Solving a system of 5 dependent non-linear equations

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I am trying to solve the system of 5 non-linear equations:
Io * exp(Voc/ (m*Vt)) + Voc/Rsh - Is = 0
(Isc - (Voc - Rs*Isc)/ Rsh) * exp(-Voc/ (m*Vt)) - Io = 0
Isc - (Vmp + Rs*Imp + Rs*Isc)/ Rsh - (Isc - (Voc - Rs*Isc)/ Rsh) * exp((Vmp + Rs*Imp - Voc)/ (m*Vt)) - Imp = 0
Imp + ((-(Rsh*Isc - Voc + Rs*Isc) * exp((Vmp + Rs*Imp - Voc)/ (m*Vt))/ (Rsh * m*Vt) - 1/Rsh)/ (1 + Rs * (Rsh*Isc - Voc + Rs*Isc) * exp((Vmp + Rs*Imp - Voc)/ (m*Vt))/ (Rsh * m*Vt) + Rs/Rsh)) * Vmp = 0
(-(Rsh*Isc - Voc + Rs*Isc) * exp((Rs*Isc - Voc)/ (m*Vt))/ (Rsh * m*Vt) - 1/Rsh)/ (1 + Rs * (Rsh*Isc - Voc + Rs*Isc) * exp((Rs*Isc - Voc)/ (m*Vt))/ (Rsh * m*Vt) + Rs/Rsh) + 1/Rsh = 0
in which m, Rsh, Rs, Io and Is are the unknown variables. The other variables have the values:
Vmp = 31.1;
Imp = 8.67;
Voc = 38.2;
Isc = 9.19;
K = 1.38 * 10^-23;
T = 298.15;
q = 1.6 * 10*-19;
Vt = K*T/q;
I know the solution must be around m=62.3, Rsh=786, Rs=0.2748, Io=407.33, Is=9.1932
I have tried using the function fsolve as shown below
syms m Rsh Rs Io Is
V = [m, Rsh, Rs, Io, Is];
F = @(V) [V(4) * exp(Voc/ (V(1)*Vt)) + Voc/V(2) - V(5);
(Isc - (Voc - V(3)*Isc)/ V(2)) * exp(-Voc/ (V(1)*Vt)) - V(4);
Isc - (Vmp + V(3)*Imp + V(3)*Isc)/ V(2) - (Isc - (Voc - V(3)*Isc)/ V(2)) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt)) - Imp;
Imp + ((-(V(2)*Isc - Voc + V(3)*Isc) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) - 1/V(2))/ (1 + V(3) * (V(2)*Isc - Voc + V(3)*Isc) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) + V(3)/V(2))) * Vmp;
(-(V(2)*Isc - Voc + V(3)*Isc) * exp((V(3)*Isc - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) - 1/V(2))/ (1 + V(3) * (V(2)*Isc - Voc + V(3)*Isc) * exp((V(3)*Isc - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) + V(3)/V(2)) + 1/V(2)];
InitialGuess = [50; 100; 1; 100; 10];
XY = fsolve(F, InitialGuess);
but I get the Error using trustnleqn (line 28). Objective function is returning undefined values at initial point. FSOLVE cannot continue. error message. Can someone tell how they would solve this problem?
  2 Comments
Frederico Lourenço
Frederico Lourenço on 2 Mar 2018
I'm sorry, I understand your frustration, I really don't know much about how most of the functions work, that's why I came here. Maybe someone would be able to explain to me what I was doing wrong.
Anyway, that unimaginably small number that you refer was due to a typo in the value of q. The real value for Vt is 0.0257, and not -1.3534e-23, which I thank you for pointing to me. I already corrected that and since then I no longer get the undefined values that crash fsolve.
My problem now, like I mentioned in your answer, is that the function just does not converge and as you said I don't understand what I'm doing wrong since I'm following the explanation in the link of the MathWorks Support Team.
You say that I should not use fsolve. Well my original question was how would someone solve the problem I presented, solve a system of 5 dependent non-linear equations.

