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Engineering
Electrical Engineering
Layering different dielectric materials to increase breakdown voltage
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[QUOTE="Baluncore, post: 5679696, member: 447632"] 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 modeled 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 [/QUOTE]
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Layering different dielectric materials to increase breakdown voltage
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