Solving for two variable.

I have:

clearclc

x=[7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];

y=[0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];

but I need do find n and I0 from:

I = I0 * e^( (q*U)/(n*k*T) )

I already know q, U, k and T.

[Alan Stevens]

What are the equivalents of x and y in your equation? Presumably, y represents I. What does x represent?

[Sebastian]

x=I and y=U

Answer 1

[Alan Stevens]

In that case one way is to take logs of both sides to get:

log(I) = log(I0) + q/(n*k*T)*U

then do a best-fit straight line to the data (use log(x)) and get log(I0) from the intercept and q/(n*k*T) from the slope, from which yoiu can then get I0 and n.

[Sebastian]

I started with matlab yesterday so I need help with how I would do that?

clear

clc

x=[7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];

y=[0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];

p=polyfit(x,y,1);

f=polyval(p,x);

plot(x,y,'o',x,f,'-');

[Alan Stevens]

Like this

x=[7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];
y=[0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];

% You said x = I and U = y so

p=polyfit(y,log10(x),1);
f=polyval(p,y);
plot(y,log10(x),'o',y,f,'-'), grid
xlabel('U'),ylabel('logI')

% Intercept is p(2), slope is p(1)
I0 = 10^p(2);
q_on_nkT = p(1);  % You need to rearrange this to get n, using
                  % your known values for q, k and T
                  
disp(I0)

   2.4861e-10

disp(q_on_nkT)

   11.0383

[Sebastian]

I have q = 1.60*10^(-19), k=1.38*10^(-23) and t=300

I get n = 1.5966

using n=(q*y)/(k*t*(log(x)-log(I0)))

the correct n is 1.521

[Torsten]

x=[7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];
y=[0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];

% You said x = I and U = y so

p=polyfit(y,log(x),1);
f=polyval(p,y);
figure(1)
plot(y,log(x),'o',y,f,'-'), grid
xlabel('U'),ylabel('logI')

% Intercept is p(2), slope is p(1)
I0 = exp(p(2));
q_on_nkT = p(1);  % You need to rearrange this to get n, using
                  % your known values for q, k and T
q = 1.60*10^(-19);
k = 1.38*10^(-23);
T = 300;
n = q/(k*T*q_on_nkT);
disp(I0)

   2.4861e-10

disp(n)

    1.5206

figure(2)
plot(y,x,'o',y,I0*exp(q/(k*T)*y/n),'-'), grid
xlabel('U'),ylabel('I')

Answer 2

[Torsten]

I = [7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];
U = [0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];
q = 1.60*10^(-19);
k = 1.38*10^(-23) ;
T = 300;
value = q/(k*T);
fun = @(I0,n) I - I0 * exp( value * U / n );
p0 = [1 ; 10];   % Initial guess for I0 and n
options = optimset('TolX',1e-10,'TolFun',1e-10,'MaxFunEvals',100000,'MaxIter',100000);
sol = lsqnonlin(@(p)fun(p(1),p(2)),p0,[],[],options);

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

format long
I0 = sol(1)

I0 = 
     4.840384083194344e-10

n = sol(2)

n = 
   1.570628124624527

hold on
plot(U,I,'o')
plot(U,I0 * exp( value * U / n ))
grid
hold off

[Sebastian]

sorry, I don't have lsqnonlin

[Torsten]

I = [7.53*10^(-5) 3.17*10^(-4) 1.07*10^(-3) 3.75*10^(-3) 1.35*10^(-2) 4.45*10^(-2) 1.75*10^(-1) 5.86*10^(-1)];
U = [0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85];
q = 1.60*10^(-19);
k = 1.38*10^(-23) ;
T = 300;
value = q/(k*T);
fun = @(I0,n) I - I0 * exp( value * U / n );
p0 = [1 ; 10];   % Initial guess for I0 and n
options = optimset('TolX',1e-10,'TolFun',1e-10,'MaxFunEvals',100000,'MaxIter',100000);
sol = fminsearch(@(p)sum(fun(p(1),p(2)).^2),p0,options);
format long
I0 = sol(1)

I0 = 
     4.840330426445296e-10

n = sol(2)

n = 
   1.570627287586739

hold on
plot(U,I,'o')
plot(U,I0 * exp( value * U / n ))
grid
hold off

[Sebastian]

I0 and n are both wrong.

I0 = 0.249*10^(-9)

and

n = 1.521

[Torsten]

No, you are wrong.

Applying log to your equation distorts the fitting.

You must fit I0*exp(value * U / n) against U to get unbiased estimates for your parameters.

Fitting log(I0) + value/n * U against log(U) only gives an approximation for I0 and n.