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Talk:OS:Magnetic Electron Pump

Lasted edited by Andrew Munsey, updated on June 15, 2016 at 1:10 am.

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Discussion page for OS:Magnetic Electron Pump

Image:Magnetic Electron Pump 95x95.jpg

Solid state (no moving parts) electromagnetic device, said to be simple to build from readily-available, inexpensive materials, allegedly puts out four times as much energy as is required to run it. FE professionals point out measurement problems and math errors.


Forum - Official group forum for those wishing to replicate the technology, discuss the theory, improve on the design, and even commercialize the instructions, kits, and final applications

New Energy Congress Comments

The following comments come from members of the New Energy Congress.

Measurement Suggestion

On Oct. 7, 2007, NEC member, Congress:Member:Robert Indech, PhD PE wrote:

The design of this device appears to be that of a a 24-volt battery

source energizing a pulse generator, feeding into a transformer ("magnetic

electron pump"), whose output is then rectified and smoothed before being

placed into a load. The input power can be measured by placing a low value

sensing resistor where the fuse is indicated, monitoring the voltage across

it, and integrating this voltage over time, multiplying by 24 and the

resistnce value to get watts in. Watts out would require sensing both the

voltage across and the current through a sample load, and then integrating

over time. This measurement is relatively simple and, as the author is an

electronics engineer, completely within his capabilities. I would be curious

to see the results.

DC in versus DC out?

On Oct. 7, 2007, NEC member with 40+ years EE experience, Earl wrote:

The biggest red flag is not measuring DC in versus DC out, using a multi-pole low-pass filter on input and on output.

I get very nervous when seeing pulse measurements used to calculate COP.

It would be very simple for [1in2out] to put multi-pole low-pass filters on input and output so he could measure DC in versus DC out. Any Engineer can build a simple filter using 3 parallel capacitors in each leg, e.g. 10nF ceramic in parallel with 100nF ceramic in parallel with 100uF. I would suggest using three capacitor legs and two coils in each pi filter, for a total of 2 inductors and 9 capacitors in each filter. Capacitors, especially the ceramics should have short lead length. Inductors should not saturate. ONLY MEASURE DC.

Other Skeptical Comments

Bad Math?

On Oct. 8, 2007, Linda Damiani wrote: [

Interesting, but looking at figure 1, we can see a large error in

his calculation of input power. He shows an input pulse measured on

a 0.01 ohm shunt and peaking at 0.3V. This is 30A.

Following the rest of the math, and giving him an advantage by

calculating the area under the curve as if it were a linear ramp

(50%), we have a pulsed power of 30A x 24V x 1/2 or 360W. Now he

says the pulse duty factor is 12.4% giving a total average input

power of 44.64W not 8W as he states!

In reality the figure would probably be more like 55W based on the

exponentially-rising (>50% probably more like 63%) curve of the

current waveform, which is clearly larger area under the curve than a

linear ramp.

This looks like a simple case of bad math. I think he used a figure

of 3A peak (off by 10X) then failed to reduce this based on less than

full area under curve and then multiplied by 24V, getting 72W and

then multiplied by .124 (12.4% duty cycle) to get his approximate 8W


50+W input for 32.8W output is not so impressive when you add up the

numbers correctly!

Good Catch

: On Oct. 8, 2007, NEC member, Congress:Advisor:Kenneth M. Rauen writes:

: Good work, Linda!

: I missed it because I did not study the details as far as you did, though I questioned the numbers. Fresh eyes and minds help keep us on track. For as many of these as I have seen over the years, admittedly I am bleary-eyed at times from fatigue. Sorting the wheat from the chaff is a tedious chore, as there is a lot of chaff out there.

My Math

On Oct. 8, 2008, the inventor, 1in2out wrote:

The formula I used to calculate the input power is:

Power Average= I average X V average

I peak =.3 divided by .01= 30 Amps X avg. 707 = 21.2 Amps avg. X duty cycle (12.4%) .124 = 2.63 amps avg. X

V avg. V avg.= 24 Volts X duty cycle .124 = 2.98 V avg.

