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Author Topic: Reactive power - Reactive Generator research from GotoLuc - discussion thread  (Read 261690 times)

Offline gotoluc

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One thing is for sure if the energy companies see people or hear of people doing this they will not be pleased, and take action. People have been put in jail for similar schemes involving the stealing of energy using various means by manipulating their grid.

What I don't get is if it is actually free energy then why can it not be done with a battery and true sine wave inverter, until extra energy is shown using a source of AC other than the grid I don't see valid gains.

Cheers

Hi Farmhand and everyone,

please note, this circuit idea is not to get a free lunch using the Grid.

I maybe attaching my test circuits to the grid (because it's convenient) to demonstrate what it can do but that is only for demonstration purposes and not the intended use of this circuit if in the end is proven to work.

Ideally the developed circuit would be fed by low impedance DC batteries as this research is a possible solution to give power to the poor people of this world who have no grid and no way to make a living or even live if the can't pump water to their crops.

So please try to help in a positive way instead of trying to find faults as I know you are quite a capable individual.

What would be needed next is a special circuit that would take high current lead acid battery DC and make AC BUT the circuit needs to handle the return (reactive circulating current) back to the batteries. An off the shelf Inverter cannot do this.
So if you can develop a DC to AC circuit that can do this you would be my hero.

Thanks for your time

Luc

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Offline gotoluc

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Hi Luc,
if you have an 12 Volts DC to 120 Volts AC inverter, just try to power your AC drive motor with the inverter
from a 12 Volt battery and measure at the battery, if  drawing load from the generator will
see any increase into the inverter at the battery.

Okay, this adds losses from the inverter also, but this way you can see quickly  if the DC input power
changes, without needing a DC driving motor !

Regards, Stefan.

Good idea Stefan

I will see if I can find one. This will also eliminate the need to do a scope test on the motor (which I already did but did not post a video)

Luc

Offline gotoluc

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You are right.

The best way would be, if LUC would try it again with a high efficiency DC motor in front of his generator and
see, how much DC power he needs to drive the generator. If the DC power into the drive motor will also
not increase, he might be able to scale it up and use the MOT output also via a FWBR rectifier to charge up a supercap
and then run from it the DC motor and close the loop....
But this would probably require at least a factor 10 in MOT size as it seems to depend on iron core.

Maybe it can be done smaller with using higher frequencies and Ferrite transformers...

Regards, Stefan.

Stefan and everyone,

my most current tests are now leading me to believe the circuit needs a LOW Impedance power source.
My new double MOT circuit attached to the grid can now deliver 50 Watts and still maintain a 90 degrees phase shift but when I attach it to my 1000 watt Alternator generator I can only get 29 watts max without affecting the prime mover.

This leads me to believe that my Alternator Impedance which is much higher than the grid is causing losses as the reactive re-circulating power is being wasted in Alternator coil resistance.

If we consider two of the supposed to be looped generator videos (Valy being the most recent) one thing they have in common is they have over size Alternators. Something in the range of 30KW or so.
Obviously these Alternators would have extremely low Impedance compared to mine and would not waste the return current and also deliver more because of the minimal losses.

So considering this, the possibility of looping my small scale setup is next to impossible. What we would need to do is come up with a solid state version of a Battery DC to AC switch (like suggested to Farmhand above post) and eliminate the generator and loses altogether.

Looking forward in the participation of the bright minds on this forum

Luc

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Offline totoalas

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Free electricity while charging battery   part 3   .yt   dreamyear     
 :)

Offline MileHigh

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Permit me to make a suggestion.

I believe the essence of this is the presumption is that the power factor from the mains is zero wile still powering a small load, implying an exact 90 degree phase shift between the mains voltage and the current.  It appears that the power factor meter may not have sufficient resolution to show a true zero power factor.

If you set up the scope in x-y mode and make a lissajous figure, you will see a circle or a near circle when you look at the voltage and current in x and y and you adjust the variable gain of one channels to get as close to a circle as possible.  The problem is that it's hard to judge by eye if it is a true circle or if there is a slight phase shift from 90 degrees.

