Quick FE by resonance

[gyulasun]

I don't quite follow this...I know that efficiency is highest when source and load impendances mach. but somehow I don't see this situation
within mains AC as source.

Hi Igor,

I was not clear enough in my previous mail but the point is trying to preserve and maintain the created parallel resonant circuit impedance which is Z=Q*XL where XL is the inductive reactance and Q is the loaded quality factor of the parallel resonant circuit.  (Say your L is 5H, then XL is 2*pi*50Hz*5H = 1570 Ohm inductive reactance, and in case of a loaded Q of 10 for the parallel circuit, this gives 10*1570=15.7 kOhm resistive impedance at 50Hz resonance with the 2.02uF capacitor. And if you connect this high impedance parallel circuit to the mains as you show in your schematics without any matching, then the result is a big shunting load across the parallel circuit, coming from the very low inner impedance of the AC mains, hence you lose the high impedance immediately. (It is like you short circuit the LC circuit from AC point of view.)

But wouldn't the nature of simple LC resonator render less amperage draw from mains (at res freq.)? and according slow down the speed of aluminium plate counter ?

Yes, but is able to render less amperage draw if the resonant higher impedance of the simple parallel LC circuit does not get shunted by the AC mains.  You can preserve the higher impedance by inductive or capacitive couplings to the LC resonator.  This is why I mentioned (welding) transformers where the secondary coil is of very low number of turns with respect to the primary coil, hence the impedance transformation can be a high ratio. If you do not wish to use transformers (the extra input transformer is needed for stepping down the mains voltage near to the low value of the coupling (i.e. the original secondary) coil of the welding trafo) then you may choose capacitive coupling too.
This latter is in the simplest case a capacitor of suitable value that connected in series with one of the AC mains connector and your parallel LC circuit receives a reduced (or divided down) AC voltage (a voltage drop develops across the series capacitive reactance).
This would make a welding trafo (and the extra trafo) unnecessary. Of course you would need a suitable inductance anyway with a L value of some Henry to form a resonant circuit and this coil should not be saturated by the 4-5 Amper or even higher reactive currents that will flow inside the resonant tank.

rgds,  Gyula

[alan]

Quick response for now:
kirchhoff current law and complex domain/plane for Z calcs.

[Paul-R]

Don't forget that if you are dealing with a hair dryer (FE researchers like to pay attention
to their grooming), then you can't separate the L and the R. They come together.
Paul.