid determination) to be on order of 1400 Ohms. This
conforms closely to the value found above, and the data reported by
RCA for Rp in a design of similar characteristics. The range of
possible values taken from RCA: 900 Ohms to 1500 Ohms.

Taking your low, the high from my range, and the higher computed
through your supplied data using Terman's formula, the average is 1500
Ohms with a variation of roughly 6% which is about the limits of
superlative accuracy for conventional bench equipment. What this
means is that all three values are identical on the basis of
accumulation of error.

Having said that, this is not the end of analysis. However, at the
first pass your data has demonstrated that the Plate serves as a real
resistor dissipating half of the available power being supplied to a
load that is conjugately matched.

The only difference is that you state as much, but dismiss the plate
dissipation as being a real resistor. Perhaps I mis-state you. If
not, I take issue with that and ask once again, as this perception is
unique to your hypothesis, do you have data that differentiates the
resistance of steel absorbing and dissipating the impact of the
electron stream as being different from carbon absorbing and
dissipating the impact of the electron stream?

Would it serve to replace the steel plate with a graphite one like the
845 (similar to the GM-70)? When I review the spec sheets for this
tube, it reports an Rp of 1700 Ohms - hardly remarkable at 100 more
Ohms than the computation above. Of course, there are compounding
application differences, but still, graphite plates or steel don't
seem to force a new conclusion. Materials don't seem to be an issue.

Certainly the mechanisms of resistance differ in the kinetics of
speed, but not in the product of heat. To my knowledge, no authority
bases the concept of resistance upon the speed of the electron, but
rather in the kinetics of its collision.

73's
Richard Clark, KB7QHC

Plate Resistance
 

On Jun 3, 8:14*pm, Richard Clark wrote:
On Thu, 3 Jun 2010 12:08:42 -0700 (PDT), walt wrote:
Data from Terman’s example on Page 449 of Radio Engineers Handbook:

(1) Eb = DC Source Voltage = 1000 v.

From your own data
Eb = DC Source Voltage = 800 v.(2) Emin = Eb - EL = 1000 - 850 = 150 v. [See Terman, Figs 76(a) &
76(b)]

You do not supply context for me to apply to your data.(3) Idc = DC Plate Current = 75.l ma. 0.0751a.
Idc = 260mA
(4) EL = Eb - Emin = 1000 -150 = 850 v. = Peak Fundamental AC Plate
Voltage

You do not supply your minimum nor maximum plate voltage swing.(5) I1 = Peak Fundamental AC Plate Current = 132.7 ma. 0.1327 a.

You do not supply your peak AC Plate current.(6) Pin = Eb x Idc = DC lnput Power = l000 x O.0751 = 75.l w.
Pin = 208W
(7) Pout (Eb - Emin)/2 = ELI1/2 = Output Power Delivered to RL =
[(1000 -150) x 0.1327]/2 = 56.4 w.

You report Pout, but we cannot use this formula for lack of data.
Pout = 100W(8) Pd = Pin - Pout = Power Dissipated in Dissipative Plate Resistance
Rpd = 18.7 w,

From your report of Pin and Pout:
Pd = 108W(9) Rpd = 18.7W/0.0751^2 = Dissipative Plate Resistance Rpd = 3315.6
ohms

From what is reported by you:
Rpd = 108W/(.260mA)²
Rpd = 1597 Ohms(10) RL = (Eb - Emin)/I1 = EL/I1 = Load Resistance = 850/0.137 = 6405
Ohms (6400 in Terman)

You do not report Emin but you report Load Resistance
RL = 50 Ohms(11) Plate Efficiency = Pout x 100/Pin = 56.4 x 100/75..1 =
75.1%

Plate Efficiency = 48%

By all reckoning according to your reference from Terman, using what
you report, it appears that you have exhibited a Conjugate basis Z
match as you claim, and that the plate Rpd by the same reckoning is
the same as we formerly arrived at Rp. *It would appear that over the
course of some dozen years between publications that Terman simplified
the term Rpd to Rp which, according to you, is not found in your
volume, and as such this migration of terms seems logical by the
numbers agreeing in both volumes for different labels. *Inasmuch as
Rpd does not appear in Terman's later work, nor in any of the Tube
specifications since that era, Rpd appears to be an orphan.

You report your own Rp (now Rpd) or its equivalent by resistor
substitution (a valid determination) to be on order of 1400 Ohms. This
conforms closely to the value found above, and the data reported by
RCA for Rp in a design of similar characteristics. *The range of
possible values taken from RCA: 900 Ohms to 1500 Ohms.

