Gerald,
thanks for the comment.
Assuming the network type:
input - C1 - L1 - L2 - C2+C3 - L3 - Cin+ R
and excluding C1-L1, I think I have solved the network as a
L-pi circuit.
It is interesting that this method is an inverse way of designing
a plate pi-L circuit where Rp, Cout, Rm, QL and load impedance
(normally 50 ohm) given initially, whereas the input L-pi network
should start from fixed Cin, a tube input (drive) impedance R and
network input impedance to deliver remaining parameters.
Totally C1+L1 gives a variable L which is added to L2;
varying C1 sets variable reactance Xs = XL1 + XL2 - XC1.
Then the Xs simply determine QL and Rm, where Xs = 50*QL
and QL = rt(Rm/50 - 1)
Since Q1 = R/XCin, Q1 is assumed as a fixed parameter.
Then Qopi can be determined by Q1 and Rm by solving the equation:
Q1 = (R*Qopi -rt(R*Rm*Qopi^2 - (R-Rm)^2))/(R-Rm)
(where Rm>R so R-Rm will produce negative value, but totally OK)
then Qo = Qopi + QL, Q2 = Qopi - Q1
and XC2 = Rm/Q2, XC3 = Rm/QL,
L3 = (Qopi*R)/((Q1^2 + 1)*2*pi*freq)
Since L3, optimum input impedance R and Cin are all fixed value,
only a center frequency should vary in conjunction with Xs.
Since XC2 = Rm/(Qopi - R/Xcin) and XC3 = Rm/rt(Rm/50 - 1),
C2+C3 will vary with Rm and Qopi set by Xs.
With a return-loss bridge I found C1 (=VC1) determines
(or dominates) a center frequency, and seemingly C2+C3 (=VC2) is
effective to cancel any reactance because VC2 determines
the depth of return-loss dip at the frequency.
Accordingly I understood how C1-L1 contribute to the network.
thanks to all,
de Han JE1BMJ
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