# Split the positive

1. Feb 16, 2004

### cala

Hello.

Here is a circuit layout based on the ideas of Edwin Gray and Stefan Hartmann.

We all know that when a difference of potential is set along a circuit, the charges goes from one terminal to the other, and finally deplete the source.

Now, I explain briefly how the proposed circuit works:

We have a DC source of low voltage and high current. This battery is conected to the first coil of a transformer through a chopper circuit, that converts the DC voltage into AC signal to the coil. This branch of the circuit tends to deplete the source as happens on all normal circuits.

But now, we have the second coil of the transformer, that transforms the low voltage-high current signal into high voltage and low current, with almost the same power.

This signal pass through a diode bridge, and is rectified to be AC possitive values only.

Now, we add a DC offset to the potential signal, the same value of the primary DC voltage source.

Now, we will connect a load between this potential signal, and the possitive terminal of the primary source (this load can be inductive, capacitive or resistive, and takes the same power the primary source is giving).

So imagine that in one end of the load we have a 220 AC potential value with an offset of 12 V DC, and on the other end of the load we have 12 V DC.

Then, we are powering the load (because there is a potential difference on its terminals), and also, this branch of the circuit is recharging the battery with the same power that was extracted before instead of depleting it!.

Finally, we have the chooper and first coil branch depleting the source, and the second coil, diode bridge, DC offset and load branch recharging it, so the primary source doesn't ever deplete, but the load is being powered!.

Can anybody explain me what is happening here??

Only physical explanations allowed.

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2. Feb 17, 2004

### Nice coder

This is odd to say the least...[?]

3. Feb 17, 2004

### Guybrush Threepwood

Why do you need a diode bridge? What is it's purpose?

4. Feb 17, 2004

### cala

without the diode bridge, the AC signal of the second coil of the transformer is sometimes positive, sometimes negative, then, if the load is connected between the second coil and the primary DC source positive terminal, we are sometimes recharging the battery (when the AC signal is positive and > 12 V) and sometimes we are discharging it (When the AC signal is less than 12 V).

With the diode bridge and the DC offset, we are asuring the signal of the second coil of the transformer is always greater than 12 V of the primary source.

With this method, it seems that we can power the load at certain amount of power with only the losses of the chopper you talked about. (Maybe 1Kw of power wasting 1w ?... I mean, 1Kw +1w of power wasted on the chopper branch, then 1Kw generated on the second coil branch, and almost that 1Kw going through the load into the battery positive terminal again.)

Last edited: Feb 17, 2004
5. Feb 17, 2004

### Guybrush Threepwood

1) as you said you have no way of obtaining more power than the one given by the first voltage source.

2) IF the first voltage source is an accumulator, I'm not sure that this is a proper way to recharge it ....

3) are you sure that using the high-voltage, low-current signal after the transformer you can open the diodes in the rectifier? Those diodes need current you know?

6. Feb 17, 2004

### cala

2) Yes, of course, the recharging cycle is not so simple, the circuit pictured here is only a "proof of the concept". Maybe the battery can't be charged like that, but surely there are proper circuits to do that, placing them between the load and the battery.

3) On the other hand, diodes doesn't need current to work, indeed they need voltage (the treshold voltage). You think the diodes need some current to start function, but what is needed is a voltage about 0.7 V to let the current (any current value) run through the circuit.
A diode is a voltage governed device, not current one, and in the circuit I pictured, the condition of voltage on diodes takes place.

Finally, the DC offset voltage is only set there to avoid the discharging of the primary source when the AC voltage is cero.

Imagine two batteries of 12 V connected with the same polarity trhough a ressistance. Would you say both batteries are doing work?

In fact, we can remove the second battery. Then we will have to add a new term for losses to the losses of the chopper, when the second coil signal drops down from 12V, but we keep doing more work on the load than the power lost on the chopper and on the little discharging cycle of the primary source.

I mean: the whole idea seems to make sense or not?

7. Feb 17, 2004

### wimms

Nope. Not at all. Learn circuits basics and you won't see PM everywhere.

