What is the significance of using U/2I in four point probe theory?

ScieneShines
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Dear Ladys and Gentlemans,

I want to measure the sheet resistance R_square on a printed product.
My measuring equipment consists a SMU to supply and measure and a four point Probe by Jandel.
As output size I get the electrical resistance R (basically the measured Voltage U divided by the supply current I)
So I have to multiply the geometric factor and a correction factor (correction factor because the width is small)

The derivation for the geometric factor I have found on :
http://www.four-point-probes.com/four-point-probe-...

And the correction factor table is shown here, in my case its on page 54
https://www.iiserkol.ac.in/~ph324/StudyMaterials/G...

Funny but true, I get twice the value than in the reference measurement system (eddy current technology)
So may we can discuss why?
And the bigger Question: why is the resistance calculated in the derivation with U/2I.
In my opinion there is no superposition of currents because they have a different sign.
Thank you in advance,
ScieneShines
 
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Well, since you didn't show any of your work, we would have no idea where any error lies.

Now my disclaimer. Not my area of expertise, so take anything I mention merely as a 'second pair of eyes.'

From doing a spot read/scan of the links you supplied, you may have mis-applied the geometry correction. See especially pgs 4 thru 7 of the PDF you referenced. That seems to answer your 'why the factor of two' question and also cover the 'factor of two' error you seem to be getting.

Hopefully the above is at least a place to start!

Cheers,
Tom
p.s. Since this is a schoolwork question, I have requested it be moved to the Homework forum.
 
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First things first : Thank you for your answer and no this is definitivly not a Schoolwork, Homework or anything else... Did you do those things in school? Respect for that.

I'm 100% sure that i used the correct correction factor.
Symmetry condition are given.
 
Welcome to the PF. :smile:
ScieneShines said:
I'm 100% sure that i used the correct correction factor.
I agree with @Tom.G that we need to see your work to be able to help you.
 

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