What We Have Learned – The Topic.

[PeterP]

What We Have Learned – The Topic.

This discussion is a continuation of the blog post of the same name. When I posted the first post on that thread I honestly believed that all the members of the forum (who have technical back grounds) understood and accepted the “brick model”. When I found that was not true I attempted to explain the model more clearly. I am afraid that I failed in that, but one thing led to another and I have kept on posting. Essentially I am reviewing points that were raised on the old site and which I think are pertinent to “How the EESU Works”. Just a few more posts and I will be done.

The usual rules apply: Please keep replies short, to the point and polite. ee-tom is banned.

I had started to review the possible polarisation mechanisms (for the EESU) so I am going to repost the two posts concerning that, then I will continue.

Regards,

Peter

[PeterP]

­­­­­Macroscopic Concepts:

Macroscopic in the context of dielectrics basically means “measureable”. Measurements can be made on homogenous dielectrics between electrodes and then the results extended to cover the heterogeneous case.

 

Macroscopic Field Strength: 

Macroscopic field strength is regarded in the texts about the matter as a volume average calculated (perhaps weighted) over a reasonably large number of cells.

It can also be calculated as a volume average over one unit cell. This certainly applies for a homogeneous dielectric where all unit cells are assumed to be the same.

It can also be calculated by evaluating the microscopic potential at two points ‘a’ metres apart along a line perpendicular to the plates and dividing the difference by ‘a’.  Where ‘a’ is the unit cell dimension.

Macroscopic field is discontinuous at interfaces. Changes in field strength take place over a distance delta where delta goes to zero.

 

Macroscopic Flux:

This concept is valid over the range of field strengths where the macroscopic approximation is valid.

 

Permittivity:

This is a macroscopic concept.

 

Macroscopic Potential: 

This is continuous through interfaces. Both Metal/dielectric and dielectric/dielectric interfaces.

It can be seen that the “macroscopic approximation” is an abstraction. To be useful for considering the EESU an abstraction for space charge needs to be included.

 

Abstract Space Charge: 

Space charge is represented by two charge planes. Defining parameters are:

 

1) The field strength Esc which is equal to the breakdown field strength of the ionised volume of dielectric which provides the space charge.

 

2) The dipole moment Msc of the above volume calculated by the usual means.

 

The space between the charge planes can be regarded as having a permittivity of 1. 

As the charge planes are abstractions then the charge density is continuous and there is no limitation on value (except as determined by analysis of the system).

 

Warning: Care is required not to accidently introduce microscopic quantities and mix them with macroscopic quantities. That way lies confusion. That is the way of the "sceptic".

Regards,

Peter.

 

 

 

 

[PeterP]

Why Does the EESU Work?

This is not the same question as “How Does the EESU Work”, this is much easier really.

Here is Why:

As the voltage across the EESU is increased (say in steps) during charging the field in the dielectric changes so that the energy in the dielectric is minimised. This is standard and is the condition for equilibrium (after each charging step charges, in the dielectric stop moving).

This is physics and as a result a fully charged EESU has about half of its dielectric “ionised” (this being the minimum energy solution). So there is space charge. The potential of the space charge and the amount is therefore known from the physics. Applying electrostatic theory to the amount of charge gives a potential which does not agree with the actual potential (I think this is well known).

 

This is because electrostatic theory is incomplete. For example when using electrostatics charges are held in position by “clamping” or are moved by “agents” which store energy. A convenient fiction.

 

Perhaps an analogy with a voltage regulating diode might help. The voltage across the diode is constant as the current is increased. With space charge the field strength is constant across the space charge region as the voltage is increased (nothing new).

Anyhow if you can accept that the charges and potentials are according to the “physics” the EESU claims are quite reasonable. To prove it you have to make one.

Also as there is very little voltage across the rest of the CMBT so the leakage current of the EESU is incredibly low.

Of course I have said all this several times before, can’t seem to stop myself.

Regards,

Peter

[PeterP]

The Battery is the Boss:

When a capacitor is charged the battery determines the voltage across the plates of the capacitor. If you pull the dielectric out current will flow back into the battery. If you disconnect the battery and pull the dielectric out the voltage across the plates will rise. If you do that to a fully charged EESU you will deserve what you get. So we are not going to do that. The potential difference between the plates depends on the battery (yes, I am repeating myself).

If you draw a straight line from the positive plate to the negative plate of a capacitor or an EESU then if you measure the voltage along that line it will always decrease as you move from positive to negative. There will be no voltage rises only voltage drops. The sum of the voltage drops equals the potential difference between the plates.

