Van der Pauw method
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The van der Pauw method is a technique for doing 4-probe resistivity and Hall effect measurements. In essence it provides an easy way to measure:
- Sheet resistivity/conductivity
- From this one can calculate the resistance/conductance provided the thickness of the sample is known
- Hall voltage
- Sheet carrier density
- From this one can calculate the carrier density provided the thickness of the sample is known
- Hall mobility
The advantages of this method include low cost and simplicity. The van der Pauw technique can be used on any thin sample of material and the four contacts can be placed anywhere on the perimeter/boundary, provided certain conditions are met:
- The contacts are on the boundary of the sample (or as close to the boundary as possible)
- The contacts are infinitely small (or as close as possible)
- The sample is thin relative to the other dimensions
- There are no isolated holes within the sample
Measure Sheet Resistance
Eight measurements need to be made in order to calculate the sheet resistivity. A current must be forced through two adjacent contacts, and a voltage must be measured across the other two contacts. Let the current passing from contact i to contact j be denoted as:
- <math>I_{ij}<math>
and the voltage measured across contacts k (positive) and l (negative) be give by:
- <math>V_{kl}<math>
The sheet resistance in this configuration is given by:
- <math>R_{ij,kl} = \frac{V_{kl}}{I_{ij}}<math>
Assuming that the contacts are numbered sequentially along the permieter of the sample, we need to measure the following resistances:
- <math>R_{12,43},R_{21,34},R_{23,14},R_{32,41},R_{34,21},R_{43,12},R_{41,32},R_{14,23}<math>
We define
- <math>R_A = \left( R_{12,43} + R_{21,34} + R_{34,21} + R_{43,12}\right) /4<math>
and
- <math>R_B = \left( R_{23,14} + R_{32,41} + R_{41,32} + R_{14,23} \right) /4<math>
Van der Pauw discovered that the sheet resistance of the sample could be determined from R_A and R_B using any arbitrary shape. The actual sheet resistance, R_S is obtained by solving:
- <math>e^{-\pi R_A/R_S}+e^{-\pi R_B/R_S}=1<math>