So the way around this problem is to shift the phase of the voltage or the current by another 90 degrees.  Now when you compare the current with say the 90 degree phase shifted voltage, you will see a diagonal line angled at 45 degrees.  If you don't see a perfectly straight diagonal line on the scope but instead see a long thin oval shape, that is telling you that the phase shift between the actual voltage and current waveforms is not exactly 90 degrees.  So this trick gives the scope display a way to show the deviation from the 90 degree phase shift very easily.

The only requirement is to create the 90 degree phase shift with a small circuit.  I am not going to get into the circuit itself, but my first thought is to use an op-amp as an input voltage integrator that is isolated from the mains power with a small step-down transformer.  It's a trivial circuit that could be powered by a pair of 9-volt batteries.  Since the circuit is isolated via the step-down transformer, there no problem connecting the op-amp ground to the scope ground.

If the whole thing worked properly, when Luc started to draw a small amount of power to go into his load, you would start to see the small phase shift on the scope display.  You also have the option to turn up the gain on both channels of the scope so you are only looking at a segment of the diagonal line.  That gives you more precision to detect any phase shift.  If you see two separate and distinct diagonal lines that would indicate that there was a tiny phase shift and thus a real power draw from the mains.

MileHigh

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Offline poynt99

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I donĀ“t think so.
Just short out the cap and place a positive voltage on the channel 1 from the grid.
Then a positive input current will flow through the transformer coil
which is registered right in the shunt as a positive value !

So no need to invert the scope channel !

Regards, Stefan.
Yes, you are absolutely correct, IF you wanted your power to come out as a positive result instead of negative.

So let me repeat:

Power sources, when measured properly, have a NEGATIVE POLARITY, and elements that dissipate power have a POSITIVE POLARITY.

So in the case of my diagram, and if you invert channel two as noted, you will not only get the correct polarity of the resulting grid power, but the phase relationship between the voltage and current will be displayed correctly. This is actually basic electronics, but most people forget. If you don't believe me, start at the bottom of the grid generator and going clockwise, use your finger and trace around the circuit in a loop, noting in your mind the polarity of the probes. You should be saying to yourself "...negative.....positive....positive....negative". Notice the inversion? To be correct, we should have the CH2 probe inverted so that we would say to ourselves as we travel around the circuit..."negative...positive...negative...positive". The reason we DON"T connect the scope this way, is because the scope gnds must be commoned. So instead, we connect it as shown, and to be 100% correct, we should invert channel 2 in the scope. I know no one does this, but they should when testing under the circumstances we are dealing with here where phase and power polarity means everything.

Offline poynt99

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Hi poynt,

sorry for staying out of the discussion. I needed to do this to stay focused on my tests in hopes to better understand the circuit with my way of understanding which is non conventional I know but this is what allowed me to come up with these ideas.

So, don't take it personally. Your knowledge is and has been very helpful.
No problem Luc.

Quote
I see you have posted the below diagram for probe position, which are my probe positions. However I see you say to invert Channel 2 (current probe) is this the correct way?... and do you wish I test it this way and post the results.

Thanks for your time

Luc
Yes thanks, I would appreciate seeing this. I am not expecting your resulting net power to remain negative. In fact I expect it will go net positive, which would indicate more power going back to the grid than what is being used by the circuit. However, the phase relationship between the current and voltage might change. If it does change and you re-adjust to get your 90 degree phase shift again, the results may be different.

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Offline poynt99

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Here are some simple diagrams I made a while ago to explain this same power polarity issue to Rosemary Ainslie, because she had (has) the same mental block about this.

Offline allcanadian

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@Gotoluc
Quote
What would be needed next is a special circuit that would take high current lead acid battery DC and make ACBUT the circuit needs to handle the return (reactive circulating current) back to the batteries. An off the shelf Inverter cannot do this. So if you can develop a DC to AC circuit that can do this you would be my hero.


I believe you already have it in your circuit but it just needs a little clarity. An AC signal is dictated by two changing DC half cycles and if a circuit elements instantaneous voltage is higher it's a generator function and if it's lower it is a motor function. In an AC machine the instantaneous voltage relates to the rate of change of the expanding/contracting field, ie.. a moving magnet. A standard inverter cannot work here because energy is dissipated on the DC primary side regardless of whether the load becomes a source or not.


