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Main >> Applications >> Sample problems >> Plane capacitor simulation
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 nonideal dielectric can be replaced by electric circuit with ideal capacitor C and resistivity R connected in parallel.
Current I has two components: active I_{A} and reactive I_{R}.
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 I_{A} to increase. It could be calculated if we run the analysis with effective value of electric conductivity, which is
g_{apparent} = 2πf·εε_{0}·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) = P_{A} / P_{R} = 0.0000695 / 0.00695 = 0.01.
View simulation report in PDF.
Download simulation files (files may be viewed using any QuickField Edition).