The bad leakage current and how was meassured

[Tecnófilo]

Hello you eestorians. I´ve been scrolling your chat and searching posts. I am surprised nobody has made comments here (well, I did not see them) about the "testing equipment" specified in the EEStor report of jan 28. And someone is calling for an explanation of how a "leakage current" is meassured.

As I think you are guessing about me. I am a spanish physicist and my speciliaty is in electronics.  And, yes, I have ZEEN shares at 0.87CAD/sh. I shall try to explain you my opinion about the recently published leakage and its measure.

First problem I see: as "Weir's style" implies, we don’t know which "testing equipment" was used; we only know that "testing equipment included"... so maybe more testing equipment were used.

"Testing equipment included: QuadTech 1715 LCR Digibridge LCR, Keyence GT2-212K, Yokogawa WT3000, Stanford Research PS350/5000B-24W"

QuadTech 1715 LCR Digibridge LCR (quadtech - specifications sheet): is useful for testing impedances (resistance, capacity and inductance). So Weir should have an analysis of the impedances at "Test Frequency: 100Hz, 120Hz, 1kHz and 10kHzv (9.6kHz)". But it is not related with current leakage. Also, for energy storage, which matters must to ZENN, it is without interest the response at 1KHz: charge and discharge will be at DC or cuasi-DC, unidirectional current.

Keyence GT2-212K (GT2 - GT2 specifications - GT2 applicationes): GT2 is a series of "contact probes". I think were used to messure the thickness and maybe wardpage (electrodinamic response with applied voltages). It also has nothing to do with the leakage current.

Leakage current is very simple concept: it is the current between plates through the dielectric insulator, even when the capacitor is disconnected. Even disconnected, the capacitor has a voltage between plates (culprits are the charges accumulated in both plates as they attract), so there will be a current that (self)discharge the capacitor: a very bad thing if you pretend to storage charge in it. The energy storaged in a capacitor is only the potential energy due electrostatic attraction between charges storaged in plates; of course you can speak about the changes in dielectric (polarization, field, fux, etc.), but at the end the storaged energy is that potential energy.

So, how is measured the leakage?: it is also very simple. The capacitor will be connected between a constant voltage source which is equal to the voltage it is intended to maximum charge the capacitor; then the capacitor is let charging until it has the intended maximum charge, which occurs when the capacitor has the same voltage than the source but with inverted polarity. 
Then the current is messured, with the same constant voltage source applied! That is so because this current should have the same value that the leakage current in the first instant when capacitor is disconected. If you disconnect the voltage source, at the first moment the leakage current is equal to the current measured as I stated here; after disconnection the leakage current decreases exponentially as it does voltage between plates, and so the storaged charge and energy (you have beautiful graphs posted in this blog).

So the "testing equipment included" that matters for leakage are: Yokogawa WT3000 and Stanford Research PS350/5000B-24W (we don’t know if they used other equipment).

Stanford Research PS350/5000B-24W (PS300Sereis Catalog - PS300Series Manuals): is the voltage source, and it can force up to 5000V (5kV) between plates. It is very sophisticated and can measure the current with "current Resolution 1uA" (micro A), it has display for current and can be controlled with computer. But the problem is obvious: preliminary EESTOR results of leakage currents are 0.0002, 0.0003 and 0.0007 mA: well out of its resolution, 1 orders of magnitude(*).

PS350 cannot meassure neither 0.2uA nor 0.7uA: it should display 1uA or 0uA.

So there is no doubt for me: leakage current was NOT measured with the Stanford Research PS350/5000B-24W.

Yokogawa WT3000 (WT3000): it is a power analyzer. An impressive meter for sure, can be computer controlled (interfaces GP-IB, Ethernet, RS-232 and USB ), frequency up to 1MHz. Of course it can measure the current; I am not familiar with it, but from its specifications I understand the current ranges that can be selected are for the 2A input element (I understand it means up to 2A):
5mA, 10mA, 20mA, 50mA, 100mA, 200mA range.

 

So, the best option I see in specifications is to select 5mA range; again I understand up to 5mA. What is the DC accuracy for that range (5mA):
"DC accuracy: 0.05% of reading+0.05% of range+2uA"

Forget the reading: 0.05% of range+2uA=5*10^-4*5*10^-3 + 2*10^-6=(2.5+2)*10^-6=4.5uA. So, at best case current could be measured with +-4.5uA: too much error margin.

