Actually, when the tube(or any other active device) is delivering
power into a load, in a class C amp, you have either high Vp and Ip cut
off, or LOW Vp and heavy Ip flowing. When the tube is turned on, the
voltage should only be the tube "bottoming" voltage. for the 6146 I
know best this will be between 75 and 100V, depending on Ip(peak).

The part of the tube curves that determines modulating waveforms,
efficiency, and maximum useable loading voltage is the part of the
curve where full(peak) Ip is flowign at only a LOW voltage. For the
6146, for instance, that woudl be currents of 400-800ma flowing with
only 75-125V. If you were drawing full tube current with all your Vp
across the tube, your efficiency and power output would be zero!

Where does the rest of the Vp go? it appears across the load. If the
tube was a hypothetical perfect switch, bottoming voltage would be
zero, and in Class C would be either all the way on or all the way off
at all times. The 120 degree conduction angle common in Class C is used
to minimize switching voltage and therefore overlap of Vp and Ip. That
overlap and bottoming voltage are the source of the heat dissipated by
the tube. Class E and F amps use different tuned circuits to get zero
switching votage and efficiency as high as 90%!

If the tube were a perfect switch, and the conduction angle were
constant with respect to Vp, the resulting amp would give perfect
linear plate modulation! This is becuase the load should be a pure
ohmic resistance at resonance. The reason it does actually work this
way is that bottoming voltrage runs up fast as Ip is increased,
cutting the efficiency of the tube and cutting the percentage of Vp
that appears across the load. If doubling the plate voltage only raises
the voltage applied to the load by 90%, it will only increase current
by 90% at most, meaning the power increase cannot be more than about
81%!

This is why a little grid modulation added to plate
modulation(ESPECIALLY WITH TETRODES/PENTODES!) does so much for the
modulation characteristic of the amp. This is also true to a very
severe extent of MOSFETS, BTW. Solid state AM is utter hell to
modulate and I suspect will never sound as good as tubes. you know,
just like transistor guitar amps sound like garbage(but for different
reasons).

To design an AM final for plate modulation, begin by figuring desired
PEP, which is four times desired carrier power. Now select a tube(s)
and operating point that will make this power at a reasonable
efficiency. If the amp can run key-down at this level, great, but this
is not necessary in any way. In teh real world, this would always be
over its thermal dissipation rating.

In other words, if you ran it without modulation at double plate
modulation, it should run with good class C efficiency unit it
overheats from being too small in dissipation rating! It must NOT be
overloaded to the point that efficiency is dropping. This sets the
loading used on the amp as built, tuned, and run.

When operated at carrier, it will be at a rather light loading compared
to the CW ratign for the same device. It could make more power, but
you'd never modulate it. Do NOT increase the loading. At carrier, it
will still make somewhat over 1/4 the PEP power, due to greater
efficiency at the lighter current. Becuase your efficiency waqs good to
begin with, this difference will not be severe.

A small amount of grid modulation is now introduced. On a triode, the
use of grid leak bias has this effect, as grid current falls with
rising plate voltage durign the switchin time of a real tube. On a
screen grid tube,you put a choke or tertiary winding in the screen
supply. If you just rran the screen all trhe time at enough voltage for
the PEP condition, you would little too much carrier power.

The point of the grid modulation is this: it reduces efficiency just
enough at carrier to be the same as the efficiency at PEP. This, in
turn, is one of the reasons the PEP efficiency must be good-otherwise
you now always have poor efficiency.

Suppose you are designing a final for the amateur limit of 1500 wats
PEP output. You are NOT going to design a 325 watt(the carrier power)
amp! Thermal loading will be on a sine wave signal about that of a 437
watt class C amp, but the amp must not clip or round off the current
pulses at 1500 watts output and double plate voltage. Essentially, your
tubes must be able to efficiently MAKE 1500 watts, even if they can
only get rid of 200 watts of heat in the process.

I learned about this the ahrd way, on MOSFETS for a MW application. Two
IRF 510's could make 50W carrier but wouldn't modulate worth garbage.
To get acceptable modulation requred backing power off all the way to
27 watts and then still needed an assymetrical modulating voltage.

Viewing the modulating voltage on the scope proved the modulator to be
adequate when wound for a normal syymetrical output waveform, leavign
the Class C amp as the culprit. To fix it right too four of those
little MOSFETS, and 2V or gate modulation from a tertiary winding. Just
forward biasing to handle the current didn't cut it. This gave 45W
carrier, with still a shortfall in PEP, but now only the shortage
predicted by the power supply voltage sagging about 3/4V out of 16V.
The heatsink runs very cool at carrier, and doesn't get too hot even
with the modulator running flat-out.

A pair of 6146B's of course, with a pair of 6550 audio tube as a
modulator, would have been able to pound out several times that much
power! Still, they would never, ever be able to make four times the CW
rating from the tube manual simply by applying twice the CW plate
voltage. Similarily, to run the same current at carrier as in a CW rig,
they would need a lot more drive, as at PEP the plate current would be
higher, as high as the current drawn in a low-impedance VHF rig.

Straydog wrote:
Since my earlier post (dealing with the question of what is peak evelope
power output in an AM transmitter), I've been doing more scrutinizing
of tube Ip/Vp characteristic curves. They are much more non-linear than
the impression you get from just looking at the curves. Also, it is rare
or almost non-existant to find Ip vs screen voltage!

Lets look at the venerable 833 (from my RCA TT-3 transmitting tube
manual). This is a KW input class C triode.

From the curve:
at zero grid volts, 1 kV on the plate gives 175 ma plate current
2 kV 500 ma
That's more than a doubling of Ip for a doubling of Vp

at minus 50 grid volts, 2 kV on the plate gives 50 ma plate current
4 kV 750 ma

looking in my RCA receiving tube manual (RC-20) I found for a 6FG6
a sharp cutoff tetrode that only at zero grid volts was there a near
linear relationship between d a straight line [which actually deviated
slightly from a straight line] with some slope. But at 100 v on plate,
current was 14 milliamps, at 200 v on the plate, plate current was 34
miliamps. Definitely NOT a linear relationship. For the 6EM7 a triode,
and at any of a wide range of grid voltages, plate current could be
doubled with only a 15-20% increase in plate voltage.

My thinking on all of this leads me to claim that anyone who can start
with a 100 watt carrier from an AM transmitter and make a few assumptions
about 100% modulation and come up with a _calculation_ of something like
400 watts of peak power and represent that as having something to do with
reality is pure conjecture.

If anyone wants to put an appropriate oscilloscope on the transmitter output
and measure the RF voltage of unmodulated carrier into an appropriate load
and then measure the peak RF voltage when the carrier is modulated, then
and only then do they have a reasonable _basis_ for making a claim about
peak (instantaneous) output power.