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I'm trying to get a better understanding of the relationship of dielectric constant (relative permitivity) to dielectric strength (breakdown voltage).I want to know what happens when you layer materials of high dielectric constant but low dielectric strength with materials of low dielectric constant but high dielectric strength. For example Barium titanate has a dielectric constant of 1200, But a dielectric strength of only 1.2 Mv/m. Mica has a dielectric strength of 118 Mv/m, but a dielectric constant of only 3.

If you put a 1mm sheet of mica between 2 1mm sheets of barium titanate, what would be the combined dielectric constant and dielectric strength?

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This turns out to not be helpful in the way you are suggesting. The capacitance of the combined structure will be limited by the lower dielectric constant materials.

The way you model this type of structure is to just treat it as three capacitors in series. You can mentally imagine that there is a thin conductive plate at each interface. So the overall capacitance adds using the normal rule for series capacitors: Ctot = 1/(1/C1 + 1/C2 + 1/C3)

Each of the three capacitors will have the same charge 'Q' as the entire capacitor. So you can figure out the maximum charge for each capacitor with the equation Qmax = Cmax*Vmax. Whichever is less will set the overall charge (and therefore voltage) limit for the compound capacitor.

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  • \$\begingroup\$ so are you saying there is no combine dielectric constant or dielectric strength, only combined capacitance? \$\endgroup\$
    – Cassie
    Commented Jan 31, 2017 at 6:50
  • \$\begingroup\$ Since series properties have the weakest link ( lowest capacitance and withstand voltage becomes the limit) it ends up making the result worse. In math the result is as I explained. \$\endgroup\$ Commented Jan 31, 2017 at 16:16
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These material properties are tradeoffs and cannot be improved by alternating materials. Each one becomes the weakest link.

Barium titanate k=1200, E = 1.2 kv/mm (max)  k*E=1440
Mica            k=3   , E = 118 kv/mm (max)  k*E=354

multiplying by cost gives you another metric kE$ /mm

When you compare every material, you find that plastic is most economical for some such as polycarbonate, polystyrene unless there is some other requirement such as mechanical strength, thermal resistance or max temp size, etc.

For example Teflon tape and epoxy coated paper are common HiV materials used inside power transformers and also mineral and/or vegetable oil purified for transformer quality voltage insulation 25 to 75kV/mm with high thermal conductance.

The purity of materials ( and processing) also affects the max. electric field E.

If C1 =1 and C2=100*C1 are in series the net C value is \$C= \frac{1}{\frac{1}{C1}+\frac{1}{C2}}\$ Which ends up being pretty close to the smaller Cap C1 or 0.99.

Similarly, if two materials with vastly different E field limits are in series, the weakest link will breakdown. Bottom line is there is no advantage to interleaving these two materials unless there are some other requirements not given.

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  • \$\begingroup\$ Economics doesn't play into my question at all. I'm only interested in understanding. My questions is completely detached from any particular real world application. \$\endgroup\$
    – Cassie
    Commented Jan 31, 2017 at 6:57
  • \$\begingroup\$ Series values do not add, rather the inverse adds then you invert the result so it gets worse. This is for both the property of series C and series breakdown voltage. Resistance adds but for Conductance or capacitance or dielectric constant or MV/m withstanding level they behave almost as it they are in parallel for voltage breakdown and in series with same current most the voltage is across the smallest dielectric constant material \$\endgroup\$ Commented Jan 31, 2017 at 16:19
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A conducting layer, like a 'field ring', can shape the field nearby so a single faulty dielectric layer does not produce a cascade failure in a multilayer capacitor structure.

Except for some multilayer HV film capacitors, where an intermediate layer is added to make the item fail-safe against single internal short circuits, this is rarely done.

Capacitors store energy in an electric field; you can't exceed breakdown E field, so the dielectric constant times E^2 sets the maximum energy density for any diectric. Layering dielectrics doesn't improve the limit, it just puts capacitors in series.

1mm of mica, with 2mm of BaTiO3 (the order doesn't matter) with area A would give Ctot = 1/(1/(3*A/1) + 1/(1200*A/2) = 2.98*A (in some arbitrary units)

When you apply AC to such a composite, most of the voltage will drop across the mica (small capacitor), so the composite only takes the voltage of the mica and has the capacitance of the mica. But, it's three times as big.

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