In this case you'd need to understand nonlinear circuits, not just plain batteries and resistors. Replace batteries with capacitors, and this circuit looks like inefficient resonant oscillator. Something remotely similar is used in pulsed power supplies.

When you close the switch, left batt is only used to kickstart pulse on rightmost coil, after that right batt gets effectively connected with wrong polarity to the circ, and it gets fast depletion with danger of exploding. Its because, if upper coil has any load, upper transformer has low impedance and resistence, it becomes the "ground" lead for right batt. Any current that you'd get from rightside of circuit would flow through closed switch to the primary coil of your lower transformer, as an avalanche effect, until it saturates and shortcircuits. After that righmost coil looses all output, and your left battery is drained like hell. Opening the switch merely stops the abuse, and any current in right half.
Anyway, main power source will be your right battery, left one is merely "control" signal. It won't be charged. Current wold flow either from both left batt and upper coil to the rightside primary, or won't flow at all.

8. Feb 18, 2004

### Guybrush Threepwood

did you ever see a transfer characteristic of a diode? it has voltage on the Oy axis and CURRENT on the Ox axis.
Try this: take a source and start putting diodes in parralel. Once there are too many diodes and the source cannot supply current for al of them they will all block.

9. Feb 18, 2004

### cala

Wimms, again your analisys is very well exposed, thank you for your explanations.

On your exposition, the main problem seems to be the timing of the switching. If we were able to recharge the battery with the chopper switch open... What will happen?

Imagine you put a capacitor on the right side, and follow this sequence:

1 - the left battery and the chopper are activated for a time. Then, the second coil of the transformer and diode bridge charges up a capacitor of high voltage. The load is disconected this time.

2 - Now, we open the chopper circuit, and close the connection of the load, letting the capacitor previously charged discharge through the load to the left battery (with a circuit to properly charge it, as exposed by guybrush).

Now, there are two separated "time frames" to the switching chopper action (primary battery discharging), and the load powering (primary battery recharging).

What are the problems on that situation? What will happen?.

Thank you Wimms.

10. Feb 18, 2004

### cala

I post a new drawing of the circuit with the capacitor, the signals on the different elements, and a simplification on the load branch (we have to add some elements to properly give the power to the battery on the recharging proccess, as guybrush pointed).

In fact, this circuit is closer to the Edwin Gray original circuit!.

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11. Feb 18, 2004

### wimms

If using your first pic, then 1) right capacitor would not charge, and 2) it could not discharge either due to diode blocking it.

On your 2nd pic (nice graphics, btw) you wisely used full diode bridge. Actully you draw a pretty working circuit of stepup power supply, but connected its output back to the source.

Question you need to answer is - what is the purpose of this? Take a fresh look at your pic and think - what is THE energy source of this whole circuit? Its the battery. Whatever you do with the electrons, when energy leaves into the load, its lost, and battery is depleted. Reordering the steps of a dance does not change the eventual outcome the slightest, its destined to end. In no way could you do any useful work and still have battery "fresh". There is no other source of energy, see?

First, the only "battery" that can be recharged as fast as discharged, is capacitor. Any other battery would take much more time to recharge, and would require limiting to much lower current. This means that your load will have little energy to it.

Second, if you connect 220V through a load to 12V battery, its not much different from connecting your battery to your mains outlet. pop quiz - what happens? (don't try this at home)
If you protect your battery from overvoltage, your have to drain excess current to the ground, or basically discharge cap through the load to the ground directly, bypassing battery - no charging.

Third, its nice that you draw some cap charging graphs. Take its 2nd part, from top volts and down to zero discharing. This shape would be the shape of current in the load - very sharp rise when cap is switched to it, and rapid exponential decrease - this is nonlinear signal, bad for most equipment. Its effective value is around 1/3 of peak (integrate over time), and the peak would be way lower than you expect (basically charging current).

By me, its pretty useless circuit. Connect your load parallel to the cap permanently, forget sw-2, never stop sw-1 alternation, and you can power 220V equipment from your car battery. At least useful.

12. Feb 18, 2004

### cala

Thank you Wimms, your analisys is again clear and contundent.

