Radiant disruptive discharge

evostars

Radiant disruptive discharge
«  »
3 bifilar pancake coils

1 being pulsed 50% duty with a mosfet.

When the mosfet opens the circuit, the inductive negative voltage spike (back EMF) is captured in a capacitor and stored until the mosfet closes the circuit again.

When the mosfet closes the circuit, the capacitor is being discharged through the mosfet and L2, giving a fat kick.

L3 is coupled (they all are) to L2 and gets to see this kick, producing a fat current from it.

L3 at the same time is resonant from L1 being pulsed. L3 is tuned to a low frequency(100kHZ) , L2 is not tuned (resonant at >  1500kHz).

The current induced in L3  (and L1) by the kick of L2 is generating current, in L3, at the same moment that it has a maximum voltage.

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Re: Radiant disruptive discharge
« Reply #1,  »
tried the circuit, but only rings.
no disruptive discharge
Re: Radiant disruptive discharge
« Reply #2,  »Last edited
next step add C2 in series between L2 and the diode.

c2 2.2uF
c1 50nF

this should also resonate.

I also will reverse the values of c1 and c2

then a diode should be added between l2/c2 and ground

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Re: Radiant disruptive discharge
« Reply #3,  »Last edited
Rebuild everything on a new board and did the tests.
L2 becomes series resonant when C1=54nF and C2=2.3uF 
When C1 and C2 values are reverse, it doesnt resonate.

When tuning, first the back emf is at the positive max of the resonant sine (newfile10)
when further tuning down, the discharge shifts towards the closing of the swithc (mosfet) and it is at the minimum of the resonant sine (newfile11)

L1 L3 L2 are stacked bifilar pancake coils. L3 is open in the center of the stack.

When a diode is added between L2/c2 and ground (V-) the C1 and C2 get a positive offset DC voltage where the sine rides on top of it.

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Re: Radiant disruptive discharge
« Reply #4,  »
next, add a capacitor parallel to L1 and see if it can resonate in the time when the mosfet is off.

Lynx

Re: Radiant disruptive discharge
« Reply #5,  »
I just LOVE your journey here Evo, it's really inspiring to see :thumbsup2:
I only wish I could have more to offer, but I try to chime in whenever I see something that which I think/hope that I can contribute with, wherever and whenever I find it :-)

evostars

Re: Radiant disruptive discharge
« Reply #6,  »
Quote from Lynx on November 2nd, 2018, 01:27 PM
I just LOVE your journey here Evo, it's really inspiring to see :thumbsup2:
I only wish I could have more to offer, but I try to chime in whenever I see something that which I think/hope that I can contribute with, wherever and whenever I find it :-)
Thx, sometimes the journey can be hard, and than its nice to get some cheers.

I just was reminded again of the joule thief. it also gives some nice spikes.
If I understand correctly, its because the 2 coils of the JT stop each other.

one coil starts conducting and opens the base to collector current, this opens up th emitter to collector current that is much larger.

the other coil that is connected to the emitter gets a sudden string current.

the core produces a magnetic field, opposite of that which is produced by the other coil (base collector current).

the base current stops flowing due to the strong  reversed magnetic field.

this turns of the transistor. everything stops working and the core demagnetises.

then all starts over again.

this creates a voltage spike.

Maybe something similar is happening in L1 and L2 and L3 of nelsons circuit.
the coils can act as a diode. the mosfet   closes the circuit and the discharge of c5 starts. then L1 is magnetised, and acts as a diode for L2. which stops the discharge.
coil polarity
« Reply #7,  »
polarity of the coils does matter
outside rim OR
inside rim= IR

L1 OR needs to be connected to source if the mosfet
L1 IR needs to be V- ground

L2 OR connects to the diode
L2 IR connects to C1 (c5 in nelsons circuit)

this gets the highest voltage resonance for the lowest current draw

If one is reversed, the magnetic fields oppose and the current draw goes way up.
not a magnetic diode
« Reply #8,  »
I remember seeing in nelson rocha's videos, the l2 being separated from l1 and still having current (disruptive discharge) so its not due to the coupling of the coils.

in the same video he used a l1 coil with high inductance, that wasn't a pancake coil

L3 wasn't a coil at all but a tin filled with water, that heats up

https://youtu.be/x5h6NI2nQoQ
Re: Radiant disruptive discharge
« Reply #9,  »
playing with some ideas see if it makes sense

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Matt Watts

Re: Radiant disruptive discharge
« Reply #10,  »Last edited
I second Lynx's sentiments.  Nothing worth having comes easy.


Let me pose a question to you evostars, something fundamental that I have been personally struggling with for quite a long time...

