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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