>
>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.
>
? amen, Steve
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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