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Old 24th May 2014, 10:05 PM   #351
popilin is offline popilin  Argentina
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Quote:
Originally Posted by tinitus View Post
Quote:
Originally Posted by avp1 View Post
But I want to listen EXACTLY to what authors (musicians, producers, engineers) intended.
good luck
Dear Lars

I'm sorry to answer so late, but wherever you are now, I want to say you that your joke is on the top five, and I always will remember you with a smile.

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Old 8th June 2014, 01:29 AM   #352
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Another rough demonstration

Part 1 – Ferromagnetic Materials

I will follow Landau-Lifshitz approach [1], with little changes [*], because it can't be more clear and easier, intuitively based on solid physical foundations without loosing on mathematics.

Let’s consider a single domain, uniaxial magnetic particle with magnetization M, under the influence of a magnetic field H

Click the image to open in full size.

Magnetic anisotropy energy is given by

anisotropy = ½ β M² sin²θ

Where θ is the angle between M and the x-axis, taken to be along the principal axis of symmetry, the so called easy axis, and β is a constant which depends on temperature.

After some thermodynamic considerations, and a few calculations we obtain

(Hy / sinθ) - (Hx / cosθ) = β M

Taking the derivative with respect to θ

(Hx / cos³θ) + (Hy / sin³θ) = 0

Eliminating θ from these two equations gives

Hx ²/₃ + Hy ²/₃ = (β M) ²/₃

In the Hx Hy plane this represents the so called astroid curve

Click the image to open in full size.

This is the clearest way to see how varies hysteresis with anisotropy. Not shown in the book because Landau-Lifshitz presupposes some basic understanding.

For φ = 0, along the easy axis, anisotropy reaches its maximum and also does hysteresis.

For φ = π/2, along the hard axis, which effectively means no anisotropy at all, also does hysteresis, i.e. no hysteresis at all.

[*] Ex profeso, I changed the axes of coordinate system, so equations and graphs are completely different as those of the book, just in case someone wants to make copyright claims, furthermore, most textbooks follow the same approach adorned with more equations, so no need to reinvent the wheel.

At macroscopic level one can use the old trick to say that, for a magnetic domain macroscopically small and microscopically large, demonstration above is also valid.

Anisotropy/hysteresis will be in between both extremes, i.e. something similar to the case for 0<φ<π/2.

The understanding of magnetic anisotropy and how it can be controlled is the key factor in order to improve the properties of magnetic materials, e.g. amorphous and nanocrystalline alloys.

But again, don't dream, isotropic magnetic materials don't exist in real world. [2]


[1] L. D. Landau, E. M. Lifshitz. Electrodynamics of Continuous Media. Course of Theoretical Physics Vol 8. Pergamon Press.

[2] Amikam Aharoni. Introduction to the Theory of Ferromagnetism. Oxford Science Publications.
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Old 8th June 2014, 01:42 AM   #353
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Another rough demonstration

Part 2 – Dielectric Materials

A more general form to write constitutive relation for a dielectric medium is

Di = D₀i + εij(ω) Ej

Where

Di are constants in order to contemplate the case

D (E=0 , ω=0) ≠ 0

εij(ω) is the rank 2 permittivity tensor, from Thermodynamics, it can be proved that in most cases it must be symmetric, i.e. εij = εji, and it is always possible to choose coordinate systems in that this tensor is diagonal, i.e. εij = 0, i ≠ j

1) The medium is Isotropic

D₀i = 0, i=1,…,3

ε11(ω) = ε22(ω) = ε33(ω) = ε(ω) ∈ ℝ

Then

D = ε(ω) E

i) The response of the medium is Linear

Then, ε(ω), must be a real linear function.

The curve D=f(E) must be an open curve, so no hysteresis.

In the particular case of an Homogeneous medium

D = ε E

With ε=constant, then the curve D=f(E) must be a straight line, so no hysteresis.

ii) The response of the medium is Nonlinear

Then, ε(ω), must be a real nonlinear function, and again, the curve D=f(E) must be an open curve, so no hysteresis.