If you want a true AC source from a periodic alternating DC source then the AC source must be a series resonant circuit which you already have to some extent with your series capacitor off the AC generator. Now if we replace the AC generator with a large inductor then we have an AC generator (kind of) with no expanding/contracting field source other than the reactive current circulating in the system. Here we should remember that all those silly magnets in our generators moving past our coils ever did was induce a voltage "higher" than the instantaneous reactive current voltage which is why we called it a "generator".


Now if we added a large inductor in place of the generator and periodically charged your series capacitor when it is at peak voltage with a higher DC voltage of the correct polarity then the alternating current oscillations in the system would be maintained. The reactive current integrity would also be maintained and the AC system would be allowed to act like a true AC system should. The most efficient way to periodically charge the series capacitor is from a discharging inductor.. ie boost converter/joule thief. A voltage source such as a capacitor/battery cannot be used to charge the series capacitor otherwise you will lose 1/2 the Energy.. ie the parallel capacitor paradox.


Note: it would be very easy to measure the true input as the input is now DC.


Good luck


AC

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Offline gotoluc

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Thanks for your input AC ;)

Sounds great!... could you please take my basic circuit below and add the components so we can see what it may look like.

Thanks for sharing

Luc

Offline lancaIV

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gotoluc:
from Barbosa&Leal to their  reference  Camel Eslam Camal to his reference http://www.google.com/patents/DE3736921A1?cl=de

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Offline allcanadian

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@Gotoluc
It's just an LC tank with an extra inductor in place of the generator inductance. The capacitor always charges first as the inductors resist changes in current so why not charge the capacitor?. I'm not sure how most charge an LC tank circuit but pulsing the capacitor with DC from an external inductance (an inductive discharge) is the way I have done it because it is the most efficient way. Think of it as a one way reactive current charging the capacitor every half-cycle periodically adding to the reactive current already present.


As a general rule nothing moves unless it is LC, inductor to cap or cap to inductor.


You can also "ping" the circuit to find it's natural resonant frequency. Charge the cap, close the circuit and note the frequency of oscillation which is the natural resonant frequency of the circuit. Now pulse it at that frequency or use a theshold detector to detect peak voltage and pulse and your there.


AC

Offline gotoluc

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No problem Luc.
Yes thanks, I would appreciate seeing this. I am not expecting your resulting net power to remain negative. In fact I expect it will go net positive, which would indicate more power going back to the grid than what is being used by the circuit. However, the phase relationship between the current and voltage might change. If it does change and you re-adjust to get your 90 degree phase shift again, the results may be different.

Hi poynt,

please find the attached scope shots below. First is the standard way I've been doing it and the second one is selecting Invert in channel 2 menu.

This is my circuit powered by my variac from the grid and with a 5 Ohm 1% 50w rated load Resistor on the Neutral leg of the primary (per CSR). Channel 3 is displaying the voltage across that load. I didn't connect channel 3 probe ground so not to affect CSR. So all ground points are standard Grid side Neutral after the 0.1 Ohm CSR

Let me know what you thing

Luc

Offline gotoluc

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@Gotoluc
It's just an LC tank with an extra inductor in place of the generator inductance. The capacitor always charges first as the inductors resist changes in current so why not charge the capacitor?. I'm not sure how most charge an LC tank circuit but pulsing the capacitor with DC from an external inductance (an inductive discharge) is the way I have done it because it is the most efficient way. Think of it as a one way reactive current charging the capacitor every half-cycle periodically adding to the reactive current already present.


As a general rule nothing moves unless it is LC, inductor to cap or cap to inductor.


You can also "ping" the circuit to find it's natural resonant frequency. Charge the cap, close the circuit and note the frequency of oscillation which is the natural resonant frequency of the circuit. Now pulse it at that frequency or use a theshold detector to detect peak voltage and pulse and your there.


AC

I see 8) ... very interesting!

Lots to try

Thanks for sharing

Luc

Offline gotoluc

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At everyone,

I've attached 2 scope shots. The only change in the circuit  between both scope shots is, the first is with the Secondary shorted and the second is with the Secondary open.

Now you can see what happens in each case and why the preferred is with the secondary shorted.

Luc

 

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