Solving for two variable.

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Sebastian
Sebastian 2022-9-10
编辑： Torsten 2022-9-10
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.
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Alan Stevens
Alan Stevens 2022-9-10
编辑：Alan Stevens 2022-9-10
What are the equivalents of x and y in your equation? Presumably, y represents I. What does x represent?

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采纳的回答

Alan Stevens
Alan Stevens 2022-9-10
编辑：Alan Stevens 2022-9-10
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.
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Sebastian
Sebastian 2022-9-10
编辑：Sebastian 2022-9-10
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
Torsten 2022-9-10
Ran in:
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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更多回答（1 个）

Torsten
Torsten 2022-9-10
编辑：Torsten 2022-9-10
Ran in:
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
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Sebastian
Sebastian 2022-9-10
编辑：Sebastian 2022-9-10
I0 and n are both wrong.
I0 = 0.249*10^(-9)
and
n = 1.521
Torsten
Torsten 2022-9-10
编辑：Torsten 2022-9-10
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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