2) Let's consider now a Nonlinear Anisotropic single domain/crystal with its more simple uniaxial symmetry, then we can write

ε11(ω) = ε22(ω) = ε∥(ω) ∈ ℂ

ε∥(ω) = ε’∥(ω) + i ε’’∥(ω) = ∣ε∥(ω)∣ [cos (δ∥) + i sin (δ∥)]

ε33(ω) = ε⊥(ω) ∈ ℂ

ε⊥(ω) = ε’⊥(ω) + i ε’’⊥(ω) = ∣ε⊥(ω)∣ [cos (δ⊥) + i sin (δ⊥)]


With chosen coordinate system, a rotation around z-axis does not change anything, i.e. within the x y plane a uniaxial crystal is isotropic.

It can be proved that, if we apply an electric field E out of the x y plane, the curve D=f(E) is a closed curve, then we have hysteresis.

Same reasoning, another formulation, applies to biaxial symmetry.

Then, anisotropy and nonlinearity are necessary conditions for hysteresis, conversely, hysteresis implies both anisotropy and nonlinearity.

********************************************
Anisotropy + Nonlinearity Hysteresis
********************************************

As an example of how capricious can be real world, let's consider now the case of an electric field E applied on the x-axis direction, and let's also suppose that really exist a dielectric material which can be perfectly isotropic along x-axis.

Contrary to all preconceptions, this is not an impediment to have hysteresis.

The reason for this cannot be simpler: Real world capacitors are not perfect infinite parallel plane plates, at least those that I can buy on this side of the world, never knows...

Click the image to open in full size.

Not a contradiction at all, even on the case where anisotropy can be ignored, on the real world it can't.

Note that W. Thomson (Lord Kelvin) had a dirty trick to approximate the behavior of an infinite plane capacitor, but I hate the idea to see in the future "gold foil snake oil Kelvin type audio grade capacitors", then I will remain silent, but maybe it is too late, never knows...

Unfortunately, real world capacitors are far to be ideal, edge effects are greater because of their construction.

Click the image to open in full size.

With other geometry, e.g. cylindrical axial capacitors, things are even worse, because equipotential surfaces are not perfectly concentric cylinders, then exists a tangential component of electric field E along the whole dielectric, not only at the edges.

Click the image to open in full size.


Additional notes

Just in case someone escaped him the turtle, we must remember that constitutive relations have the same mathematical form both for magnetic media as for dielectric media, then analogies are valid and the problem admits a general demonstration using Tensor Geometry.

Furthermore, as permittivity tensor can't know if we are analyzing a single microscopic domain, or a macroscopic sample, demonstration applies on both cases, anyway, this is a rough demonstration, not a treatise on condensed matter physics.

However, passing from one or two symmetry axes (uniaxial and biaxial anisotropy), to an infinite number of symmetry axes (isotropy), is a lot difficult even to imagine it.
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Last edited by popilin; 8th June 2014 at 01:56 AM.
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Old 8th June 2014, 01:51 AM   #354
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Another rough demonstration

Part 3 – More Anisotropy

Isotropy is a difficult thing in the real world, to the point that even the vacuum isotropy is discussed today.

As well as Schrödinger's cat can not be, nor half-alive, neither half-dead, Schrödinger's mistress can not be half-pregnant.

If a material is infinitesimally anisotropic, then it cannot be isotropic.

