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ACElec1: Plane capacitor

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

Problem Type:
Plane problem of AC conduction.

Geometry:

Plane capacitor model

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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Download icon Download simulation files (files may be viewed using any QuickField Edition).

There are no restrictions applied to the QuickField Student Edition postprocessors.
You can view field maps, make plots, calculate integrals and print pictures in the same way that the Professional Edition users do.