Antonio Vernucci wrote:

The higher the grid resistor, the higher the bias voltage that must be
overcome by the drive. Hence, higher drive, more power lost in the grid
resistor, and lower conduction angle.

So, too high a grid resistor and you'll need to beef up your drive
stage. Plus (as mentioned), your conduction angle decreases, and your
final-stage efficiency may suffer.

Get the grid resistor too low, your conduction angle will increase, and
your final-stage efficiency may suffer.

Note that I say "may" -- there's an optimum conduction angle. There's
handbook values for it (which I can't remember!) but I'll bet that no one
amplifier works best right at the handbook value.

If you _really_ want to be scientific about it then for each grid
resistance value monitor your final stage input power, the amplifier
output power, and calculate the grid resistance dissipation. If nothing
else, that'll help you make an informed choice.

Otherwise, if it's given, calculate the grid resistor value to get you
both the desired current and the RF p-p voltage, or the rated bias
voltage, whichever is listed for your tube in that service.

--
www.wescottdesign.com

Thanks for your comments. I agree that there should be an optimum grid
resistance value (even if rather dull), but in my case the optimum
occurs at zero grid resistance.

Let me report you some tests I have made, increasing the grid resistance
in steps (starting from R=0) and then re-adjusting the drive power each
time (and also re-optimizing the Pi-network controls):

- increasing the grid resistance and then adjusting the drive power so
as to keep the GRID current constant, the plate current - and hence the
output power - decreases. Therefore, to obtain maximum output power, the
grid resistance must be zero

- conversely, increasing the grid resistance and then adjusting the
drive power so as to keep the PLATE current constant, the output power
remains about the same for a quite wide range of grid resistance values
(except when resistance becomes very high). It should be noted that,
increasing the grid resistance at constant plate current, the grid
current increases significantly, to the extent that, for fairly high
grid resistance values, the grid current gets beyond the allowable limit.

In conclusion, it looks like the final stage operates best at zero grid
resistance:

- no efficiency loss
- minimum grid current for a given ouptut power.

In such conditions, the tube operates in class B (the fixed -33V bias
causes an idling plate current of about 10 mA), with a circulation angle
of more than 180 degrees. Increasing the grid resistor causes a
reduction of the circulation angle, with no practical benefit and some
drawbacks.

Where has the class-C efficiency advantage gone?

73

Tony I0JX

The advantage of class C isn't necessary greater efficiency. By
reducing the conduction angle the tube is drawing current for a short
period of time and therefor can run cooler. It also means that the tube
can be run at a bit higher power level than it could in class B since
the AVERAGE power dissipated is the same. HOWEVER the duty cycle of
both the time transmitting vs not transmitting and that of the signal
also play a role. In other words a class C CW transmitter in theory
could be run at higher power than a class C FM phone transmitter (even
though both are usually run at the same typical parameters) since the
tube can cool between elements on CW, while FM is key down forever.
Also class B audio has a different duty cycle than a class B RF linear
amplifier running FM (don't need to be linear for FM 'thou).

In the 30's there was an article in QST on how someone ran a 200 watt
tube at a KW CW. It worked because of CW's short duty cycle, but the
editor suspected 'short dashes'.