But surprise!!! I have found some measurements!!! [1]

The dielectric anisotropy can be expressed as normalized ratios

∆Aε = 2 ∣ε’∥(ω) – ε’⊥(ω)∣ / [ε’∥(ω) + ε’⊥(ω)]

∆Atan(δ) = 2 ∣tan (δ∥) – tan (δ⊥)∣ / [tan (δ∥) + tan (δ⊥)]

Curiously, "practically isotropic" materials exhibit some degree of anisotropy, e.g.

i) PTFE, 0.945 mm thickness

∆Aε ≈ - 0.1 % ∆Atan(δ) ≈ - 4.0 %

ii) Polyolefin, 0.7725 mm thickness

∆Aε ≈ - 0.6 % ∆Atan(δ) ≈ 3.7 %

iii) Polycarbonate, 1.000 mm thickness

∆Aε ≈ 0.2 % ∆Atan(δ) ≈ - 4.0 %

iv) Polycarbonate Lexan® D-sheet, 0.51 mm thickness

∆Aε < 3 % ∆Atan(δ) < 11 %

Although author's effort on discredit his own measurements, this gives us a clue about what really happens with real world dielectrics. [2], [3], [4]

A picture is worth a thousand words

Click the image to open in full size.

A bulk material can be "practically isotropic", "almost isotropic" or "isotropic by definition" or whatever, as can be seen on fig (a), unfortunately, industrial processes to obtain thin films leaves us on the situation shown on fig (b) or fig (c), i.e. biaxial anisotropy or uniaxial anisotropy, maybe here it can be seen more clearly

Click the image to open in full size.

A very understandable explanation [6]

Thin film = Size reduction of bulk material in one dimension

Change of polymer molecule from "isotropic" Gaussian coil (spherical "ball") to elliptical form.


To put things even worse, electrostriction is a property of all dielectrics, regardless of their symmetry [5]. It produces a relatively small mechanical deformation under an applied electric field. Reversal of electric field does not reverse the direction of the deformation. If an electric field E is applied on a material, the electrostrictive strain x is defined by

xij = Mijkl Ek El

Where Mijkl are components of the fourth-rank tensor and are called electrostrictive coefficients.

As its contribution to anisotropy is very difficult to formulate correctly, it is quite often ignored.

Ignoring also the effects of temperature and phase changes, another ignored little detail: Real world capacitor dielectrics are useless without a conductive foil or a metalized surface, at the interface between the dielectric and the conductor, can have air, among many other impurities, air tends to form bubbles and this is a great contributor to anisotropy.

However this should not be an issue for paper dielectrics, we already know that paper is permeable, but we also already know that paper is anisotropic. [7], [8]

But, what about polymers? Are they also permeable? It seems that yes they are. [9], [10]

Essential factors for permeation
. Free volume: In order for any molecules to move in a material there must be holes available.
. Continuous path through the polymer.

Polymer experts say that polymers look like a pile of tangles of spaghetti, following this food analogy, seems to me that a decent plate of spaghetti needs some cheese, with a lot of holes, maybe impurities would be the sauce, and the whole picture it is anything but isotropic.


Conclusion: Real world isotropic dielectrics are as real as Santa Klaus…



[1] Plamen I. Dankov (2010). Dielectric Anisotropy of Modern Microwave Substrates, Microwave and Millimeter Wave Technologies from Photonic Bandgap Devices to Antenna and Applications, Igor Minin (Ed.), ISBN: 978-953-7619-66-4, InTech, DOI: 10.5772/9061

[2] Senden, D.J.A. (2013). Strain Hardening and Anisotropy in Solid Polymers. Eindhoven: Technische Universiteit Eindhoven. ((Co-)promot.: prof.dr.ir. H.E.H. Meijer, dr.ir. L.E. Govaert & dr.ir. J.A.W. van Dommelen).

[3] Ellen M. Arruda, Mary C. Boyce. Evolution of Plastic Anisotropy in Amorphous Polymers during Finite Straining. Mechanical Engineering Department. Massachusetts Institute of Technology.

[4] T.B. v. Erp. Anisotropic plasticity in oriented semi-crystalline polymer systems.
Eindhoven University of Technology.

[5] Marko Budmir. Piezoelectric Anisotropy and Free Energy Instability in Classic Perovskites. École Polytechnique Fédérale de Lausanne.
(The relevant section about electrostriction was taken from another paper written about eight years before by Dragan Damjanovic.)

