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Plane capacitor dielectric loss

QuickField simulation example

This is an example of the plane capacitor simulation, performed with QuickField software.

Problem Type
Plane-parallel problem of AC conduction.

Geometry
Plane capacitor Capacitor with non-ideal dielectric can be replaced by electric circuit with ideal capacitor and resistivity connected in parallel. 10 mm 0.1 mm 10 mm GND U Dielectric

Given
Relative permittivity of dielectric ε = 10.
Voltage U = 5 V, frequency f = 1000 kHz.
Loss tangent tan(δ) = 0.01.

Task
Calculate dielectric loss in the plane capacitor.

Solution
To calculate dielectric losses we specify effective value of electric conductivity for dielectric, that is a function of loss tangent, frequency and relative permittivity:
g = 2πf·εε0·tan(δ) = 2·3.142·1000000·10·(8.854·10-12)·0.01 = 5.56·10-6 S/m.

Results
Active power loss is PA = 0.0695 mW per 10 mm³.
Electric current vectors and electric potential distribution in plane capacitor
Electric current vectors and electric potential distribution in plane capacitor