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Layering different dielectric materials to increase breakdown voltage

  1. Jan 30, 2017 #1
    This is a general question, not a homework question, and @berkeman said I should post it in the EE forum...

    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?


    I came up with 4 possible scenerios:
    1. The barium titanate being next to the plates would determine the dielectric constant, with no effect from the mica. Mica having the highest dielectric strength, the combined dielectric strength would be determined the percentage of mica. Thus the combined dielectric strength would be 39.33 MV/m and the combined dielectric constant would be 1200.

    2. The dielectric constant would be limited by the mica. Thus the combined dielectric strength would be 39.33 MV/m and the combined dielectric constant would be 3.

    3. The dielectric constant and dielectric strength would be would be limited by the lowest common denominator. Thus the combined dielectric strength would be 1.2 MV/m and the combined dielectric constant would be 3.

    4. Both materials detrmine the dielectric constant and dielectric strength, which probably could only be determined experimentally.

    Are any of these Correct, and if not what is correct and why?
     
  2. jcsd
  3. Jan 30, 2017 #2

    Baluncore

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    Welcome to PF.
    The fact that you are discussing dielectric constant suggests you are considering AC insulation.

    Each different material layer will make a parallel plate capacitor with a particular thickness between it's surfaces. You need to know the dielectric constant and the dielectric strength in V/mm of the materials used before you can put numbers on a multilayer insulation model.

    For DC it is the leakage current or resistance across each layer that sets the voltage drop. But with layers of capacitance in series, AC voltage is shared in proportion to the reciprocal of the capacitance. So if you double the thickness of a layer you double the breakdown voltage, but halve the capacitance, so twice the AC voltage appears across that layer. For that reason, when considering different dielectric material layers you must pay attention to the ratio of dielectric strength to dielectric constant.

    Your question can only be answered by putting numbers on the layers, and then operating some layers at voltages below their dielectric strength.

    Metallising the floating equipotential surfaces within multilayer insulation was called “chroming”. Layers of foil are used to make more reliable insulators by averaging out variations of the voltage drop across layers.
     
  4. Jan 31, 2017 #3
    hi balancore thanks for replying,

    Actually, I'm only thinking about a DC capacitor by itself.

    I give all of that in my original post.
    Barium Titanate: dielectric constant=1,200 dielectric strength=1.2 MV/m (1.2 V/µm)
    Mica: dielectric constant=3 dielectric strength=118 Mv/m (118 V/µm)
    (If these seem like high numbers. it doesn't matter. This is only a hypothetical example to explore the concept, not a real world specification).

    There's alot of good information here, but I'm having difficulty seeing how this answers my question.

    Thanks again and I look forward to your next reply
     
  5. Jan 31, 2017 #4

    Baluncore

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    For a long term DC voltage, the dielectric constant is irrelevant as the voltage will be shared across the layers in proportion to their resistance. You have not specified the resistivity of the materials or the series leakage current that will flow through all layers.

    For a DC turn-on transient, the dielectric constant is a critical voltage division parameter as the circuit must then be modelled as an AC circuit.
    Analysing the DC step response is the same as analysing the AC situation.

    Given; Barium titanate, dielectric constant = 1200, dielectric strength = 1.2 Mv/m. Mica, dielectric constant = 3, dielectric strength = 118 Mv/m. We can hypothesise the layer capacitors as;
    C1 = Barium titanate. A 1 mm layer gives a voltage rating of V1 = 1.2 kV, with say C1 = 1200 pF.
    C2 = Mica. A 1 mm layer gives a voltage rating of V2 = 118 kV, with say relative C2 = 3 pF.
    C3 = C1.

    We know that voltage will be shared in inverse proportion to capacitance, so; V1 / V2 = C2 / C1. If we operate C2 at the breakdown voltage V2, the voltage across C1 will then be a trial value of V1;
    V1 = V2 * C2 / C1. So; V1 = 118 kV * 3 / 1200 = 0.295 kV which is OK as it is below the maximum 1.2 kV calculated earlier.

    To survive a single step turn-on transient, the answer to your question is that;
    1. The breakdown voltage of the combined sandwich is 0.295 kV + 118 kV + 0.295 kV = 118.59 kV. But since it is now 3 mm thick, the effective dielectric strength has become 39.53 kV/mm = 39.53MV/m.
    2. The capacitance of the series sandwich is Ct = 1/ (2/C1 + 1/C2 ) = 2.985 pF. But since it is now 3 mm thick, that makes the equivalent dielectric constant = 8.955
     
    Last edited: Jan 31, 2017
  6. Jan 31, 2017 #5
    Thank you, this is exactly what I was looking for. Though, I don't know where the capacitance figures were derived from, since no area was specified, but if we assume an area of 112.95 cm2, then the math works.
     
  7. Jan 31, 2017 #6

    Baluncore

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    Sorry, I used a capacitance equal to dielectric constant in pF so as to reduce the arithmetic and keep it realistic.
    I could avoid the defined value of Eo and the area of the hypothetical capacitors because only relative capacitance was needed.
     
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