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Author Topic: Cap to Cap dump  (Read 10045 times)

Offline poynt99

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Re: Cap to Cap dump
« Reply #45 on: June 14, 2016, 01:19:43 AM »

Could it be due to maintaining the electric field between the plates,that is created due to the displacement current,as i mentioned earlier?.


Brad
I don't think so. Once the polarization takes place, no more energy is required to maintain it.

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Re: Cap to Cap dump
« Reply #45 on: June 14, 2016, 01:19:43 AM »

Offline dieter

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Re: Cap to Cap dump
« Reply #46 on: June 14, 2016, 02:18:12 AM »
Brad, in your video you're getting more than 50% remaining  left over after the transfer. Actually I don't see how you can end up with two diffrent voltages when the caps where identical, unless the current could not really flow. However, I noticed there was a FWBR connected. What I have observed is when you connect a diode and a coil or antenna to one plate of a cap, you'll get a steady current of up to several hundred mV, surprisingly regardless of the caps' capacitance. So this probably had an impact as well. The cap then acts almost as a pseudo ground.


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Offline allcanadian

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Re: Cap to Cap dump
« Reply #47 on: June 14, 2016, 03:37:43 AM »
@Poynt99
Quote
Charging a cap from a voltage source has pretty much the same effect as charging a cap from another cap; half the energy is lost (dissipated as heat) in the connecting conductors.


It may be that a cap to cap discharge invokes I2R losses because of the current however there is something else to consider. The capacity to store energy is a function of 1)surface area, 2)the dielectric and 3)the distance between the conductive elements. How much Energy can be stored is a function of the capacity to store energy and charge density or differential charge density which we call voltage.


Note in a cap to cap scenario we are discharging the source cap at a high charge density into a circuit having twice the surface area... two capacitors. Thus it would seem obvious the final charge density or voltage would be one half as the surface area has doubled. Thus the charge distribution or density has changed has it not?, over twice the surface area.
Quote
Like verpies said, we can charge a capacitor from a constant current source too, and if we choose the constant current value (which affects the timing) carefully we can reduce the loss almost to zero.


We should remember a constant current source is just an ambiguous term for a current limiting device reducing current thus I2R losses just like an inductance and if the current limiting is due to any resistive element then that becomes the focal point for the dissipation of energy for reasons too numerous to explain here.
Quote
Why does an inductor help with the efficiency when charging from a voltage source? Inserting a big value, high L/R ratio inductor between the voltage source (supply or capacitor), the inductor acts as a constant current source, thus reducing the loss.


An inductor impedes a large current rise from the source cap through self-induction or a Counter Electro-motive force. As the source cap discharges and the current rises the magnetic field expands inducing a  Cemf limiting the current, reducing I2R losses. When the source cap is exactly one half empty, equilibrium, the collapsing field induces an Emf which forces a current forward which near fully charges the destination cap minus ohmic resistance losses.


Do you not find it curious that an inductive element which literally induces itself, self-inductance a Cemf, has almost no associated losses while a resistive element incurs great losses?. When I found the root cause of why energy dissipates I actually learned a few few things... which was odd for an old dog such as I am.


Now if we could miraculously lose one half our energy in a circuit due to the dissipation of charges within a system I can only imagine what might be possible if we learned how to make them accumulate. We are very good at dissipating energy however accumulation seems to be a problem.


AC


Offline dieter

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Re: Cap to Cap dump
« Reply #48 on: June 14, 2016, 04:04:36 AM »
Quote
a circuit having twice the surface area... two capacitors. Thus it would seem obvious the final charge density or voltage would be one half as the surface area has doubled.

I may have misunderstood you, but, twice the surface means twice the capacitance, but twice the capacitance at half the voltage is not equal twice the voltage at half the capacitance, because the energy in the cap is:

capacitance x ((voltage ^2)/2)

I actually said that already in this thread, but most likely was ignored as usual. In fact, that is the whole point: 2 caps at half voltage together have only 50% of the energy of one cap at full voltage.

Well, I think I misunderstood you. If so, pls excuse.