Taking your low, the high from my range, and the higher computed
through your supplied data using Terman's formula, the average is 1500
Ohms with a variation of roughly 6% which is about the limits of
superlative accuracy for conventional bench equipment. *What this
means is that all three values are identical on the basis of
accumulation of error.

Having said that, this is not the end of analysis. *However, at the
first pass your data has demonstrated that the Plate serves as a real
resistor dissipating half of the available power being supplied to a
load that is conjugately matched. *

The only difference is that you state as much, but dismiss the plate
dissipation as being a real resistor. *Perhaps I mis-state you. *If
not, I take issue with that and ask once again, as this perception is
unique to your hypothesis, do you have data that differentiates the
resistance of steel absorbing and dissipating the impact of the
electron stream as being different from carbon absorbing and
dissipating the impact of the electron stream?

Would it serve to replace the steel plate with a graphite one like the
845 (similar to the GM-70)? *When I review the spec sheets for this
tube, it reports an Rp of 1700 Ohms - hardly remarkable at 100 more
Ohms than the computation above. *Of course, there are compounding
application differences, but still, graphite plates or steel don't
seem to force a new conclusion. *Materials don't seem to be an issue.

Certainly the mechanisms of resistance differ in the kinetics of
speed, but not in the product of heat. *To my knowledge, no authority
bases the concept of resistance upon the speed of the electron, but
rather in the kinetics of its collision.

73's
Richard Clark, KB7QHC

Richard, I'm totally shocked by what I've read above. I can't believe
it! I can't believe you've distorted what I've written in my last post
above to the extent that I can't possibly clarify or correct it--it's
more than just misunderstanding. In addition, it's totally misleading
to other readers of this thread, and makes me appear as a moron and an
idiot.

Sorry, Richard, I'm through with this thread. There's nothing I can do
now to fix the situation.

Walt, W2DU

Plate Resistance
 

On Jun 2, 10:01*pm, Roy Lewallen wrote:
This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.
....
The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Roy Lewallen, W7EL

I'm happy to see that Roy posted this clarification about dynamic
versus static resistance. I haven't read all the posts here, but
those others that I have seemed to completely skirt the issue of the
meaning of "plate resistance."

Further to Roy's posting, though, you must realize that a device with
more than two terminals is complicated by the fact that you need to
specify that the dynamic resistance is measured with other terminals
held constant. So, for example, plate resistance of a tube is
normally defined as the partial derivative of plate voltage with
respect to plate current, with the grid-to-cathode voltage held
constant, and of course at some particular plate voltage (and
corresponding current). For a tetrode, the screen-to-cathode voltage
must also be constant.

The reason for this is that you will get completely different answers
if you don't hold the voltage of the grids constant with respect to
the cathode. For example, if you set up a circuit with a tube with
the grid grounded, and an appropriate resistor between the cathode and
ground (not bypassed), to establish a desired bias point, then a
change in plate current will be accompanied by an essentially
identical change in cathode current, and a corresponding change in
drop across the cathode resistor, which represents a change in grid-to-
cathode voltage. Measuring dV/dI at the plate in such a circuit will
yield a much higher resistance than if the grid-to-cathode voltage is
kept constant. It is, in fact, a good way to make a constant current
source.

There is no particular relationship that must be maintained between
the load resistance you present to an amplifier element -- a tube or a
bipolar transistor or a FET -- and the plate/collector/drain (dynamic)
resistance of the device. For example, it's common to operate a push-
pull pair of 6146's in audio service with a load around 5000 ohms
(plate to plate) at 500V plate supply voltage, 185V screen voltage,
and perhaps 30mA plate current. That's effectively 2500 ohms to each
plate of the pair. But at that operating point, the plate resistance
of a 6146 is around 600V/30mA, or 20k ohms (roughly, from plate
characteristic curves in my old RCA transmitting tubes manual). On
the other hand, a triode will have a much lower plate resistance,
likely similar to the load resistance--but again, not really related
to it.

Another rather interesting thing happens, though, when you connect an
RF load through a tank circuit such as a PI network. It's quite
possible for a pi network to transform a 50 ohm RF load so it presents
a reasonable (say) 4k ohm RF load to the plate of a tube like a 6146.
But that same pi network will in turn transform the plate resistance--
or rather, the net impedance of the tube in its particular circuit
configuration (grounded grid or grounded cathode, and invariable some
amount of feedback whether you wanted it or not), along with the DC
feed (RF choke) in parallel, to quite possibly an impedance not very
close to 50 ohms, and likely rather reactive. I can, without much
difficulty, design a PI network that will yield a source resistance at
the output connector that's a lot less than 50 ohms, using a tetrode
amplifier tube, while giving a very decent transformation of a 50 ohm
load to a desired load at the plate of the tube.

