Electric current distribution in the conductive laminated polymer films

QuickField simulation example

Assembly made of two conductive films and cylindrical copper electrodes is used to deliver the electric power to two light fixtures. We need to see the current distribution in the conductive films.

Problem Type
A plane-parallel problem of DC conduction.

Geometry
Films size 1x1 m. Cylindrical electrode diameters 1 mm. Hole diameters 2 mm.

Given
Surface electric resistance of films R_s = 10 Ω/square, film thickness is t = 200 nm.
Copper resistivity is 56 MS/m
Light fixture with connecting electrodes has electrical resistance of 800 Ohm.
DC voltage source 24 V.

Task
Find the current density distribution in the conductive films.

Solution
For the film we should specify the electrical conductivity, which is reciprocal to the product of the surface resistance and the film thickness: σ = 1/(R_s*t) = 1/ (10 * 200e-9) = 0.5 MS/m.

For the electrodes connected to the DC voltage source we specify the electric potentials 0 and 24 V respectively. For the light fixture electrodes we specify the rated current 24V / 800 Ohm = 0.03 А defined as an external current in the central node.

Results
Current density distribution in the grounded electrode film (G) and powered electrode film (V).

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