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

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

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

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

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 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

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

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

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