Offline Magluvin

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Re: Cap to Cap dump
« Reply #49 on: June 14, 2016, 05:46:15 AM »

Could it be due to maintaining the electric field between the plates,that is created due to the displacement current,as i mentioned earlier?.


Brad


I had been talking about this on and off for some years now.  I was just seeing it in my mind.  It was only till recently that I was able to put it down on paper as electron count imbalance of the capacitors that I was able to convey the idea as proof.

The energy we have stored in the capacitor is basically pressure.  Like an air tank, if it is 100psi, and we connect it to a tanks at 0psi, we should end with 50psi each. We lost 50% of the total pressure by letting the high pressure container depressurize into a container that is twice the size.

So my theory on the loss is that we lost pressure by doing so. It wasnt heat loss, or converted to another form. Thats why I say we just lost it stupidly. And it should put a kink in the conservation of energy ideas.  Unless we can figure out where we lost the energy into another form. Im at a loss there so far. But is on my mind daily.

But the most important thing we have learned so far is that resistance was not the cost of the loss, in this case. So I think we need to examine other cases such as power source to empty cap. Running a sim on that seems to show big resistance losses.  But the sim is only giving a trace with a peak values. Would need to be graphed out over time to get solid results. The reason I bring that up is the case of a very large cap charged to 10v, into a very small cap, say 10uf 10v onto 1uf at 0v, we end with 5v each. But the loss of doing so is not as great as 2 caps of the same value. Mostly it is because we didnt dump from the full cap any energy levels that would bring it down much. So the total energy we began with is only minus a small portion of the whole. But I think we still suffered a loss in that transfer by not using the energy of the transfer.  Mostly because of stupid losses of doing so.  The inductor between the caps takes advantage of the transfer from cap to cap. Of which I call a smart move, as it avoids the stupid loss of not making use of the transfer from cap to cap. Same as the air tanks.  We can just connect the 2 air tanks and 'watch' the energy go by from tank to tank, or we could use an air motor to do something with the transfer. In both cases in the end, the 2 tanks will result in half the energy left once they are equalized.

Mags

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Re: Cap to Cap dump
« Reply #49 on: June 14, 2016, 05:46:15 AM »
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Offline allcanadian

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Re: Cap to Cap dump
« Reply #50 on: June 14, 2016, 06:41:58 AM »
@Dieter
Quote
I may have misunderstood you, but, twice the surface means twice the capacitance, but twice the capacitance at half the voltage is not equal twice the voltage at half the capacitance, because the energy in the cap is:

capacitance x ((voltage ^2)/2)


The sequence of logic here is why did the energy change in our final two caps versus the original one?...because the voltage dropped. Why did the voltage drop?... because the original charges now rest on twice the surface area thus the charge density in now 1/2 of what it was originally. I simply look at it differently because I don't normally use equations I use reality... what is actually happening in reality and why is it happening?.


Or


Energy=1/2 CV^2... a 1F cap at 10v(E=1/2*1*10^2)=50. The one cap discharges into another identical cap and we now have two caps at 2F at 5v(E=1/2*2*5^2)=25. We lost 1/2 the Energy because it takes progressively more work to force charges together due to Coulomb forces and we also lose more energy when the charges move apart occupying a greater surface area. Coulombs law states the magnitude of the electrostatic force between the charges is inversely proportional to the square of the distance between them. A greater surface area (twice the surface area as the capacitance has doubled) means more distance between the charges and energy falls following the inverse square of the distance between the charges.


I always work the problem from the particle/field level outwards to my observable level, the electrostatic force between charges in Coulombs Law is dependent on r^2(a distance) thus V must also be squared in E=1/2 CV^2 because the Voltage is a function of charge distribution(a distance which is r^2)... the number of charges and the distance between charges relating to surface area. Again twice the capacitance means twice the surface area = more distance between charges = less voltage = less energy falling by the square of the distance between the charges.


This isn't water flowing in a tube under pressure or buckets full of water... this is electrodynamics. The force between the charges dependent on the distance between them is the real Energy in the capacitor and it falls by "the square" of the distance which is dependent on the surface area. Double the surface area and we lose one half the energy.



Quote
I actually said that already in this thread, but most likely was ignored as usual. In fact, that is the whole point: 2 caps at half voltage together have only 50% of the energy of one cap at full voltage.


