Joule Thief 101

[MileHigh]

Magluvin:

So with that said, if we can associate and determine the voltage charge in the cap if we're able to count the electron differential between the positive and negative plates, then we will always have half the differential in each cap as we did in the 10.0000v source cap to begin with doing a direct cap to cap transfer.   So how, how is it possible to end up with 7.07v in each cap by direct transfer in theoretical superconducting capacitors, connections and switch, etc?  Wouldnt there need to be some electrons added to the circuit in order for that to happen?(http://overunity.com/Smileys/default/huh.gif) 7.07 million differential???   

Isnt it odd when you think of it that way?  Thats why I like the air tank analogy because the psi (at particular temp) can be determined by how many air atoms, to say, are pumped into the tank in the same way we can look at caps.  So if the air tanks were to do a direct tank to tank transfer till equalized, how could we start with 100psi in the source tank and end up with 70.7psi in each when done? Eliminate losses, and how did we get the extra air atoms/molecules that we didnt start with?

There is an answer to the conundrum with ideal electrical components.  We do have to add the ideal diode, but what you missed was the impossibly infinitely fast switching function.  I will explain but first let's go back to the air tanks.

In the air tank example, and in fact using an unreal model where we ignore temperature for a second and have idealized components, when one tank discharges into the other tank, that spins up an ideal flywheel pump.   So you can set it up such that when the first tank drops to 70.7 psi, you then stop the air flow out of the first tank, and then the ideal flywheel pump takes over and pumps extra air into the second tank so that it also reaches 70.7 psi and therefore no energy is lost.  The net result is no energy was lost and extra air was pumped into the second tank by the ideal flywheel pump.

So, let's do that with ideal capacitors and ideal diodes and an ideal switch.

As Poynt said, an ideal capacitor connecting to an ideal capacitor is a no-no.  You were absolutely right though about the conundrum of "missing" electrons to get to 7.07 volts in each cap.

If you have two ideal caps, and an ideal inductor connecting between them, that is a manageable situation with no ripping of space-time.  In this case, Cap A discharges into Cap B via the coil and the charge goes back and forth forever.  There are no extra electrons and there is never a condition with 7.07 volts in each cap.  The best you can get is this in terms of equal voltage something like 5 volts in each cap, and current flowing through the coil.  I am not saying it has to be 5 volts in each cap either, just an equal voltage in each cap.  The coil discharges into Cap B and stops discharging when Cap A has zero volts and Cap B has 10 volts.  Then the whole process reverses.  During that infinite back-and-forth cycle there will be an instant in time where there is an equal voltage in each cap and the "missing" energy is in the coil.

I will do another post to explain the two caps with 7.07 volts each.

MileHigh

[Magneticitist]

@Mag

you know it's definitely interesting thinking about it, I just don't get how it could mathematically be worked out that there is some gain when placing a charged capacitor in parallel with a discharged capacitor. The air tank analogy is definitely similar, but I don't think we can say that a battery, or capacitor, or any electrical power 'source' is under the same kind of pressure. it's actually a good analogy for voltage I guess, like water pressure in pipes, but when the charged caps plates coalesce with the neutral cap isn't it just a balancing-out transference at electron speed, and not the result of some form of actual pressure like that of air? when it comes to counting electrons, isn't the 'empty' cap more like a container that already has electrons in it just like the empty tank has air in it before being fed pressurized air from another tank? In ideal terms where we have a 100% transference of electrical energy from one place to another then it seems like we answered our own question. If we are talking about a current flow happening between one cap to another, and assuming there is 0 resistance, then if we're going with the flow and also assuming current will actually pass over 0 resistance in the real world then I don't think a 10v cap would have to balance out to 5v to have 'equalized the pressure' so to speak.

[shylo]

Hi Poynt, "If you insert a high Q inductor between them, you can approach a lossless transfer."
I don't know what "high Q" is, but I have 10 mot coils wired in series, that when I discharge one cap through them to turn a mag rotor , which puts voltage back in , it also sends almost all of its voltage to the cap at the other end.
Then discharge that cap and it sends almost all of its voltage back to the first one.
The numbers go down slowly, but that is only using two points, with those ten coils there are eleven points that can be used.

Hi Mags,"I hope that what Im saying here makes sense. I reread it a couple times.  Its as good as I can put it to ya."
To me your halfway there you just have to go a little deeper.
artv

[MileHigh]

Okay, so now let's discuss an ideal situation where we end up with 7.07 volts in each cap.  It's going to be essentially identical to the case with the simplified model of the air tanks and the ideal flywheel pump pumping in extra air molecules.

Assume Cap A has discharged to 7.07 volts.  You know that current is still flowing, and therefore there is energy in the ideal coil connected between the two ideal caps, and there is also some energy in Cap B.  Like I said before, we don't need to know the size of the coil, or the amount of current flow, or the voltage in Cap B.  The most critical thing is that we are at the instant in time were Cap A is at 7.07 volts.

So what we have to do is instantly change the configuration of the circuit at this instant in time to realize the goal of 7.07 volts in each cap.

So the new configuration is this:  Cap A is completely disconnected from the circuit, and the inductor is across Cap B only with an ideal diode.

So what will happen is that the inductor will discharge all of its energy into Cap B and you end up with 7.07 volts in Cap B.   That is the "ideal flywheel air pump."  The ideal inductor supplies the extra electrons.  You have to remember that an ideal inductor can supply an infinite amount of electrons into a zero ohm load.  So in effect the ideal inductor "pumps up" Cap B to 7.07 volts.

Of course, we did an "impossible switching event" where we reconfigured the circuit and put the inductor across Cap B to make this all happen.  Such is the beauty of a thought experiment using ideal components.  The critical point being that you need extra electrons to reach 7.07 volts and the inductor supplies them by sucking them off of one plate and depositing them on the other plate.

It's also worth noting that without this "trick" then you can never get 7.07 volts across each cap.  Even in the ideal case we are discussing, charge has to be conserved.

MileHigh

[minnie]

   MH,
        thought it interesting to know MarkE's take on air tank.
Can't do quotes, you'll have to look at Re Rosch that I posted
earlier.
      Tinman struggles with fundamentals!
            John.