A "pure" LC circuit is lossless... both the L and the C are pure
reactances, and neither will (or can) dissipate any power.
Not dissipate but surely radiate in a closed box!
It will not radiate efficiently.
The radiation resistance of a physically-small coil is going to be
very small... a small fraction of 1 ohm. What your transmitter is
going to "see" will consist of:
- The radiation resistance of the circuit (probably less than 1 ohm),
in series with
- The loss resistance (mostly in the coil)
in series with
- Whatever reactance remains after the series L and C resonate against
one another.
If the coil is "perfect" (lossless) and the L and C reactances are
equal and opposite, then all you'll be left with is the radiation
resistance.
If it's in a closed (shielded) box, matters get *worse*. The presence
of the reflections from the walls will reduce the radiation resistance
even further.
Will your transmitter be happy with a load which looks almost exactly
like a short circuit to ground.
Yes, it's for a single frequency. Since an antenne is also a LC-circuit
that doesn't get warm but radiates in fields.
And it radiates with reasonable efficiency, precisely because it's
physically large - it has a substantial radiation resistance.
This will not be the case for a physically-small LC circuit.
I had it all open built so i suspect mantelwaves etc. But on paper a series
LC circuit forms a loading of a particular Zo. This statement is still right?
No, I don't think so... at least, not in the way that you think.
If you show us the specific circuit you're thinking of (components and
their values, wires, and desired operating frequency) we can analyze it.
Lets say an input on a semiconductor p.a. stage with an LC series circuit..
A series LC circuit is not, by itself, going to present a resistive
impedance.
--
Dave Platt AE6EO
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