Main >>
Applications >>
Sample problems >>
Dielectric losses

# Dielectric losses

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

**Problem Type:**

Plane problem of AC conduction.

**Geometry:**

*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* = εε_{0}·*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*

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) = *X*_{C}/*R*, where

*X*_{C} - capacitor reactance, *X*_{C} = 1 / 2π*f·C*,

*R* - capacitor resistance.

After substitution we have:

tg(δ) = 1/ 2π*f**C**R* = 1 / 2π*f*·εε_{0}·*S*/*d*·(1/γ)·*d*/*S*

Electric conductivity is γ = 2π*f*·εε_{0}·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 *P*_{A} = 0.46 mW.

Download video.

View simulation report in PDF.

Download simulation files