# Graphing the Van Der Pauw Equation

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Snow 2023 年 1 月 31 日

I am trying to write a code to solve the Van Der Pauw Equation (https://en.wikipedia.org/wiki/Van_der_Pauw_method) and then evaulate what happens when R_horizontal= R + R_vertical.
The Van Der Pauw equation is defined as
VDP = exp(-pi*R_h/R_s)+exp(-pi*R_v/R_s)=1 , where R_s is the sheet resistance.
I am not sure how to approach this problem. The only real unknown quantiity would be R_s, but the thing is R_h and R_v can take any value?
I came across this post, but this solves for specific values?
I thought maybe defining
R_h = linspace(1,100)
R_v = linspace(1,100)
then writing a for loop for when R_h ~=R_v I input those values into VDP. But then I still wouldnt have R_s?
Does anyone have some insight on how I should approach this?
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Rik 2023 年 1 月 31 日
What exactly do you want to solve? It sounds like you want to plot Rs versus some other variable, but I don't fully understand which.
Another point of clarification: can you confirm the R in your equation is Rs?

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### 回答 (1 件)

Sai 2023 年 2 月 27 日
I understand that you wanted to solve for sheet resistance (‘R_s’) in Van Der Pauw equation.
If there is only one equation, there should be only one unknown variable in it for which we have to solve.
I am not sure of the values ‘R_h’ and ‘R_v’ can take, but you can use range of values using ‘linspace()’ so that ‘R_s’ results in a vector.
R_h = linspace(1,100);
R_v = linspace(1,100);
k = length(R_h);
for i = 1:k
vdp = @(R_s) (exp(-pi*R_h(i)/R_s) + exp(-pi*R_v(i)/R_s));
R_s(i) = fsolve(vdp,1);
end
disp(R_s)
The above code works and results in ‘R_s’ as a vector
Please refer to fsolve and fzero for future reference

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