ACElec1: Plane Capacitor

AC current in the plane capacitor

Problem Type:
A plane-parallel problem of AC conduction.

Geometry:

Due to symmetry only a small part of 1 mm height is used. The length of capacitor in z direction is L = 10 mm.

Given:
Relative permittivity of substrate ε = 10;
Conductivity of substrate g = 10^-8 S/m.

Voltage U = 5 V,
Frequency f = 50 Hz.

Problem:
Find current and dissipation factor tg(δ) of the plane capacitor with non-ideal dielectric inside.

Solution:
Capacitor with non-ideal dielectric can be replaced by electric scheme with ideal capacitor C and resistivity R connected in parallel.

The capacitance of the plane capacitor is calculated by the equation
C = εε₀S/d, where S is the plate area S = h·l.

Resistance of the substrate is calculated by the equation
R = ρ·d/S.

Current I has two components: active I_A and reactive I_R.
For parallel scheme
I_A = U / R, I_R = U / X_C.

tg(δ) = |P_A / P_R| = |U·I_A / U·I_R| = |I_A / I_R|

Results:
S = 100 mm².
R = (1/10^-8)·0.1e-3/100e-6 = 10⁹ Ohm.
C = 10·8.854e-12·100e-6/0.1e-3 = 8.854·10^-11 F.
X_C = 1 / 2·3.1416·50·8.854·10^-11 = 0.3595·10⁹ Ohm.
I_A = 5 / 10⁹ = 5·10^-9 A
I_R = 5 / 0.3595·10⁹ = 13.908·10^-9 A

tg(δ) = 5·10^-9 / 13.908·10^-9 = 0.36

See the ACElec1.pbm problem in the Examples folder.