[6] Vorlesung/Lecture: Hans-Ulrich Krebs Polymer thin films. (Prof. Dr. Hans-Ulrich Krebs. Georg-August-Universität Göttingen. Institut für Materialphysik.)

[7] A. Kallmes, J. Scharcanski and C.T.J. Dodson. Uniformity and Anisotropy in Nonwoven Fibrous Materials.

[8] Yong-Joo Sung and Ramin Farnood. Characterizing Anisotropy of the Deterministic Features in Paper Structure with Wavelet Transforms. Department of Chemical and Applied Chemistry.University of Toronto. Toronto, Canada.

[9] Polymer permeability. Edited by J. Comyn, Elsevier Applied Science Publishers, London, 1985.

[10] P.M. Wood-Adams. Diffusion And Permeability In Polymers - Concordia University.
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Last edited by popilin; 8th June 2014 at 02:10 AM.
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Old 9th June 2014, 01:35 AM   #355
popilin is offline popilin  Argentina
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Dear SY

I respect and admire you, more than you can imagine, but this time you screwed up.

Please don't take this as anything personal against you, but only as a serious technical discussion.

Quote:
Originally Posted by SY View Post
Any amorphous polymer containing polar groups will be isotropic (by definition) yet show hysteresis to a greater or lesser degree.
Fortunately the laws of physics can't be broken; anisotropy and hysteresis are not unrelated phenomena.

Then, your "isotropic by definition" polymer, if really it is isotropic, hence cannot show hysteresis, conversely, if it shows hysteresis it can’t be isotropic.

Even more, I must answer by myself my own questions

Quote:
Originally Posted by popilin View Post
Can you please explain to me how?
Because, so far, you didn't explained the hysteresis phenomenon for isotropic materials, although result quite clear from my own rough demonstration(s), that this would be physically impossible.

Quote:
Originally Posted by SY View Post
edit: And I should add that the rolling direction is normal to the applied field, so doesn't have much relevance to capacitor action.
Well, that is some progress, now you seems to accept some kind of anisotropy, but you are still ignoring real world capacitors.

That is for your ideal edge effects-free capacitors, filled into vacuum with an ideal dielectric which is perfectly isotropic along one axis.

Quote:
Originally Posted by SY View Post
Indeed, I agree that you would say that. Nonetheless, it's incorrect. Take a block of semicrystalline polymer. Measure the properties from face to face in all three directions. Get back to me once you've done that.

Take a “block” of vacuum. Measure the properties “from face to face” in all three directions. Get back to me once you've done that.

I warn you that you will need an outstanding lab and struggle with gravitational field among other minor details.

Quote:
Originally Posted by SY View Post
Your confusion, I think, is separating what happens on a microscopic, single domain level with what happens at a macroscopic level. You're also, I believe, not thinking through the consequences of the symmetry aspect- we only apply a field in the z direction.
No, I think it is the other way around, is you who confuse what happens on a bulk sample, "isotropic by definition" or rather "slightly anisotropic" with what happens in a real world thin film capacitor dielectric, under strain, pressure, temperature changes, phase changes, permeation, etc. that it is anisotropic by forming, among other things.

You're also, I believe, thinking through the consequences of the symmetry aspect, but ignoring elementary electrostatics.

You also forget, I believe, letting aside the different nature of fields, the fact that dielectric permittivity for decent dielectrics, is various orders of magnitude lower than magnetic permeability for ferromagnetic materials, hence the difficulties to measure dielectric anisotropy.

Finally, as I know that you love measurements, take your oscilloscope, your signal generator, and a capacitor, with a very simple circuit you can measure hysteresis, if you find hysteresis, not me, but physics, assures you that you are also measuring dielectric anisotropy.

If you can't find hysteresis, even with a large electric field, please tell us about such amazing isotropic hysteresis-free capacitor, I will be the first on the checkout line.
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Old 16th June 2014, 03:48 AM   #356
popilin is offline popilin  Argentina
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“Science is the belief in the ignorance of experts.” (Richard P. Feynman)

Quote:
Originally Posted by DF96 View Post
You need to understand a bit more basic physics before you pontificate about capacitor size etc.
I must recognize that you are right, learning process never ends, and this forum is a clear example, every day one can learn something new.

