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ACElec1: Plane Capacitor

AC current in the plane capacitor

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

Geometry:

Capacitor

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.

Capacitor

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 IA and reactive IR.
For parallel scheme
IA = U / R, IR = U / XC.

tg(δ) = |PA / PR| = |U·IA / U·IR| = |IA / IR|

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.
XC = 1 / 2·3.1416·50·8.854·10-11 = 0.3595·109 Ohm.
IA = 5 / 109 = 5·10-9 A
IR = 5 / 0.3595·109 = 13.908·10-9 A

tg(δ) = 5·10-9 / 13.908·10-9 = 0.36

QuickField

Theoretical result

IA, A

5·10-9

5·10-9

IR, A

13.908·10-9

13.908·10-9

tg(δ)

0.36

0.36

See the ACElec1.pbm problem in the Examples folder.