Mr. Dollard states that a dielectric field exists between conductors.  Let's just assume that is an actual fact without debate.

If we have an open circuit with a switch, then there should exist a dielectric field between the two conductors.  We can measure voltage between the conductors, so we know something is there.

If we close the switch, the two conductors effectively become one.  The dielectric field now has to reorient itself between this one conductor and some other conductor.  My question then is:  Where does the dielectric field go, what path does it take to get there and does this change in orientation create a current flow?

It is my understanding the dielectric and magnetic field lines always exist perpendicular to each other, so when we close the switch, the magnetic field must also reorient.

If you can visualize and explain this with enough detail, it should become quite clear how Nelson's circuit actually functions.
Re: Radiant disruptive discharge
« Reply #11,  »
It's just a hunch, but I feel the word "current" is improperly associated with amperage, or the magnetic component of electricity.  I think "electric current" implies both dielectric and magnetic.  The "electric current" is what we notice changing when the dielectric & magnetic fields are forced to reorient.  Someplace I do recall Mr. Steinmetz defining this term.

evostars

Re: Radiant disruptive discharge
« Reply #12,  »
Quote from Matt Watts on November 3rd, 2018, 10:45 AM
It's just a hunch, but I feel the word "current" is improperly associated with amperage, or the magnetic component of electricity.  I think "electric current" implies both dielectric and magnetic.  The "electric current" is what we notice changing when the dielectric & magnetic fields are forced to reorient.  Someplace I do recall Mr. Steinmetz defining this term.
the dielectric is normally terminated on both ends, on conductors. Maybe, a neutrino is an open ended dielectric line if force. but thats speculation.
but if so, there are plenty of them coming from the sun trying to terminate on earth (conductor).

the coil with its magnetic vortex has the dielectric field terminated on different parts of the wire (between windings) due to the voltage difference (one is v+ the other is v-).

when the circuit is opened, the magnetic vortex (that is a loop, self terminated on itself) is opened up. then the open lines of force terminate on the conductor again as dielectric lines of force (bothe ends on different windings). In the process of reconnecting, neutrinos might also connect (speculation).

more importantly, that same aether can  be pulsed by a violent capacitor discharge over a coil(l2) . the pressure this creates is seen by the  secondary L3, and produces current from it.

Current is a tricky thing.
current is the loss of the dielectric field component, said Dollard.
I agree, but is it lost? No not when its a magnetic field, because it can transform back into a dielectric field (back emf or resonance).

when dielectric energy is transformed into heat (resistance) then it is lost (tranformed into heat, and not recoverable).

this disruptive discharge is a temporal field event. all of the energy in a split energy.
the collapse gives infinite current or voltage, if its a capacitor or inductor, said Steinmetz.
it only depends in the speed of the transmutation, and how low(capacitor closing circuit ) or high (inductor open circuit) you can get it.

the switch opens, and closes, super fast, with very low and very high resistance. thats why I like the mosfet.

the c5 capacitor discharges when the mosfet closes the circuit again. but... it needs to discharge faster than it resonates...
here lies a key somewhere
DC offset resonant
« Reply #13,  »
the AC resonant sine wave of L2 can ride on top of a negative DC bias.

if that negative DC is provided by a rather large capacitor, it would also provide a stable ground for the LC series resonance.

When the mosfet closes the circuit, the capacitors discharge, making L2 resonant, but also, removing the dc offset. This is the rising part of the spike.

than the discharge stops again. at the same time, the circuit is still closed by the mosfet.

Why does it stop discharging so quickly?

Here the mosfets could play a role.
Is the capacitor that provides the dc offset, discharged by a diode, that opens up by an event when the mosfet closes the circuit?

A quick opening and closing of the diode?

Maybe because another small capacitor is discharged, and once its empty, closes the diode again?

I cant stop thinking about what makes the capacitor discharge within a resonant system, producing a spike on top of a sine
how to disruptive discharge in a coil
« Reply #14,  »
in nelsons circuit, C5 and C4 are at each end of L2.

the mosfet acts as a sparkgap, and its a bit like a hairpin circuit.

But, L2 is a coil, and becomes resonant. If the cap and the frequency match.

the only way to get a disruptive discharge is to use a large enough cap with a high enough frequency.

the large cap would not discharge completely. but only a little bit, very shortly.

this brings in a problem, the large cap should first be charged, to a high voltage. with a charge pump.

the other problem is the discharge, the 50% duty cycle of the mosfet is much to long. It should only have a small time to discharge and then, charge again.

