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An approximate solution to the Van der Pauw equation

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    question: approximate the function f if the ratio R_1 / R_2 is about 1000.

    given:
    the van der pauw equation for resistivity:
    \rho = ( pi * d / ln[2] ) * (R_1 + R_2})/2 * f(R_1 / R_2),

    where f is a function of the ratio R_1/R_2 only and satisfies the relation:

    (R_1- R_2)/(R_1+R_2) = f arccosh{exp(ln[2]/f) / 2}.

    from van der pauw's 1958 paper:if R_1 and R_2 are almost equal, f can be approximated by the formula

    f ~ 1 - [ (R_1 - R_2) / (R_1 + R_2) ]^2 * ln[2] / 2 - [(R_1 - R_2)/(R_1 + R_2)]^4 * {(ln[2])^2 / 4 - (ln[2])^3 / 12}.

    2. Relevant equations
    -
    3. The attempt at a solution
    I have no idea how to start this.
     
  2. jcsd
  3. Oct 12, 2009 #2

    gabbagabbahey

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    That approximation doesn't seem like it would be very accurate when [itex]R_1=1000R_2[/itex]!:wink:


    What does [itex]\frac{R_1-R_2}{R_1+R_2}[/itex] equal when [itex]\frac{R_1}{R_2}=1000[/itex]?
     
  4. Oct 12, 2009 #3
    What does [itex]\frac{R_1-R_2}{R_1+R_2}[/itex] equal when [itex]\frac{R_1}{R_2}=1000[/itex]?[/QUOTE]

    It's 999/1001. But I don't know how to apply that to find an approximate solution. How do you go about finding an approximate solution to that, or any other function?
     
  5. Oct 12, 2009 #4
    I can't seem to solve the equation explicitly for f (I think it's transcendental). Can I still use the Taylor series method?
     
  6. Oct 12, 2009 #5

    gabbagabbahey

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    Yes, it is transcendental, but I think you can still use a Taylor series. However, at the moment I can't think of exactly what expression your going to want to expand to make things easiest....I'll give it some more thought in the morning, but in the meantime hopefully someone else will jump in.
     
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