Solving for two variable.

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Sebastian
Sebastian on 10 Sep 2022
Edited: Torsten on 10 Sep 2022
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.
  2 Comments
Sebastian
Sebastian on 10 Sep 2022
x=I and y=U

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Accepted Answer

Alan Stevens
Alan Stevens on 10 Sep 2022
Edited: Alan Stevens on 10 Sep 2022
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.
  4 Comments
Torsten
Torsten on 10 Sep 2022
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')

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More Answers (1)

Torsten
Torsten on 10 Sep 2022
Edited: Torsten on 10 Sep 2022
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
  4 Comments
Torsten
Torsten on 10 Sep 2022
Edited: Torsten on 10 Sep 2022
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.

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