Sent: Monday, February 21, 2005 9:11 AM
Subject: Re: [Amps] Fwd: Pi-L In-circuit Adjustment Question
>
> 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.
Okay boys and girls, IMO, there are several miss-statements in the above.
Your homework for today is to identify them and correct them lest some of our
newer readers take this as gospel and loose a beer to someone later on.
(((73)))
Phil, K5PC
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