Joule Thief 101

[Magluvin]

@Mag 
lol definitely not books =). I'm just not understanding the electron differential you speak of not out of disagreement, but sheer ignorance. I don't see it as counting electrons so to speak, but measuring imbalance in charge. but maybe this is what you are saying by differential. the electron ratio to positive charge which produces a voltage. Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water. and we apparently don't. we do lose energy though, rather have to expend some more in the process. so long as those electrons are moving and there is a current, that voltage will not only drop due to resistance, but due to the 'pressure' equalization so to speak when the joules are spread out like jelly long the capacitance of this extra capacitor introduced, which is like another way of saying the electrons are rearranging yet the differential remains the same.

it's like, yea. naturally I want to say it in essence we are basically saying, what happens when I cut this in half? does it form 2 equal halves? sure, why not? I'm just not necessarily saying the voltage unit would be cut in half imo. I'd assume the same total energy, or joules, or whatever you want to call it that was stored, was cut in half.  so basically I think it COULD end up as 5v/5v so long as stored current was adjusted. how would we measure that? electrons I assume. I just don't necessarily think it 10v has to end up as 5v/5v after the dump using equal caps because it seems to work out that way in our other analogies in nature. it most certainly could though and probably does.

You raise important details as I go on.......

"but maybe this is what you are saying by differential. the electron ratio to positive charge which produces a voltage."

Yes, I probably should use the word ratio instead of differential. 


"Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water"

Water is not a good analogy for pressures like air and charge. Water is not compressible. Air and electrical charge are compressible.


"we shouldn't lose any electrons like we shouldn't lose any air or water."

Ok, here is where I have not went into detail......

If we have 2 caps, one 10uf and the other 100uf, the 10uf cap plates will have less total atoms than the 100uf.  With no charge, there are no excess electrons on the neg plate and no electrons taken from the pos plate.  When we add an electron to the neg plate and take an electron from the pos plate of each cap, the 10uf cap will have more voltage/pressure than the 100uf cap. And if it were a 1uf cap the voltage/pressure would be even higher, just with the 1 electron difference for each plate.  So if we had a theoretical pair of tiny caps and the smaller cap had say 100 atoms and the larger had 1000 atoms for their plates, when we take one electron from one plate and put it into the other plate, the smaller cap will have more pressure than the larger cap. Similar to a small air tank and a larger one, it would take less atoms of oxygen pumped into the smaller tank to reach say 100psi than it would to get the larger tank to 100psi. In both examples, the containers determine how much is needed to be pumped in to get to a particular pressure or voltage.

So when we take that ratio of electrons that are in the 10uf cap at 10v, we are letting that same ratio of the same number of electrons to be in a larger cap/container by doing the cap to cap connection. So now we have the same 'total' number of excess electrons on the neg plates and the same number of electrons missing from the pos plates, but they are divided between the to caps, which if they are now in parallel equals a larger capacitor/container. So pressure/voltage is reduced. It is not resistance that caused the loss, it was the fact that we put our initial pressure into a larger container, there by reducing the total pressure. The only loss is that we didnt use the action of the transfer that happened to do work.



" it's like, yea. naturally I want to say it in essence we are basically saying, what happens when I cut this in half?"

Ah. Thats where you are stuck.  We are not cutting the capacitor or air container in half. We are dividing that pressure into a larger container of twice the size. that is why we have reduced pressure/voltage. 10uf cap in parallel with a 10uf cap equals a 20uf cap. A 10lb nitrous oxide bottle and connected it to another empty bottle, there would be 5lb in each. But we lost total pressure, even though we still have our total amount of gas in weight. We lost the energy of pressure. Resistance had nothing to do with the loss. I cannot use the 10lb of nitous in my car if the pressure were 0psi. As a further example, if we were to have a large enough container that if we sucked out all gas to a perfect vacuum and put in that 10lb of nitrous till it was 16lb, equal to ambient sea level air pressure, the gas will not come out when we open the nozzle. That 10lb of nitrous(about $35) is not usable.



"Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water"

This is the point Im making...  If we did the cap to cap transfer and we start with 10v in the first cap and 0v in the other, the only way we can end up with the total amount of initial energy in the 2 caps total would be to end up with 7.07v each.  So that would mean that we would need a different ratio/number of electrons on the cap plates.  We would have to introduce more electrons in the system, or pull more from the pos plates and put them into the neg plates through some mechanism of doing so. So the cap to cap dump can only end up as half, .5, of the total electron ratio once spread between a container of twice the original size. How could we ever see a .707 in each of 2 caps when we started with 10 doing a cap to cap equalization?

If we have a 10uf cap at 10v and another cap 10uf at 0v, and instead of a direct cap to cap dump, we use an inductor between the caps and watched it in slow motion, we would see the field build in the inductor as current flows. The inductor acts as a flywheel. It stores energy as the current increases. It will pump the remaining electron ratio from the source cap into the receiving cap till the source is 0v and the receiving cap is 10v.  So now if we do the same and we cut off the source cap at 7.07v, the inductor is still charged, and we switch the inductor to be across the receiving cap when we disconnected the source cap, the inductor will pump electrons from the pos plate of the receiving cap into the neg plate till the receiving cap is at 7.07v and we disconnect the inductor. We will now have the total amount of energy in the 2 caps as we did in the source cap to begin with. Thats because we used the energy of the transfer to do work, where the direct cap to cap we did nothing with that work potential. Resistance has nothing to do with the loss because there is no other outcome no mater the value of resistance introduced whether it be 1Mohm or 0ohm, we will still end up with half of the electron ratio in each cap. The total number electron imbalance will be the same, just divided between the 2 caps as reduced pressure.


