ENERGY AMPLIFICATION

[totoalas]

My finished coil
Primary  bifilar  14 awg  40 t
Secondary  23 awg  160 t
Pick up   23 awg   600t
Tomorrow for termination test
N

[gyulasun]

Hi all,
For this test i have 2 multimeters connected to the two caps for dc voltage measurement.Ok here's my quick test today, i pre charged c1 with 10vdc, c2 is zero volts, i close sw1, induction occurs, c1 voltage drops to almost zero, then c2 generates 2.5vdc, i did it vice versa....almost the same readings...
conclusion; i need to wind more for the secondary coils to generate more voltage, for the 300 turns = to 2.5 volts, maybe i need to wind 1000 turns to get to the 10v input range... Thats all for now....

Hi crazycut06,

Before you wind 1000 turns let's discuss how quickly or slowly did you discharge the capacitor C1 when you close SW1.  I do think that you have to be very quick if you do the switching by hand and make sure you just switch off SW1 before C1 fully discharges.  So starting with the 10V precharge voltage, try to switch SW1 off when coil L1 consumed a few volts only from the 10V, say 4V and this would mean that current (kept on by the remaining 6V in that moment) is still able flow in the series LC circuit when you interrupt it: this means that the coil is also able to give the kick-back spike any inductance gives when current is cut in it and the amplitude of this spike is (Vi=L*dI/dt)  can be several times of ten volts, even above 100V, depending on the suddeness of the switch-off, coil inductance and the current itself. Question is and this is the relevant question: is this spike able to recharge capacitor C1 via the diode bridge?   (In your above test I assume you did not switch sw1 off quick enough and there was no chance to interrupt any meaningful current flowing in the coil just because C1 was discharged by then.)

Of course there can be circumstances when it is quasy impossible to perform as quick switch-off by hand as would be needed to keep on some or certain current flow in the coil BEFORE the charge in the capacitor diminishes to zero or near zero where a negligible current can only flow in the coil in the very moment of switch-off.  In this case the only solution is electronically controlled switching.
Here comes into picture the L/R time constant you wish to consider.  For this you need to know the DC resistance and the L inductance of coil L1,  (L2).
For those not familiar with L/R time constant, there are several links like this: http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php 
You can see from the graph that probably the best time for switching off current is at T=1*(L/R) second or just below that.  Say you have 20mH coil for L1 and its DC resistance is say 4 Ohm. Its time constant is T=20mH/4 Ohm=5msec (millisecond) this is the time duration you wish to keep current flowing into this coil after the switch-on moment, and 5msec later you switch it off very quickly (at least under some microsecond or at some hundred nanosecond speed).   (in this calculation I neglected the switch series ON resistance and the capacitor C1 inner resistance (ESR=equivalent series resistance), these two resistances increase the R value in the time constant calculation.)

rgds, Gyula

PS1: What Magluvin says on the Tesla electromagnet patent can also be of significance, for Tito mentioned it indeed but one step at a time...

PS2: crazycat06, please charge up C2 to also 10V, even if you do not discharge it first because its uncharged presence in the circuit means a big load to any spike (any capacitor is almost a short circuit for the first moment when you start charging it), and anyway Tito said charge up both caps to a certain DC level.

[crazycut06]

Hey Crazy

Im talking in general. Knowing transformers a bit, there has to be some 'twist' or something that is not common.

There has to be something that produces more out than in. Something that tricks the secondary into thinking it has been induced by a higher power that what is actually consumed.

I was going to dig out the pat. and read it over again to see if there is anything more to it than what I got before, but Im busy with another project at the moment.

This is the only thing I can think of that would be amiss from the circuit. Unless it is the individual windings shown and they are all wound at the same time. Not sure.

Ive read several patents that claim the use of a bifilar winding gives the transformer better efficiency.  Easy to search for. When they claim better efficiency, its not clear whether it is a greater COP or just closer to 100%.

Mags

Oh okay,    Glad to have you back in this thread,    My coil is just single wound, i think you are correct about bifilars or multifilars in transformers, better efficiencies, i've seen many transformers in our daily appliances that has multifilar winds old and new ones, mostly in pulsed dc, will try that as we go along....
Regards
Cc

[crazycut06]

Hi crazycut06,

Before you wind 1000 turns let's discuss how quickly or slowly did you discharge the capacitor C1 when you close SW1.  I do think that you have to be very quick if you do the switching by hand and make sure you just switch off SW1 before C1 fully discharges.  So starting with the 10V precharge voltage, try to switch SW1 off when coil L1 consumed a few volts only from the 10V, say 4V and this would mean that current (kept on by the remaining 6V in that moment) is still able flow in the series LC circuit when you interrupt it: this means that the coil is also able to give the kick-back spike any inductance gives when current is cut in it and the amplitude of this spike is (Vi=L*dI/dt)  can be several times of ten volts, even above 100V, depending on the suddeness of the switch-off, coil inductance and the current itself. Question is and this is the relevant question: is this spike able to recharge capacitor C1 via the diode bridge?   (In your above test I assume you did not switch sw1 off quick enough and there was no chance to interrupt any meaningful current flowing in the coil just because C1 was discharged by then.)

Of course there can be circumstances when it is quasy impossible to perform as quick switch-off by hand as would be needed to keep on some or certain current flow in the coil BEFORE the charge in the capacitor diminishes to zero or near zero where a negligible current can only flow in the coil in the very moment of switch-off.  In this case the only solution is electronically controlled switching.
Here comes into picture the L/R time constant you wish to consider.  For this you need to know the DC resistance and the L inductance of coil L1,  (L2).
For those not familiar with L/R time constant, there are several links like this: http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php 
You can see from the graph that probably the best time for switching off current is at T=1*(L/R) second or just below that.  Say you have 20mH coil for L1 and its DC resistance is say 4 Ohm. Its time constant is T=20mH/4 Ohm=5msec (millisecond) this is the time duration you wish to keep current flowing into this coil after the switch-on moment, and 5msec later you switch it off very quickly (at least under some microsecond or at some hundred nanosecond speed).   (in this calculation I neglected the switch series ON resistance and the capacitor C1 inner resistance (ESR=equivalent series resistance), these two resistances increase the R value in the time constant calculation.)

rgds, Gyula

PS1: What Magluvin says on the Tesla electromagnet patent can also be of significance, for Tito mentioned it indeed but one step at a time...

PS2: crazycat06, please charge up C2 to also 10V, even if you do not discharge it first because its uncharged presence in the circuit means a big load to any spike (any capacitor is almost a short circuit for the first moment when you start charging it), and anyway Tito said charge up both caps to a certain DC level.

Hi Gyula,
I did not do the switching quickly, and let the cap to fully discharge, manual switching is hard to do, im preparing my relay pulser to do the job, regarding inductance, i do not have an L meter    And don't have       
Any quality multimeter to get perfect measurements.
I tried charging both caps to 10v and closed sw1 but no noticeable voltage rise on c2, but if i connect the 10v battery to c1 then switch it, i get 13v on c2 with one switch, and if i continue switching it rose to 25v maybe more, ill try it again with the relay for more controled switching....thanks for the heads up....
Regards
Cc

[totoalas]

Tito dsif he used pc fan
Magmet on the blades
Reed switch for relay pulsing
I tried to charge the cap to 12 v but easily dischargef with multimeter
Will try ur test with battery on one cap