This is what happens at the macroscopic level. At the microscopic level the voltage may go up and down like a yo-yo but ultimately it must be consistent with the macroscopic voltage (through an averaging process, if you like). In the EESU the macroscopic voltage is known from the “physics”.

The separated charge is also known from the “physics”. So the amount of charge on the plates can be calculated. I have done this by assuming that ionisation of the dielectric allows 1 electron to escape from each unit cell in the ionised volume. This gives 1xEESU energy density.

The only question remaining is “where is the energy stored”?

The energy is stored as charge on the plates. The energy is stored in the field.

These two statements are equivalent. This may not be obvious so I will attempt an explanation in my next post.

Regards,

Peter

[PeterP]

Levels of Abstraction for the EESU:

 

1 Level n (Top Level):

k= 18500, Max working E = 350 v/micron, Max working P = 60 C/m^2

 

2 Level n-1 (Macroscopic Representation):

Potential distribution from plate to plate and charge distribution in dielectric determined from material characteristic and “physics” (see previous posts).

 

3 Level n-2 (Charge Sheet Representation):

 

Step 1: Charge Sheets:

Space Charge is represented by two charge sheets of opposite sign. The potential difference between the charge sheets is slightly less than the potential difference between the plates, say Vs. The charge density on the sheets is that required for the “state of charge”. Say for full charge the charge sheets are separated by .5xplate separation then charge density = 2xplate charge density, say Ds. The potential difference calculated from Ds does not equal Vs.

 

This enigma is resolved by including  “pseudo charge sheets” which takes into account the “short range forces” or the “molecular potentials”. These are similar to the Lennard Jones potential for inert gases, but are for CMBT.

 

Step 2: Pseudo Charge Sheets: (Short range forces).

These represent the interface to the “not ionised” portion of the CMBT. The potential at a distance d from these sheets has a value Vc = A/d^na.

A and na are determined by the geometry and the requirement that the combined potential at the interface (due to charge sheet and pseudo charge sheet) should be Vs.

I expect that na will be quite high (by examining cases where the values are known) (possibly >8 or higher) so the pseudo charge sheets will have no (or very little) effect on the plate potentials.

I have no doubt that Lucas successfully described the lower levels, finishing with a Quantum Electrodynamics description of the EESU, but do we need it?

 

Regards,

Peter   

[PeterP]

Electrostatic Energy in a Collection of Charges:

 

For 2 Charges:

U = kc*q1*q2/r12

 

Where q1, q2 are charges, r12 is the distance separating them and kc is Coulomb’s constant. The formula is derived by considering an “agent” moving one charge slowly to “infinity” and calculating the work done (or energy gained, depending on the relative polarity of the charges) by the agent.

Consider N charges of value qi. By moving the charges, one by one to “infinity” the energy for the distribution is obtained.

 

U = kc*Sum(qi*qj/rij)………………………………………………………………………………1

 

Where rij is the absolute distance between charges i and j. The sum is taken for all values of i and j except when i equals j.

 

The equation is based on Coulomb’s Law as are Gauss’s Law and Maxwell’s First Law (Poisson’s and Laplace’s equations). Other expressions for energy are derived from it using mathematics. For example:

 

U =.5*e0*Volume_Integral( E^2)…………………………………………………………………….2

 

Where the charge distribution is continuous and the integral is over all space. Point charges not allowed because the integral includes i=j, which is excluded in equation 1. However this difficulty can easily be worked around.

I used equation 1 to calculate the distribution of charges on a disk on another thread.

Equation 1 can be used as the starting point for calculating the “self-energy” of a lattice. A term for the short range force has to be included. Given the lattice parameters (x_ray diffraction) and the “ionising energy” the constants for the short range force term can be evaluated. The standard example is the NaCl lattice.

 

Let us say we have a thin slab of NaCl. We can work out the self-energy. If we attach electrodes (perhaps “gold leaf”) we can apply a field, this requires energy. This changes the lattice dimensions slightly and hence the total energy. The effect is reversible so the energy can be recovered. This is capacitor dielectric operation. Two points to note:

 

1) The plate charges must be included (directly or indirectly) in the calculation of energy when there is an applied field.

2) The applied field is pulling the lattice apart which is the opposite to the self-energy. You might imagine that total energy could be reduced to zero when the lattice would fly apart. Usually this does not happen, instead electrons escape from the molecules and the dielectric starts to conduct. If uncontrolled this leads to avalanche breakdown which if uncontrolled causes heating and destruction of the dielectric.

 

I think that covers the basics of the energy relationships in a conventional dielectric. There are refinements, techniques, conclusions and there are questions; for somebody like Carl Nelson there may be a path.

 

Regards,

Peter.

 

 

 

[PeterP]

Paradox?

