Electric heating film for glass

QuickField simulation example

Electric current is passing through the conductive thin film with two electrodes attached.

Problem Type
Plane-parallel problem of DC conduction.

Geometry

Given
Indium tin oxide film thickness t = 170 nm, sheet resistance R_s = 12.6 Ω/sq.
Electrode thickness d = 2 μm, copper electric conductivity σ = 56 MS/m.
Voltage V+ = 220 V.

Task
Calculate the current and the Joule heat power.

Solution
Film resistivity* ρ = R_s * t = 12.6 * 170e-9 = 2.142 μΩ*m. In QuickField DC conduction analysis we specify the electrical conductivity, that is reciprocal to the resistivity σ = 1 / ρ = 1/2.124e-6 = 467 kS/m.

Electrodes and film thickness are different. To be able to simulate this model in 2D we reduce the electrodes thickness to that of the film and proportionally increase the conductivity σ*d/t = 56 MS/m * 2μm/170nm = 659 MS/m.

Heat power distribution in the film is not uniform, so is the temperature. To get accurate results that take into account the heat redistribution inside the glass we should simulate a 3D heat-transfer problem.
Alternatively we can ignore the fact that the heat is redistributed in the glass and use this example to remap volume heating power Q to temperatures.
First we convert volume power to a surface power q[W/m²] = Q [W/m³] * t [m]
Then we can use the data from the table to get temperature values.

Result
Electric current is I = 3.218 A. Total power is 220 V * 3.218 A = 708 W.

References:
* Wikipedia, Sheet resistance.

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