Jamie said:
Hi,
Is a capacitors dielectric breakdown voltage similar to an inductors
core saturation, in that no more field energy can be stored at that
point?
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.
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?
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.
In there any case where a capacitor dielectric can saturate without
voltage breakdown?
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.
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.
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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