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Author Topic: Skycollection's "Pentafilar Pancake" inductively coupled "Overunity Potential".  (Read 185319 times)

Offline skycollection 1

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Yes, it is correct, I have been working with my circuit for many years with very good results, it has a COP>, <or = 1 this is possible, with this circuit I can move any type of motor that has magnets,
The advantage is that this circuit  can be connected  of all kinds of motors, as an example you can see my MAGNETIC GYROSCOPE, with three circuits and with three coils, with low power consumption, I prefer my circuit compared to John's Bedini circuit.
I myself designed this circuit in 2012 with 12 VOLTS IN, my question was how to protect my transistors with 24 volts IN, I already have the answer that is only to change the 1K resistor for a 2.2K resistor.
In a few days y will post a new video with a new generator and battery charger, in this motor i have two similar circuits connected in parallel with very low consumption of energy.

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Offline gyulasun

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...
I myself designed this circuit in 2012 with 12 VOLTS IN, my question was how to protect my transistors with 24 volts IN, I already have the answer that is only to change the 1K resistor for a 2.2K resistor.
...


Dear Jorge,

Please allow me to note that the goal of changing the 1 k resistor to 2.2 kOhm as I suggested was to reduce any unnecessary increase in input current draw from the 24 VDC input, this resistor does not directly protect the transistors as it did not protect them at the 12 V input.  You can run your circuit with the 1 k resistor too, just check with a DC Ampermeter how the average input current draw increases and whether a higher input current draw gives any advantage in the operation of your setup. 

IF you do not use the BEMF output for anything (i.e. you leave it unloaded), then the transistors are protected only by the neon bulb which limits the collector-emitter peak voltage to 80 - 90 V, hopefully the neon bulb will not be damaged by the higher kickback spikes the 24 V input creates versus the earlier 12 V input.
I mean the neon bulb may get overloaded and may get burnt just like any light bulb which receives extra high voltage during operation.
And the moment the neon bulb fails and becomes an open circuit, the transistors will have no protection any more, and in case the kickback spikes have higher than 250 V peak amplitudes, the transistors will be toasted too.

This is why it is advisable to load the BEMF output as a precaution.  Or use two identical type neon bulbs in parallel, to reduce possible overload for a single bulb. If you can see a starting dark colouring on the inside glass wall of the neon bulb, it would be an indication of a starting overloaded operation.

Note also that the neon bulb dissipates some part of the coil kickback energy because it conducts current whenever the kickback pulse amplitude is higher than trigger voltage for the bulb, 80-90 V, this is how it protects the transistors. I mention this to realize that eventually the load for the BEMF is the neon bulb itself whenever the kickback spikes across the coil exceed 80-90 V peak values (could be seen by an oscilloscope).

Hopefully these notes are not confusing you but help better understand your circuit behaviour. 
   
Greetings,  Gyula

Offline skycollection 1

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Gyula, thank you very much for your valuable help, it is very important to me since I did not study electronics and this will help a lot my development in future projects.
You are really right to mention the neon bulbs, I usually see how my neon bulbs burns little by little, it turns black and after some time the transistor is damaged, (I am talking about 12 volts with a 1K resistor) now to apply 24 volts and I know that I should change the 1K resistor for a 2.2K one AND AT THE TIME PLACE RL a led bulb in the BEMF.
And thirdly, is to place two neon bulbs in parallel, I am already doing it and I am going to do several tests.
I AM USING THIS CIRCUIT ONLY TO ACTIVATE THE MOTOR, MY COILS ARE bifilar, ONE COIL GOES TO THE CIRCUIT AND THE OTHER IS PICKUP COIL, WITH A DIODE RECTIFIER AND A CAPACITOR, WITH THIS I CAN CHARGE BATTERIES.
In a few days i will post a new video, i have a new generator and i hope you can see on YOUTUBE
Thanks for your help

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Offline skycollection 1

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SPECIAL THANKS TO GYULA, THIS IS MY NEW PROJECT, I HOPE YOU HAVE A TIME TO SEE IT

https://www.youtube.com/watch?v=kgINt2_68P8&t=48s

Saludos desde Mexico

Offline gyulasun

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Hi Jorge, 

Very nice job, thanks for showing the setup running.   

Will you run the setup from 15 or 24 V DC input too? No problem if you do not wish to do so of course. 

Do you happen to have a third LED bulb (120 V, 6 W, same type as the other two LEDs) ?  If you have one more, you could feed it directly from the 120 V AC mains and place it close to the other two LED bulbs to see the brightnesses of all the 3.

This way you could see whether the consumption of the two LED bulbs operated from the BACKEMF outputs approaches the nominal 6 W power the 3rd bulb is supposed to consume from the AC mains. 

Also, if you load each 66 V DC output by say a 2 kOhm resistor, then you would have two more useful loads besides the two LED bulbs and could see how the overall efficiency comes out. 