Furthermore, several current sensors are proposed in specifications sheet, with its accuracy:
"751521....DC to 100 kHz (-3 dB). -600 A to 0 A to +600 A (DC) Basic accuracy:(0.05% of rdg* + 40 mA)"
751523... idem
CT1000...DC~300 kHz, ±(0.05% of reading +30uA)
CT200...DC~500 kHz, ±(0.05% of reading +30uA)

What do you think about: "basic accuracy 40mA"? "30uA"?

With those specifications I'll give you my opinion:

The Yokogawa WT3000 has a margin error non less than 4.5uA, probably much more. Neither a scientist nor an engineer should claim a 0.2, 0.3 or 0.7uA +- 4.5uA or +-(much more uA)!

So, the specified 0.7uA was not actually measured.

I am not saying Weir is lying; he is not stupid, only a foolish could lie about that data.

Another possibilities must be explored.

1.- I am plain wrong.
2.- There is an error in published date. I think it is highly improbable: leakage has been specified for 4 samples (and also in previous public reports).
3.- Could be 0.7mA (0.7*10^-3A) correct? Impossible, 0.7uA is too much leakage, but 0.7mA is impossible (for one layer only).
4.- A multi layer capacitor was actually measured.
5.- Another one.

I think the most probable possibility is the 4th.

Even within the 4th case there are two possibilities:
a.- The specified leakage current is for whole multilayer capacitor. I would like this one, and it is possible, but it not appears to be the real case. “Layer” is explicit three times in table data. Futhermore, there is a conclusive prove it is not the real case, at least not with a lot of layers: if the published leakage was for a whole multilayer capacitor with a lot of layers we must had testified a much more powerful “leakage”: the share price should be now above the sky.
b.- The specified leakage current is the mean leakage for its layers: total leakage divided by the number of layers.

So I think most probable possibility is that was measured a multilayer capacitor and then published the mean data of leakage current.

I it is true, the bad leakage data is more representative: it is not an only one bad layer of 0.7uA; it is a statistically significative data, because is the mean of multiple layers.

There is no question this leakage is huge. With this leakage there is not possible EESU, the multilayer capacitor cannot storage energy for a long time (only a few thousand of seconds, as has been posted in this blog); Weir, EESTOR, ZEEN and us have a problem of several orders of magnitude, three. If it is not solved the Technology Agreement is not worth too much.

 

Disclosure:

I am not familiariced with specified equipment, muy opinion is based only in my interpretation of the specifications sheets.

I was lucky and had a good entry point, and have not too much money, so I shall not sell (now) :)

 

[ee-tom]

PeterP - not is not dishonest to view impossible measurements as impossible. I never said faked up. You did.

You can agree with me about impossibility. Indeed you did that, I have annotated your disagreement to show which part is wrong.

You only think I'm dishonest because you cannot imagine having that clear a view of this matter. That's OK. And of course I may be mistaken - though I've shown you precisely where your criticism of the key argument goes wrong. But I'm not dishonest.

You however are supremely unimaginative if you cannot see how it is possible for somone to have a clear idea, right or wrong, in this matter. I've stated my argument. I've shown you where your refutation of it goes wrong. I'll do the same to any other refutation, or admit it is wrong.

My position about the current tests is different. There is enough wiggle room for them to exist but it is very highly improbable. And that makes me see measurement error of whatever kind as being less improbable - though this is a balance of improbabilities rather than a strong statement.

PeterP said:

It is dishonest to claim that your speculations about the EESU are true. It is dishonest to claim that you know Carl and Dick faked up the patent measurements.

It is perfectly honest to speculate about possible polarisation mechanisms for the EESU and say "this won't work" or "that won't work".

 

Just because you can't understand the EESU does not mean that others can't or that it does not exist.

 

That is "egomania".

 

[PeterP]

ee-tom,

Anything Ulrich and/or Capman say that is not signed off on by Eestor is heresay. You are entitled to make deductions but they are not proof or evidence. Saying measurements are unlikely due to theories you understand is okay. Saying you know they are not real (due to your deductions) is a lie.

I will continue to regard your statements about the EESU as lies unless they can be traced back to Eestor supplied information.

I will reply to your comments about the "killer flux" in another post.

[PeterP]

ee-tom Wrote:

You are mixing two levels of description. QM replaces point charges by charge distributions. Either of these cases can then be analysed in a number of ways. Local field is one tool in the analysis.

Nit picking.

ee-tom Wrote:

How about we don't use the word flux…

You are the guy that likes to talk about flux, not me.

ee-tom Wrote:

It is the equations that are real, not the words.

No. There is reality, dimly perceived and poorly described by words or equations, but we are slowly getting better.

ee-tom Wrote:

That is true, in the case that the macroscopic approximation holds. I've said on my other thread that in the system we are interested in, it holds for ionic and (normal) electronic polarization, but not for polarization due to macroscopic movement of charge over long distances. This macroscopic movement means we'd have to average over a similar spatial scale for the macroscopic approx to work. However that averages out the killer field!

