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bmulyanto1-Visitor
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January 11, 2019
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Definite integral evaluation (Fuchs-Sondheimer Resistivity model for Nanowires)

  • January 11, 2019
  • 1 reply
  • 3886 views

I really appreciate it if someone can help me to double check the following equation by running my sheet in their Mathcad.

It is because the result that I got is the opposite with the result in the paper where the equation comes from.

 

The decreasing result at small w comes from this part.

image.png

This is my sheet.

image.png

 

The original paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.61.14215

The equation from the paper (phi and psi should be the same): 

image.png

the expected result (solid line):

image.png

Best answer by LucMeekes

Where did you find the value for Rho0 (to be 2.44 uOhm.cm?) and that phi should be dependent on w (and h) as phi=arctan(w/h) ? I think Rho0 would be rather 6.5 uOhm or so (from Fig 4). It's a bad article because it doesn't specify what phi is, and if phi is really atan(w/h), it should have been spelled out.

I constructed the formula for rho0/rho from the article, resulting in:

LM_20190112_Resistivity.png

For values of w<43 nm Mathcad (11) cannot find a converging solution (TOL=10^-7). And you see I did use phi=atan(w/h). Setting phi to a fixed value (somewhere between atan(1) and atan(10)) doesn't change the graph significantly, at least not such that it turns the hockeystick upside down.

So this gives rho0/rho. This means that with (a constant) rho0, then rho as a function of w etc.. will be:

LM_20190112_Resistivity1.png

Sort of what you found.

 

Success!

Luc

1 reply

L
LucMeekes23-Emerald IVAnswer
23-Emerald IV
January 12, 2019

Where did you find the value for Rho0 (to be 2.44 uOhm.cm?) and that phi should be dependent on w (and h) as phi=arctan(w/h) ? I think Rho0 would be rather 6.5 uOhm or so (from Fig 4). It's a bad article because it doesn't specify what phi is, and if phi is really atan(w/h), it should have been spelled out.

I constructed the formula for rho0/rho from the article, resulting in:

LM_20190112_Resistivity.png

For values of w<43 nm Mathcad (11) cannot find a converging solution (TOL=10^-7). And you see I did use phi=atan(w/h). Setting phi to a fixed value (somewhere between atan(1) and atan(10)) doesn't change the graph significantly, at least not such that it turns the hockeystick upside down.

So this gives rho0/rho. This means that with (a constant) rho0, then rho as a function of w etc.. will be:

LM_20190112_Resistivity1.png

Sort of what you found.

 

Success!

Luc

    B
    bmulyanto1-VisitorAuthor
    1-Visitor
    January 12, 2019

    I would like to thank you for helping me double checking the calculation.
    Today I finished the calculation using matlab. The result is the same with mathcad.
    So I can pretty much be sure about the integration and my mathcad installation. 

     

    Rho0 = 2.44 uOhm.cm for Gold. This is a bulk resistivity.
    I got from this.  

    https://aip.scitation.org/doi/10.1063/1.1473868

     image.png

     

    It's a bad article because it doesn't specify what phi is, and if phi is really atan(w/h)

    Yes, I am also confuse. but from the original paper, where the formula comes from, it is said it is "the azimuthal around z axis". I am not sure if I am able to derive the formula from its general form, so I cannot guarantee whether they refer to the same angle. z axis is parallel with cross sectional area vector A.

    https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1950.0107

     image.png

     

    image.png

      

     

    For values of w<43 nm Mathcad (11) cannot find a converging solution (TOL=10^-7). And you see I did use phi=atan(w/h).

    I guess, the formula is valid for w/h>1.xx.

     

     

    This is my matlab result.

    Thank you once again for your help.

     

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