Sure you can help me on this:

Making an analogy with water, water tanks and water pipes, to me it seems that the electric potential is comparable to the height of water on a tank. The surface of the tank is comparable to the current, and the volume of the water on a determined tank comparable to power (height multiplied by surface equal to volume, as V*I=P).

You can see you can store the same volume of water on different tanks with different height to surface proportions. (also, the same power on different V to I proportions).

When you connect two different tanks, the water will flow always from the greater height, until rising the same levels of water on both tanks. It doesn,t matter if one tank is thiner. The surface only will affect the duration of this transference. When this transference happen, we can use a turbine to extract work.

Back to the electric systems, given a quantity of power, the transfer of charges only will happen when there is a potential difference, it doesn't matter if the power stored on one place is less than the other. The transfer happen if the potential is greater. The transfer duration depends on the current(until the levels of potential equalyzes). And also, as this transfer happen, we can extract work.

Now, back to the water analogy. Now, you have the same levels on both tanks, and want to put the levels as they were before. Then, you should use a water pump, making some work to extract the water from one tank, and increase the height of the other tank.

What amazes me is that transformers on electric systems are like water pumps on water systems, but they do not expend work to do their job!, i mean, they can change the potential of the electrical containers, extracting (or better said, wasting) the power from one container, and increasing (or better said, generating)power on the other side, without expending work on doing this proccess!

Once we have different potential levels, they will tend to be equal again, and we can use this natural transfer of power to extract work.

Can you explain me this a little better?

Just another thing:
Do you know any circuit layout to transform the high voltage-short pulse of a high voltage discharging capacitor into low DC voltage of a given value?

13. Feb 19, 2004

### Guybrush Threepwood

what??? the transformer does not increase the power on the other side. The power is the same in both sides IF AND ONLY IF you have an ideal transformer. If you have anormal transformer you have losses in coils and in the magnetic core. More (as you probably noticed because you used a chopper) the transformer would not work with a DC power source.
A personal opinion - I always find water analogies more difficult to understand than the actual circuit, and I'm sure they never work on nonlinear devices....

14. Feb 19, 2004

### cala

Guybrush, I'm not saying the transformer increase the power, i'm saying the transformer transfers the same power from one side to the other, but at another V to I ratio.

Imagine you have 200 W at 20 V and 10 A AC running from one coil, then, on the other hand you could have the same 200 W, but at 200 V and 1 A AC.

Making the water analogy, it's like if with some circulating water through the pump, you could lift the same amount of water on another tank the height you wanted, freely!.

You know that to lift water from a lower tank to a thiner but higher level tank, you must do work in order to increase the water potential energy. Then, if liberated, the water will fall down naturally, and we can extract the potential energy as work.

But, in the electrical system, we can change the electric potential on both sides without expending work on the transformer, and then, let the both sides transfer the energy naturally to get the same potential level on both sides again, extracting also work.

15. Feb 19, 2004

### cala

I post another animated diagram, it represents the water analogy.

The water is the power, the height of water is the voltage or potential, and the width is the current.

The left tank is the battery, the right tank is the capacitor, and the animation represents the switching times, the action of the transformer as a free water pump and the turbine as the recharging battery through the load.

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16. Feb 19, 2004

### wimms

No. Power is work done over unit time. Memorise this well. V*I=P is only convenient formula.
Bringing height into the picture is wrong. You get impression that gravity is your source of free energy.
If you really want analogy with water, which is not good as Guybrush pointed out, then you can, given you restrict this only to purely resistive circuits (ie no caps nor inductances), think of Voltage as of pump Pressure, and current as of water flow speed (like in litres/second at specific point). Then power is how many litres you displace some specific distance per second (litre meters/sec)

Transformers are not pumps. Only power source is pump, like battery. Transformers trade current for voltage, so that power remains same, minus losses. Battery is your only source of energy, to do any work (ie. move water). There is no good water analogy for transformers without going too much off track. You'd better drop that, electricity is too much different from water. Definitely forget about lifting water to any heights and hoping gravity to do the work while bringing it down.

