Why am I getting the error "Too many output arguments"?

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Aftab Mohammed
Aftab Mohammed am 3 Mär. 2024
Beantwortet: Walter Roberson am 3 Mär. 2024
%initialise paramteres
A = 1.7;
k = 1.380658*(10^-23);
q = 1.6 * (10^-19);
Eg = 1.1;
Ior = 19.9693*(10^-6);
Iscr = 3.3;
ki = 1.7 * (10^-3);
ns = 40;
np = 2;
Rs = 5*(10^-5);
Rp = 5*(10^5);
Tr = 301.18;
Ta = 25;
%array error = 1;
amb = [25 40 60];
irr = [0.3 0.5 1];
%Evaluate
hold on
for i=1:3
Ipv = [];
Vpv = [];
Ppv = [];
G= irr(i);
Tc = Ta + (0.2*G) + 273.18;
Isc = (Iscr + ki*(Tc-Tr))*G;
Is = Ior * ((Tc/Tr)^3) * exp(q*Eg*(1/Tr-1/Tc)/(k*A));
I = Isc;
V = 0;
while I>0
while error>0.0001
f1= I-(np*Isc)+(np*Is)*(exp(q*((V/ns)+(I*Rs/np))/(A*k*Tc))-1)+(((V*np/ns)+(I*Rs))/(Rp));
df1= 1+(Rs/Rp)+((q*Is*Rs)/(A*k*Tc))*(exp(q*((V/ns)+(I*Rs/np))/(A*k*Tc)));
I= I-(f1/df1);
error=abs(f1/df1);
end
Ipv= [Ipv I];
Vpv = [Vpv V];
Ppv= [Ppv I*V];
V = V+0.1;
error = 1;
end
plot(Vpv, Ipv);
% plot(Vpv, Ppv);
end
Error using error
Too many output arguments.

Akzeptierte Antwort

the cyclist
the cyclist am 3 Mär. 2024
You are confusing MATLAB by using error (a MATLAB function) as a variable name.
I changed that variable name to errorVal, and initialized it to -Inf.
%initialise paramteres
A = 1.7;
k = 1.380658*(10^-23);
q = 1.6 * (10^-19);
Eg = 1.1;
Ior = 19.9693*(10^-6);
Iscr = 3.3;
ki = 1.7 * (10^-3);
ns = 40;
np = 2;
Rs = 5*(10^-5);
Rp = 5*(10^5);
Tr = 301.18;
Ta = 25;
%array error = 1;
amb = [25 40 60];
irr = [0.3 0.5 1];
%Evaluate
errorVal = -Inf;
hold on
for i=1:3
Ipv = [];
Vpv = [];
Ppv = [];
G= irr(i);
Tc = Ta + (0.2*G) + 273.18;
Isc = (Iscr + ki*(Tc-Tr))*G;
Is = Ior * ((Tc/Tr)^3) * exp(q*Eg*(1/Tr-1/Tc)/(k*A));
I = Isc;
V = 0;
while I>0
while errorVal>0.0001
f1= I-(np*Isc)+(np*Is)*(exp(q*((V/ns)+(I*Rs/np))/(A*k*Tc))-1)+(((V*np/ns)+(I*Rs))/(Rp));
df1= 1+(Rs/Rp)+((q*Is*Rs)/(A*k*Tc))*(exp(q*((V/ns)+(I*Rs/np))/(A*k*Tc)));
I= I-(f1/df1);
errorVal=abs(f1/df1);
end
Ipv= [Ipv I];
Vpv = [Vpv V];
Ppv= [Ppv I*V];
V = V+0.1;
errorVal = 1;
end
plot(Vpv, Ipv);
% plot(Vpv, Ppv);
end

Weitere Antworten (2)

Torsten
Torsten am 3 Mär. 2024
Bearbeitet: Torsten am 3 Mär. 2024
You don't initialize "error" in your code by giving it a value before you use it in the while loop.
That's why MATLAB assumes error to be the built-in "error" command:

Walter Roberson
Walter Roberson am 3 Mär. 2024
You do not initialize error before you use it.
error happens to be the name of an important Mathworks function -- a function that is not permitted to return a value and so cannot exist in a form
while error>0.0001
Best would be if you used a different variable name (and made sure to initialize it). Compromise would be if you assigned to error before you use it.

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