[Amps] Fwd: Pi-L In-circuit Adjustment Question

[Bill Fuqua]

	The L networks are resonant. All three of them in a Pi-L network. Two step 
down and one steps impedance)
	Looking at a single L network ( impedance step down) with a resistive load 
you will see that it is a capacitor to ground and a series  inductor and 
resistance across it. This forms a parallel resonant circuit.
    	The Q is easily calculated and so it it's effective impedance as seen 
at the "input" of the L network. The HI-z side is across the capacitor and 
low-z  side is at the end of the inductor to ground. The impedance 
transformation ratio is the square-root of the Q.
The L-network is the easiest to solve because it is a basic parallel tuned 
circuit with a resistor in the inductive leg.
           If the L-network has a Q of 16 and the inductor is connected to 
50 ohms the input impedance, looking into the capacitor end, is 4x50 or 200 
ohms. And  resonance is the standard formula where XL =XC
	The problem with the PI-L network is that it will not match as large of an 
impedance range as a PI network alone due to the fixed output inductance. 
But it does attenuate harmonics just as well while at a lower Q allowing 
for less expensive less expensive band switches and coils. But the down 
side of it is that the intermediate impedance is higher than the output 
impedance so the "loading" capacitor has to be of higher voltage rating and 
larger capacitance as well.  Some old tube type AM broadcast transmitters 
often had 3 ,4 or 5  sections to assure harmonic suppression. This was 
particularly hard to tune correctly.

73
Bill wa4lav