But Rich you misunderstand what I am trying to tell you and the others!!!!
The anode voltage swing WILL be the 80% or so of the dc supply voltage -
but it is now developed across a much higher impedance of 22800 ohms.
Power is the voltage squared divided by the RL, which you can surely see
from the formula!
If people ignore basic formula and principles they are putting the hobby
back into the dark ages. The one thing about this wonderful hobby is that
it is supposed to contain a "self education" aspect. How can you
blatantly ignore facts!
As far as the other aspect of the argument regarding the linearity versus
anode voltage - it is a well documented fact that the greater the anode
supply available the more linear the operation will be for a given output
power. Why else do we use such high voltages if we could use lower and
safer voltages? Look again at the formula - here you can plainly see that
the predominant factor in obtaining high output power is the voltage term!
By running the tube over its most linear portion of the anode current curve
the linearity MUST be improved. This of course entails moving the RL to a
higher value and a greater transformation ratio between the RL and the
antenna load of 50 ohms. After a certain point it becomes impractical to
provide high transformation ratios due to the unrealisable circuit values
and a move to a more complex network - such as the Pi-L is often necessary.
John ZS5JF
----------
> From: Rich Measures <measures@vc.net>
> To: John Fielding <johnf@futurenet.co.za>; Amps Contesting
<amps@contesting.com>
> Subject: Re: Fw: [AMPS] Anode load line calculations.
To: <amps@contesting.com>
> Date: 20 December 1998 01:41
>
>
> >In response to various comments made on my original posting.
> >
> >I think that some hams misunderstand the basic principles of of rf
> >amplifiers!
> >
> >The load impedance of a tube is defined by the anode voltage and the
output
> >power and is calculated using the formula: (assuming that 80% of the dc
> >supply is available to develop the rf voltage swing, which is pretty
close
> >to what happens in practice)
> >
> >
> >RL = (0.57 x Va2)
> > -------------------
> > Po
> >
> >Taking an amplifier with, say, 2kV anode supply and an ouput power of
100W
> >gives RL as 22800 ohms.
> >
>
> >Using the same anode voltage but now running 1kW output the tube load
line
> >(RL) is 2758 ohms.
> >
> >Obviously the Pi tank network will not be able to provide matching at
both
> >of these load impedances without some drastic changes to the L & C's.
> >
> >This is an over simplified example .......
>
> ? This presumes that the anode's voltage swing at 100w output in the
> 1000w amplifier is approx. 80% of the supply potential, which is
> undoubtedly not the case. It seems to me that RL at any power level is
> closely related to dV/dI
> - later, John
>
>
> Rich...
>
> R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
>
>
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