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Main >> Applications >> Sample problems >> Plane capacitor simulation

# Plane capacitor

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

Problem Type:
Plane problem of AC conduction.

Geometry:

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.

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

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

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