Questions & Answers Regarding Sample Size and Correction Factors

Before the questions and answers shown below, here are two information packed technical documents that can be downloaded/viewed:

1) Tables showing correction factors for various material shapes can be found at this link:

Haldor Topsoe

2) Tables showing correction factors from the 1964 National Bureau of Standards Technical Note 199, “Correction Factor Tables for Four-Point Probe Resistivity Measurements on Thin, Circular Semiconductor Samples”, can be found at this web page link:

National Bureau of Standards Technical Note 199

Also, before the questions and answers section shown below, the following table shows one of the most commonly needed correction factors, i.e., Correction Factors for Measuring a “thin, circular slice”.

The table shown directly below pertains to sheet resistance measurements made in the center of a circular slice. d/s = diameter of sample divided by probe spacing (probe spacing being the distance between any two adjacent probes). For example, a 4mm diameter sample probed with a four point probe with 1mm tip spacing would have a correction factor of 0.6462. A 100mm wafer measured with a four point probe head that has 1mm tip spacing would have a correction factor of 0.9991. A result better than 0.1% can be obtained by measuring in the center of a circle with diameter greater than 100 x s., better than 1% is obtained with 40 x s.:

Correction factors for measuring a thin circular slice, measured in the center
d/s
Correction
3
0.5
3.448
0.5734
4
0.6462
5
0.7419
6.061
0.8089
7.5
0.8665
8.696
0.8972
10
0.9204
12.5
0.9475
15
0.9628
20
0.9788
28.57
0.9895
40
0.9945
100
0.9991
infinite
1

 

The following information represents emailed questions from customers
with responses from the founder and managing director of Jandel Engineering Ltd.,
John Clark, C. Eng, M.I.Mech.E., F.B.H.I.

 
Question: What is the need for correction factors in four point probe measurements?

Answer: The geometry of the sample determines the correction factors that must be used, additionally the position of the probes on the sample and the spacings between the probes. The need for correction factors is caused by the proximity of a boundary which limits the possible current paths in the sample. The most basic sample would be semi-infinite in extent i.e., it extends to infinity in all directions below the plane in which the four probes are located. All other cases would restrict the current paths available, eg., an infinite plane sample of finite thickness requires a correction factor based on the thickness.


Question: We can control the sample size, the thickness of the coating is approximately 30 nano meters. For example if I had a rectangular specimen and the pins on my probe were linear and 1.59mm apart, what is the smallest specimen size I could measure? I have seen probes with a variety of pin spacing, does that affect what size specimen I need? I may also want to measure odd shaped specimens, is there a minimum amount that the specimen must extend beyond the pins on the probe that would ensure accurate measurements for any shape specimen?

Answer: A spacing of 1.59mm would give an overall separation (outer two pins) of 4.77mm.

When it comes to the size of a sample there are a number of things to bear in mind. To obtain a reading where no correction factor is needed, there would need to be a sample of sufficient size to incorporate a circle of diameter 100s (where s is the spacing of the probe), i.e., for a probe with 1 mm spaced needles, a 10cm circle would be needed. This drops to 6.35cm for a 0.635mm spacing probe or 5cm for a 0.5mm spacing probe. However, at 40 times the spacing there is a 0.9945 correction factor (i.e., less than 1% error) and even at 10 times there is less than 10 % error.

With a rectangular sample there are similar correction factors. The pins should contact parallel to the longer of the sides. Again, the correction factors show that if the length of the rectangle is over 100s then there will be no error. At 40 times the spacing there is less than 1% error and 10s less than 10% error.

As for irregular shapes I would consider what size circle/rectangle could be included within its perimeters and work from there.

These correction factors (if you are working more or less on these limits mentioned) will only be valid for measurements at the center of the sample.

Generally the larger the sample used, the more accurate any measurement of resistivity is going to be.


Question: I have a question about the sample size requirements. Does the sample that I want to measure need to be larger than the head of the probe, or will I get accurate measurements as long as all the pins make contact with the sample?

Answer: There are correction factors available to deal with a number of possible situations. Basically there are two distinct possibilities;

1) Where the sample is of semi-infinite volume

2) Where the sample is an infinitely large plane of finite thickness

Considering practical samples such as wafers and parts of wafers corrections may need to be made. Generally the formula for sheet resistance is used.

Sheet resistance = 4.5324 x ( V/I ) ohms per square and multiplying this figure by the thickness in cm gives the bulk resistivity in Ohm cm. Provided the thickness does not exceed half the spacing of the probes there is not too much error in using this formula. The situation where the sample is small in diameter or width is more complex. For example, with a circular sample 3mm in diameter and a spacing of 1mm the 4 probes would only just be on the wafer and the correction factor would be 0.5. This rises to 0.991 at 100mm diameter and obviously any intermediate diameter the factor could be between 0.5 and 1. We can provide some charts to arrive at a suitable figure. The correction is more complex with rectangular samples. From a practical point of view, too, the probes ought to be presented at right angles to the sample, so probably the rear of the holder should be rested on a piece of material equal in thickness to the sample. Please tell us more about the sample thicknesses and size so that we can be more helpful.


Question: I am looking for some information concerning your four point probes. I am looking to measure the bulk resistivity of Bars with a dimension of 3mm x 4mm x 50-60mm.

