A new approach to field modelling

Language:
de dk es fr it cz ru cn


RSS Twitter Facebook Linkedin YouTube

>> >> >>

Dielectric losses

Simulation of the dielectric losses in the capacitor with known loss tangent.

Problem Type:
Plane problem of AC conduction.

Geometry:

Plane capacitor model
d=1 mm, h=10 mm.
The length of capacitor in z direction is l = 10 mm.

Given:
Relative permittivity of substrate ε = 2.3.
Loss tangent tgδ = 0.75%.
Voltage U = 220 V,
Frequency f = 100 kHz.

Task:
Calculate dielectric losses in capacitor.

Solution:
Capacitor with nonideal dielectric can be represented by electric circuit with ideal capacitor C and resistivity R connected in parallel, so that R corresponds to the insulation resistance.
For plane capacitor the capacitance can be calculated as C = εε0·S/d, where S=l·h is a plate surface area.
For plane capacitor the resistance can be calculated as R = (1/γ)·d/S

dielectric losses in capacitor equivalent circuit

In the AC conduction problems electric conductivity (γ) of materials should be set. We can derivate the electric conductivity value using the formulae:
tg(δ) = XC/R, where
XC - capacitor reactance, XC = 1 / 2πf·C,
R - capacitor resistance.

After substitution we have:
tg(δ) = 1/ 2πfCR = 1 / 2πf·εε0·S/d·(1/γ)·d/S

Electric conductivity is γ = 2πf·εε0·tg(δ)

Results:
Electric conductivity is γ = 2·3.142·100000·8.854e-12·2.3·0.0075 = 0.096 uS/m.

Active power dissipated in dielectric is PA = 0.46 mW.

Download Download simulation files: dielectric_losses.zip

View movie Video.