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Old June 12th 10, 02:13 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.

However the information and measured data provided in the text
excerpts below is not very supportive of that theory.

These excerpts discuss and show the effects of the energy entering a
transmitter at its output connector by frequencies offset from the
transmitter frequency.

There is direct applicability of the conclusions of the paper showing
that the operational source impedance of the transmitter near/at the
carrier frequency is much different than 50 ohms.

If it WAS a functional 50 ohms, then the termination provided to the
transmission line for signals entering the transmitter by its output
connector (whether on or off frequency) would not be present at the
plate of the PA tube to react with the power being generated by the PA
tube.

Rather the data leads to a logical conclusion that the operational
source impedance of this configuration at the carrier frequency will
be very low (approaching zero), when it is optimally tuned/adjusted to
produce its rated output power.

Further discussion or comment is invited.

RF

From:

A STUDY OF RF INTERMODULATION BETWEEN FM BROADCAST TRANSMITTERS
SHARING FILTERPLEXED OR CO-LOCATED ANTENNA SYSTEMS, by Geoffrey N.
Mendenhall, P.E.*

II. INTERMODULATION AS A FUNCTION OF "TURN-AROUND-LOSS".
"Turn-Around-Loss" or "Mixing Loss" describes the phenomenon whereby
the interfering signal mixes with the fundamental and its harmonics
within the non-linear output device. This mixing occurs with a net
conversion loss, hence the term "Turn-Around-Loss" has become widely
used to quantify the ratio of the interfering level to the resulting
IM level. A "Turn-Around-Loss" of 10dB means that the IM product fed
back to the antenna system will be 10dB below the interfering signal
fed into the transmitter's output stage.

"Turn-Around-Loss" will increase if the interfering signal falls
outside the passband of the transmitter's output circuit, varying with
the frequency separation of the desired signal and the interfering
signal. This is because the interfering signal is first attenuated by
the selectivity going into the non-linear device and then the IM
product is further attenuated as it comes back out through the
frequency selective circuit.

"Turn-Around-Loss" can actually be broken down into the sum of three
individual parts:
(1) The basic in-band conversion loss of the non-linear device.
(2) The attenuation of the out-of-band interfering signal due to the
selectivity of the output stage.
(3) The attenuation of the resulting out-of-band IM products due to
the selectivity of the output stage.

Of course, as the "Turn-Around-Loss" increases, the level of
undesirable intermodulation products is reduced and the amount of
isolation required between transmitters is also reduced.

The small portion of the interfering signal that is not reflected is
what causes intermodulation products to be generated. Obviously the
lower the output source impedance, the more complete the reflection
(lower return loss), with the result being less production of
intermodulation products.

III. EQUIPMENT PARAMETERS THAT AFFECT INTERMODULATION LEVELS.
The interfering signal must be coupled into the transmitter's output
stage before the IM products are produced and the output level of the
intermodulation products will be related to the interfering signal
level.

The two parameters (outside of the filterplexing equipment) that most
affect the interfering signal level into the transmitter's output
circuit are the output loading and the circuit's frequency selectivity
(loaded "Q"). These two parameters are interrelated because the degree
of output loading will change the loaded "Q" of the output circuit
while also affecting the return loss of the interfering signal looking
into the output circuit.

"Output Return Loss" is a measure of the amount of interfering signal
that is coupled into the output circuit versus the amount that is
reflected back from the output circuit without interacting with the
non linear device.

To understand this concept more clearly, we must remember
that although the output circuit of the transmitter is designed to
work into a fifty ohm load, the output source impedance of the
transmitter is not fifty ohms. If the source impedance were equal to
the fifty ohm transmission line impedance, half of the transmitter's
output power would be dissipated in its internal output source
impedance. The transmitter's output source impedance must be low
compared to the load impedance in order to achieve good efficiency.

The transmitter therefore looks like a voltage source driving a fifty
ohm resistive load. While the transmission line is correctly
terminated looking toward the antenna (high return loss), THE
TRANSMISSION LINE IS GREATLY MISMATCHED LOOKING TOWARD THE OUTPUT
CIRCUIT OF THE TRANSMITTER (LOW RETURN LOSS). THIS MEANS THAT POWER
COMING OUT OF THE TRANSMITTER IS COMPLETELY ABSORBED BY THE LOAD WHILE
INTERFERING SIGNALS FED INTO THE TRANSMITTER ARE ALMOST COMPLETELY
REFLECTED BY THE OUTPUT CIRCUIT.

