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Author Topic: The Bedini Holy Grail  (Read 3468 times)
Group: Guest
For all you Bedini enthusiasts:

The user Callanan (a.k.a. "Ossie") from overunity.com posted an amazing circuit that allows you to recycle your back-EMF from your Bedini coil back into the source battery.  It is a deceptively simple circuit and I know that many of you have been looking for this for years.  (Why didn't I thinka it?)

The attached image is not a direct plug-in for a Bedini setup but let me suggest what can be done to make it fit.  Again we are talking about making a Bedini setup without a target battery, only a source battery.  The Holy Grail!!!

I am going to suggest a solution that uses a single power coil without a second trigger coil.  In the schematic (see below), remove Toroid 2 so that you are just left with Toroid 1.  This is your main power coil.

[EDIT] --- I just realized that at this point you can just plug this circuit into your Bedini setup and run!!!  The rest of the discussion involves using CMOS logic to control the two transistors.  You can still use a reed switch to do this function, OR, you could use any other trigger/CMOS logic circuit to get more precise timing than you can get by moving a reed switch around.  So the discussion below is "overkill" if you are just going to use a reed switch, but does include a trick or two nonetheless.  My suggestion would be to use an optical trigger and a dual CMOS 555 timer setup to have complete control over your pulse timing once you get your optical trigger signal. --- [EDIT OUT]

Remove the reed switch and the 1K resistor.

The PNP and the NPN transistors must be driven with complimentary signals, i.e.; "0 1" to energize the coil and "1 0" to switch off the coil.  (At the switch-off the energy in the coil will go back to the source battery.  Yay!)

The complimentary signals can be generated with CMOS logic because it will work at the 12-volt battery voltage with an added bonus that the CMOS chip will consume almost no power.

The triggering can be done with a standard reed switch wired the following way:

Assume that when the reed switch is open-circuit you want the coil to be off, and when the reed switch is short-circuit you want the the coil to be energized.

Make a resistor divider network with a 100K resistor connected to +12 volts and a 10K resistor connected to ground.  The voltage at the point where the two resistors connect together will be at about 1.2 volts.

Connect one side of the reed switch to +12 volts and the other side of the reed switch to the junction between the two resistors.

When the reed switch is open, the voltage at the junction will be about 1.2 volts, and this will be seen by the CMOS logic as a logical zero/false.

When the reed switch is closed, the voltage at the junction will be at 12 volts and this will be seen by the CMOS logic as a logical one/true.

You use this logical true/false signal to drive the CMOS logic to generate the complimentary "1 0" or "0 1" signals to control the two transistors.  Naturally you need base resistors between the complimentary CMOS outputs and the transistor base inputs.

My suggestion would be to do this with a standard CMOS 4009 hex inverter chip  (there is probably a more modern equivalent).  A hex inverter gives you all that you need to implement this function.  Remember to tie all unused CMOS inputs to +12 volts or ground.

Note that your reed switch will never get burnt because it is performing a pure logic function.

Have fun - the Holy Grail is finally here thanks to Callanan/Ossie!!!!!!!   John Bedini eat your heart out...

MileHigh

« Last Edit: 2009-12-29, 08:14:11 by MileHigh »
   
Group: Guest
OK, I'm willing to be the token dummy here...

And believe me, I want to be wrong.

But I was always of the understanding
that the type of electricity coming out,
was not the type going in.

That the type of charge on the battery was odd and different
than what we normally accept as traditional battery chemistry.

And that the type of charge it exibited
to the target batteries did not lend itself well.
to the function performed by the source battery.

Any attempt to utilize the type of output it makes
by feeding it back to the source battery
would be no more than a slowly loosing battle.

Please, somebody, prove me wrong, OK?
   
Group: Guest
For all you Bedini enthusiasts:

The user Callanan (a.k.a. "Ossie") from overunity.com posted an amazing circuit that allows you to recycle your back-EMF from your Bedini coil back into the source battery.  It is a deceptively simple circuit and I know that many of you have been looking for this for years.  (Why didn't I thinka it?)

The attached image is not a direct plug-in for a Bedini setup but let me suggest what can be done to make it fit.  Again we are talking about making a Bedini setup without a target battery, only a source battery.  The Holy Grail!!!

I am going to suggest a solution that uses a single power coil without a second trigger coil.  In the schematic (see below), remove Toroid 2 so that you are just left with Toroid 1.  This is your main power coil.

