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Re: [Amps] Plate Load Calc.

To: amps@contesting.com
Subject: Re: [Amps] Plate Load Calc.
From: Manfred Mornhinweg <mmornhin@gmx.net>
Date: Sat, 03 Mar 2007 16:30:02 -0300
List-post: <mailto:amps@contesting.com>
Hi Steve,

> For calculating Resonant Plate Load, Orr's Handbook suggests 1.8 for
> the K-factor for Class B and C. Tonne's PI-EL software uses a
> K-factor of 1.8 for Class B and 2.0 for Class C. This gives quite a
> difference in the Resonant Plate Load.  Let me guess...the answer is
> "both are correct".

I prefer not using any such "fudge factors", but instead calculate the 
real situation, based on the actual tube and operating condition.

I will explain this with an example: Let's assume that you want to 
calculate the plate load resistance for a tube that's delivering 1500W, 
while fed from 3kV. And the specific tube you are using, under those 
conditions will saturate at 300V (this is an information the tube data 
sheet often doesn't give directly, so you have to infer it, or actually 
measure it. And it is partly a function of acceptable distortion).

Under these conditions, the peak RF voltage at the plate is 
3000-300=2700V. The peak-to-peak is of course twice that, but we don't 
need that value. What we do need is the RMS voltage, which is 0.71 times 
the peak. So we have about 1900V RMS.

Power is voltage squared divided by resistance. So the resistance is the 
voltage squared divided by power. 1900^2 / 1500 = 2400 Ohm. There you 
have the plate load resistance for that tube under those operating 
conditions.

Simple, isn't it?

In class C you would typically overdrive the tube to bring it deeply 
into saturation. Distortion is no matter in this case. So the saturation 
happens at a somewhat lower voltage, but specially the RMS is 
significantly more than 0.71 times the peak. So, running this tube in 
class C might land us with about 2400V RMS, which would give 3840 Ohm 
load resistance. But then, bias has an influence too. The more negative 
you set it, the lower the RMS again gets, and this will in turn reduce 
load resistance.

As you can see, using a fudge factor and applying it without regard for 
the exact conditions, will never lead to a precise result, and the 
difference between 1.8 and 2 hardly matters at all. The unknown factors 
in the tube's characteristics and its operating conditions are much more 
important! And anyway the PI circuit fortunately provides enough 
adjustment range to render exact calculation a platonic exercise!

Manfred.

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