VI. CONCLUSIONS
1. "Turn-Around-Loss" is a function of the particular non-linear
device and the amount of loading on its output circuit.
2. "Turn-Around-Loss" increases as the interfering signal and the
resulting IM products are moved away from the carrier and out of the
output circuit passband.
3. "Turn-Around-Loss" will be least when the interfering signal is
within the transmitter's passband.

The figure posted at the link below shows the measured data supplied
with this paper.

http://i62.photobucket.com/albums/h8.../TAL_Chart.gif


* Geoffrey Mendenhall presently is Vice President, RF Engineering at
Harris Corporation Broadcast Division, and a recognized authority on
transmitter system design. Harris Broadcast is one of the largest
manufacturers in the world of AM/FM/TV broadcast transmitters, rated
for power outputs up to 2,000 kW.
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Old June 12th 10, 03:22 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Jun 12, 1:13*pm, Richard Fry wrote:
It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.

However the information and measured data provided in the text
excerpts below is not very supportive of that theory.


yeah, lots of words that say things kind of related, but more
confusing than useful for what is really being discussed.

what is so hard about stating:
1. tubes and transistors are non-linear devices
2. tank circuits and other output matching systems TRY to make them
appear more linear to the outside world, but are not perfect.

any circuit that has non-linear characteristics will have a response
that is not like a simple resistance. such systems must be modeled
using appropriate non-linear techniques to get the full and proper
response characteristics. An APPROXIMATION may be used in many
systems that allows an equivalent simple impedance representation, but
often only over SMALL ranges of voltage/current or frequency. The
problem appears to be in this discussion that some people have used
this small signal approximation to derive results in a domain where it
is not applicable and are drawing the wrong conclusions.
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Old June 12th 10, 03:44 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On 12 jun, 15:13, Richard Fry wrote:
It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.

However the information and measured data provided in the text
excerpts below is not very supportive of that theory.

[deleted]


Hello Richard Fry,

First, I use your last name also to avoid confusion with Richard
Clark.

As you may know, I have also doubts about the conjugated match issue.
So I am happy with this posting. It is not that I dispute the
conjugated match condition itself, but whether or not the conditions
are met. In my opinion, required conditions aren't met in many cases,
so output impedance of many amplifiers is not equal to the load
resistor.

To avoid discussions about measurements and traceable measurement
instruments/procedures, I did some simulations on a valve amplifier
and a class C mosfet circuit (that actually isn't operating in class C
mode). Everyone can do these simulations at their own PC en get an
opinion. I also described the difference frequency method that is
easily implemented in simulation and gives the output VSWR almost
directly without complex calculus. The link is: www.tetech.nl/divers/PA_impedance.pdf.

Except for the condition of matching to maximum power output given
certain drive, all output VSWR of the amplifiers is far from expected
load impedance. Just a little voltage saturation (for example to get
better efficiency) did drop the plate impedance below the required
load impedance (voltage source behavior). When reducing the drive, the
plate impedance rises rapidly above the load impedance (current source
behavior).

So power entering the output of the amplifier will mostly be reflected
back into the cable, unless you specially design for output impedance.
I know, this is disputed by some members of this group, but therefore
I did the simulations and presented the results.

I specially took "amateur" examples to avoid comment that I only use
"exotic" examples. I did not copy results from professional
activities, so I will not run into problems. The circuits are simple
to ease reproduction in any spice based simulator.

I dispute:
"If the source impedance were equal to
the fifty ohm transmission line impedance, half of the transmitter's
output power would be dissipated in its internal output source
impedance."

A real class C amplifier with very small conduction angle has
efficiency 50% when optimally tuned. When it operates at the
transition of current/voltage saturation is can show 50 Ohms output
impedance for very small change in load impedance. But as soon as the
load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of
the amplifier changes rapidly. So practically spoken, most high
efficient amplifiers have high output VSWR (or bad Return Loss), but
theoretically it can be 50 Ohms (for example). If there are doubts I
will add a special simulation example for this.

Best regards,


Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me.

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Old June 12th 10, 04:17 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Jun 12, 9:44*am, Wimpie wrote:

A real class C amplifier with very small conduction angle has
efficiency 50% when optimally tuned.


Would that not be evidence that the amplifier source impedance
necessarily is lower than the load impedance?

When it operates at the
transition of current/voltage saturation is can show 50 Ohms output
impedance for very small change in load impedance. But as soon as the
load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of
the amplifier changes rapidly.