Now attend carefully:

1) The local field is the field "at the molecule"; it is a microscopic quantity.

2) The “killer field” is due to the plate charge, no averaging is required.

3) The field due to the macroscopic displacement of charge can always be evaluated at a point (at the molecule) by Coulombs Law.

What are you on about?

Okay:

Macroscopic field 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.

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’. ‘a’ is the unit cell dimension.

No doubt you are familiar with the result from calculus which makes that statement valid.

For anything other than the homogeneous case a clear picture of the charge distribution must be drawn.

If we can come to agreement about the above we may be able to continue.

One step at a time.

[ee-tom]

One step at a time. PeterP as you say we agree on much.

The issue is not the field Ed but =the dipole moment which gives rise to it. I agree absolutely we can claulate this microscopically and all will work. But we need to do this, and take into account edge effects.

So, let us consider, microscopically, the plate/dielectric interface. You claim Ed can cancel the killer field P/e0 at this interface, I think. Shall we evaluate it and see?

Now to do this you have to consider the microscopic charge distribution in the dielectric. What moves, how far does it move. You then need to consider what happens to this distribution at the edge. Finally you need to work out (integrating over all charge in dielectric and plate, what is the field at this intrface.

Oh, the field at the interface will also be near the interface, as you know from your simulations if they are accurate.

Now, I think your point is that because of Ed, at this interface, the overall filed will be small. Ed ~ Q/e0

It looks like the issue then will be doing the calculation to see if this is  true. Would you like to say what charge distribution around the edge, and in the dielectric, you expect? You can express the dielectric interior charge distribution as a dipole and an overall charge density but you need to justify the value of the dipole - it is easy to make mistakes with dipole when electrons move more than one lattice cell.

Remember, dipole density (and any Ed derived from this) is a macroscopic approximation. It may not apply properly to edges, so we much be careful. If each electron to a positive charge and sum the dipole moment of all, microscopically, then that is OK but again we have to consider where do the edge electrons move relative to their parent charge (there is not so much room as we get closer to the edge!)

PeterP said:

ee-tom Wrote:

You are mixing two levels of description. QM replaces point charges by charge distributions. Either of these cases can then be analysed in a number of ways. Local field is one tool in the analysis.

Nit picking.

ee-tom Wrote:

How about we don't use the word flux…

You are the guy that likes to talk about flux, not me.

ee-tom Wrote:

It is the equations that are real, not the words.

No. There is reality, dimly perceived and poorly described by words or equations, but we are slowly getting better.

ee-tom Wrote:

That is true, in the case that the macroscopic approximation holds. I've said on my other thread that in the system we are interested in, it holds for ionic and (normal) electronic polarization, but not for polarization due to macroscopic movement of charge over long distances. This macroscopic movement means we'd have to average over a similar spatial scale for the macroscopic approx to work. However that averages out the killer field!

Now attend carefully:

1) The local field is the field "at the molecule"; it is a microscopic quantity.

2) The “killer field” is due to the plate charge, no averaging is required.

3) The field due to the macroscopic displacement of charge can always be evaluated at a point (at the molecule) by Coulombs Law.

What are you on about?

Okay:

Macroscopic field 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.

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’. ‘a’ is the unit cell dimension.

No doubt you are familiar with the result from calculus which makes that statement valid.

For anything other than the homogeneous case a clear picture of the charge distribution must be drawn.

If we can come to agreement about the above we may be able to continue.

One step at a time.

[Y_Po]

PeterP said:

ee-tom,

Anything Ulrich and/or Capman say that is not signed off on by Eestor is heresay.

Yeah, laws of physics are hearsay unless Dick Weir signed it off.

[ee-tom]

And just to be clear: I agree with all of this: and suggest that we do the calculation microscopically based on an appropriate point charge distribution & movement that will provide 50C/m^2 polarisation. Since you claim this disproves my statements, I guess you have such a distribution in mind. Lets look at it and do the calculation. I don't mind a continuous distribution, but that makes edges more difficult because we don't know which + charge each - charge comes from originally.

Now attend carefully:

1) The local field is the field "at the molecule"; it is a microscopic quantity.

2) The “killer field” is due to the plate charge, no averaging is required.

3) The field due to the macroscopic displacement of charge can always be evaluated at a point (at the molecule) by Coulombs Law.

What are you on about?

Okay:

Macroscopic field 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.

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’. ‘a’ is the unit cell dimension.