However, I was right from the beginning; anisotropy and hysteresis are not unrelated phenomena.

The root of your confusion, I think, is that to study/understand physics, it is divided into “compartments”, e.g. Mechanics, Electrodynamics, Relativity, and a long etc.

However, Nature does not need to take classes at university, and physical phenomena occur in a more complex, regardless epistemological issue, reality flows as it is.

Quote:
Originally Posted by DF96 View Post
I am trying to correct errors put forth in a public forum, so others are not mislead.
That is very laudable, but for the next time, please you try to not mislead yourself before confusing people with your own false assumptions/misunderstandings.

Quote:
Originally Posted by DF96 View Post
A standard technique which seems to be used on here by some, including you, is to make an incorrect statement based on misunderstanding and then demand that others prove that it is untrue. This is not a helpful way to advance knowledge, or to acquire existing knowledge. I have told you, and others have told you, that anisotropy and hysteresis are unrelated phenomena. These are facts. It is for you to defend your speculation, not for us to prove you wrong. So far your defence has serious holes.
This is a serious accusation, but lacking any support.

Your mistake is almost as big as your obstinacy on insisting about it, if being a modest TV repairman removes me credibility, please talk with your colleagues, anybody who knows about condensed matter physics will tell you that those “facts” are wrong.

“Beware of false knowledge; it is more dangerous than ignorance.” (George B. Shaw)

Quote:
Originally Posted by DF96 View Post
I am not re-entering the 'debate', but merely reducing potential confusion for others who may read this thread.
Please, do not underestimate the readers, if you really want reducing potential confusion for them, I would say that, with all due respect, you should follow your own advice and try to understand a bit more basic physics before you pontificate about things that obviously you don't understand.



Quote:
Originally Posted by SY View Post
Let's see this evidence. It's odd that you can't afford basic distortion measurements, but you can afford the rather specialized equipment it takes to do measurements of mechanical, thermal, and optical anisotropy.
I must confess that I always enjoy your jokes and sarcasm, to the point of repeating by myself some of your jokes.

But desperation to repair a TV to put a meal on the table is not fun, poverty is not funny, so no need to be sarcastic with mine.

Just to clarify, I found such evidence on the web, as I already said on post#217, I thought it was understood, sorry.

Quote:
Originally Posted by SY View Post
The problem is, if I may be blunt, that you have to be willing to discard most of what you know that isn't so in order to build the knowledge base that it will take to understand what the correct answers are.
The problem is, if I may be blunt, that on this issue you are not able to give all the correct answers, nobody can, and I don't want to be unkind with you, but your physics is almost as pathetic as my chemistry.



Finally, my favorite philosopher, said
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Old 28th June 2014, 04:33 AM   #357
popilin is offline popilin  Argentina
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A Short Digression on Linearity

Constitutive relation for dielectric media

Di = Doi + εij(ω) Ej

For a perfectly isotropic and homogeneous dielectric becomes

D = ε E

With ε=constant, i.e. D=f(E) is a straight line, and the dielectric is said to be perfectly linear, so a capacitor will introduce no distortion at all.

Unfortunately, real world capacitors are not perfectly linear and they need a hysteretic (anisotropic) model for permittivity.

Using the results of post#301 we can write

D = ε(ω) Eo cos (ω t - δ)

E = Eo cos (ω t)

ε(ω) = ε’(ω) + i ε’’(ω)

tan (δ) = ε”(ω) / ε’(ω)

ε(ω) =ε(ω) [ cos (δ) + i sin (δ) ]

A good indicator of linearity is to evaluate the ratio

b/a

The closer to zero, the closer is the ellipse to a straight line, then the greater linearity.

Click the image to open in full size.

