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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
Tags:
Interlude:
In a few hours I will be heading to the airport. I am visiting my daughter who because she is a civil engineer and works in the mining industry lives in Western Australia.
Therefore I will probably not be posting much for a few weeks.
I am not blind; I see the walls of text that are posted in response to my posts by a member I have banned. I don’t read them anymore, that is why I banned him, I hoped to save him the trouble. That was not always the case, once I spent hours preparing careful responses. No matter how hard I tried I was never able to establish communication.
When you understand that a person is quite willing to tell lies in order to support his case then it becomes too much trouble to bother with.
There are many interesting puzzles to consider without bothering about an idiot ( Actually that is probably not quite correct, “smartarse” is probably more accurate).
For example:
A particle of CMBT will exhibit ferroelectricity if the size is large enough. Can I relate the size to the polarisability of the CMBT?
Or
I have tried to calculate the permittivity of an idealised simple solid. I call it MySulfur, it has atoms of sulphur but the packing is that of close packed spheres. The answer I get differs from real sulphur. I know why but how can I improve the estimate?
Or
Triboelectricity itself, how does it really work?
Some time ago I calculated 60 C/m^2 polarisation could be achieved with about 3000 volts across the EESU. This required ionising about 30% of the dielectric. There was also a formula for effective permittivity as I recall. When I return I will dig that out, check it and repost it.
This thread is supposed to be revision after all.
Regards,
Peter
pp said :
Leave it to Technopete, I believe he can handle eight atoms.
That is misinformation!
Current DFT runs using CASTEP are on tetragonal BT (and BZ) using an infinite slab which is 4 unit cells thick. It makes sense to use a symmetric slab, and at the moment we are looking at a [1 0 0] slab boundary direction. Currently both edge terminations are TiO2 rather than BaO, so the modelling uses 23 atoms (5 x TiO2 + 4 x BaO) in a "supercell" which is infinitely replicated in all three directions to represent a BT slab which is 18 Angstroms thick, infinitely wide and infinitely high.
This model is currently running on 3 cores out of 8 on my college workstation, rather than needing the college supercomputer (which is available). There's no point in wasting computer time until it is clear you are modelling the right scenario. Clearly the college CX1 supercomputer will be able to cope with a slab which is significantly thicker, or a mixed BT/BZ slab when the right time comes.
Regards,
TP
pp said:
Leave it to Technopete, I believe he can handle eight atoms.
tp said:
That is misinformation!
Sorry Peter, I merely intended to emphasise the difficulty of doing detailed simulations at the microscopic level.
I think the subject is worthy of a thread of its own. It is not clear to me that DFT is the correct (or even satisfactory) method of analysing the EESU as conceived by Carl Nelson. I would need to use it to evaluate known dielectrics as a first step. ( Sulphur, Sodium Chloride, Barium Titanate, Rutile, D/A doped rutile, the Chung dielectric, the Hansen dielectrics). Call it benchmarking.
It may be satisfactory for what you are doing. However as far as I know DFT does not take into account the Van der Waals forces and these may be important.
My view is that compared to Carl we are all novices and amateurs. Of course the patent does give insight to what Carl was doing, so we have a head start providing we use it (the patent) to guide our thinking. So I say let us keep the alumina until we know for sure that it has been replaced and by what.
The field from 60 c/m^2 (or 27 c/m^2 for that matter) on the plates is much larger than that normally encountered in standard capacitors. The charge movement in the dielectric required to satisfy the minimum energy requirement may not affect just the outer electrons.
So the micro scale behaviour of the EESU dielectric is probably not clear to anyone, no matter how they evaluate their own capabilities. It was/is probably quite clear to Carl though.
The macroscopic scale behavior is a different matter.
Regards,
Peter
The Effective Permittivity due to Space Charge:
This is a repost of an equation I posted on the old site, it elicited one comment. I have evaluated it for the device described in the Eestor 2007 patent.
kef = kv*kd*t*e/( e0*Ei* a^3)
= 26700
Where:
kv .5 (volumetric efficiency; ref Nanocarbons)
kd (difference between centre of mass of positive space charge and centre of mass of negative space charge normalised to width of space charge region.) A figure of 1/6 has been used.
e electron charge
a lattice dimension
e0 permittivity of free space.
