NegentropicInductor

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Pulse Charging Capacitors Through Inductors

Introduction

In this pdf I will go through the details of a series of experiments Iperformed on the 5th of August 2008 and the 22nd November2008, the first two of which were posted on youtube under thename Overunity Demonstration : 12v + 0v = 15V.

There has been some controversy over the video as it does notshow an overall gain in energy. However, I used it as an

experiment to demonstrate overunity in terms of Conservation ofCharge and as an indication that energy is entering the systemfrom outside of the circuit.

It is commonly accepted that pulsing a capacitor into anothercapacitor through an inductor increases the efficiency of theenergy transfer between the capacitors. In this PDF I suggest itdoesn't as I believe it increases the COP, not the efficiency.Although these experiments don't provide conclusive evidence, I

believe they are a strong indication that this theory is correct.

EXPERIMENT 1

Im sure many of you have tried this experiment.

We start with two equal capacitors. Any size will do but lets use

two 10,000uf caps and lets say the C1 is charged to 10 volts andC2 is at 0 volts. This gives us 0.5 joules and 0.1 coulombs in the


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system.

If we close the switch the charge from C1 will discharge into C2until the voltage in each cap is equal, in this case 5 volts.

So now we have two caps charged to 5 volts and the total chargein the system is still 0.1 coulombs, but half the energy hasdisappeared. There is now only 0.25 joules left in the system.

But this doesn't mean the transfer was 50% efficient. As I said,there was originally 0.5 joules in C1 before it was discharged.After the discharge there was 0.125 joules, so in actual fact theefficiency of the energy transfer was only 25%! Only 25% of theenergy that left the capacitor was recovered by the charging

capacitor.

So where did this energy go? The theory is that it is dissipitatedas heat through resistance in the circuit. Even though we didntuse a resistor in the circuit, the circuit itself will have someresistance and that is where the energy is lost.

EXPERIMENT 2

It is widely accepted that pulsing a capacitor into another capacitorthrough an inductor can distribute the energy between thecapacitors with very high efficiency. A simple experiment to seethis will be the circuit below which is very similar to the circuit Iused in the video.

This is a basic solid state oscillator. C1 is charged to 12.45v(0.7250125 joules) and C2 is sitting on 0 volts. When the switch is


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closed the inductor energizes and discharges in very rapid pulsesthrough C2 and the flyback from the collapsing field is channeledinto C2 as well.

Using this method you could theoretically achieve 100% efficiencyin energy distribution between the two capacitors which would giveyou about 8.5 volts in each capacitor. In my circuits the best I canachieve is something like 8.47 volts in the C1 and 7.24 volts in C2.

C1 = 0.381938 joules , 0.0874 CoulsC2 = 0.262088 joules . 0.0724 Couls

Total Combined = 0.644026 Joules , 0.1598 Couls(Originally in the System = 0.7250125 joules , 0.1245 Couls)

So 0.366308 joules left the original capacitor (C1) and 0.262088joules was recovered by the charging capacitor (C2) which is a71% efficient transfer of energy (compared to 25% efficencywithout the inductor).

This time we have saved most of our joules and gained somecharge. So at first glance it appears that pulsing one capacitorinto another capacitor through an inductor cangive us far greater

efficiency in energy distribution than simply discharging onecapacitor into another. I have been told by some rather learnedchaps on a popular mainstream physics forum that this is becausean inductor holds back the current while it is energizing so there isless energy being lost through resistance as heat.

Let's test that.

EXPERIMENT 3

To test the theory posted above, I rewired my oscillator so that C1discharges into C2 through the pulsed inductor, but the flybackfrom the collapsing field is being channelled into a third capacitorcalled C3.


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If the increased efficiency is due to current being held back by the

inductance during the coil charging, then we should see asignificantly improved efficiency in the distribution of energybetween capacitor C1 and C2 even though C2 is not collecting theflyback from the collapsing field.

This time we start with C1 charged to 10.4 volts. After we closethe switch, C1 is left with 5.75 volts and C2 has charged to 4.57volts which converts into these figures:

C1 = 0.54553125 Joules , 0.18975 CoulsC2 = 0.34460085 Joules . 0.15081 Couls

Total Combined = 0.8901321 Joules , 0.34056 Couls(Originally in the System = 1.78464 Joules , 0.3432 Couls)

This time 1.23910875 joules left C1 and only 0.34460085 Jouleswas recovered by the charging capacitor (C2) which is a 27.8%efficient transfer of energy which is almost exactly the samefigures we would expect to see without the inductor in the circuit!In my opinion it is highly unlikely that this is a coincidence and itleads me to believe that 72.2% of the energy is still lost throughheat in this experiment.

Or to put it another way, approximately 75% of the energy willalways be lost through resistance as heat when transfering energybetween one capacitor and another, regardless of whether youplace an inductor or a resistor in series.

But in this experiment we still have the energy being collectedfrom the collapsing field by C3 which was left with 3.95 volts


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C3 = 0.31205 Joules , 0.158 Couls

I believe this is the energy that makes it appear that the efficiency

of the transfer of energy between two capacitors can be increasedusing an inductor. And if we believed that this circuit is a closedsystem, then this would be a logical conclusion.

However, I believe that this energy has come from the collapsingfield alone, and is independent of the energy input into the system.The figures above indicate that 75% of the energy that wastransfered through the system had already left the circuit as heat,so an explanation for the energy in C3 is that the energy of amagnetic field formed by an inductor is not a storage of the input

energy, but is formed by the space surrounding the inductormaking it an open system.

Of course, the only way to confirm this conclusion would be toperform the same series of tests in a calorimeter to find out theactual amount of heat that is released by these circuits.Unfortunately, I don't have a calorimeter available to do thesetests, but seeing as how we end up with almost the same figuresin Experiment 1 and 3 for Capacitors C1 and C2, I feel that this is

a very likely conclusion.

Sephiroth24th November 2008

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