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Originally posted by Harvey
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Hi Mike,
This really is extraordinary. I don't know what type of circuit your PWM incorporates, but for it to work below 3V would really be interesting to evaluate. I wasn't aware that you were running the PWM circuit off the the battery as well.
So, if I understand this correctly, you ran the circuit down to zero battery volts and then the 9120µF capacitor began to reverse charge...is that correct? Or was that a different circuit configuration?
In the configuration shown, it would appear that your LED's would drain the battery at a rate of 105mA / hr. All batteries are different, but most 12V lead acid batteries are considered discharged at about 9V as by the time they reach that voltage there just isn't much usable charge left in them. Your battery will deliver 4.2 average Amps for one hour and then it is considered depleted of charge. With just the LED's as a load, we would expect your battery to be discharged in 40 hours (40 * .105 = 4.2). But since your circuit also powers the PWM (I would need to look back at its power requirements) that too would discharge the battery by its value. And then lastly, the circuit resistance itself (copper wire and interconnects and diode junctions) also dissipates power when operated. Interestingly, most of your circuit is reactive and would not result in a great drain factor as most of its operation would entail 'apparent' power. The FET naturally will have an amount of power dropped across it when on and the gate resistors also will exhibit a resistive power loss during the on time.
The beautiful thing about your circuit is that if the PWM is simply set to zero Hz with a fully charged battery, your LED's will effect a natural draw-down curve that can be monitored until the battery can no longer support a current through them and they turn off. That would set a baseline for you. Of course it would take a couple of days, but you may be able to take a reading every 4 hours. Then, after recharging the battery, you could do the exact same test with the PWM at 4.33KHz. This would be a very worthwhile endeavor. If the scalar function of the circuit does produce the gain it appears to, this will be very evident from the tests.
If this is the circuit that produces the negative charge on the capacitors, then we will be looking much closer at the scalar potential and how it may continue to source energy when the battery is drained. This is an important aspect of Rosemary's thesis as well, which states that a secondary event occurs which causes the inductors to function as a power source in addition to the returning energy of the magnetic field collapse. In her model, the additional power is derivative of the atomic structure of the inductor which loses its bonds and gives up the energy of those bonds resulting in the material decay of the inductor. Glen's and Aaron's work in the COP>17 thread are the first attempts at documenting the evidence in a scientific manner to help prove her thesis. For whatever reason, the previous records produced by qualified researchers was not made available to her.
Your work here, may actually provide proof. Of course Koontz, Bearden and Tesla may argue that the energy arrives from a different pool, but at this stage of the 'game' I would say that where it comes from is irrelevant as long as the energy is free to be put to work. And in your case, you are doing just that by lighting the LED's.
This really is extraordinary. I don't know what type of circuit your PWM incorporates, but for it to work below 3V would really be interesting to evaluate. I wasn't aware that you were running the PWM circuit off the the battery as well.
So, if I understand this correctly, you ran the circuit down to zero battery volts and then the 9120µF capacitor began to reverse charge...is that correct? Or was that a different circuit configuration?
In the configuration shown, it would appear that your LED's would drain the battery at a rate of 105mA / hr. All batteries are different, but most 12V lead acid batteries are considered discharged at about 9V as by the time they reach that voltage there just isn't much usable charge left in them. Your battery will deliver 4.2 average Amps for one hour and then it is considered depleted of charge. With just the LED's as a load, we would expect your battery to be discharged in 40 hours (40 * .105 = 4.2). But since your circuit also powers the PWM (I would need to look back at its power requirements) that too would discharge the battery by its value. And then lastly, the circuit resistance itself (copper wire and interconnects and diode junctions) also dissipates power when operated. Interestingly, most of your circuit is reactive and would not result in a great drain factor as most of its operation would entail 'apparent' power. The FET naturally will have an amount of power dropped across it when on and the gate resistors also will exhibit a resistive power loss during the on time.
The beautiful thing about your circuit is that if the PWM is simply set to zero Hz with a fully charged battery, your LED's will effect a natural draw-down curve that can be monitored until the battery can no longer support a current through them and they turn off. That would set a baseline for you. Of course it would take a couple of days, but you may be able to take a reading every 4 hours. Then, after recharging the battery, you could do the exact same test with the PWM at 4.33KHz. This would be a very worthwhile endeavor. If the scalar function of the circuit does produce the gain it appears to, this will be very evident from the tests.
If this is the circuit that produces the negative charge on the capacitors, then we will be looking much closer at the scalar potential and how it may continue to source energy when the battery is drained. This is an important aspect of Rosemary's thesis as well, which states that a secondary event occurs which causes the inductors to function as a power source in addition to the returning energy of the magnetic field collapse. In her model, the additional power is derivative of the atomic structure of the inductor which loses its bonds and gives up the energy of those bonds resulting in the material decay of the inductor. Glen's and Aaron's work in the COP>17 thread are the first attempts at documenting the evidence in a scientific manner to help prove her thesis. For whatever reason, the previous records produced by qualified researchers was not made available to her.
Your work here, may actually provide proof. Of course Koontz, Bearden and Tesla may argue that the energy arrives from a different pool, but at this stage of the 'game' I would say that where it comes from is irrelevant as long as the energy is free to be put to work. And in your case, you are doing just that by lighting the LED's.
Thanks for looking at this as my electronics is very basic, I am a HAM operator, or I was, don't have much time now, and my training and Bsc is in industrial engineering which does just about cover everything, but electronics pure is a field on its own and you are up there with me as one of the tops
Now back to the serious stuff
.When the caps started to negative charge it was the circuit which was self oscillating with the coil and reed switch.The last test was with the PWM that I have which is the Lawton circuit, and I was only using the last half of the circuit to give me duty and frequency. The test was done with the one battery only, the current draw for the PWM is 20ma. By my calculations I am getting more out than putting in when you add the 20ma to the 105ma = 125ma, the 24hrs and the voltage drop on the battery. BUT I do understand that I should have charged up the battery to its max first, but I was in a hurry and I was thinking that maybe the battery voltage would have risen up a bit and in which case it would be proof in itself.
But still with the drop, which was very small, it has shown that the circuit has produced more than the consumption on the battery.
Now to something that I have found and which is part of my HAM days.
When I change the resistance across the output of the phasing trans: the consumption goes down. E.G. put my 5w 12v bulb across the output which has a resistance cold of 3.6 ohms and the coil has a resistance of 3.85 ohms and so the resistance has dropped. Now if we take the frequency at this point, the change in resistance would have a markable effect on the frequency in the coil and the secondary coil.
Now I think that we need to drop the frequency on the output transformer so as to create more current on the output with less voltage "change voltage for current". Now if we can drop the frequency down to 50 or 60Hz we will have good usable current. I am not saying that we drop the input frequency as this is part of the way there is a gain, but the final output should be at a useable frequency
I am thinking on the line of a resistor and capacitor in parrallel, what do you think of this? and if you think it might be a good idea could you draw your thoughts so as we could compare before putting into practice.Going back to the caps and negative charge, when I removed the battery the voltage on the caps went down then up then down then up etc. untill 0v and then started to charge in reverse. Now we might have a fight on here inside the caps, taking both negative and positive charge depending on the cycle
Now when it is connected with both battery and caps, may be the caps are taking the positive charge and the battery the negative!!!!!!!!!!! thinking aloud here, so if I am rambling, but I think we can uncover a lot of things here. I await your commentsMike
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