Ei ionising field strength, 600 v/micron from 2007 Eestor patent.
The equation was derived using stored energy.
U = 1/2 *e0*kef*Ep^2
Where U is energy density and Ep is the plate field.
The calculated figure is reasonably close to the patent figure. Exact correspondence is not expected as the figures kv and kd depend on geometry and will vary during manufacture.
Regards,
Peter
Edit: t is the particle diameter. I used 680 nm.
Moving Along / The Chung Dielectric / Plod, Plod, Plod:
The formula given in the previous post may be applied to the Chung dielectric by making some substitutions.
Assume that the dielectric has been doped so that there are Nd electrons/m^3 in the conduction band when the field strength exceeds Ea at room temperature. Then replace a^3 by 1/Nd and Ei by Ea
kef = kv*kd*t*Nd*e/ ( e0*Ea)
In the Chung dielectric Ea is chosen to be quite low (giving high kef) so the dielectric works for very low applied voltages. Because of this it does not store much energy. Do Chung et al know this? It seems very unlikely that they would not.
Okay the equation above may be regarded as valid for a case when Poisson’s equation is needed.
The other equation is:
kef = kg*kb*t/d
where:
kg is a factor depending on geometry.
kb is (definitely) the barrier permittivity
t is the thickness
d is the barrier (silica/alumina) thickness
This is valid when Laplace’s equation is appropriate.
The fact that the Chung dielectric has high permittivity due to space charge has been obvious to some forum members for a long time. For example DRP and myself, I expect that BigMig has always known too and there may be others.
Nevertheless there have been many pages of rambling bullshit written about it. Makes you think, does it not?
Regards,
Peter
Hindsight:
An Eestor PR contained the following:
The aluminum oxide particle coating material purification has been certified to be in the parts-per-trillion level.
EEStor, Inc. has certification data from outside sources that purified
aluminum oxide, in the range that EEStor, Inc. has certified, can have a
voltage breakdown of 1,100 volts per micron.
This means that the “bulk breakdown” for alumina thicknesses of about one micron will be as stated. Thicknesses which are greater will have the breakdown reduced according to Tautscher’s Rule. Thicknesses which are less will have the breakdown increased due to “thin film” effects.
In fact, as BigMig explained on another thread, breakdown does not actually occur for high bandgap dielectric films which are thin and pure (without defects).
You do get tunnelling though and this is important for the EESU operation. It is important that the alumina not “breakdown” before the CMBT ionises. Once the CMBT does ionise (at the interface) the field strength in the alumina is limited by that (a bit like Zener protection). This is a necessary condition for EESU operation.
It seems quite clear with hindsight, but I have to admit I did wonder at first why Dick commented on the alumina and not the CMBT.
It is because the alumina barrier and its characteristics are vital to the operation of the EESU. I am sure that with time, the purpose of other PR’s will also become clear.
Regards,
Peter
Thank you Peter. Tunneling w/o BD and able to polarize the CMBT (which IMO must also not conduct). Assembly is not trivial. The solution for that may take weeks or eons.
Hello devotEE,
The terms “breakdown” and “conduction” describe macroscopic phenomena. At the microscopic level we have tunnelling, ionisation and the avalanche effect. Irreversible breakdown is due to heating caused by the scattering of electrons as they travel through the lattice.
At the microscopic level “conduction” is the movement of electrons and in an insulator this is a transient phenomenon.
So, yes, I agree with you the CMBT does not conduct except for leakage. Note that the field is not uniform and there are regions of CMBT with very low field. Of course very low current flows because of this low field and as these regions are “in series” the extremely low leakage currents measured and listed in the patent result.
I don’t know what is holding Eestor up at this stage; I guess that there may be difficulties in achieving consistent quality product in volume. Achieving this could certainly be called “fine tuning”.
Regards,
Peter
What We Have Learned From The Chung Dielectric:
1 Permittivity:
For a dielectric consisting of silica coated CMBT particles; space charge polarisation gives high permittivity. This is not new. The permittivity is high over a wide frequency range. This is new, check the books.
2 Flux:
The experimental results can only be explained in terms of flux by saying the space charge flux is an independent flux (a parallel flow) to the flux which flows through the CMBT.
