Quantum Energy Generator

This looks very promising interesting*:

Download the open source specs here:


*Update May 10, 2014:

This technology is not home free, in that the over-unity self-powering loop has not been closed. Instead, what is claimed is that the output energy is greater than the input energy. The question is, how is the output energy measured?

In electrical systems, the power P (energy) is equal to the voltage (V) times current (I):

 P = V \times I

Therefore if one measures a high output voltage V and high output current I and multiplies them together to derive a calculated power P which is greater than the input power, it is then tempting to claim over-unity has been achieved. However, with alternating currents, as is the case with the QEG, there is a critical third variable, which is the phase difference \phi between the current and voltage:


The phase between voltage and current determines whether there is usable power.

Since both the current and voltage are oscillating in an alternating current, it is important that they be sufficiently synchronized in order to have usable energy. The phase \phi is an indicator of the degree of synchronicity. Whether or not the QEG satisfies this requirement is still an open question.

Updated May 21, 2014

A few days ago HopeGirl posted the video

which report the following numbers:

Output voltage = 1.9 kV
Output current = 1 Amp
Input power = 655 Watts

James commented that the output measurements are "peak readings".

Taking into consideration the phase \phi between output voltage and current, the formula, assuming a perfect sine wave for the alternating voltage and current, is

 \mbox{Output power} = 1.9{\rm kV}\times1{\rm Amp}\times\frac{1}{2}\cos\phi = 950{\rm W}\times\cos\phi

In the best case, in which output voltage and current are in-phase, \phi=0 and \cos\phi=1 then we have the efficiency = \frac{950{\rm W}}{655{\rm W}}=145\% which is over-unity.

In the worst case, in which the output voltage and current are out of phase by \phi=90^\circ, then \cos\phi=0 in which case there is no real energy output.

We need to know the relative phase between the output voltage and current before the claim of over-unity is made.