LCR Over Unity Generator

Do you think it is possible that the load is receiving extra energy from somewhere outside of the system? It is impossible to get more energy out of a system than what goes into it. Unless of course an outside energy is being drawn into the system then this becomes perfectly feasible that absolutely does not break any laws of physics. Solar cells and wind generators do this all the time all day long in which the environment supplies the energy. We no longer have to put any energy into the device in order to receive the free environmental energy.

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If the load resistor is directly connected to the battery there is a lower current because the battery voltage is lower. The meter corresponds exactly with an Ohm's law calucation.

When the load resistor is placed back on the capacitor that has a much higher voltage than the battery then the load's current becomes higher, because the voltage is higher and again the Ohm's law shows this to be exactly true and so the meters are indeed measuring correctly. The output wattage is larger than the input wattage. It is believed that the higher voltage across the load resistor is originating from the collapsing magnetic field from the coil and because there is no diode to capture this energy then a vibratory oscillating energy between the coil and capacitor is supplying energy into the load resistor, which should not be over unity, but it is.

Here are two versions of the same schematic for this circuit:

[drchocolat]

dude this channel is dope! I'm so happy I found you. it's not super confusing. I enjoy your smooth approach to over unity device's. appreciate your time and effort!

[smokyatgroups]

Hi Marc the electrolytic at the output (C3) can look high impedance to fast transients & high frequencies. Your meter's true rms frequency range and shape factor, likely doesn't like these impulses on the output so it's not as accurate as we might think. I suggest bypass C3 with 0.01 disc ceramic.

[DS-oldv]

Hey dude, do you sell any of your completed devices? I'd like to purchase one! Thanks

[area]

Interesting...just how much energy can a caducious ferrite hold and how long will it keep the voltage high...since it cancels the magnetic field you would think the fly back would be low and act like a resistor...I suggest taking voltage drop reads from the ten k resistor and calculate amps from it but you might want to install another resistor like a one ohm in series and take direct voltage reads from it since ohm laws says one amp through one ohm a read of .00123 volts will be exactly that in amps I use the ten watt 1 ohm for current anything over ten volt drop would probably ruin the resistor.

[smokyatgroups]

It's hard to make accurate measurements without a fast oscilloscope. The idea you are exploiting is "Bearden's" mass-less displacement current, whereby the capacitor "sees" the voltage when FET turns ON for preferably < 100uSec (which is less than enough time for electrons to form skin effect current). So before electron current can flow.. the FET turns off. Capacitor acquires electrostatic charge but needs to pull the electrons from somewhere else ... presumably the ambient. The suggested 0.01 disc capacitor might interfere with this process even if it makes the meter read more accurately.

[tomgeorge]

Are your DMMs rated at 145kHz? I don't think so, your meters cannot measure AC voltage or current at that frequency, hence no AC meter reading.

The capacitor is in a pulsed current circuit, it will contribute to the current flow, read what I have posted in the previous video comments.
You have to measure the resistor ONLY current to to work out the power dissipated in the resistor.
Your DMMs are trying to measuring the pulsed DC current that flows through both capacitor and the resistor, this is an average current and NOT the resistor current.

The instant you introduce pulsed DC current, which you have, you cannot use the DC version of OHMS law, you have to consider the capacitor and inductor IMPEDANCE.

When you take the capacitor out of circuit, you still have a pulsed DC current and the coil in circuit.

When you demonstrate "under unity" you are working with DC, different ball game.

All because of AC and your components IMPEDANCE.

[paolomarani]

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