Comments:
Formally, Poisson’s equation can be used to analyse the dielectric.
The view on “flux” is simply a way of modelling the measured results. I remarked as much “way back when”. Only DRP noticed, discussed and agreed with me. He also noted that the two fluxes existed in the same space. This is true; there is only one kind of space.
Nevertheless the measured behaviour is what it is. Once this fact is accepted then a theoretical explanation can be derived.
However the engineering approach is to accept the fact and use it. This is possibly what Carl did when he invented the EESU. Possibly he also derived the theory at the same time. Possibly Lucas refined the theory.
Whatever happened, it was a remarkable achievement.
Dick’s achievement is the automatic factory and that has been a bastard to get going. Not long now.
Regards,
Peter
Lattice Energy (Part 1):
This is the energy required to move all the ions of the crystal far enough apart so they no longer inter-react and to ensure charge neutrality of each ion, the ions are in the gaseous state.
To determine the value is a Chemistry exercise. The energy for the reactions involved is determined by experiment. It is not a direct measurement. I think it is instructive to consider an example of the results so for Alumina:
Experimentally determined Lattice energy: 15269.9 kj/mole
Molar mass: 101.96 g/mol
Density: 4g/cm3
Therefore:
Ulattice = 15269.9/101.96 /4 *1000 J/cc
= 37441 J/cc
Now calculating the energy stored in alumina used as a capacitor dielectric:
k = 10, Emax = 1000 v/micron
Umax = 44.3 J/cc (In an aluminium electrolytic this will be reduced by the infrastructure required)
The ratio is
Umax/Ulattice = ru = 0.0012
This is quite small but that is the limit due to the alumina “ionising”. If the increase in thickness of the alumina due to this energy is worked out it comes to .43 per cent which seems reasonable.
So for a conventional capacitor energy stored is a very small fraction of the lattice energy.
What about barium titanate? Well I couldn’t find any figures but alumina is a metal oxide and so is BT. The energy figures will be similar but because both barium and titanium atoms are bigger than aluminium atoms the energy will be reduced, say 30000 J/cc as a guess.
Now the lattice energy can be calculated by assuming the lattice is a collection of point charges (hard balls). This is interesting and leads to some interesting ideas so I am going to write a second post on that.
If anybody has more accurate information on BT lattice energy I would appreciate if they could post it (non-banned members only).
Regards,
Peter
Lattice Energy (Part 2):
The equation used is attributed to Born and Lande (e acute); BigMig posted it on another thread not long ago. DRP posted some results quite a while ago and I have referred to it when I talk about short range forces. Here it is again (kind of):
U = A/a – (B/a)^N + Uinfinity
A depends on the Madelung constants for each of the ions in the crystal. This is simple geometry, so A can be evaluated. a is the unit cell dimension, for two ions the separation of the ions is used but for 3 or more the distances are conveniently normalised to the unit cell dimension (a). Initially a is regarded as a variable so that when a is large the energy is Uinfinity.
Uinfinity is zero for electrostatic charges. For ions it is the sum of the ionising energy of the cations and the electron affinity of the anions.
The term –(B/a)^N is really a suggestion made by Born. The constants B and N can be worked out if you know the unit cell dimension (x-ray diffraction) and the lattice energy (chemistry). Basically a kind of curve fitting.
The lattice equilibrium condition is:
dUda = -A/a^2 –N(B/a)^(N-1).(-B/a^2) = 0 at a = a0
A/(N.a0) = (B/a0)^N
Ulattice = A/a0.(1-1/N)
A/a0 is due to electrostatic forces only; N is probably in the range 8 to 12. It can be seen that the difference in lattice energy due to changing N is not that much.
Now considering a to be a variable:
a -> a0.x
U = A/a0.( 1/x -1/(N.x^N) )
For convenience I normalise the lattice energy to have the value 1 when x = 1. Also I am interested in energy difference from a reference condition which is at x = 1.
So I plot u (little U) as a function of x, here are some graphs for typical values of N:
The value of N does not make a significant difference for x values greater than 1. For values of x<1 the difference is significant.
I have some further comments which will have to wait for part 3.
Regards,
Peter
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