Mags

[Magluvin]

Does a visual help any?

can you 'visualize' the electron ratio that would occur if a charged cap is directly connected to an empty cap?

Mags

[Magluvin]

What if the other cap is adjustable??

So you start it out at 0.0MFD,, connect the caps together and then adjust the cap slowly up to the same value as the other one,, while doing this the current flow could be used in some sort of inductive fashion so that when current flow stops the collapsing induced field returns the energy it has taken,, that would be the energy not used but would be wasted,, and tops up the adjustable cap.

What if that slow induced field had a secondary interaction with something that allowed for the collapsing field to have a faster rate of change in the flux density??

If the cap is adjustable say 1uf to 10uf, then when you charge it at 10uf to 10v, the voltage goes up when you adjust the cap to a lower value. the electron ratio and count number remains the same. We have just increased the pressure by changing the value of the cap. The energy would remain the same, only the voltage value and capacitance value would change. Good question, as it helps my explanations.

Mags

[Magluvin]

The cap below. Would you consider it to be a charged cap? If so, could you understand that if we took some of the electrons from the bottom plate and put them in the top plate that the voltage would go down?. And in the reverse, if we took some of the electrons from the top plate and pumped them to the bottom plate that the voltage would be higher?

So to what extent 'can't' we say that the number of electrons ratio is directly related to the voltage of a determined cap value of capacitance? Same number of electrons, same ratio, but the value of the capacitance is what will determine the voltage. So I say that we can determine the voltage on that caps capacitance value if we could count the electron number ratio and be very accurate at doing so.

All this is just to help understand that when we do a cap to cap dump, there is still the same total amount of extra electrons on the neg plates that there was in the source cap neg plate, and the same total number of missing electrons from the pos plates as there was in the source cap positive plates. Its just now that the numbers are divided between the 2 caps, thus less pressure, less voltage. Resistance will not change those numbers or ratios. And 0ohms will not change those numbers either. Heck put 10 10ohm resistors in series with the cap to cap transfer. All those actual voltage drops will not change the outcome from 10v in one cap to 5v in 2 caps.

So if thats the case, the only thing we lost in the cap to cap transfer was pressure. If we didnt do anything with that pressure release from one cap to the other, then we just lost it stupidly. Resistance had nothing to do with it.

So the big question is if we have ideal caps, no resistance, and we do a cap to cap transfer, would we not still end up with the 'same' electron ratio and electron number count in either system, real world or fantasy ideal world? So with no resistance we still end up with 5v in each cap from the single 10v cap. What am I missing here??? If the 2 caps ended up at 7.07v each, total energy of both caps equal the energy of the original source cap at 10v, then something must have changed in the electron count and ratios. What mechanism would have occured to create that 'theoretical' situation?

The energy was lost as pressure. If we use that pressure without wasting it, we can then do a full cap to cap transfer, as described by me and MH several times here already. No matter what we do, cap to cap direct connect, we lost 'pressure' if we didnt use it during the 'action' transfer. Ideal world or not. 

Do you get it now that resistance has nothing to do with the 50% loss by doing the cap to cap deal?? 

Now if we did the transfer with the inductor and diode, resistance will affect that for sure due to the voltage division. It will allow less voltage across the inductor throughout the transfer process and the inductor will never build up enough to do the full transfer from cap to cap and we would end up with some left in the source and not all in the receiving cap. The higher the resistance, the less is transferred.  but direct cap to cap, no mater the resistance, there will ALWAYS be a leveling out of pressure and the resistance will only delay that transfer over time.

MH said that the super conducting, ideal caps would have instantaneous transfer.  Well maybe at the speed of light, which is not instantaneous in the least. Its just incredibly fast. Even with superconductors there must be a limit to how much current can flow through the stuff. Thus the cooling I suppose. Is it the cooling that takes away the heat that is not suppose to occur?

Anyway......


Mags

[Magluvin]

Here I have edited one of the previous pics to describe what Im saying....

The first pic is of 2 caps, same value, both at 0v.

The second pic is of the left cap charged to a particular voltage.

The third pic is after cap to cap connection.

This is to show that the 'total number of electrons in both caps' is the same throughout the process and just shows the ratio when charged and then cap to cap connection. We never lose or gain electrons in the system. They are only moved or say pumped from one plate to the other, but while charging the total remains equal. One in and one out at a time to say as an example. So if we do a ratio electron count, and we know the other variables as in the caps capacitance value and the electron ratio change in the left caps plates when charged, then we should be able to calculate the voltage in the left cap when charged. Every time. Then when we do the cap to cap deal, we again should be able to calculate the voltage in each when they are equalized. And each time we do the count, we should be able to do the math to find the voltage.

So in an ideal system, how could 0ohm vs resistance(any value) affect the electron count values changing the voltage outcome for each cap? Where did the extra electrons come from to have 7.07v in each cap from 10v initial charge in the ideal model vs 5v on each cap in the resistance energy losing model? Thats the big question.

If we get the same outcome, ideal or real world, how could we say we lost half the energy in heat, if there wasnt any heat in the ideal model??

Ok. Im done for now.  I cannot further explain it. Let me know where I am going wrong, if you can.

Mags