Help needed for below error

Kindly help with the below error for the below mentioned code. All other values are constant
scale1 = 100;
T1 = linspace(10, 2000, scale1);
Ta = rdivide(1,T1);
ND1 = 1e17;
NA1 = 1e5;
%Computing n and p values to be plotted
hole1 = ones(scale1,1)*0;
Eflevel1 = ones(scale1,1)*0;
electron1 = ones(scale1,1)*0;
NC1 = 2*Mc*(((mde*kb*T1(1:100))/(2*pi*(hbar^2))).^(3/2));
NV1 = 2*(((mdh*kb*T1(1:100))/(2*pi*(hbar^2))).^(3/2));
kbT1 = (kb*T1(1:100))/e;
ev1 = 0;
ea1 = ev1 + Eai;
eg1 = 1.17 - (4.73e-4*((T1(1:100)).^2))./(T1(1:100) + 636);
ec1 = eg1;
ed1 = ec1 + Edi;
x2 = [0 1.73];
for range1 = 1:prod(size(T1))
eq2 = @(EF1) ((NC1(range1))*exp(-(ec1(range1)-EF1)/kbT1(range1))) + ((NA1)/(1+4*exp(-(EF1-ea1(range1))/kbT1(range1)))) - (((NV1(range1))*exp(-(EF1-ev1(range1))/kbT1(range1))) + ((ND1)/(1+2*exp(-(ed1(range1)-EF1)/kbT1(range1)))));
Eflevel1(range1) = fsolve(eq2, x2);
y1 = Eflevel1(range1);
x2 = y1.';
s11 = Eflevel1(:,1,:);
electron1(range1) = (NC1(range1))*exp(-(ec1(range1)-Eflevel1(range1))/(kbT1(range1)));
s12 = electron1(:,1,:);
hole1(range1) = (NV1(range1))*exp(-(Eflevel1(range1)-ev1(range1))/(kbT1(range1)));
s13 = hole1(:,1,:);
end

3 Kommentare

John D'Errico
John D'Errico am 16 Mär. 2022
Amy Topaz
Amy Topaz am 16 Mär. 2022
Sorry about that, I have modified it. I didn't mean in that sense.
Rik
Rik am 29 Mär. 2022
I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.
The page has now been archived.

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Antworten (2)

Walter Roberson
Walter Roberson am 16 Mär. 2022

0 Stimmen

x2 = [0 1.73];
row vector
Eflevel1(range1) = fsolve(eq2, x2);
vector is used as the example parameters for fsolve, so eq2 is going to be passed something the same length as x2.
eq2 = @(EF1) ((NC1(range1))*exp(-(ec1(range1)-EF1)/kbT1(range1))) + ((NA1)/(1+4*exp(-(EF1-ea1(range1))/kbT1(range1)))) - (((NV1(range1))*exp(-(EF1-ev1(range1))/kbT1(range1))) + ((ND1)/(1+2*exp(-(ed1(range1)-EF1)/kbT1(range1)))));
That vector becomes EF1 on input. You use that vector EF1 in several places, including in
/(1+4*exp(-(EF1-ea1(range1))/kbT1(range1)))
so the right hand side of the / is a row vector.
But remember that the / operator is matrix-right-divide, mrdivide, with A/B being similar to A*pinv(B) except with some size constraints. And the size constraints are such that row vector / row vector is not permitted .

6 Kommentare

Amy Topaz
Amy Topaz am 16 Mär. 2022
oh got it, thank you for your explanation. Do you know how can I modify it so that it is corrected?
Walter Roberson
Walter Roberson am 16 Mär. 2022
No... you haven't defined how it should work, so I cannot recommend a correction.
If you switch to element-by-element processing using ./ instead of / then your eq2 would return a vector of two elements. fsolve() can deal with a vector of two elements being returned: it tries to adjust the values so that both values become 0. However, since nothing else in your expression modifies the expression differently for the two elements, it follows that the same value (0) would be returned if you managed to find a 0 for one of the elements and then set the other element to the same value. So at present, the calculations cannot differentiate between the two values passed in to the function.
Notice you are assigning the result of fsolve() to the scalar location Eflevel1(range1), implying that you expect fsolve() to return a scalar. But fsolve() is always going to return a non-scalar whenever the example input (x2) is non-scalar, since fsolve() returns the array of model coefficients that lead the function to have all-zero output.
Amy Topaz
Amy Topaz am 16 Mär. 2022
oh okay.
Can we use fzero instead of fsolve?
Amy Topaz
Amy Topaz am 16 Mär. 2022
I basically need all the values of EF for the given range so that I can plot a graphs between Ta and s11
Walter Roberson
Walter Roberson am 16 Mär. 2022
fzero can only deal with a single variable, not a vector of variables, so it would be an error to use fzero with that x2.
Amy Topaz
Amy Topaz am 16 Mär. 2022
okay, so how shall I proceed?

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Torsten
Torsten am 16 Mär. 2022

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Since it seems that you only solve for a single unknown, change
x2 = [0 1.73];
to
x2 = 1.73;
Then your code should be
for range1 = 1:prod(size(T1))
eq2 = @(EF1) NC1(range1)*exp(-(ec1(range1)-EF1)/kbT1(range1)) + NA1/(1+4*exp(-(EF1-ea1(range1))/kbT1(range1))) - (NV1(range1)*exp(-(EF1-ev1(range1))/kbT1(range1)) + (ND1/(1+2*exp(-(ed1(range1)-EF1)/kbT1(range1)))));
Eflevel1(range1) = fsolve(eq2, x2);
x2 = Eflevel1(range1);
end
y1 = Eflevel1;
x2 = y1.';
s11 = Eflevel1;
electron1 = NC1.*exp(-(ec1-Eflevel1)./kbT1);
s12 = electron1;
hole1= NV1.*exp(-(Eflevel1-ev1)./kbT1);
s13 = hole1;

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Gefragt:

am 16 Mär. 2022

Kommentiert:

Rik
am 29 Mär. 2022

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