For some additional input -- I have taken part in factory tests of
high-power Class C single-tube/tuned cavity FM broadcast transmitters
driving 50-ohm test loads measured to have 1.03VSWR, showing a DC
input to r-f output efficiency of the PA to be in excess of 80% --
including the loss in the harmonic filter. Load power was measured
using calorimetric methods. In fact, 80% PA efficiency is the
published spec for this transmitter line as long as load VSWR relative
to 50 ohms is 1.7:1 or less (any phase angle).

Those results don't appear to be fitting very well with the idea that
the tx source impedance can be ~50 ohms for loads with low VSWR.

RF
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Old June 12th 10, 04:56 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Jun 12, 8:17*am, Richard Fry wrote:
On Jun 12, 9:44*am, Wimpie wrote:

A real class C amplifier with very small conduction angle has
efficiency 50% when optimally tuned.


Would that not be evidence that the amplifier source impedance
necessarily is lower than the load impedance?


No, not at all. It is important to realize that a real source is NOT
necessarily like either the Thevenin or the Norton equivalent. The
Thevenin and Norton equivalents, when discussed in any respectable
text, will be noted to behave exactly like any other real linear
source with respect to the load, but NOT necessarily with respect to
the source itself, and in particular with respect to source efficiency
and dissipation.

For example: I can, at least theoretically, build a switching power
supply (a source) which measures its output voltage and current and
adjusts the voltage down by 50 volts for each amp that's drawn from
it. Because it is always operating as a switching supply with very
high efficiency, even when it's loaded so the output voltage is half
the zero-current output voltage, it's not behaving internally like a
Thevenin equivalent, even though it has a 50 ohm output resistance.

When it operates at the
transition of current/voltage saturation is can show 50 Ohms output
impedance for very small change in load impedance. But as soon as the
load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of
the amplifier changes rapidly.


For some additional input -- I have taken part in factory tests of
high-power Class C single-tube/tuned cavity FM broadcast transmitters
driving 50-ohm test loads measured to have 1.03VSWR, *showing a DC
input to r-f output efficiency of the PA to be in excess of 80% --
including the loss in the harmonic filter. *Load power was measured
using calorimetric methods. *In fact, 80% PA efficiency is the
published spec for this transmitter line as long as load VSWR relative
to 50 ohms is 1.7:1 or less (any phase angle).

Those results don't appear to be fitting very well with the idea that
the tx source impedance can be ~50 ohms for loads with low VSWR.

RF


As Wim says, the conjugate-matched output impedance is guaranteed only
over a very narrow range of conditions. Specifically, it's for the
condition where you can't adjust the load impedance in any direction
without lowering the power delivered to the load. For a power output
stage with a tank that can tune both resistance and reactance (such as
a pi network), that should be equivalent to adjusting the pi network
to maximize the power delivered to the load with a particular fixed
set of bias/drive conditions for that output stage.

But why would I necessarily want to operate the output stage that
way? What if I'm providing enough grid drive and plate voltage to my
6146 that it's capable of outputting 150 watts (class C), but if I
were to tune up the output matching network to get that much output
power, the tube's life would be significantly shortened because the
plate dissipation would be too high? What if I'm operating it class
AB, with low enough drive so that the plate voltage and current change
only by small percentages?

Or to me, more to the point: why would I care what the source
impedance of my transmitter is? With the power at the wall outlet in
my house, I expect a nearly constant voltage, and have no intention of
loading it with a "conjugate match." I don't know WHAT its source
resistance is, exactly, but I know it's much lower than the loads I
put on it. With an electric motor, I have no intention of putting a
load in excess of the motor's rating on it, for fear of burning up the
motor, even though I know that for short periods the motor can deliver
quite a bit more output power than its nameplate rating. With a
switching power supply, same thing: I don't load it to the point of
significant drop in output voltage. All I want from ANY of these,
including the RF power amplifier, is the ability to deliver the
expected power to the rated load.

In the instrumentation systems I deal with professionally, there are
times I care very much about source impedance. With respect to ham
transmitters and power amplifiers, though, I just can't get excited
about caring that much what the source impedance is. The point of the
amplifier or transmitter is to deliver power to my antenna system, and
the source impedance it represents is irrelevant to that task.