The unloaded 66 V DC outputs will drop to a lower voltage level with the 2 kOHm loads and the voltmeters would measure how many volts will be across them. Suppose for example the voltmeters will show about 20 V each, then the dissipated power in the 2 kOhm resistors would be (20V x 20V) / 2000 Ohm = 0.2 Watt each. Use 1 Watt rated resistors.

It is possible the input power to your circuits would be higher than 12 V x 0.23 A = 2.76 W when the 2 kOhm loads will be connected, you will see this.

It is also possible that the brightness of the two LED bulbs will also change (reduce a little) when the two 2 kOhm resistors are connected, you will see it.

Thanks,
Gyula

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Offline skycollection 1

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Thanks Gyula, this project is 100% interesting and in the following days I will do some experiments as you describe in your comments, with 1K resistors in the current outputs and see what results it gives through those resistors, thank you very much for your support which will be very useful in the future.

Offline skycollection 1

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Hi Gyula, I did the experiment by placing a 2.2K resistor in the current output with the following result, from 66 volts the voltage drops to 40 volts through that resistance, the rotor did not stop, it did not affect its rotation, I donĀ“t know how to interpret this result, maybe it is necessary to try other resistance values, what do you think ...? With a resistance IK the rotor slow down with only 4.5 volts.

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Offline gyulasun

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Dear Jorge, 

Many thanks for the tests! I assume you connected the 2.2 kOhm resistor in parallel with the + and - outputs of the R rectifier diode bridge, right?  i.e. to the same pins where you discharged the puffer capacitors in the video. 
 
So in this case the 40 V DC across a 2.2 kOhm dissipates 40 x 40 / 2200 = 0.72 Watt power. If you use a second 2.2 kOhm resistor in parallel with the other + and - outputs of the other R rectifier bridge, then it will also dissipate about 0.72 W power. 

Question is whether the input current increases from the 0.23 A (at the 12 V input) when you connect the first 2.2 kOhm resistor and then the second 2.2 kOhm in parallel with the outputs? 

This would be good to know, especially because you wrote the rotor did not stop, the 2.2 k did not affect its rotation. 
 
You mentioned that with 1 kOhm load in parallel with the output, the rotor slows down and the 66 V drops to 4.5 V. Question would be how the brightness of the two LED bulbs running from the BACKEMF outputs changes during the 1 k and during the 2.2 kOhm loads?   

Thanks
Gyula

Offline skycollection 1

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Ok thanks, i have a new test now with both diode rectifiers: 

NUMBER ONE RESISTANCE 2.2K= 66 VOLTS DROPS TO 38.7 VOLTS
NUMBER TWO RESISTANCE 2.2K= 66 VOLTS DROPS TO 35.3 VOLTS

INPUT VOLTAGE WITHOUT RESISTANCES 12 VOLTS X 0.23 = 2.76 W

INPUT VOLTAGE WITH ONE RESISTANCE 2.2K BOTH SIDES  12 VOLTS X 0.30 = 3.6 W

LEDS BULBS BOTH BEMF OFF
what does this result mean?

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Offline gyulasun

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Hi Jorge, 

Thanks for the new tests. To understand the results, first some calculations:
power dissipated in one of the 2.2 k resistors is 38.7 x 38.7 / 2200 = 0.68 Watt
power dissipated in the other  2.2 k resistor   is 35.3 x 35.3 / 2200 = 0.566 Watt
Summing these two numbers gives 1.246 Watts output power.

Because you experienced the LED bulbs across the BEMF outputs go off when the 2.2 k resistors are used, this means the following:

The two 2.2 kOhm resistors impose already such a high load for both bifilar coils that the peak amplitude of the BEMF is not enough any more to operate the 120 V LED bulbs. You know there is a minimum input voltage specified for LED bulbs by the manufacturer and in your setup the peak voltage of the BEMF was so far enough to light up both LED bulbs when the 2.2 k loads were not present. 

You could use for instance 3.3 kOhm or 4.7 kOhm load resistors instead of the 2.2 kOhm, so the LED bulbs may remain lit at a reduced brightness,  this depends on the trade-off how much part of the total BEMF you use up by the resistors and the rest of the total BEMF may light up the LED bulbs, albeit at a reduced brightness.  Let me note (although you may know also) that LED bulbs have a nonlinear voltage-current characteristic, this means that near their threshold ON voltage range a smal change in the voltage across them causes a high current change hence brigthness change. This is why we cannot really estimate actual power a LED bulb may consume by our naked eye. 

You know that the total output power comes from two sources: the rotor magnets induce current in the coils (normal Faraday motional induction) and the input current from the 12 V source charges up the coils and this latter appears at the BEMF outputs when the transistors switch this current off and the coils can release their stored energy. 

This is how I see this. 

