Let's suppose you have a vertical antenna with a base feedpoint
impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300
ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A
network consisting of 1275 ohms in *parallel* with 319 ohms of
capacitive reactance has the same impedance. (Note how the Xc isn't much
different from the Xc of the series circuit in this case.) If we put an
inductor with 319 ohms of inductive reactance in parallel with the
antenna (that is, from the base feedpoint to ground), the reactance of
the inductor cancels out the capacitive reactance of the antenna, and
we're left with 1275 ohms of pure resistance from the antenna base to
ground (that is, across the inductor).

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see at
that tap will be very nearly 1275 * k^2, and it'll be purely resistive
(no reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2 =
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of
the way up the coil.

Roy Lewallen, W7EL

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt