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) :)

 

[PeterP]

Hello ee-tom,

You Wrote:

..could take back your allegations about my competence/mental state/truthfulness on this issue?

I have no real knowledge of your competence/mental state/truthfulness.

You might be incompetent and truthful.

You might be competent and lying.

I suppose you might even be mad.

I don't know.

All I see is your behaviour.

You are the Denigrator.

I don't know what motivates you either, you may be a "good guy" working for KPCB.

I just don't know.

 

It is possible that you like to debate and convince people of things that are not true.

 

Beats me.

However nothing but Denigration of the EESU is important to you. That is clear.

Regards,

Peter

[PeterP]

ee-tom Wrote:

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

Because I don’t want to feel any obligation to have to read and parse your nonsense.

[ee-tom]

PeterP, you are under no obligation to defend the strong and indefensible (I claim) statements you've made here. Your last attempt was logically very poor.

PeterP said:

ee-tom Wrote:

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

Because I don’t want to feel any obligation to have to read and parse your nonsense.

[ee-tom]

PeterP,

That is honest. But if you don't understand my arguments why spend your time saying you know I'm incompetent? And if they are correct I'm not denigrating, but bearing true witness. My not respectful remarks about you would then also be a natural reaction to your bad behaviour.

PeterP said:

Hello ee-tom,

You Wrote:

..could take back your allegations about my competence/mental state/truthfulness on this issue?

I have no real knowledge of your competence/mental state/truthfulness.

You might be incompetent and truthful.

You might be competent and lying.

I suppose you might even be mad.

I don't know.

All I see is your behaviour.

You are the Denigrator.

I don't know what motivates you either, you may be a "good guy" working for KPCB.

I just don't know.

 

It is possible that you like to debate and convince people of things that are not true.

 

Beats me.

However nothing but Denigration of the EESU is important to you. That is clear.

Regards,

Peter

[PeterP]

Tecnofilo's Last Post:

 

I suppose he got tired of us:

 

He Wrote:

 

For me, it has much more sense to test the "leakage current" in a multilayer capacitor: more statistically signficative, it solves accuracy problems and it is more convincing for possible investors (more "real").

 

And it is time for Weir to have a real multilayer capacitor. The problem is the published leakage is huge, a very bad notice for ZENN unless Weir has all the necessary tricks to solve it. In such case he is a genious.

And don't you remember previous pr Capman saying "leakage current" was under resolution of 1uA?

 

PeterP said:

Hello Tecnófilo,

You Wrote:

I don't see in the pr that "measurements were done on a single layer".

Nor do I after looking at the PR once again. Thank you for pointing that out. I can’t even see that the PR is deliberately misleading it is just an assumption that I have made based on the way it is written.

The number of layers has been deliberately withheld then.

It makes a lot more sense to make more than one layer and I have always said that he had that capability. So how many layers?

A 6 mm square capacitor would probably be no more than 2 mm thick. Say 40 layers. But actually we have no idea.

So the effect of the intrinsic leakage could be reduced by as much as 40 but we just don’t know.

By the way I see no reason why Eestor should stick to 6mm square once the production process is under control. Thick layers and larger area will make production easier, maybe much easier.

Thanks and Regards,

Peter

[PeterP]

ee-tom Wrote:

And just to be clear:I agree with all of this:

I Wrote:

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.

That is a big step forward for him. Good.

Then he Wrote:

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.

 

ee-tom wants to lead me on a Merry Dance starting at the wrong place and leading nowhere.

No thanks Tom.

Po must Go.

[ee-tom]

We are clear about this then. We agree with the math here. But you don't want to prove your point by applying it (I notice you have lots of simulations, so you could do one) to show how surface charge (or whatever) can give 50C/m^2 without killer fields...

I've shown why it does give this killer field.

PeterP said:

ee-tom Wrote:

And just to be clear:I agree with all of this:

I Wrote:

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.

That is a big step forward for him. Good.

Then he Wrote:

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.

 

ee-tom wants to lead me on a Merry Dance starting at the wrong place and leading nowhere.

No thanks Tom.

Po must Go.

[PeterP]

ee-tom Wrote:

We are clear about this then. We agree with the math here. But you don't want to prove your point by applying it (I notice you have lots of simulations, so you could do one) to show how surface charge (or whatever) can give 50C/m^2 without killer fields...

My point is that the correct method of analysis is to start with zero applied volts and increase it until breakdown. Then the maximum polarization (whatever it is) will be known.

Why would you think I have not done the simulation, seeing that you recognise that I can?

Why are you talking about 50 c/m^2 when for years we have been talking about 60 C/m^2. Is ee-tom a group entity and “you” are a new member of the group?

 

[PeterP]

ee-tom Wrote:

We are clear about this then. We agree with the math here.

 

ee-tom is agreeing to my correction of his (quite wrong) maths. It is a matter of record, and more than I hoped for. He has now moved on to his new thread, thank God.

 

Regards,

Peter

[PeterP]

I thought it might be appropriate to include this excerpt from a JPL report on this thread:

                       

A 1000 times improvement might reasonably expected in the leakage measurements made in the PR if “proper” measurements were made. This information would have been available to Dick when he designed the EESU assembly.

I am expecting some great results soon.

Regards,

Peter