I wonder if this circuit would work, or perhaps a variation of it.

I know simulations will NOT reveal FE since the codes are not built that way, and usually the interesting stuff happens at the singularities which codes fail at or assume (and add some resistance etc..)

Has anybody tried anything like this before? I haven't yet, but will play with it soon.

EM

Hi EM and all,

I found such sharp notches in frequency response of different filters when the unloaded Q (figure of merit) of inductances used in the filter circuits had a very high value, higher than 1000 or more in the simulator, which are already not practical. There are simulators in which the built-in inductance model fails to consider the copper (or core) losses and in this case the quality factor is unreasonably taken as a high value just like for the capacitors.

Basically you have a resonant parallel tank circuit, L1-C1, and the tank circuit impedance at the resonant frequency of 159.15kHz increases to the highest value so the input current at this frequency gets minimized.

You included a 1 Ohm resistance (R1) inside the tank circuit. This adds up in series to the coil L1 own loss resistance because in practice you cannot access the loss resistance itself to take out power from across it.

In practice the quality factor for a 100uH coil can be around 300-400 (with pot or ring cores and using Litz wire) at around 100-200kHz and this gives an "r" loss resistance for L1: r = X

_{L}/Q so for your L1 the series loss resistance is 0.249 Ohm if the assumed Q of L1 is 400. (X

_{L}=2*3.14*0.159*100=99.8 Ohm divided by 400 gives 0.249 Ohm.)

Your 1 Ohm of R1 adds up with this to 1.249 Ohm. But now the resultant Q for the tank circuit will be less than 400, namely, Q= (X

_{L})/1.249= 79.9 only, lets round it to 80! and assuming infine Q for C1.

A tank circuit like this will have a resonant impedance Z=Q*X

_{L} where X

_{L} is the inductive reactance like before.

Calculation gives Z=80*2*3.14*0.159*100=7.988kOhm impedance at resonance (0.159MHz) and the input current taken from the generator should be around I=V/Z=1Volt/7.988kOhm=125.18uA (microAmper). Better circuit simulators should give out this amount of current too as a result of an exact simulation, if their built-in inductance model can be instructed to calculate with Q=400 and you include R1 of 1 Ohm into the circuit too.

So inside the tank circuit the current should be Q times this 125uA, i.e. 80*125uA=10mA in my example. This current makes a voltage drop of 10mV across the R1 (1 Ohm) resistor at resonance.

The power dissipation in R1 is .01*.01*1=0.1mW at resonance and the power taken from the generator is 0.125mW at resonance {(V*V/Z), (1*1)/7988=.125} so efficiency is 80% [.1/.125=.8]

Basically this example is exactly the same problem than the case of a rotoverter: how to draw power from a resonant tank circuit (that is constituted by a windings of a 3-phase motor and run capacitors) without loading the source and taking advantage of the coils quality factor? Peter Lindemann may shed some light on this later on because he started to examine the rotoverter principle and how to take out real power from a volt amper reactive power that usually circulates inside a parallel resonant circuit, see here Esa's mail:

http://www.overunity.com/index.php/topic,203.msg82319.html#msg82319 rgds, Gyula