Fw: [AMPS] Anode load line calculations.

[Steve Thompson]

In message <19981221185213.AAA303@[209.239.238.117]>, Rich Measures
<measures@vc.net> writes
>
>>Both - but what's the difference?  An amplifier set up to give 100W using a
>>2000V supply or a 1kW amp using the same ht have the same requirement.  
>
>€  The requirement is seemingly set by the conditions under which the 
>amplifier is tuned up.  Assuming 60% effiency, if a 1666w input, w. 2000v 
>anode supply, 1000w-out amplifier is, , tuned up at 1666w input, approx  
>0.833a of anode current is required.  Therefore RL is roughly. 2000v/(2 x 
>0.833a) = 1200 ohms and the Pi-network tank circuit must be tuned to 
>match 50 ohms to 1200 ohms.  Assuming that the drive level is susequently 
>reduced until the output is 100w, the anode-V swing is reduced and the 
>anode-I swing is also reduced.  In my opinion, when the output of this 
>amplifier is 100w, RL does not become 22k-ohms. 

It definitely does not. Very simplistically, reducing the drive reduces
the grid voltage swing which reduces the anode current swing (Ia is
related to gm x Vg); the reduced anode current swing through RL means
less anode voltage swing.

> .  However, if it were 
>decided to tune up the amplifier for operation at 100w out, RL would 
>increase to roughly 22k-ohms.  
>
>>The point I was trying to make, and which you are conveniently ignoring, is
>>that the RL in either case is radically different if you use the formula I
>>posted previously.
>>
>Semi-agreed.  .  However, the RL formula is an approximation that applies 
>for the power level at which an amplifier is tuned.  When drive is 
>reduced to a tenth, RL does not  increase ten-fold.  .    

RL is set by the antenna impedance and matching network settings. If you
don't retune the matching network, RL can't change.

Steve

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