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Accepted Answer

John BG
John BG on 2 Mar 2018
Hi Frederico
1.
Actually, command fsolve works pretty well here, let me explain:
Since V(5) unknown only shows up in the 1st equation I decided to put off the 1st equation, let's solve the remaining system with fsolve
clear all;clc;close all
syms m Rsh Rs Io Is
Vmp = 31.1;
Imp = 8.67;
Voc = 38.2;
Isc = 9.19;
K = 1.38 * 10^-23;
T = 298.15;
q = -1.6 * 10^-19;
Vt = K*T/q;
V = [m, Rsh, Rs, Io];
F = @(V) [ (Isc - (Voc - V(3)*Isc)/ V(2)) * exp(-Voc/ (V(1)*Vt)) - V(4);
Isc - (Vmp + V(3)*Imp + V(3)*Isc)/ V(2) - (Isc - (Voc - V(3)*Isc)/ V(2)) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt)) - Imp;
Imp + ((-(V(2)*Isc - Voc + V(3)*Isc) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) - 1/V(2))/ (1 + V(3) * (V(2)*Isc - Voc + V(3)*Isc) * exp((Vmp + V(3)*Imp - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) + V(3)/V(2))) * Vmp;
(-(V(2)*Isc - Voc + V(3)*Isc) * exp((V(3)*Isc - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) - 1/V(2))/ (1 + V(3) * (V(2)*Isc - Voc + V(3)*Isc) * exp((V(3)*Isc - Voc)/ (V(1)*Vt))/ (V(2) * V(1)*Vt) + V(3)/V(2)) + 1/V(2)];
InitialGuess = [50; 0.1; .01; 1];
options = optimoptions('fsolve','Display','none','PlotFcn',@optimplotfirstorderopt,'MaxFunctionEvaluations',2000);
XY = fsolve(F, InitialGuess,options);
XY =
51.755945628868020
3.071518542695716
1.085173513887406
-1.062922141399835
2.
These are semiconductor equations, I like reading what is what, so I am putting back the initial variable names
m=XY(1)
Rsh=XY(2)
Rs=XY(3)
Io=XY(4)
3.
Now the initial equation is trivial
Is= Io * exp(Voc/ (m*Vt) )+ Voc/Rsh
Is =
12.436844990189343
4.
fsolve even produces the optimality graph ending up in 3.5e9 quite high.
5.
I was wondering, should your 1st equation have an inverted polarity, then Is would be
Is= Io * exp(-Voc/ (m*Vt) )+ Voc/Rsh
Is =
-3.101249033198816e+12
6.
Please note that function fsolve has a default amount if iterations equal to 400. If not changed, for this equations system, fsolve completes all the loops, without a solution, but warns that it may have a solution if increasing such default value.
To so, just change in advance the options field
'MaxFunctionEvaluations'
to for instance
2000
.
Frederico
Cosi fan tutte :)
If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?
To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link
thanks in advance
John BG
  3 Comments
Frederico Lourenço
Frederico Lourenço on 3 Mar 2018
Sorry, I said stable and I meant convergent. It is an equivalent model of an electrical circuit with a diode and two resistors, no oscillatory behavior

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More Answers (3)

John D'Errico
John D'Errico on 2 Mar 2018
Maybe you THINK you know where the solution lies. But if you are correct, you will essentially NEVER be able to solve this problem using tools like fsolve.
I'll use symbolic tools here.
syms m Rsh Rs Io Is
E(1) = Io * exp(Voc/ (m*Vt)) + Voc/Rsh - Is == 0;
E(2) = (Isc - (Voc - Rs*Isc)/ Rsh) * exp(-Voc/ (m*Vt)) - Io == 0;
Etc.
We can stop at equation 1 in fact.
pretty(vpa(E(1),5))
/ 24 \
| 2.8224 10 | 38.2
Io exp| - ----------- | - 1.0 Is + ---- == 0.0
\ m / Rsh
You claim that m is around 62.3.
So that would have you compute exp(-2.8224e24/62.3). The number is so small that it is effectively zero, even when using symbolic or my own HPF tools.
In fact, every place where you have m, you divide it into a number on the order of 1e23 or so, then you exponentiate. So m is irrelevant. It results in an exponential that is zero.
pretty(vpa(E(2),5))
/ 24 \
| 2.8224 10 | / 9.19 Rs - 38.2 \
exp| ----------- | | -------------- + 9.19 | - 1.0 Io == 0.0
\ m / \ Rsh /
>> pretty(vpa(E(3),5))
/ 22 \
(17.86 Rs + 31.1) 1.0 | (8.67 Rs - 7.1) 7.3886 10 | / 9.19 Rs - 38.2 \
0.52 - --------------------- - exp| - --------------------------- | | -------------- + 9.19 | 1.0 == 0.0
Rsh \ m / \ Rsh /
So, even if you are calling fsolve wrongly, you are wasting your time. In fact, you are wasting your time trying to solve this even with symbolic tools, IF m is on the order of 60. Unless m was on the order of 1e20 or more, you will have no chance of solving this. And even then, fsolve would be a complete waste of time.
  1 Comment
Frederico Lourenço
Frederico Lourenço on 2 Mar 2018
First of all, thanks for your fast answer! What you say is indeed true, I had a typo in the value of q. It was suppose to be 1.6 * 10^(-19), and not 1.6 * 10*-19.
The error went away when I changed that but unfortunately I believe I still have a mistake because when I do the line
ShouldBeZero = F(XY)
ShouldBeZero =
-2.2483
0.0165
-3.0388
0.8064
-0.0109
The program only stops because of this
fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 500 (the default value).
and not because it has reached results within the tolerance of the function. Do you know why this happens?

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Alex Sha
Alex Sha on 2 Feb 2020
Edited: Alex Sha on 2 Feb 2020
Hi, if vt = 0.0257, the solution should be:
m: 512.891368076595
rsh: -1.08505661410065
rs: 3.40678689852078
io: 0.856812866408915
is: -19.6641364483436
Fevl:
3.80140363631654E-13
1.59949831157746E-12
4.13891143580258E-13
-2.0961010704923E-13
3.41060513164848E-13
while, if Vt = K*T/q=-1.35344407894737E-23, the solution will be:
m: -9.73908591023589E23
rsh: -1.08505661410068
rs: 3.40678689852086
io: 0.856812866410473
is: -19.6641364483141
Fevl:
1.4210854715202E-14
1.11022302462516E-16
-8.34887714518118E-14
-1.24344978758018E-14
-1.55431223447522E-15
  1 Comment
Frederico Lourenço
Frederico Lourenço on 4 Feb 2020
It's been two years since I posted this and got the answer, but thank you for your comment.

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Alex Sha
Alex Sha on 4 Feb 2020
My pleasure, it is actually very challenging problem, hard to be solved by common software or algorithms.

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