The area within hysteresis curve is

A = D dE = (D dE/dt) dt = πε(ω)Eo² ω sin (δ)

We must remember that for a capacitor, it is valid the approximation

E ≈ V/d

Now we can see that a negligible voltage does not mean necessarily a negligible electric field across the capacitor, and it depends on its geometry, also we can write

A ≈ π ε(ω) (V/d)² ω sin (δ)

It can be proved that, for the ellipse

A = π a b

Where a and b are the semi-major and semi-minor axes

a = Eo ω ≈ (V/d) ω

b = Eoε(ω)sin (δ) ≈ (V/d) ε(ω)sin (δ)

For good dielectrics, dissipation factor is very low, and it is valid the approximation

sin (δ) ≈ tan (δ)

Then
b ≈ (V/d) ε(ω)tan (δ)

b/a ≈ [ε(ω)tan (δ)] / ω

We can also compare linearity of two capacitors at the same frequency

b1/a1 ≈ {[ε1(ω)tan (δ1)] / [ε2(ω)tan (δ2)]} (b2/a2) (*)

Due to “difficulties conservation theorem”

tan (δ) = tan (δP) + tan (δS) + tan (δD)

Where
tan (δP) = 1 / (RP ω C)

tan (δS) = ESR ω C

tan (δD) = Characteristic of the dielectric

The parallel component, tan (δP) depends on the insulation resistance RP, as it is extremely high, this component is negligible.

The dielectric component, tan (δD) depends on the particular dielectric, in polypropylene capacitors it remains constant with frequency and will typically result in a value of about 10⁻⁴

The series component, tan (δS) is determined by the series resistance ESR, this component increases rapidly with frequency until it becomes dominant.

Click the image to open in full size.

This illustrates the behavior of tan (δ) vs. measuring frequency, using two polypropylene capacitors of different capacitance as examples, and usually it is not shown on datasheets.

Then
a ≈ (V/d) ω

b ≈ (V/d) ε(ω)ESR ω C

b/a ≈ ε(ω)ESR C

We can compare different capacitors at the same frequency as

a1 ≈ [(d2 C2) / (d1 C1)] a2

b1 ≈ [(d2 ε1 ESR1) / (d1 ε2 ESR2)] b2

b1/a1 ≈ [(ε1 C1 ESR1) / (ε2 C2 ESR2)] (b2/a2)

In order to compare parallel plate capacitors we must taking into account that, calling AD the dielectric area

C ≈ (ε AD) / (4πd)

Parallelepiped shape capacitors are made with many paralleled substructures, being n the number of paralleled substructures of thickness d

Ci ≈ ni (εi ADi) / (4πdi)

For metallized film capacitors, under the supposition that metallized thickness mi << di, calling the with of capacitor as WD, is valid the approximation

WDi ≈ ni di

Then
(d2/d1)² ≈ (ε2 AD2 WD2 C1) / (ε1 AD1 WD1 C2)
Or
d2/d1 ≈ [(ε2 VD2 C1) / (ε1 VD1 C2)]½

Now, finally we can write

a1 ≈ [(ε2 VD2 C2) / (ε1 VD1 C1)]½ a2

b1 ≈ [(ε1 VD2 C1) / (ε2 VD1 C2)]½ (ESR1 / ESR2) b2

b1/a1 ≈ [(ε1 C1 ESR1) / (ε2 C2 ESR2)] (b2/a2)


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Last edited by popilin; 28th June 2014 at 04:36 AM.
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Old 28th June 2014, 05:04 AM   #358
popilin is offline popilin  Argentina
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Some loose ends

1) Just as an example, with some MKP capacitors like those I use in my amps, they are: C1=3.3 μF 450 VDC (Epcos B32674D4335 15.0 mm.x24.5 mm.x31.5 mm.) and C2=6.8 μF 450 VDC (Epcos B32674D4685 22.0 mm.x36.5 mm.x31.5 mm.)

a1 ≈ 2.12 a2

b1 ≈ 0.88 b2

This is illustrated on the following graph, with coordinate system placed to show the point more clearly.

