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[AMPS] basic question about dipping the plate

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Subject: [AMPS] basic question about dipping the plate
From: W8JI@contesting.com (Tom Rauch)
Date: Mon, 23 Apr 2001 10:15:51 -0400
> Now we add the output load C and it becomes a Pi-net.  Traditionally,
> this circuit is viewed at two L networks back to back, primarily
> because that view simplifies the computation of values (in the days
> before computers).  But, I personally think its reasonable to think of

We view it that way because that is how the network behaves. If 
you moved along the inductor measuring voltage to ground with a 
RF voltmeter, you would see voltage at some point in the network 
is LOWER than either the input or output voltage.

We have two L networks, BOTH stepping impedance down when 
looking from each end towards some point in the network. Phase 
shift is more than 90 degrees, and system Q is more than the 
square root of the ratio of impedances. 

If the Pi-network has inadequate Q, it behaves like a single L 
network and one of the capacitors behaves "backwards" or has 
very little effect when adjusted. Some "pi-network" antenna tuners 
are actually L networks, because the second capacitor is far too 
small and Q is far to low to make them behave like Pi-
networks....even though they have three components.       

> what's going on differently.  The Tune C and L create a tuned circuit
> with some Q (higher than what's needed for the L network)...which
> creates circulating currents.  The load C shunts some of the
> circulating current back to the tank but lets some of the current flow
> in the output.  The less load C, the higher its impedance and the more
> circulating current flows in the output load

Only when the network has enough Q to really be a Pi.

Also, the input and output areas each have a Q...and that Q is 
usually different. The network Q is higher than the Q at one end, so 
when we are using Q's of 12 in simple tank circuit calculations the 
Q is really higher.


73, Tom W8JI
W8JI@contesting.com 

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