Electric circuits are closed loops, that energy which you apply to the primary coil and which pops up on secondary coil, is lost from primary half of circuit, like pumping water in circles.
Potential is nothing by itself. It won't do work if not supplied by sufficient current. And that current is supplied by primary coil, in proportions you mentioned for a transformer. To get 1A out, you need to put 10A in. Zero net effect.

17. Feb 19, 2004

### Guybrush Threepwood

yes you said exactly that , it's in your quote...... but nevermind

as I said earlyer water analogy does not work for AC current. The transformer works only for AC current. If you do not agree with that please provide a clear water analogy for a RC circuit and than we can move on to the transformer....

yes IF everythnig in your system is ideal. But the first law of thermodinamics assures you there are losses whenever you convers power from one form to another....
basically you want to get power from the battery, pass it through a transformer (which only converts this power) to charge a condensator and then not only recharge the battery with the same power but also use it to do something. Even assuming no losses you couldn't do that.

I think you make a confusion between electric potential and energy.

Last edited: Feb 19, 2004
18. Feb 19, 2004

### cala

I agree that potential is not energy, and agree that power is not energy.

But potential is the key for energy transfers, it doesn't matter if the energy amount on one side is greater or fewer than on the other. If you connect two places with different electric energy amounts, but equal electric potential, the transfer will not happen.

In the first switching sequence, the battery is the energy source.

Some of the energy of the battery is wasted to activate the transformer. The transformer transfers this same power quantity to the capacitor with another V to I ratio. So finally, on this cycle, almost the same energy wasted from the battery source is stored on the capacitor.

But when we connect the second switch, the capacitor is the new source of energy, because it has more potential than the battery now, and potential indicates the direction of the transfer of energy. It doesn't matter if the capacitor has less energy stored than the battery, the potential indicates the direction the charges will follow on the current path.

So finally, in this cycle, almost all the energy stored on the capacitor could pass again to the battery, until the potential levels become equal.

Then, we could start the sequence again.

I mean, on first cycle, battery is energy source, energy of the battery is wasted, but transformer makes capacitor storage that same energy amount at higher V value, and less I value.

Then, the capacitor becomes the energy source, due to the higher potential value (not for the real amount of energy stored.). The energy stored will transfer from the capacitor to the battery until the potentials on capacitor and battery becomes equal. The time of this transfer will depend on the load caracteristics (more or less current), but the quantity of energy tranferred will be always the same (near all the energy stored on capacitor, and almost the same wasted on the primary activation).

I have no other way to explain myself, please, forgive me, but i still don't see some of the problems you talk about.

Wimms: to me, the battery is the energy source on one half of the cycle. The capacitor is the source of energy on the other half.

Guybrush: to me, potential is not energy, but is energy transfer key. Increased potential on capacitor is what causes that this capacitor becomes the energy source and can power a load (current flowing, so charges passing from capacitor to battery) meanwhile it also recharges the battery with the charges that run through the load, until the potential levels on battery and capacitor becomes equal.

19. Feb 19, 2004

### Guybrush Threepwood

and the capacitor gets its energy from where???
do you agree with that?

I suggest the following experiment (be careful though). The usual car batery has 12 V. Take one of those and then link it to 2 small 9V batteries (that's 18V > 12V). See if the car battery is charging.....

20. Feb 19, 2004

### wimms

Okay. Here is key to your confusion. You believe that energy that returns to the battery through the load is _all_ the energy that is stored on the capacitor. This is NOT true. Work on load reduces the amount of energy that reaches battery. Learn about resistence.

As an analogy, consider bullet that kicks a wall, goes through it and then kicks you. By going through the wall it looses energy, and you won't get killed. Work its energy does is a hole in the wall, and that energy is now not available to make a hole into you. Same above, energy returning to battery is that stored in cap, minus energy spent on working with the load, minus losses as heat. Get that?

In effect, all this circuit does is simply a fancy and inefficient way to dischard battery into the load, nothing else. As I said above, you could as well connect your load in parallel to the cap, and have 12V->220V transforming unit to power some 220V equipment.
There is NO way to power a load for useful work without spending energy, battery charge or whatnot. Turning the cap back into the battery accomplishes absolutely nothing besides making it very inefficient - it won't charge battery, it won't do work either.

