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

Hi Walt,

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

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

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

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

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

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

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

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

....Keith

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

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

Hi Keith,

You want to try that again with actual data?

Start with Walt's:
2 Finals 6146B

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

73's
Richard Clark, KB7QHC

On Jun 2, 9:52*am, Richard Clark wrote:
Start with Walt's:

2 Finals 6146B

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


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

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

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

....Keith

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

On Jun 2, 9:52*am, Richard Clark wrote:
Start with Walt's:

2 Finals 6146B

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


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

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

Hi Keith,

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

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

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

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

*********

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

73's
Richard Clark, KB7QHC

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

wrote:
On Jun 2, 9:52 am, Richard Clark wrote:
Start with Walt's:

2 Finals 6146B

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

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

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

Hi Keith,

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

Those 108 W did have to go somewhere.

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

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

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

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

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

*********

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

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

....Keith

On 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

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

Start with Walt's:

2 Finals 6146B

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

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

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

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

...Keith

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

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

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

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

Walt, W2DU

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



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

Start with Walt's:

2 Finals 6146B

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

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

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

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

...Keith

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

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

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

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

Walt, W2DU

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

Walt

On Wed, 2 Jun 2010 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