Can we make use of the body diode, to discharge for just a fraction of the period, until the mosfet turns in again?
C1 needs to be smaller
« Reply #15,  »Last edited
When the mosfet is turned off, D3 is blocking the positive supply.
This could make C1 part of a resonant system.
L1 would still be grounded on the supply, but C2 makes it resonant, when the mosfet is off.
So if it swings more positive than the powersupply, It would open up the body diode.

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Re: Radiant disruptive discharge
« Reply #16,  »
long ago I did some tests with discharging 2 capacitors with a bifilar in between.
this resulted in a spike in the center tap of the coil.

don't know for sure but its on the forum here somewhere.

this could be done with c5 and c4

the body diode can charge c5, by making l2 resonant with c2 in the "off" time. which opens the body diode. passing the higher than v+ voltage of c2 into c5/c1 as d3 is blocked.
 
need to make c5 c1 small enough to charge and see what happens, by tuning c2 to resonate with l1. giving a positive

Re: Radiant disruptive discharge
« Reply #17,  »
I looked back into my own research as I remember doeing something the same, I found it here:
http://open-source-energy.org/?topic=2952.msg44094#msg44094

It isnt clear If it produced the spike in the center tap of the coil, and there are no screaan shots of the scope, exept for the trigger signals.

I do remember it worked. So ... I need to find out if this works

Radomir

Re: Radiant disruptive discharge
« Reply #18,  »
Quote from Matt Watts on November 3rd, 2018, 10:33 AM
I second Lynx's sentiments.  Nothing worth having comes easy.


Let me pose a question to you evostars, something fundamental that I have been personally struggling with for quite a long time...

Mr. Dollard states that a dielectric field exists between conductors.  Let's just assume that is an actual fact without debate.

If we have an open circuit with a switch, then there should exist a dielectric field between the two conductors.  We can measure voltage between the conductors, so we know something is there.

If we close the switch, the two conductors effectively become one.  The dielectric field now has to reorient itself between this one conductor and some other conductor.  My question then is:  Where does the dielectric field go, what path does it take to get there and does this change in orientation create a current flow?

It is my understanding the dielectric and magnetic field lines always exist perpendicular to each other, so when we close the switch, the magnetic field must also reorient.

If you can visualize and explain this with enough detail, it should become quite clear how Nelson's circuit actually functions.
Hi Matt

Maybe these books could help you your understanding of Ether:
https://archive.org/details/WilliamLyneOccultEtherPhysicsTeslasHiddenSpacePropulsionSystem2ndEdition1997
https://archive.org/details/PentagonAliens1993-WilliamLyne
https://archive.org/details/LyneOccultDictatorship_201710

Regards
Radomir

Re: Radiant disruptive discharge
« Reply #19,  »
Quote from evostars on November 7th, 2018, 06:32 AM
I looked back into my own research as I remember doeing something the same, I found it here:
http://open-source-energy.org/?topic=2952.msg44094#msg44094

It isnt clear If it produced the spike in the center tap of the coil, and there are no screaan shots of the scope, exept for the trigger signals.

I do remember it worked. So ... I need to find out if this works
Hi Evostars

How did you wound your bifilar pancake coils as Tesla bifilar or Hopper bifilar ?

O, i have got one simple idea ! Why don't try making asymetric bifilar whatever Tesla or Hooper bifilar configuration. Idea is to make one wire longer than other exactly at Phi-ratio or golden section ratio. Lenght of L1 would be 1,618 x lenght of L2. Hooper os non inductive coil, on the contrary Tesla bifilar have inductance so you can make wind all 3 Tesla pancake coils in golden inductance ratio. 1st L1 would be some xxx uH. Second 1,618 x L1 while the third would be 1,618 x L2 or Phi square L1. I think it is natural proportion for Tesla coil and not only for coils worth. It's natural propotion.

Another variant is golden lenght ratio. So lenght of longer wire in 1st coil would be 1,618 x lenght of 2nd wire. Than meassure inductance for complete pancake and use it as inductance setup for 2nd coil. So simple Phi square of 1st coil lenght would be your 2nd. And 3rd (output or load coil) would be a Phi cube of your initial coil setup.

In any case i think nobody has played with asymetric pancakes coil so far. Maybe you be a first after Tesla, because he had used it for sure. But the purpose is still unknown for me , same as the real purpose of his bifilar coil.

And one more note for me. Maybe your main problem trying is breaking treashold value of Ether or simply said you need to increase voltages and curents in your experimet to make Ether answer you in form of Radiant pulse. Keep in mind Tesla had done experiments in Colorado Springs with 12 000 000 Volts and couple of tenth thousand ampers in peak envelope power.

Same Ed Gray in his CSET used 7kV and several kA. Think about why you still haven't achived real Radiant Impulse.

Best Regards
Radomir