P = 2.63 amps avg. X 2.98 volts avg. = 7.84 Watts I calculated that I was losing about 1 to 2 watts in the power transistor which would bring the total input power to about 10 watts max.

The power supply I used to supply input power is a 24 Volt 12 Amp linear supply. The heat sink on it barely gets warm in the configuration I released to you. So there is very little heat generated anywhere except in the load.

IF my calculation is wrong then I am wrong

Calculations Wrong

: On Oct. 8, 2007, NEC member, Congress:Advisor:Kenneth M. Rauen writes:

: Duty cycle multiplication should be done for POWER, not for VOLTAGE and AMPERAGE individually! This is reducing the actual power by its square of the duty cycle, which is an error! Power is INSTANTANEOUS VxI. Again, the duty factor is only used to multiply the power by, not (duty cycle x V)(duty cycle x I). The calculated power input is far too low. Then there is the correction pointed out by the student, Linda, that the 0.707 coefficient for average does not apply 0.707 is the correction for the sinusoidal crest factor, not for an asymptotically exponential type of peak seen, so a higher average coefficient is needed. The end result of this is a greatly increased input power.

: Also in response to 1in2out, on Oct. 8, 2007, Linda Damiani wrote:

: Two errors glare forth here:

: First, I don't know where your .707 constant comes from, although it might be close for the waveshape you drew. If half the square root of two was what you were figuring, that only applies to sine wave shapes. This probably represents only a few percent error.

: The huge error comes from multiplying both the current and the voltage by the duty cycle! This was your mistake. That is equivalent to using the square of the duty cycle, a much smaller number than reality.

: The correct calculation involves figuring the pulse power V x I and then, after that is multiplied, and only once, multiply the pulse power by the duty cycle to get the average power. No fair doing the duty cycle multiplication twice!

: On Oct. 8, 2007, Directory:Peter Lindemann wrote:

: Here, [1in2out]'s math is definitely wrong. You can't factor in the "duty cycle" twice. You can average the current, but the applied voltage remains at 24 volts at all times.

: Great eye, to catch the error in [1in2out]'s own numbers, Linda.

Inventor Relents

On Oct. 8, 2007, "1in2out" writes:

I did not intentionally try and mislead anyone, I really thought I had something that worked, If I am wrong, as apparently I am, I sincerely apologize to all those involved and thank you for the time, out of I'm sure busy schedules, to correct me. I did this for neither money or fame, but only to try and do the right thing for the world. I am grateful for sites such as PESwiki that try and bring new energy technologies to us. Please keep up the good work.

Questionable Input Pulse Measurement

The following comment was received on Oct. 8, 2007

I have to question the measurement methods used here for the input pulse. Typically, that's what is called into question. It is not as easy as is suggested to expel magnet flux from an iron core. The area under the curve of the "power" in the input pulse is what is required to know for sure. I see no mention of that here other than a small hint that it might be wrong. The quoted COP of 4 could easily come from an error in measurement of the input power. If the output power is measured as DC from a FW rectified that's probably accurate.

I can easily check how much "current" is required to expel a given amount of magnet flux from a surrounding iron core. That's a no-brainer. But what comes into play is the inductance, time constant and resulting CEMF that will set the required voltage to set up that current. The problem is the CEMF voltage that sets the POWER required for a given current. It seems to me that's not being taken into account here.

So I doubt that this actually works at COP=4 with direct power supply drive.

Magnetic Bias

On Oct. 8, 2007, Linda Damiani wrote

It is interesting to see yet another project based on the idea of

putting permanent magnets within the flux path of a standard

transformer. Hitachi Metals Magnetics Division has made certain high-

tech ferrite cores for DC/DC converters and DC-biased inductors that

had built-in high-power neo magnets.

The purpose and effect was simple: to push the BH curve to one side so

that the net DC flow in the winding(s) would bring it back to the

center of the curve and thus not cause premature saturation of the core

under AC ripple.

In this way, these cores could be about half the size as would have

been needed to stay out of saturation with DC flowing in them. I don't

know if Hitachi Metals still makes these or not. There was no

suggestion of overunity or extra power, of course. Just the ability to

get more unsaturated inductance out of a small core under heavy DC bias.