Greetings
Gyula

Offline skycollection 1

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Ha Ok I am going to get other resistors and continue experimenting with this value 4.7 Kohm, on the other hand I did another experiment without using the capacitor gave me the following result:
COIL NO. 1.- 11.38 VOLTS OUT
COIL NO. 2.- 11.75 VOLTS OUT
INPUT VOLTAGE 12 VOLTS X 0.20 = 2.4 W

My e-mail is: skycollection@hotmail.com
In case you have some time and you want to send me some advices

Thanks a lot

Free Energy | searching for free energy and discussing free energy


Offline gyulasun

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Dear Jorge, 

Please give some more details to better understand your latest experiment. 

So you removed the puffer capacitors from the output of the R full wave rectifiers if I understand you correctly. Is this correct?

And what about the LED bulbs? Did they remain at a decent brightness like they were with the capacitors in place as you showed in the video earlier? 

Thanks for your email address but I would prefer communicating openly as we have been doing now.  If you have special question, then please send a personal message via this forum, ok? 

Thanks
Gyula

Offline skycollection 1

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For my last experiment I removed the diode rectifier and the capacitor as well, and i connected a new diode rectifier without capacitor, both output voltages give almost 12 volts each coil, related to the BEMF and the led bulbs kept their brightness the same as before , it only decreased the engine consumption a little.
My second experiment was to connect a 7 amp acid battery and it began to charge very well, the BEMF bulb went out and the motor consumption remained the same, without changes, 2.7 Watts.
I think that the capacitor is not necessary to charge a battery, my motor continued to rotate even with the charging battery connected.

Offline gyulasun

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Hi Jorge, 

Okay, thanks for the further explanations on your latest tests. 

The diode bridge you labeled as R in your video schematic surely rectified both the voltage induced by the rotating magnets and the BEMF (coming via magnetic coupling from the switched-out winding) and the puffer capacitor stored both of these as a filtered and averaged DC voltage. 

When you remove the capacitor, then an unfiltered DC appears across the output and the voltmeter cannot show the voltage values correctly, this is what I think. 
Okay that the LED bulbs kept their brightness the same as before during this test. 

I assume that both the earlier and the new diode rectifiers have had at least 250 V or higher reverse breakdown voltage ratings. For such bridge rectifiers the best is to assemble the bridge from individual fast switching diodes like the 1000 V, 1 A rated UF4007 or similar types. Off-the-shelf diode bridges with the 4 built-in diodes could be used but these are not fast switching types and good for the mains frequencies but not so good for pulse rectifiers. 

Regarding your second experiment charging a 7 Ah battery and you found the BEMF bulb went out while the battery was being charged: This is normal and expected because the 12 V battery virtually appears across the output winding (and via the magnetic coupling it also appears across the switched winding) and will keep all the induced and the BEMF voltage levels at its 12-13 V amplitude while being charged. Batteries are very good voltage limiters. 
So this max 13 volts level is not enough any more to light up the LED bulbs which may be able to emit light when the peak voltage across them is at least as high as 75-80 V, maybe little higher (manufacturer's specification clarifies this lower limit).
You nicely experienced the trade-off in voltage hence power levels between the diode outputs and the BEMF outputs by observing the effect of the 1 kOhm and the 2.2 kOhm resistors on the brightness of the LED bulbs: 

- with the outputs unloaded the output voltage across the capacitors was 66 V, LED bulbs were bright and rotor speed was at maximum, 
- with  1 kOhm load the output voltage was 4.5 V only and the rotor slowed down, LED bulbs went off, 
- with 2.2 kOhm load the output voltage was around 40 V, the rotor speed was not affected and LED bulbs went also off as you wrote- with the charging battery the output voltage was surely around the battery voltage, the rotor speed was unaffected and the LED bulbs were off.
 
This shows that when the averaged voltage level was the unloaded 66 V DC across the outputs, the peak voltage spikes across the switched coil was surely around the 100-110 V voltage for the LED bulbs, these spikes gave the brightness for them in a pulsed operational mode. And the moment these spikes amplitude was reduced i.e. attenuated by the resistor loads (or by the charging battery) below the lower threshold voltage for the 110 V LED bulbs, the bulbs went off and the induced and the BEMF power went into the resistors (or into the charging battery).

Hopefully these additional comments help understand better the operation of your interesting circuit setup.

Gyula

Offline skycollection 1

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Gyula, very interesting, i really appreciate your time in this project, i am experimenting with the circuit, now i have some fast diodes, i am going to build a new rectifier and i get some capacitors 2200 Uf 16 volts, i will make new experiments with this components.
In my new circuits I am incorporating a green terminal where the 1K resistor is connected, to make the conversion to 24 volts I am changing that resistor for a 2.2k , so easy with a screwdriver and I can now apply 24 volts.
Regarding the BEMF I am connecting a 120 volt 6 watt led bulb in order to absorb the voltage peaks, which I have been very successful, the neon bulbs (a pair) are kept off and in case the bulb of led fails, the two neon bulbs will come into operation while I replace the damaged led bulb, so far I have not lost any transistors.

 

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