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 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

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

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

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.

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

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

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

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

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