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, 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

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, 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

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

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.