Cheers,
Tom


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Old June 12th 10, 08:00 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Jun 12, 11:56*am, K7ITM wrote:
On Jun 12, 8:17*am, Richard Fry wrote:

On Jun 12, 9:44*am, Wimpie wrote:


A real class C amplifier with very small conduction angle has
efficiency 50% when optimally tuned.


Would that not be evidence that the amplifier source impedance
necessarily is lower than the load impedance?


No, not at all. *It is important to realize that a real source is NOT
necessarily like either the Thevenin or the Norton equivalent. *The
Thevenin and Norton equivalents, when discussed in any respectable
text, will be noted to behave exactly like any other real linear
source with respect to the load, but NOT necessarily with respect to
the source itself, and in particular with respect to source efficiency
and dissipation.

For example: *I can, at least theoretically, build a switching power
supply (a source) which measures its output voltage and current and
adjusts the voltage down by 50 volts for each amp that's drawn from
it. *Because it is always operating as a switching supply with very
high efficiency, even when it's loaded so the output voltage is half
the zero-current output voltage, it's not behaving internally like a
Thevenin equivalent, even though it has a 50 ohm output resistance.





When it operates at the
transition of current/voltage saturation is can show 50 Ohms output
impedance for very small change in load impedance. But as soon as the
load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of
the amplifier changes rapidly.


For some additional input -- I have taken part in factory tests of
high-power Class C single-tube/tuned cavity FM broadcast transmitters
driving 50-ohm test loads measured to have 1.03VSWR, *showing a DC
input to r-f output efficiency of the PA to be in excess of 80% --
including the loss in the harmonic filter. *Load power was measured
using calorimetric methods. *In fact, 80% PA efficiency is the
published spec for this transmitter line as long as load VSWR relative
to 50 ohms is 1.7:1 or less (any phase angle).


Those results don't appear to be fitting very well with the idea that
the tx source impedance can be ~50 ohms for loads with low VSWR.


RF


As Wim says, the conjugate-matched output impedance is guaranteed only
over a very narrow range of conditions. *Specifically, it's for the
condition where you can't adjust the load impedance in any direction
without lowering the power delivered to the load. *For a power output
stage with a tank that can tune both resistance and reactance (such as
a pi network), that should be equivalent to adjusting the pi network
to maximize the power delivered to the load with a particular fixed
set of bias/drive conditions for that output stage.

But why would I necessarily want to operate the output stage that
way? *What if I'm providing enough grid drive and plate voltage to my
6146 that it's capable of outputting 150 watts (class C), but if I
were to tune up the output matching network to get that much output
power, the tube's life would be significantly shortened because the
plate dissipation would be too high? *What if I'm operating it class
AB, with low enough drive so that the plate voltage and current change
only by small percentages?

Or to me, more to the point: *why would I care what the source
impedance of my transmitter is? *With the power at the wall outlet in
my house, I expect a nearly constant voltage, and have no intention of
loading it with a "conjugate match." *I don't know WHAT its source
resistance is, exactly, but I know it's much lower than the loads I
put on it. *With an electric motor, I have no intention of putting a
load in excess of the motor's rating on it, for fear of burning up the
motor, even though I know that for short periods the motor can deliver
quite a bit more output power than its nameplate rating. *With a
switching power supply, same thing: *I don't load it to the point of
significant drop in output voltage. *All I want from ANY of these,
including the RF power amplifier, is the ability to deliver the
expected power to the rated load.

In the instrumentation systems I deal with professionally, there are
times I care very much about source impedance. *With respect to ham
transmitters and power amplifiers, though, I just can't get excited
about caring that much what the source impedance is. *The point of the
amplifier or transmitter is to deliver power to my antenna system, and
the source impedance it represents is irrelevant to that task.

Cheers,
Tom


Hello Richard Fry,

I quote a sentence from your previous post:

"Would that not be evidence that the amplifier source impedance
necessarily is lower than the load impedance?"

What you have said above is the key to the concern over the output
resistance of a Clsss C amplifier being non-dissipative. What seems to
be universally misunderstood is that there are really two separate
resistances in the operation of these amps; one, the cathode-to-plate
resistance, which is the dissipative resistance Rpd that accounts for
all the heat, due to the electrons striking the plate; and two, the
real output resistance that appears at the output of the pi-network,
the resistance comprised of the voltage-current ratio E/R. I am
speaking only of tube amps using an adjustable pi-network in the
output. Due to the energy storage in the pi-network the output of the
amp is inherently isolated from the upstream non-linear portion, thus
the output is linear. One of the myths concerning the RF amp is that
it cannot have efficiency greater than 50% because half of the power
is dissipated in the source resistance. This would be correct if the
output source resistance was dissipative, like in a classical
generator. But in the real RF xmtr this is not true.

