Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[Grumage]

Are you sure about your value of A_(RMS) = (8V*0.707=5.657V) ?   Take a look at the attachment below...
Also  0.8 / 1099000 = 0.000000728 and that means 728nH  (not 0.07μH)

Anyway, the reactance of a 728nH coil at @175kHz is 0.8Ω  (+ any DC resistance of the coil) and you would be hard pressed for an amplifier that can handle a load around ~1Ω.
Most of the commercial amplifiers are designed for 50Ω loads.

The relevant equation for the impedance of your coil at a given frequency is:
Z = 6.28*L*f + R_DC
where, Z: AC impedance of the coil in Ohms, L: inductance in Henrys, f: frequency in Hertz, R_DC: The DC resistance of the coil in Ohms

Also, the previous equation amended to account for R_DC is:
L = (A² – P*R_DC) / (6.28*P*f)

Dear Verpies,

Many thanks for your responce, and yes that should have been 14 V p/p not 8. I am using an audio amp right on it's top edge of performance. It is rated at 4 to 8 ohm impeedance.

Cheers Grum.

[verpies]

I saw different equation for current flowing through inductor  Not much more complex but exponential one. Am I wrong ?

The formula with time in the exponent is for aperiodic rectangular voltage waveform applied to an inductor.  For a sinusoidal periodic waveform, the inductive reactance is X=2πLf

[verpies]

Many thanks for your response, and yes that should have been 14 V p/p not 8. I am using an audio amp right on it's top edge of performance. It is rated at 4 to 8 ohm impeedance.

So if your sine waveform has 14V peak-to-peak then its amplitude +A_MAX = +7V and its RMS value is A_RMS= 7V * 0.707 = 4.949V

Assuming that your R_DC=0.2Ω and using the equation L = (A_RMS² – P*R_DC) / (6.28*P*f)
the inductance of your coil should be 375nH to transfer 40W to it at 175kHz and your amplifier should be stable with a 0.4Ω load (I don't think your amp can handle that!)

To have an 8Ω impedance @ 175kHz the coil should have the inductance of 7μH.
Unfortunately at 4.949V_RMS and 8Ω impedance only 3W will be transfered to the coil

To get 40W into an 8Ω load, the RMS value of the sinewave would have to be A_RMS = 17.9V, because:
A_RMS = SQRT(P*Z)
where P: the average power in Watts, Z: is the AC+DC impedance of the load in Ohms (e.g. 8Ω)

[Grumage]

So if your sine waveform has 14V peak-to-peak then its amplitude +A_MAX = +7V and its RMS value is A_RMS= 7V * 0.707 = 4.949V

Assuming that your R_DC=0.2Ω and using the equation L = (A_RMS² – P*R_DC) / (6.28*P*f)
the inductance of your coil should be 375nH to transfer 40W to it at 175kHz and your amplifier should be stable with a 0.4Ω load (I don't think your amp can handle that!)

To have an 8Ω impedance @ 175kHz the coil should have the inductance of 7μH.
Unfortunately at 4.949V_RMS and 8Ω impedance only 3W will be transfered to the coil

Cheers Verpies,

I think I need some "Dilithium crystals"!! That should do the trick

MORE POWER

Cheers Grum.

@ Zeitmaschine,

Good to see you back, I think you should have mirrored the second pics, they would look really cool

[verpies]

I think I need some "Dilithium crystals"!! That should do the trick

Yeah, Scotty, now you have only ~5V_RMS at the output of your amplifier and you should have ~18V_RMS in order to push 40W into an 8Ω load.

A_RMS = SQRT(P*Z)
where P is the average power in Watts and Z is the DC+AC impedance of the load in Ohms (e.g. 8Ω)