And as if that weren't enough, I can modify the output impedance
further by application of feedback; this is more commonly done at
audio frequencies, where amplifiers designed to drive loads like 4
ohms an 8 ohms have output impedances in the area of a small fraction
of an ohm. But it's also done with RF amplifiers, often for reasons
unrelated to the output impedance, but also to control the output
impedance so that it IS very close to 50 ohms (for example in
instrumentation, where it may be important).

Cheers,
Tom

Plate Resistance
 

On Thu, 3 Jun 2010 18:15:12 -0700 (PDT), walt wrote:

I can't believe you've distorted what I've written in my last post

Hi Walt,

Can you give an example?

73's
Richard Clark, KB7QHC

Plate Resistance
 

On Jun 2, 3:43*pm, Keith Dysart wrote:
....
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

Indeed, Keith...nor is it much use, by itself, for computing the
source impedance seen looking back into amplifier's output RF
connector, at the output end of the network that transforms a (50 ohm,
or whatever) load to the impedance seen by the tubes driving that
load.

If the tube is operated with cathode (RF) grounded and driven with a
stiff voltage source, the impedance looking back into the plate side
will be the plate resistance, shunted by stray and tube capacitances;
but put a bit of unbypassed cathode resistance in there and the
resistive part seen at the plate will be higher, possibly quite a bit
higher. Use a triode instead of a pentode, and the plate resistance
will be much lower, even though the optimal load presented to the
plate may well be quite similar to the optimal load presented to a
tetrode. Run the tube grounded-grid and the impedance seen looking
back into the plate will be much higher than if it's run grounded-
cathode.

I think it helps to understand that the "optimal" load presented to
the tube depends on what you want to accomplish: you want a
reasonable amount of RF output power with reasonable efficiency and
tube operating conditions that will yield good tube life (normally, in
our circles). If it's supposed to be a linear amplifier, you want to
keep (intermodulation) distortion products below some level. All that
has very little to do with the impedance seen by the load looking back
into the tube.

Of course, essentially the same situations exist with transistors and
FETs as with tubes.

Cheers,
Tom

Plate Resistance
 

There was a calculation error in my recent posting. Thanks to Tom, K7ITM
for spotting it and letting me know. It doesn't affect the conclusions.

Roy Lewallen wrote:
. . .
Now connect another current source across the biased diode, but make it
an AC source and connect it through a capacitor. Make the AC source
current 0.1 mA RMS, which is much less than the bias current of 10 mA.
What AC voltage should we expect? Well, it's not 0.1 mA * 70 ohms,
because 70 ohms is the DC resistance V/I, and the AC current is
encountering a resistance of dV/dI, where dV/dI is the ratio of the
*change* in voltage to the *change* in current. This is the dynamic, or
AC resistance, and it equals the slope of the V/I curve of the diode.
The AC and DC resistance are the same for a simple resistor, but not the
nonlinear diode. The dynamic resistance actually changes for each
instantaneous value of the AC waveform, but as long as the AC current
waveform is much less than the DC bias current, the change won't be
much. It turns out that the AC resistance for a diode is about 0.026/Idc
where Idc is the bias current. So in our case, the AC resistance is
about 0.026/0.01, or about 2.6 ohms. Therefore we'd see an AC voltage
across the diode of 0.1 mA * 2.6 = 0.26 volt.

The AC voltage across the diode is 0.26 mV, not 0.26 V.

The AC voltage is in phase with the AC current through the capacitor, so
the AC current source is supplying power to the diode, an amount equal
to Iac * Vac = 0.1 mA * 0.26 V = 0.026 mW. . .

The power supplied by the AC current source to the diode is 0.1 mA *
0.26 mV = 0.026 uW = 26 nW, not 0.026 mW. This is, of course, also Iac^2
* Rac = (0.1 mA)^2 * 2.6.

Roy Lewallen, W7EL

Plate Resistance
 

On Jun 4, 12:38*am, K7ITM wrote:
On Jun 2, 10:01*pm, Roy Lewallen wrote:



This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.
...
The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Roy Lewallen, W7EL

I'm happy to see that Roy posted this clarification about dynamic
versus static resistance. *I haven't read all the posts here, but
those others that I have seemed to completely skirt the issue of the
meaning of "plate resistance."

