Is a capacitors dielectric breakdown voltage similar to an inductors core saturation, in that no more field energy can be stored at that point?
If the dielectric is composed of different materials with different breakdown voltages, will each material only store energy up to its own inherent breakdown voltage, even though the dielectric as a whole is still below its breakdown voltage? If so will there be any energy wasted by the materials that are above their breakdown voltages?
In there any case where a capacitor dielectric can saturate without voltage breakdown?
I am wondering about this as I was thinking of a capacitor dielectric with barium titanate powder in a matrix of epoxy or water ice at one of barium titanate's dielectric constant peaks (~ -5C or ~+120C), and was wondering if the barium titanate would store energy beyond its characteristic breakdown voltage.
cheers, Jamie
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J
John Popelish
No. It is more like winding overheating due to inductor over current.
Core saturation is more like the dielectric nonlinearity in the high-K ceramic dielectrics, where the capacitance falls as the voltage rises. Those dielectrics are called ferro-electric, and the core materials are called ferro-magnetic.
Materials are usually permanently damaged by arching when their breakdown voltage is exceeded. Zener diodes are an exception, if the current is limited.
Google high-K ceramic dielectrics.
Once the dielectric breaks down from over voltage, it is no longer a dielectric but some kind of nonlinear resistor/plasma/flame/explosion.
T
Tim Williams
Well, yes but no, one is destructive, the other isn't. Well, usually. I actually cracked a black toroid ferrite core when testing it to saturation. Weird stuff huh.
But seriously, John's on the ball for equivalence here.
Tim
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Bert Hickman
Not quite. As a ferromagnetic core saturates, the B-H curve flattens out. But further increases in the applied magnetic field do not result in catastrophic "breakdown" of the core material. For ferroelectric materials, the D-E curve is similar in shape to the B-H curve of ferromagnetic materials, but it is not nearly as "flat" at high E-fields. And, the electrical stress cannot be increased indefinitely since the dielectric will break down long before dielectric "saturation" is reached. And, for most solid dielectrics, there often very little warning before the dielectric suddenly fails.
While a capacitor with a vacuum or air dielectric can usually recover from a flashover, and certain "self healing" metallized film capacitors are designed to electrically isolate a failing region of dielectric. However, most other capacitors catastrophically fail. The resulting arc over within the dielectric causes irreversible physical and chemical damage. The most obvious external signs are excessive leakage current/short circuiting or a markedly reduced breakdown voltage, but high energy density capacitors may actually explode.
You would normally never want to stress the composite dielectric system so that EITHER dielectric is at the limit of its breakdown strength. Assuming you use good dielectrics, the voltage stress across each dielectric will be a function of the relative permittivities of each dielectric, their respective thicknesses, and their bulk resistivities. The system behavior similar to a pair of capacitors connected in series, each shunted by a resistor (representing the resistivity of the respective dielectric material). With a composite dielectric, either transient (pulsed/AC) or steady state (DC) conditions may result in excessive electrical stress across one of the dielectrics in the system. This can ultimately lead to the failure of both dielectrics.
The dielectric with the lowest permittivity will have the highest transient E-field stress (in volts/mil), while the dielectric with the highest resistivity will have the highest DC stress. Most capacitor manufacturers design high voltage capacitors using composite dielectric systems (oil-polymer film or oil-kraft paper-film)) so that most of the actual electrical stress appears across the best insulating material (i.e., the film). A properly designed capacitor, when operated within design specifications, has a voltage stress across each dielectric element that is always below the respective breakdown voltage during either transient or DC conditions. Most of the stored energy ends up residing in the most highly stressed, lowest dielectric constant material.
No... ultimately all dielectrics (including even a vacuum) will break down under a sufficiently high electric field. The actual breakdown mechanisms will differ depending on the dielectric system(s), but the end result is the same. Higher k ferroelectrics tend to have lower breakdown thresholds. For example, a relatively low k ferroelectric material (k~30) may withstand uniform E-field of 15 kv/mm, while a higher k (4000-6000) material may only support a stress of 2.5 kv/mm.
This is analogous to a "loosely packed" ferrite with a distributed air gap - most of the magnetic energy resides in the "gap". If you separate high-permittivity grains of ferroelectric material within a matrix of another material (with a much lower relative dielectric constant), most of the electrical stress, and most of the stored electrostatic energy, will be within the lower dielectric constant material. This is one reason why silver electrodes are evaporated so as to make intimate contact with the ceramic. Ceramic capacitors do not use composite dielectric systems. If you introduce small gaps between the ceramic and electrodes, or between grains of the ferroelectric material, you will likely see excessive E-fields within the gaps, leading to ionization/breakdown of the gaps and, ultimately, the destruction of the cap. Lower k gaps will also markedly reduce the effective capacitance of the composite system. The exception might be a lower k (~30 ceramic in a water matrix (k~80) in a pulsed power application, but you'll still never "saturate" the dielectrics...
Best regards,
Bert
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Terry Given
specifically, magnetisation involves re-orienting of domains, of which there are a finite number. Once all domains point the same way, the material is saturated. no matter how much more H you throw at it, no more than 100% of domains can be aligned.
So the magnetic material effectively "goes away" once it saturates, rather than breaking, and you end up with an air-cored inductor, IOW the permeability plummets, as does (by definition) the inductance.
It is not uncommon for inductor saturation to then break other things.
Ignoring, of course, several things:
- the physical forces exerted can break ferrite (hard but brittle) (ever seen a picture of a BIG line transformer thats had an output short? smashed/twisted bus bars is the norm.)
- square-loop ferrite can break if excited at its mechanical resonant frequency (ferroxcube have a note to this effect in their square-loop ferrite datsheets)
"Ferromagnetism," Bozorth, IEEE press, is a bloody good read. If you can be bothered.
For ferroelectric
J
joseph2k
Alas, i disagree some with the comments. I agree that the fundamental physics of core saturation is very unlike breakdown in dielectrics; the first is recoverable, predictable, repeatable, and not normally destructive; dielectric breakdown is none of these. Non-linear magnetics have gotten to the point where they are well characterized. Only very few nonlinear capacitances are characterized, an then not very well. Piezoelectric substances are still a fertile field of study.
JosephKK
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