Last edited: Feb 20, 2004
21. Feb 20, 2004

### cala

Well Wimms, now I see my error in this case. I always thought that the current was the carrier of the energy, but with your example of the bullet, it seems that on energy transfers, the current direction is only a manifestation of the energy transfer, not the flow direction of the energy.

Ok, then I came to the final try. Now, talking about the actual Edwin Gray circuit, se uses the capacitor to store the energy as stated, then he discharges the capacitor through an air gap, that is to say a very quick discharge. The path of this discharge has no load, and the energy is suposed to go back to the battery.(This part is something like my previous try, but without the load).

But there is a load on the Edwin Gray circuit. He uses another plate near the capacitor that charges with electrons by electrostatic atraction as the capacitor goes charging from the source.

Then, when the spark fires, the electrons on this plate are released at the same speed, passing through the load, and powering it (induction load) He also uses the backEMF of this inductive load. Then, finally, this load is powered, and the primary battery restored in some way.

22. Feb 20, 2004

### Cliff_J

Cala, ESR (equivalent series resistance) is something you should factor into your diagrams and calculations as well. Think of it as a simple resistor that is always inline with a battery or capacitor. And it will always be converting your electrical energy into heat energy, even if it is a small amount.

A great transformer would be 80% efficient. You're converting electrical energy to magenetic and back to electrical, with losses in the form of heat.

Your switcher is what, a FET or a bi-polar and not a simple mechanical switch? So now you have those losses to account for as well, all going to the heatsink...

Now you jump that air gap with maybe 2KV. As it feeds back into the battery you have the ESR to deal with and since its happening quickly the ESL (inductance) may need to be factored back in again. Not much left going back into the battery.

A leaky pump for the battery with leaky fittings on the pipes with hydraulics used for the transformer would be a slightly better analogy, although it still not totally in-check with AC behavior.

Regardless losses abound....

Cliff

23. Feb 20, 2004

### wimms

I think that there is such a game: someone says it works, and someone is supposed to buy one, because its obviously too complex to reproduce.

With highenergy and short pulses it might overcome high impedance of both load and battery. The charge is banged into the battery, some of it going to the load.

I think primary battery is messed up in some way. It might seem like being restored, but it'd probably be boiling and'll fail soon. That part obviously belongs to "needs some more minor development".

Anyway, I don't see where the point of this is. Free energy? What is the source of the added energy?
As someone said, there is only that much your can do with ropes and shouts.

24. Feb 21, 2004

### cala

If i understand the complete Edwin Gray circuit (i'm not sure now), i think the free energy cames from the transformation of the electrostatic charge of the load branch into usable power through the load, and recharging the battery with a backEMF effect on this load.

Basically, the charge-discharge cycles of the high-voltage capacitor is used not to power the load directly, only to cause a compresion, then repulsion of the electrostatic charge on the load branch (something like to compress, then release the water on a pipe).

When the capacitor is charging, the electrons of the load branch goes near the high-voltage capacitor, then, as the capacitor discharges again into the battery through a spark gap, the electrons on the load branch are repelled or released.

The high-energy pulse generated on the electrostatic charges converts into usable power on the inductive load, and also, as the pulse ceases, the backEMF of this inductive load is used to recharge the battery.

Do you think it will work?

25. Feb 21, 2004

### wimms

Nope. Listen, you know by now, that energy spent on doing work with load is not available to recharge the battery. What backEMF?? Transformations of electrostatic charge as source of free energy? Lets grow up and not 'hope' that if there is something mystical that we don't yet understand, then that it could be the source of free woodoo. Its not. If that static charge contains energy, then its was put there. Then something does not have this energy. What? Battery. Its all running circles.

The only thing containing usable energy in this circuit is battery. Everything else is frankly just a bunch of ropes. The only way to explain free energy is to resort to some kind of mystical fields, 4th dimensions, torsion 3rd order effects, vibrations in spacetime fabric, stuff like that. And all that of course due to just reordering of ordinary ropes. \$1500 and you'll get your eternal happiness.