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Overunity Machines Forum



WHAT DO YOU THINK

Started by magnetman12003, May 18, 2010, 02:40:12 PM

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gyulasun

#5
May 19, 2010, 02:39:55 PM
I think the output voltage is not a pulse but sinusoidal shape because the rotating cylinder magnet is diametrically magnetised, hence changing poles in half turns, there must be zero voltage values between the positive and negative maximums. So the diode bridge is justified, with one diode he would lose output (due half wave rectification only).

I am not sure on his text under his video where he wrote his load resistor is in series with the output, is it a mistype or he measure the voltage via the resistor?

rgds,  Gyula

gyulasun

#6
May 19, 2010, 04:47:45 PM
Ok, Tom answered my question on youtube. so he placed the load resistor in paralell with the diode bridge output. This is good news.

Now I think he would need a DC/DC stepdown converter to convert the high voltage into a stabilized 12V DC output for looping back. His input power need is 12V @ 80mA, this is what the converter should be able to produce from his present 315.9V DC output which would feed the DC/DC converter input.


Bruce_TPU

#7
May 19, 2010, 10:10:52 PM
Hi Tom,

I sent you a message on Youtube.  Great work!  I love your setup.  If I understand you right, your input current is unaffected by the proximity of your drive magnet to the flux field of the coils.  Is that correct?

If that is true, why not build two more of your "drive generator" setups, and place one behind your flux coils and another above your flux coils.  Rectify all three outputs and wire in parallel.  This would give you the same voltage at three times the amperage.  I would say you could even add more, based on what I saw of the strength, and breadth of that flux field.   :o

Cheers,

Bruce
1.  Lindsay's Stack TPU Posted Picture.  All Wound CCW  Collectors three turns and HORIZONTAL, not vertical.

2.  3 Tube amps, sending three frequency's, each having two signals, one in-phase & one inverted 180 deg, opposing signals in each collector (via control wires). 

3.  Collector is Magnetic Loop Antenna, made of lamp chord wire, wound flat.  Inside loop is antenna, outside loop is for output.  First collector is tuned via tuned tank, to the fundamental.  Second collector is tuned tank to the second harmonic (component).  Third collector is tuned tank to the third harmonic (component)  Frequency is determined by taking the circumference frequency, reducing the size by .88 inches.  Divide this frequency by 1000, and you have your second harmonic.  Divide this by 2 and you have your fundamental.  Multiply that by 3 and you have your third harmonic component.  Tune the collectors to each of these.  Input the fundamental and two modulation frequencies, made to create replicas of the fundamental, second harmonic and the third.

4.  The three frequency's circulating in the collectors, both in phase and inverted, begin to create hundreds of thousands of created frequency's, via intermodulation, that subtract to the fundamental and its harmonics.  This is called "Catalyst".

5.  The three AC PURE sine signals, travel through the amplification stage, Nonlinear, producing the second harmonic and third.  (distortion)

6.  These signals then travel the control coils, are rectified by a full wave bridge, and then sent into the output outer loop as all positive pulsed DC.  This then becomes the output and "collects" the current.

P.S.  The Kicks are harmonic distortion with passive intermodulation.  Can't see it without a spectrum analyzer, normally unless trained to see it on a scope.

gyulasun

#8
May 21, 2010, 02:14:20 PM
Hi Tom, 

It is very good you will have a circuit for converting the high voltage output to 12V DC, as you wrote to me in your youtube answer.

I think this link also has a solution for this task:

http://www.powerint.com/sites/default/files/product-docs/lnk302_304-306.pdf 

The IC is the LNK302-304-306 LinkSwitch family from Power Integrations, see Figure 5 schematics in page 4 of the above link.  Hopefully you can obtain such IC, and the recommended printed circuit board is shown too. 

The circuit in Fig. 5 can work from any AC input voltage from 85V to 265V, at the input there is the two 1N4007 diodes in series, you may use two of your fast diodes instead if you wish. Diode D1 must be a fast diode however. The 12V DC output voltage is stabilized for up to 120mA max load current so your present 80mA input current need is well covered.

Of course there are several other types too, I show this just to inform you or others here.

rgds, Gyula
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