Graphing the Van Der Pauw Equation

Question

Snow 2023-1-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1903565-graphing-the-van-der-pauw-equation

回答： Sai 2023-2-27

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?

https://www.mathworks.com/matlabcentral/answers/462681-van-der-pauw-equation

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 2023-1-31

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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Answer 1

Sai 2023-2-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1903565-graphing-the-van-der-pauw-equation#answer_1180530

在 MATLAB Online 中打开

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)