# 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?

### Accepted Answer

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

### More Answers (3)

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.

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

Alex Sha
on 4 Feb 2020

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