Fwd: RE: [AMPS] basic question about dipping the plate

[measures]

>
>View the L-network as a parallel LC with the load Rs in series with the L.
>At resonance this circuit is equivalent to a parallel LCRp (all in 
>parallel) with the Rp equal to (X^2)/Rs . The Q is always in this case equal 
to the square 
>root of the ratio of Rs ( the load)
>  and Rp which should match the output impedance of the tube.
>So if you have a plate impedance of 5000 and an output impedance of 50 the 
>Q must be
>10 to match the two. This makes sense because for the same power 10 times 
>the current must flow
>thru 50 ohms as  thru 5000 ohms. And the circulating current, all of which 
>flows thru Rs(the load)
>is Q times the input current applied to the junction of C and L.
>A L-network should be used to match large plate and load resistance ratios 
>if you want a reasonable Q.
>     In some cases reactances are not equal and those are when matching a 
>impedance that has a reactive component
>such as a short antenna. Some of the output reactance is used to 
>cancel  out the reactive component of the antenna.
>Otherwise the rest of the circuit should resonant.
>
>73
>Bill wa4lav
>
>
>--
>FAQ on WWW:               http://www.contesting.com/FAQ/amps
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>Problems:                 owner-amps@contesting.com
>
>


-  Rich..., 805.386.3734, www.vcnet.com/measures.  
end


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