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Four-probe sheet resistance measurement - QuickField simulation example

Four-probe method is used to measure the thin sheet resistance. This method is similar to a Wenner array method. But here we deal with a limited in size specimen. So the corrections to the equation should be made [1]. We use a finite element model to verify these equations.

Engineering question

How to simulate four-probe sheet resistance measurement?

Answer
Convert sheet resistance and film thickness to conductivity, inject the specified current through the outer probes, keep the voltage probes floating, and measure their potential difference.

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Typical applications
four-probe resistivity samples, thin conductive sheets, semiconductor wafers

Four-probe sheet resistance measurement

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Simulation problem

Problem Type
Plane-parallel problem of DC conduction.

Geometry
d = 100 mm, a = 200 mm, s = 10 mm.
Four-probe sheet resistance measurement Wenner array is used to measure electrical surface resistivity of concrete specimen. Film I+ 2 1 I- s s s a d V I

Given
Surface electric resistance of the film is Rs = 10 Ω/square, film thickness is t = 200 nm.
Electrodes spacing s = 20 mm, injected current I = 4.5 mA.

Task
Calculate the electric potential difference between electrodes 1 and 2.

Solution
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.
The electrode conductivity is much higher than that of the film. We model electrodes by specifying their electric potentials and current density on the electrode surfaces.
Electrode tip diameter is 1 mm. In QuickField we specify current density at (I+) electrode 0.0045 / (PI*0.001/2) / 200e-9. At the (I-) electrode we specify zero electric potential. Electrodes (1) and (2) are floating conductors.

Four-point probe equation for surface resistivity is ρs = (V / I) * C(a/d;d/s), where C(2;10) = 4.2357 in accordance with [1].

Results
Potential difference between electrodes 1 and 2 is 0.0356-0.0249 = 0.0107 V. Following the analytical equation the resistivity is ρs = (0.0107/0.0045)*4.2357 = 10.07 Ω/square. This shows that the finite element simulation result agrees with the analytical approach with 0.7% tolerance.
Electrical resistivity of concrete

Reference
F. M. Smits Measurement of sheet resistivities with the four-point probe, The Bell System Technical Journal, vol. 37, no. 3, pp. 711-718, May 1958.

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