The current limiting resistor can even be removed (set to 1 m Ohm). The circuit will draw as much power as needed for the "losses". If the losses kept low (high Q, low Lrser, high Lrpar), the voltage drop-down on the source can be minimized. Then the current through the coil and voltage will rise. At the moment roughly 0.7 V drops down.
See attachement for coil voltage and current.
It's amazing.

Aziz
See attachement for coil voltage and current.
It's amazing.

Aziz


I'm using physics intuition instead of calculations so far, I may change my thinking later.
I am very interested in studying this multipole feed idea - excellent idea like your others.
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