Inter electrode resistances calculation for conductive films assembly - QuickField simulation example
Assembly made of two conductive sheets and cylindrical copper electrodes is used to deliver electric power to two light fixtures. We need to see the current distribution in the conductive sheets.
How to calculate inter-electrode resistances in a conductive film assembly?
Answer Typical applications Geometry
Given
Task
Solution
Taking into account that conductive sheet conductivity is much lower than conductivity of the cylindrical electrodes, we may consider the contact surfaces between the sheet and electrodes A, B, C, D, V, G to be equipotential.
Results
Engineering question
Set up a plane-parallel DC Conduction problem for the conductive sheets and electrodes, assign the film conductivity from its surface resistance and thickness, and determine the current density distribution and voltage drop between electrodes from the computed field results.
conductive film assemblies, multi-electrode resistance networks, LED lighting fixture wiring
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Simulation problem
Problem Type
A plane-parallel problem of DC conduction.
Sheets size 1x1 m. Cylindrical electrode diameter 1 mm. Holes diameter 2 mm.
Surface electric resistance of sheets Rs = 10 Ω/square, film thickness is t = 200 nm.
Copper resistivity is 56 MS/m
Light fixture current 30 mA.
DC voltage source 24 V.
Current density distribution in the conductive sheets.
For the film we should specify electrical conductivity, which is reciprocal to the product of surface resistance and the film thickness: σ = 1/(Rs*t) = 1/ (10 * 200e-9) = 0.5 MS/m.
Volgate drop in each conducting film is 0.5 V.
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