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Plane capacitor

plane capacitor loss factor, dielectric loss angle tangent, dielectric AC loss, AC loss in dielectric

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 Substrate

Given
Relative permittivity of substrate ε = 10.
Conductivity of substrate g = 10-10 S/m;

Voltage U = 5 V,
Frequency f = 1000 kHz.
Loss tangent tan(δ) = 0.01.

Solution
Capacitor with non-ideal dielectric can be replaced by electric circuit with ideal capacitor C and resistivity R connected in parallel.
Current I has two components: active IA and reactive IR.

plane capacitor equivalent circuit

Active current due to dielectric conductivity g is very low. It could be calculated if we run the analysis with actual value of electric conductivity g.

Dielectric loss cause the electric current active component IA to increase. It could be calculated if we run the analysis with effective value of electric conductivity, which is
gapparent = 2πf·εε0·tan(δ) = 2·3.142·1000000·10·(8.854·10-12)·0.01 = 5.56·10-6 S/m.

Results
Electric current vectors and electric potential distribution in plane capacitor

Electric current vectors and electric potential distribution in plane capacitor

Loss tangent tan(δ) = |PA / PR| = 0.0000695 / 0.00695 = 0.01.

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