I know you said it and that is why I posted because yes I think you were ignored, lol. I think these concepts are pretty important for people to learn and these water pipe/bucket analogies are ridiculous.


AC

Offline verpies

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Re: Cap to Cap dump
« Reply #51 on: June 14, 2016, 10:33:03 AM »
Or ask yourself the question if the dielectric can be polarized by some other means,, like a change in the pressure on the dielectric.
Yes, but not efficiently.
Too bad you ignored my question with OU potential.

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Re: Cap to Cap dump
« Reply #51 on: June 14, 2016, 10:33:03 AM »
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Offline verpies

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Re: Cap to Cap dump
« Reply #52 on: June 14, 2016, 10:42:56 AM »
We should remember a constant current source is just an ambiguous term for a current limiting device reducing current
It is not an ambiguous term.
A current limiting device is not the same as a current source.

Offline webby1

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Re: Cap to Cap dump
« Reply #53 on: June 14, 2016, 02:04:08 PM »
Yes, but not efficiently.
Too bad you ignored my question with OU potential.

Why do you think I ignored it?  I answered with something not expected,, does not mean I am not thinking about it.

The bottom line is that there is more than one storage system in play with a cap.

My view is that what happens one way should be able to be reversed,, that is in non-used up, transformed or destroyed kind of ways,,,

Offline webby1

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Re: Cap to Cap dump
« Reply #54 on: June 14, 2016, 03:45:08 PM »
The plates are attracted together as they decrease the distance, thus they perform work.
Also, the capacitance of the capacitor increases and the voltage across that capacitor decreases, while the electric flux stays constant.
The energy stored in the capacitor after the plates move closer together is smaller than the energy which was stored when the plates were further apart.  The difference is exactly the amount of mechanical work that the attracted plates have performed.

The above is pretty elementary, what is more interesting is when the distance between these plates stays constant, but there is a high electric susceptibility slug attracted in between these plates (from far away).
In that scenario, the capacitance also increases and the voltage across the capacitor also decreases while the electric flux stays constant, but the approaching dielectric becomes internally polarized.  That polarization represents a part of the stored energy and contributes a part of the total electric flux.

Ask yourself the question, whether the energy used to polarize the dielectric slug can be recovered, once the slug becomes attracted all the way in between the plates ?

With this,, I agree that the plates will pull together and express the mechanical work that is equal to the decrease in stored energy,, basic stuff.

If what you propose is considered,, then there might be an interesting problem.

If the slug were to exert a mechanical work out while it is being attracted into the space between the plates then that mechanical work should also be the same amount of energy that is reduced within the stored capacitor,, discharge the cap at the end point and there is less energy out of the cap.

If the polarized condition could be conserved while the cap is discharging, and you have the mechanical work out from attraction plus the discharged energy,, then you allow the polarized condition to relax through another out,, would that be a gain?

Is a piezoelectric crystal something that could do this?

or do you have another material in mind?

Free Energy | searching for free energy and discussing free energy

Re: Cap to Cap dump
« Reply #54 on: June 14, 2016, 03:45:08 PM »
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Offline allcanadian

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Re: Cap to Cap dump
« Reply #55 on: June 14, 2016, 06:02:58 PM »
@verpies
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It is not an ambiguous term.A current limiting device is not the same as a current source.


The term current source sounds easy enough and I have designed and built many. The last one for my Arduino pro mini controlled, variable output, 200w/19,000 lumen LED flashlight. You know it sounds so easy, it's just a current source so you pop in a big freaking resistor is it starts hemorrhaging energy like it's going out of style. Then you start switching caps and not that much has changed. Finally we are left with the only real component which works which is the inductor in a buck/boost converter configuration. So really when we speak of an efficient current source in reality we are speaking of a switched inductor circuit.


Theory and speculation are fine, using near meaningless terms which never actually describe the events which need to take place works for a while... until you have to build something. At which point builders understand theory seldom works as planned in reality. The fact is we will never truly understand what happens and why until we build it and test it...that's the way it works. This is what gives the terms we use real meaning because under the surface lies real understanding.