Answer: The problem with samples of this shape and size is correction factors for the results you obtain. We use the factors contained in a book by Haldor Topsoe Semiconductor Division. There is a section dealing with samples of the shape and size you describe, but it is rather complex. It seems to me that if you used a 25 mil spacing probe the results of measuring as shown in the diagram would be valid.

http://www.four-point-probes.com/haldor-topsoe-geometric-factors-in-four-point-resistivity-measurement/


Question: What are the limitations when probing very small geometries? I would like to probe a layer of material that covers the entire wafer, but the exposed area which I can probe may be very very small — on the order of millimeters.

Answer: Our smallest spacing probe is 0.5mm so the window needed could be 2mm. Clearly the actual “span” of four probes is 3x the spacing plus a small allowance because of the width of the probe.


Question: How close to the edge of sample can we probe? Can you just add a correction factor?

Answer: The general rule is that you can probe up to (20 x probe spacing) of the edge, eg 20mm with a 1mm spacing probe without applying any factor beyond the basic one of rho=4.5324 x (V/I). Further factors are available from literature.


Question: How small are the signals coming from the probe head?

Answer: The signals can be from microvolts to hundreds of mV dependant on the material.


Question: Can we tell what the material is from the sheet resistance and the bulk resistivity?

Answer: The sheet resistance is of no help, but if you know the thickness you can calculate the resistivity (resistivity=sheet res x t) where t is the thickness in cm. You can’t really recognize what the material is, because the raw material could be ‘doped’ to give almost any value of resistivity.


Question: Will the four point probe technique work for the samples that I sent to you to test? Will I need to use correction factors for my samples?

Answer: There are basically two possibilities:

1) Semi-infinite volume – where the sample extends to infinity in all directions below a plane on which the four probes are located. In this situation with equidistant probes resistivity (rho) = 2 x pi x s x V/I ohm.cm where s is the spacing in cm, I the current flowing between the outer two probes and V the measured voltage between the centre two probes. In practice provided the semi-infinite volume thickness is greater than five times the probe spacing the correction factor 2 x pi x s is correct within 0.7%. Your samples are of course not infinitely wide but 100mm diameter which is 100 x s. It can be shown that the correction factor for this is near unity, when measuring at the centre, and that the formula remains valid for measurements up to within 20 x s of the periphery.

2) Thin slice infinite in extent. ‘Thin’ in this context means that the ratio of thickness to spacing t/s is <1 and sheet resistance R=pi/(log(n)2) x V/I = 4.5324 x V/I ohms/square. Again, the formula is applicable to a thin circular slice, the correction factor for the limited diameter of 100mm gives near unity for a ratio d/s=100 but for the thickness, much greater than the limit t < s/2. (in fact t/s=4) the correction factor is about 0.32. To derive the volume resistivity from the sheet resistance it is necessary to multiply by the thickness t in cm.


Question: Why do the correction factors shown in the first table on your web page, “Correction factors for measuring a thin circular slice, measured in the center”: http://www.fourpointprobes.com/haldor-topsoe-geometric-factors-in-four-point-resistivity-measurement/
….have different values from the correction factors shown in the F.M. Smits document, “Measurements of Sheet Resistivity with the Four-Point Probe”, BSTJ, 37, p. 371 (1958).?

Answer: There is no error with our correction factors table. In our table we are saying that the measurement should be made as usual and the end result multiplied by the factor given. The F.M. Smits table which has been referred to is effectively the figure of 4.5324 multiplied by the correction factor in the table we use – this is why it increases as the sample gets larger. We use the figures we give to correct the sheet resistivity value calculated by (4.5324 x V/I). Therefore the value we would get for a sample which is D/S = 3 is equal to (4.5324 x V/I) x 0.5

The value in the F.M. Smits chart is the figure to multiply the V/I by, for example, for D/S= they have 2.662 x V/I, which of course gives the same result.

To simplify, the sheet resistance formula is:

4.5324 x V/I

4.5324 is a constant. When we make measurements we work out the sheet resistance using this equation. We then multiply the result by a correction factor given in the table we supplied.

The F.M. Smits table has actually already multiplied the constant, 4.5324, by the figure in the chart we use – so in the example of D/S = 3, we use a correction factor of 0.5 whereas the customers table already has 4.5324 x 0.5 = 2.2662 – so they multiply their V/I reading by 2.2662. Our table corrects the sheet resistance, whereas the customer’s table gives a figure to multiply against V/I.


Question: At what point, as the silicon chunks get smaller, does a correction factor become necessary. Can this be expressed as a volume in cubic centimeters, or does the shape of the pot scrap chuck become important?

Answer: Measurements on a sample for bulk are within 1% if the sample is greater than 5 times the probes spacing in thickness – eg 5mm for a 1mm spacing probe. Sheet resistance measurements would not be valid. The sample should also extend on a plane away from the probe as well, I am not sure at what point correction would be needed, but the pieces obviously do not have to be very big at all to get a reasonable measurement.

 


Four-Point-Probes is a division of Bridge Technology. To request further information please call Bridge Technology at (480) 988-2256 or send e-mail to Larry Bridge at: sales@bridgetec.com