Further to Roy's posting, though, you must realize that a device with
more than two terminals is complicated by the fact that you need to
specify that the dynamic resistance is measured with other terminals
held constant. *So, for example, plate resistance of a tube is
normally defined as the partial derivative of plate voltage with
respect to plate current, with the grid-to-cathode voltage held
constant, and of course at some particular plate voltage (and
corresponding current). *For a tetrode, the screen-to-cathode voltage
must also be constant.

The reason for this is that you will get completely different answers
if you don't hold the voltage of the grids constant with respect to
the cathode. *For example, if you set up a circuit with a tube with
the grid grounded, and an appropriate resistor between the cathode and
ground (not bypassed), to establish a desired bias point, then a
change in plate current will be accompanied by an essentially
identical change in cathode current, and a corresponding change in
drop across the cathode resistor, which represents a change in grid-to-
cathode voltage. *Measuring dV/dI at the plate in such a circuit will
yield a much higher resistance than if the grid-to-cathode voltage is
kept constant. *It is, in fact, a good way to make a constant current
source.

There is no particular relationship that must be maintained between
the load resistance you present to an amplifier element -- a tube or a
bipolar transistor or a FET -- and the plate/collector/drain (dynamic)
resistance of the device. *For example, it's common to operate a push-
pull pair of 6146's in audio service with a load around 5000 ohms
(plate to plate) at 500V plate supply voltage, 185V screen voltage,
and perhaps 30mA plate current. *That's effectively 2500 ohms to each
plate of the pair. *But at that operating point, the plate resistance
of a 6146 is around 600V/30mA, or 20k ohms (roughly, from plate
characteristic curves in my old RCA transmitting tubes manual). *On
the other hand, a triode will have a much lower plate resistance,
likely similar to the load resistance--but again, not really related
to it.

Another rather interesting thing happens, though, when you connect an
RF load through a tank circuit such as a PI network. *It's quite
possible for a pi network to transform a 50 ohm RF load so it presents
a reasonable (say) 4k ohm RF load to the plate of a tube like a 6146.
But that same pi network will in turn transform the plate resistance--
or rather, the net impedance of the tube in its particular circuit
configuration (grounded grid or grounded cathode, and invariable some
amount of feedback whether you wanted it or not), along with the DC
feed (RF choke) in parallel, to quite possibly an impedance not very
close to 50 ohms, and likely rather reactive. *I can, without much
difficulty, design a PI network that will yield a source resistance at
the output connector that's a lot less than 50 ohms, using a tetrode
amplifier tube, while giving a very decent transformation of a 50 ohm
load to a desired load at the plate of the tube.

And as if that weren't enough, I can modify the output impedance
further by application of feedback; this is more commonly done at
audio frequencies, where amplifiers designed to drive loads like 4
ohms an 8 ohms have output impedances in the area of a small fraction
of an ohm. *But it's also done with RF amplifiers, often for reasons
unrelated to the output impedance, but also to control the output
impedance so that it IS very close to 50 ohms (for example in
instrumentation, where it may be important).

Cheers,
Tom

I'm going to break with my earlier decision to no longer respond to
this thread in order to respond to Tom, K7ITM.

Tom, as you can see from my earlier posts I agree with your assessment
of plate resistance, Rp, in that it is the partial derivative of the
Ep/Ip relationship. Therefore, while Rp is definitely a factor in
determining the value of load resistance RL, it is by no means the
total source resistance of the RF power amplifier, as others insist.

Walt Maxwell, W2DU

Plate Resistance
 

On Jun 4, 12:38*am, K7ITM wrote:
On Jun 2, 10:01*pm, Roy Lewallen wrote:



This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.
...
The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Roy Lewallen, W7EL

I'm happy to see that Roy posted this clarification about dynamic
versus static resistance. *I haven't read all the posts here, but
those others that I have seemed to completely skirt the issue of the
meaning of "plate resistance."

Further to Roy's posting, though, you must realize that a device with
more than two terminals is complicated by the fact that you need to
specify that the dynamic resistance is measured with other terminals
held constant. *So, for example, plate resistance of a tube is
normally defined as the partial derivative of plate voltage with
respect to plate current, with the grid-to-cathode voltage held
constant, and of course at some particular plate voltage (and
corresponding current). *For a tetrode, the screen-to-cathode voltage
must also be constant.