[EDIT] --- I just realized that at this point you can just plug this circuit into your Bedini setup and run!!!  The rest of the discussion involves using CMOS logic to control the two transistors.  You can still use a reed switch to do this function, OR, you could use any other trigger/CMOS logic circuit to get more precise timing than you can get by moving a reed switch around.  So the discussion below is "overkill" if you are just going to use a reed switch, but does include a trick or two nonetheless.  My suggestion would be to use an optical trigger and a dual CMOS 555 timer setup to have complete control over your pulse timing once you get your optical trigger signal. --- [EDIT OUT]

Remove the reed switch and the 1K resistor.

The PNP and the NPN transistors must be driven with complimentary signals, i.e.; "0 1" to energize the coil and "1 0" to switch off the coil.  (At the switch-off the energy in the coil will go back to the source battery.  Yay!)

The complimentary signals can be generated with CMOS logic because it will work at the 12-volt battery voltage with an added bonus that the CMOS chip will consume almost no power.

The triggering can be done with a standard reed switch wired the following way:

Assume that when the reed switch is open-circuit you want the coil to be off, and when the reed switch is short-circuit you want the the coil to be energized.

Make a resistor divider network with a 100K resistor connected to +12 volts and a 10K resistor connected to ground.  The voltage at the point where the two resistors connect together will be at about 1.2 volts.

Connect one side of the reed switch to +12 volts and the other side of the reed switch to the junction between the two resistors.

When the reed switch is open, the voltage at the junction will be about 1.2 volts, and this will be seen by the CMOS logic as a logical zero/false.

When the reed switch is closed, the voltage at the junction will be at 12 volts and this will be seen by the CMOS logic as a logical one/true.

You use this logical true/false signal to drive the CMOS logic to generate the complimentary "1 0" or "0 1" signals to control the two transistors.  Naturally you need base resistors between the complimentary CMOS outputs and the transistor base inputs.

My suggestion would be to do this with a standard CMOS 4009 hex inverter chip  (there is probably a more modern equivalent).  A hex inverter gives you all that you need to implement this function.  Remember to tie all unused CMOS inputs to +12 volts or ground.

Note that your reed switch will never get burnt because it is performing a pure logic function.

Have fun - the Holy Grail is finally here thanks to Callanan/Ossie!!!!!!!   John Bedini eat your heart out...

MileHigh




Yawn.....

This circuit is near identical to the half bipolar circuit MileHigh, which has been around since the '80s. It allows feedback to the source, by isolating the coils from the source through the use of PNP and NPN transistors. Callahan has neglected to place a large capacitor over the batteries terminals, which would help in this configuration. Either that, or the diodes need to feed a cap which is pulsed back to the source between motor pulses.

Old hat...
   
Group: Guest
All that I can say is that when I used to read Bedini related stuff on YouTube and in the forums (going back about two years ago), I must have read hundreds and hundreds of comments by replicators wishing and longing that they could eliminate the charging battery and feed the spike directly back to the source battery.  They were always frustrated that they seemingly couldn't do it so hopefully some of them will try the dual-transistor + diode implementation.

MileHigh
   
Group: Guest
All that I can say is that when I used to read Bedini related stuff on YouTube and in the forums (going back about two years ago), I must have read hundreds and hundreds of comments by replicators wishing and longing that they could eliminate the charging battery and feed the spike directly back to the source battery.  They were always frustrated that they seemingly couldn't do it so hopefully some of them will try the dual-transistor + diode implementation.

MileHigh

Hi MileHigh,

Well this is the circuit to try. I found that the inductive discharge on this circuit doesnt seem to be as strong as the SG circuit, Im still not sure as to why, I have my theories, but nothing solid as yet. Once again, that is what I found, and should not necessarily be considered all encompassing. Ive had friends get different results with slightly different configurations.

The video that Bedini showed recently used the half bipolar switch, which is basically half of his circuit he designed with Ron Cole to switch AC over a coil from a DC source. In some ways it is similar to a "H-Bridge" or Push/Pull oscillator.

Also note, that in his video there was no attempt to collect any of the input energy. You can see the spikes clearly on his scope, and he does say that all of this can be captured, as we already know. It makes one wonder, if you can get amp draw down real low (I too wish he had used a more appropriate gauge) and still charge a capacitor above the source voltage, which can be returned to the input when the motor is off, then how much further must one go to collect power off the rotor with a separate coil?

I have seen a bipolar rotor switching on both north and south poles with a single power winding drawing 3 milliamps. Once again there was no attempt on this particular configuration to store or capture the inductive discharge presented, and no generator coils either.

Take a look at DadHavs video here if you are interested.

http://www.youtube.com/watch?v=Fw9H5p4McZA

Regards
   
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