Click the image to open in full size.

Differences on linearity are small, but still favoring the smaller capacitor.

b1/a1 ≈ 0.41 (b2/a2)

2) Same type MKP capacitors with almost the same volume: C1=4.7 μF 300 VDC (Epcos B32674F3475 14.0 mm.x24.5 mm.x31.5 mm.) and C2=4.7 μF 300 VDC (Epcos B32676F3475 12.0 mm.x22.0 mm.x42 mm.)

a1 ≈ a2

b1 ≈ 0.52 b2

Click the image to open in full size.

C1 is slightly more linear than C2.

b1/a1 ≈ 0.52 (b2/a2)

Parallelepiped shape capacitors are made with many paralleled substructures and this tends to reduce ESR, and even it can vary substantially up to a factor 2 or more with pin diameter and material, as capacitance increases, pin diameter often increases too and dielectric thickness often decreases, then can occur that ESR decreases with capacitance, so it is needed an analysis for each particular case.

If you want to use big capacitors, this is the way you go, cheap industrial grade MKP capacitors.

3) I found these cylindrical MKP capacitors that seems to have similar d, C1=35 μF 700 VDC (Vishay GLI 700-35) and C2=230 μF 700 VDC (Vishay GLI 700-230)

a1 ≈ 6.67 a2

b1 ≈ 0.5 b2

Click the image to open in full size.

As can be seen, C1 is more linear than C2 by far.

b1/a1 ≈ 0.075 (b2/a2)

No surprises here, no paralleled substructures, and connectors are very large, then, this capacitor geometry is more predictable.

Conclusion

On previous post Eq.(*) roughly proved that capacitor linearity depends on the dielectric and its losses, due to dielectric anisotropy, i.e. real world capacitors are far from being ideal, and statements like this just sit the basis for another audio myth

Quote:
Originally Posted by Merlinb View Post
In a given circuit, using a bigger capacitor will result in less distortion, not more, because the signal voltage across the cap will be reduced.
This is wrong, it is not the voltage across the capacitor that matters, but the electric field E, even more, doing the math you can see for yourself with almost any datasheet, that my original guess is quite correct.

Quote:
Originally Posted by Merlinb View Post
Armchair theorizing is all very well, but your claims simply do not stand up again the real world evidence, which conclusively shows that the bigger the capacitor, the lower distortion. Take a 100nF polystyrene capacitor and put it up against a 10000uF electrolytic, and you won't be able to measure the distortion either way.
Let me continue "armchair theorizing", but as I have no data for polystyrene capacitor, I will use those from a 100 nF MKP, if you don’t mind.

C1=100 nF 630 VDC (Solen SM010)

tan (δ1) ≈ 9 x 10⁻⁵, ε1 ≈ 2.1, d1 ≈ 6 μm, V1 = V, ω = 100 Hz

C2=10000 μF 10 VDC (Vishay 54103E3)

tan (δ2) ≈ 0.31, ε2 ≈ 8, d2 ≈ 0.015 μm, V2 ≈ V x 10⁻⁵, ω = 100 Hz

Then
a1 ≈ 250 a2

b1 ≈ 0.019 b2

This is so horrible that I don’t know how draw it…

Click the image to open in full size.

Hysteresis loop for C2 is so nonlinear that seems a rotated ellipse, it should be rotated a little because ε2 > ε1, but for clarity it isn’t, both loops start at the same point, and "major" axes of both ellipses lie on the same line, then “rotation” is an effect of extreme nonlinearity, be in mind that both V1 and V2 have the same phase.

b1/a1 ≈ 7.6 x 10⁻⁵ (b2/a2)

To my regret, I must admit that audiofools, sometimes can be right, i.e. capacitor quality does matter, and the bad reputation of electrolytic capacitors is well deserved, and if you can't measure the difference, your instruments must be worse than mine.
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Last edited by popilin; 28th June 2014 at 05:29 AM.
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