AC

Offline dieter

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Re: Cap to Cap dump
« Reply #56 on: June 14, 2016, 06:47:24 PM »
AC - ok thank you. I see... somehow... but still puzzling: if you increase the surface there will be more capacitance, hence the voltage drops, versus: when you connect them in series, the inner plates become phantom plates, resulting in the outer plates forming a cap with twice the distance, equals half the capacitance, resulting yet in half the voltage. Probably we got to deal with some sort of fuzzy "yes and no" logic here, because, whatever you do, the caps are both, connected in series and in parallel at the same time. Additionally they form a complete circuit.

From the subjective POV of each of the two caps the other one is just a diffrently charged conductor, so, as current always flows from a higher to a lower voltage, +5 flows to 0 Volt at the anodes, and zero to -5 at the cathode, where of course the negative voltage is relative. Regardless of the kind of supply, current will flow as long as A) there is a charge and B) there is a diffrence in Voltage.

Pressure is a good term for electrostatic charge, also tension (actually the germans say tension, not voltage, like 230 Volt tension), as it suggests a need for a nonlinear amount of energy to reach a certain tension, but, except for the "loss-mystery", this energy can also be extracted.

This is particulary significant in high capacitance-low voltage caps, whereas in low capacitance-high voltage caps the "loss" is proportionally much higher; at least if I haven't completely lost my mind, which I cannot exclude entirely ATM.

Even if in this special dumping example both outcomes would be the same, it would be diffrent in a real world application in which a cap must deliver a certain amount of energy: the low voltage cap with high capacitance will drop only a few dozen millivolts, while the high voltage-low capacitance cap would drop significantly in voltage to provide the same amount of energy.

If that makes any sense.


Offline webby1

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Re: Cap to Cap dump
« Reply #57 on: June 14, 2016, 08:02:13 PM »
A cap to cap dump is like pulling a cart up a ramp using a soft spring,, when you get there you know how hard you pulled on the spring and how far you pulled the spring,, now you are going to let go of the spring and collect the output from the cart rolling back down the ramp,, and then saying "look at all the energy I lost".

It is not so much of a loss if you were to also collect the stored energy within the spring that you let go of.

Offline verpies

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Re: Cap to Cap dump
« Reply #58 on: June 15, 2016, 01:45:27 PM »
If the slug were to exert a mechanical work out while it is being attracted into the space between the plates then that mechanical work should also be the same amount of energy that is reduced within the stored capacitor,, discharge the cap at the end point and there is less energy out of the cap.
But if the dielectric slug were always between the capacitor plates, especially while the capacitor was being charged, then this slug would become polarized, storing energy internally.
In theory all of the energy input into such a capacitor could be recovered, because upon capacitor's discharge, the dielectric slug would relax and give up its internal energy.  In fact, most real capacitors store and release energy that way.

What's interesting is, that the amount of energy recovered is proportional to the level of dielectric polarization, when all other things consider are being equal (e.g. plate area & spacing). 

The final level of dielectric slug's polarization can be the same when the slug is attracted into between the plates from afar and when it is always located between the plates (while the cap is charged). 
In both cases, the energy released by discharging the capacitor when the slug is in between the plates, is the same - proportional to slug's dielectric polarization.

Notice, that it does not matter whether the slug was moving just a moment ago (before the discharge process had started).

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Offline webby1

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Re: Cap to Cap dump
« Reply #59 on: June 15, 2016, 02:08:33 PM »
If the slug does not move at all,, then its influence within the cap will not change,,

And I agree that the stress on the dielectric is where some of the energy is stored,,

Since your view on the changing dielectric polarization by means of pressure applied is that of not very efficient,, how is this any more so?

You asked if the polarization could be saved once the slug is attracted all the way,, I took that to mean that it was external to start with and was attracted into the region between the plates.

So are you suggesting then,, we start with the slug external to the plates,, but on a mechanical system so that any movement of the slug can be converted into a mechanical work out,, slowly charge the cap and watch the slug get pulled into the space between the plates,, when fully inserted,, discharge the cap.

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Re: Cap to Cap dump
« Reply #59 on: June 15, 2016, 02:08:33 PM »

 

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