Sorry - for some reason my Outlook Express does not add
attribution marks to some messages.
>In normal use, you don't have resistors loading the circuit
>at both ends - you put a load on one end and use the
circuit
>to transform the impedance so that looking into the network
>you see the desired impedance.
However, the tube is a time-varying resistance.
In which case one or more of resonance, impedance and Q will
change in sympathy. In practice when tuning an amplifier we
work with time averaged values.
> If you remove one of the
>resistors, I think the resonant frequency will decrease.
Sure, but the tube is there.
On the input side, (in g-g) the tube is the load on one end
of the network - we want the other end to look like 50 ohms,
but we don't (normally) terminate it in 50 ohms. The
resonant frequency of the pi network in isolation will be
different from when it's connected to the tube, even after
allowing for tube capacitance.
On the anode side, doesn't the tube approximate to a
constant current source, which is a high resistance. Ok.
it's a variable constant current source so it's a variable
high resistance. In any case, it's my guess that the tube
resistance is somewhat higher than the load resistance
presented to it by the pi network so there's limited or
minimal effect.
We desperately need an interactive blackboard capability -
realtime paintbox perhaps?
Steve
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