No doubt you are familiar with the result from calculus which makes that statement valid.

For anything other than the homogeneous case a clear picture of the charge distribution must be drawn.

If we can come to agreement about the above we may be able to continue.

One step at a time.

[PeterP]

Y_Po Wrote:

Yeah, laws of physics are hearsay unless Dick Weir signed it off.

Hello Y_Po,

I wonder, seeing that you have such a low level of comprehension, if you actually do have tertiary qualifications.  Perhaps they can be purchased on-line these days.

Anyhow it is only genuine information about the EESU that Dick controls.

If somebody needs information on on-line character assassination, buffoonery, foul language or abuse you are the man.

Regards,

Peter

 

[PeterP]

Daniel R Plante Wrote:

If I was contracted to get to the bottom of this, I would insist on any info that had any direct bearing on the safety of individuals or the safety of the samples. Period. Beyond that, I might be provided with other info from the client such as energy density, max working voltage or even breakdown voltage, leakage @ voltage, etc.

And while I might find that extraneous info interesting I would nevertheless be ethically/statutorily bound to dismiss/ignore it.

Dan is a professional, he gets paid, he has expertise and he accepts responsibility.

The evidence he ignores is classified as hearsay. He has to ignore it.

When you are considering evidence about the EESU you should ignore hearsay too. Anything not vouched for by Eestor in writing is hearsay.

That is just the way it is.

Regards,

Peter

[Y_Po]

PeterP said:

ee-tom,

Anything Ulrich and/or Capman say that is not signed off on by Eestor is heresay.

Yeah, laws of physics are hearsay unless Dick Weir signed it off.

[PeterP]

Y_Po Wrote:

Yeah, laws of physics are hearsay unless Dick Weir signed it off.

Hello Y_Po,

I wonder, seeing that you have such a low level of comprehension, if you actually do have tertiary qualifications.  Perhaps they can be purchased on-line these days.

Anyhow it is only genuine information about the EESU that Dick controls.

If somebody needs information on on-line character assassination, buffoonery, foul language or abuse you are the man.

Regards,

Peter

[ee-tom]

PeterP - I'm still awaiting your demonstration of a possible charge distribution that breaks the "killer flux" argument? Or you could take back your allegations about my competence/mental state/truthfulness on this issue?

Notice I make no accusation about you. Mistakes are the stuff of life.

ee-tom said:

And just to be clear: I agree with all of this: and suggest that we do the calculation microscopically based on an appropriate point charge distribution & movement that will provide 50C/m^2 polarisation. Since you claim this disproves my statements, I guess you have such a distribution in mind. Lets look at it and do the calculation. I don't mind a continuous distribution, but that makes edges more difficult because we don't know which + charge each - charge comes from originally.

Now attend carefully:

1) The local field is the field "at the molecule"; it is a microscopic quantity.

2) The “killer field” is due to the plate charge, no averaging is required.

3) The field due to the macroscopic displacement of charge can always be evaluated at a point (at the molecule) by Coulombs Law.

What are you on about?

Okay:

Macroscopic field 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.

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’. ‘a’ is the unit cell dimension.

No doubt you are familiar with the result from calculus which makes that statement valid.

For anything other than the homogeneous case a clear picture of the charge distribution must be drawn.

If we can come to agreement about the above we may be able to continue.

One step at a time.

[ee-tom]

PeterP's reply is below on a thread I can't post. I wonder why he did not want to put it here?

It is logically incorrect. In order for a 60C/m^2 capacitor to work the field at every possible stage of charge including full charge must be feasible (ie not cause capacitor insulation to break down).

I've shown insulation does break down at full charge (and, actually, anything near this).

I don't think this is misdirection: I think peterP is unable to work out logically any charge distribution that would work at full charge, so he resorts to "it is so complex we don't know". That is however not a refutation of a valid argument which is simple and gives the answer.

However peterP's obsessive wish to call me dishonest is sad.

The Killer Flux Fallacy:

The argument:

Put 60 C/m^2 on the plates of a capacitor then a “gigantic field results”.

This is true. But charge is not “put” on the plates of a capacitor. The capacitor “charges up”.

The correct approach:

In order that the potential difference between the plates of a capacitor is 3500 volts when charged to 60 C/m^2 there must be a reverse field to cancel the “gigantic field”.

To generate the correct field the dielectric must be polarised to 60 C.m/m^3.

In order to evaluate the possibility of this, the charging process must be analysed in detail starting at zero volts with zero charge on the capacitor.

Starting with full charge is bad logic and bad physics.

It is excellent misdirection though.

Regards,

Peter