The reason for this is that you will get completely different answers
if you don't hold the voltage of the grids constant with respect to
the cathode. *For example, if you set up a circuit with a tube with
the grid grounded, and an appropriate resistor between the cathode and
ground (not bypassed), to establish a desired bias point, then a
change in plate current will be accompanied by an essentially
identical change in cathode current, and a corresponding change in
drop across the cathode resistor, which represents a change in grid-to-
cathode voltage. *Measuring dV/dI at the plate in such a circuit will
yield a much higher resistance than if the grid-to-cathode voltage is
kept constant. *It is, in fact, a good way to make a constant current
source.

There is no particular relationship that must be maintained between
the load resistance you present to an amplifier element -- a tube or a
bipolar transistor or a FET -- and the plate/collector/drain (dynamic)
resistance of the device. *For example, it's common to operate a push-
pull pair of 6146's in audio service with a load around 5000 ohms
(plate to plate) at 500V plate supply voltage, 185V screen voltage,
and perhaps 30mA plate current. *That's effectively 2500 ohms to each
plate of the pair. *But at that operating point, the plate resistance
of a 6146 is around 600V/30mA, or 20k ohms (roughly, from plate
characteristic curves in my old RCA transmitting tubes manual). *On
the other hand, a triode will have a much lower plate resistance,
likely similar to the load resistance--but again, not really related
to it.

Another rather interesting thing happens, though, when you connect an
RF load through a tank circuit such as a PI network. *It's quite
possible for a pi network to transform a 50 ohm RF load so it presents
a reasonable (say) 4k ohm RF load to the plate of a tube like a 6146.
But that same pi network will in turn transform the plate resistance--
or rather, the net impedance of the tube in its particular circuit
configuration (grounded grid or grounded cathode, and invariable some
amount of feedback whether you wanted it or not), along with the DC
feed (RF choke) in parallel, to quite possibly an impedance not very
close to 50 ohms, and likely rather reactive. *I can, without much
difficulty, design a PI network that will yield a source resistance at
the output connector that's a lot less than 50 ohms, using a tetrode
amplifier tube, while giving a very decent transformation of a 50 ohm
load to a desired load at the plate of the tube.

And as if that weren't enough, I can modify the output impedance
further by application of feedback; this is more commonly done at
audio frequencies, where amplifiers designed to drive loads like 4
ohms an 8 ohms have output impedances in the area of a small fraction
of an ohm. *But it's also done with RF amplifiers, often for reasons
unrelated to the output impedance, but also to control the output
impedance so that it IS very close to 50 ohms (for example in
instrumentation, where it may be important).

Cheers,
Tom

I'm going to break with my earlier decision to no longer respond to
this thread in order to respond to Tom, K7ITM.

Tom, if you review my earlier posts, you'll find that I agree with you
concerning plate resistance Rp as the partial derivative of the Ep/Ip
relationship when both the grid and screen voltages are held constant.
Although Rp is definitely a factor in determining the value of load
resistance RL, it is by no means the total source resistance RL, as
others have insisted.

Walt Maxwell, W2DU

Plate Resistance
 

On Fri, 4 Jun 2010 11:11:02 -0700 (PDT), walt wrote:

Although Rp is definitely a factor in determining the value of load
resistance RL, it is by no means the total source resistance RL, as
others have insisted.

Hi Walt,

Do you have other Data to replace into your copy of Terman's equation?
I got the impression you were satisfied with his work and his
conclusions. Myself, I would say Terman's are suitable but incomplete
and additional data wouldn't change anything as he presented a classic
case.

Hemenway, Henry, and Caulton go miles beyond merely suggesting that
there are partial derivative forms (you can choose to offer your own
from Terman if you can find them), and they give the full treatment
from soup to nuts in chapters outside of the one I have distributed.

As for the comparison of Rp to RL, yes, it is not complete and in fact
although they closely agree, that is only a first cut. Refinement may
go as far as 10s of percent shifts and with the accumulation of
possible errors, that shift might still resolve to Rp = RL.

However, if Rp (or its distant cousin Rpd) slips in value away from
RL, then the match is going to migrate away from the Conjugate basis Z
match towards some other solution. I cannot see how it could be
otherwise unless there were some undocumented and unmeasurable R (not
real) were to combine to balance the books.

If there's a match, the Data and the numerous counter arguments to you
seem to prefer an image basis Z match. I don't hold much confidence
in that outcome either.

73's
Richard Clark, KB7QHC