Returning now to your quote above, your reference to the source
impedance (resistance Rpd) is actually the plate-to-cathode
resistance, the dissipative resistance, which is less than the output
resistance R = E/I appearing at the terminals of the pi-network. I
must add that when the pi-network has been adjusted to deliver all the
available power at any particular grid-drive level, the output
resistance of the network equals the load resistance, but not
necessarily 50 ohms, i.e., whatever the load resistance might be.

If you have doubts concerning the conditions I stated above, I invite
you to review Chapter 19 in Reflections 3, which is available in two
parts on my web page at www.w2du.com, as the first part appears in
Reflections 2 and the second part appears in 'Preview of Chapters in
Reflections 3'. In that chapter I use an example taken directly from
Terman's 'Radio Engineers Handbook'. You will note that the
dissipative resistance Rpd (3315.6 ohms) is less than the output
resistance (6405 ohms) obtained from the E/I relationship appearing at
the output

This example verifies the explanation in the text. In addition, I also
invite you to review the portion of the chapter appearing in the
'Preview Chapters in Reflections 3', in which I report additional
measurements that offer additional proof that the output source
resistance is non-dissipative. The numbers prove it to be true.

Walt, W2DU
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Old June 12th 10, 10:42 PM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

walt wrote in
:

....
What you have said above is the key to the concern over the output
resistance of a Clsss C amplifier being non-dissipative. What seems to
be universally misunderstood is that there are really two separate
resistances in the operation of these amps; one, the cathode-to-plate
resistance, which is the dissipative resistance Rpd that accounts for
all the heat, due to the electrons striking the plate; and two, the


It is my understanding that the average power (heat) generated at the anode
of a triode can be found by averaging the product of the instantaneous
anode current and anode-cathode voltage over time. In a Class C amplifier,
the voltage and current are not linearly related to each other, ie there is
no constant of proportionality, no constant or fixed resistance.

I don't understand why then, that people try to explain the anode
dissipation in terms of some value of resistance.

Owen
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Old June 13th 10, 01:30 AM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote:

It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.

However the information and measured data provided in the text
excerpts below is not very supportive of that theory.


Hi Richard,

Deja Vu all over again. I suppose you posted this to inspire me to,
once again, remind you from Mendenhall's own notes about Class C
amplifier construction - and so I will:

"It was thus necessary to determine the plate load impedance
= (Eb - Emin) / I1
~= Eb/ Idc
~= 1000 Ohms
[Not very far from Walt's determination of 1400 Ohms from his own
bench.]
"Since this was to be coupled into a Z output of 50 Ohms, a impedance
transformation of 20:1 was needed. This was accomplished by using a
voltage divider of two series capacitors to series tune the plate
circuit."

The tube is an EIMAC 4CX300A running no where near capacity.

All details (and more) written in Mendenhall's own hand.

73's
Richard Clark, KB7QHC
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Old June 13th 10, 06:03 AM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote:

It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.


From Mendenhall:

"VHF amplifiers often exhibit a somewhat unusual characteristic when
tuning for maximum efficiency. ... If the amplifier is tuned exactly
to resonance, the plate load impedance will be purely resistive and
teh load line will be linear."

73's
Richard Clark, KB7QHC
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Old June 13th 10, 06:21 AM posted to rec.radio.amateur.antenna
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Default "Non-dissipative Source Resistance"

On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote:

It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network
provides an operational “non-dissipative source resistance” of 50 ohms
for energy present at the output connector of the transmitter.


From "Care and Feeding of Power Grid Tubes:"

"The first step in designing the output circuit is to specify the
resonant load impedance of the tube, the loaded Q of the circuit and
the desired output impedance of the network. The resonant plate load
impedance for the 4065A is determined by dividing the plate peak RF
voltage swing by the plate peak fundamental RF current.
[work shows Resonant load impedance = 7600 Ohms]
"if it is assumed that the output impedance of the network is to be 50
Ohms, and the loaded Q is to be 15, the output tuned circuit may now
be designed. The output impedance of 50 Ohms will match into a
properly terminated 50 Ohm transmission line."

This document provides plenty of math and charts for the design of
efficient output networks (finals' plate load).

73's
Richard Clark, KB7QHC
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