We have seen that a parallel plate capacitor can have a stable region of heterocharge which is centrally located. The formation of this heterocharge is initiated and controlled by the presence of the barrier.

Let us represent the space charge by two charge sheets of opposite polarity.

Consider that the plate potential difference is V then the potential difference between the charge sheets is less than V. Call it V1

 

where v1 < V.

 

There is no way out of that V1 must be less than V.

Now consider a charge density on the charge plains of D =+/- Q/A

Using Gauss’s Law the potential difference between the charge plains is:

 

V2 = D*t/e0……………….where t is the separation of the charge plains.

 

Quite moderate assumptions about the value of t and D give tremendous values of V2.

 

But V2 must equal V1 and V1 < V.

 

This is the paradox.

 

It is resolved by realising that V1 is a macroscopic value and that it is incorrect to use Gauss’s Law to calculate macroscopic voltage. Gauss’s Law is always correct but it can be misused. Anyhow it can be checked by doing the calculation correctly.

 

The correct way to calculate the macroscopic voltage for the space charge region is to calculate the volume average of potential over that region.

 

The calculation of this potential from microscopic values is an interesting task.

 

Regards,

Peter

[PeterP]

The Last Step?

This is the reconciliation of macroscopic potential and microscopic potential in the EESU.

For an homogenous dielectric when the applied voltage is V; the macroscopic potential in the dielectric can be represented by a straight line starting at 0 and rising to V ( A graph of V/x where x goes from 0 to t, where t is the thickness).

If the microscopic potential is plotted on the same graph it will oscillate around the straight line. It may swing between plus and minus infinity. I call the points where the macroscopic potential and the microscopic potential are equal the “zeroes”. If there are points where the potential goes to +/-infinity I would call those the poles.

For a homogenous dielectric the zeroes will be uniformly spaced except for a few near the metal/dielectric interface. The effects of these are usually (reasonably) ignored.

The slope of the line joining the zeroes is the field strength and if V is increased the breakdown field strength will be reached and the dielectric will breakdown.

The EESU may be represented by three layers of dielectric (as previously explained), the middle layer is alumina.

The graph of macroscopic potential has three lines the slope of the CMBT lines is very low and the alumna line is high.

The microscopic potential again oscillates about these lines and again there are zeroes.

When space charge forms at the dielectric interfaces (as it inevitably will) there is a field generated by this space charge. This causes charge to flow on/off the plates and this charge has a field. The combination of the two fields shifts the zeroes.

The effect is non-uniform (analyse it and see). The zeroes near the plates are not shifted much but near the interface the shift is significant.

This means that an increase in plate voltage will cause the dielectric near the interface to breakdown thus generating more space charge (and maintaining the equilibrium).

The shift in the “zeroes” is an indication of distortion of the lattice. The lattice is compressed. This results in a reduction of permittivity when the distortion is “high” and this in turn increases the field which again aids the ionisation process.

I hope that is clear. If any eligible member has a question, please ask. An eligible member is one that hasn’t been banned.

 

Regards,

Peter

[Sydd]

Peter,

 

thank you very much for your interesting posts on this and other topics.

 

Even if this is way over my head I want to ask some questions:

 

1. Would you be so kind and post some graphs maybe even some of your simulation runs

2. Can you tell us what the ED limitation of such a dielectric would be

3. Would you say DW's actions/patens are inline with your "model" (glass/PET/epoxy, paraelectric/paramagnetic, alumina optional, k-in-the-millions, etc.).

 

Thanks again! Keep up your good work!

 

Regards,

sydd

 

[PeterP]

Hi Sydd,

Thanks for your comments.

1 Okay, I will post some graphs/drawings, it might take a few days though. I have posted some simulations;  “A Disk of Charge”, “A Line of Charge” and the “Field at the Interface” (on this thread) for example.

 

2 Energy density depends on the amount of space charge generated, the Ebd of the CMBT and the elastic limit of the CMBT (for a start). An early calculation of mine indicated that 1xESU could easily be obtained.

 

3 My model is in line with the CMBT/alumina/PET patent (I hope).  I see no reason why low field k of CMBT should not be very high, but note that k is not a very useful parameter for describing the behaviour of CMBT. It is the stability of the measured “k” that is impressive. I don’t see any significance in paramagnetism for the EESU. Yes, I think it is important that the CMBT be in the paraelectric phase.

I don’t know if epoxy is used, I don’t know if alumina is the best barrier and I don’t think that k in the millions means any more than 1/k > 0. (k in the millions was for high k polymer and I don’t have any information that Dick is using this in the EESU.)

 

I hope that answers your questions.

Regards,

Peter