polynomial fit for a schottky diode and evaluation of its characteristics (ideality factor, barrier height and I0)

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federico
federico 2023-8-29
评论： Star Strider 2023-9-15
i have a series of data taken in lab of a Schottky type of diode and i have to evaluate them neglecting the series resistance even though it can't be ignored.
i'm using this code to plot the I vs V specs of the sample at each temperature
clear
%Importare dati, indicare percorso file dati completo
load ST_n_293K.txt;
xdata=ST_n_293K(:,1); % v (riga, colonna)
ydata=ST_n_293K(:,2); % i (riga, colonna)
%plot dati
plot(xdata,ydata,'ro');
title('fit I vs V a 293K')
xlabel('V [V]');
ylabel('I [mA]');
then i proceed with the fitting app using y=a*(exp(x/b)-1) as equation but it seems like the fit is not good enough.
i'm using this a s reference https://zenodo.org/record/1085918 and in particular the first method at page 2, if i do as the article says i'm getting values way higher, so my question is:
am i applying it wrong or because of the diode manufacture the values has to be this high?
  1 个评论
federico
federico 2023-8-29
i've now tried using this equation a*log(1+exp(b*x))-c and the fit is clearly better than before, but i still can't manage how to get them value from the semilog plot of I vs V.
i got the a,b,c parameters from iv fit, but stil can't figure out how to find the other parameters, does somebody have suggestions?
my previous fit led me to n close to ..*10^2 now should be way lesser, it should be higher than standards but not lower than 1

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采纳的回答

Star Strider
Star Strider 2023-8-29
Ran in:
It is relatively straightforward to do a nonlinear parameter estimate of the required parameters in MATLAB.
Try this —
T1 = readtable('ST_n_293K.txt');
T1.Properties.VariableNames = {'V','I'}
T1 = 53×2 table
V I ____ ________ -4 -0.16481 -3.9 -0.16405 -3.8 -0.16281 -3.7 -0.16177 -3.6 -0.16063 -3.5 -0.15949 -3.4 -0.15867 -3.3 -0.15721 -3.2 -0.15607 -3.1 -0.1549 -3 -0.15379 -2.9 -0.15227 -2.8 -0.15114 -2.7 -0.14962 -2.6 -0.14848 -2.5 -0.14696
xdata = T1.V;
ydata = T1.I;
Lv = xdata > 0;
% y=a*(exp(x/b)-1)
objfcn = @(b,x) b(1).*(exp(x./b(2))-1);
B = fminsearch(@(b) norm(ydata(Lv) - objfcn(b,xdata(Lv))), rand(2,1)) % Use 'fminsearch' To Estimate The Parameters
B = 2×1
381.8461 39.1460
xv = linspace(min(xdata(Lv)), max(xdata(Lv)));
figure
semilogy(xdata(Lv), ydata(Lv), '.')
hold on
plot(xv, objfcn(B, xv), '-r')
hold off
grid
xlabel('V')
ylabel('I')
text(0.5, 2, sprintf('$I = %.2f (e^{\\frac{V}{%.2f}}-1)$',B), 'Interpreter','latex', 'FontSize',14)
To get statistics on the fit, use fitnlm instead. Also see the documentation on NonLinearModel for details.
.
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显示 7更早的评论
federico
federico 2023-9-15
the thing i'm trying to do is to plot Ln(I) vs V and then evaluate the parameters in the linear fit of values between V=0 and V=1 to demonstrate that this model doesn't work neglecting series resistance.
Star Strider
Star Strider 2023-9-15
Ran in:
I am still not certain that I understand what you want.
Perhaps this —
T1 = readtable('ST_n_293K.txt');
T1.Properties.VariableNames = {'V','I'}
T1 = 53×2 table
V I ____ ________ -4 -0.16481 -3.9 -0.16405 -3.8 -0.16281 -3.7 -0.16177 -3.6 -0.16063 -3.5 -0.15949 -3.4 -0.15867 -3.3 -0.15721 -3.2 -0.15607 -3.1 -0.1549 -3 -0.15379 -2.9 -0.15227 -2.8 -0.15114 -2.7 -0.14962 -2.6 -0.14848 -2.5 -0.14696
xdata = T1.V;
ydata = T1.I;
Lv = xdata > 0 & xdata <= 1;
xv = linspace(min(xdata(Lv)), max(xdata(Lv)));
BL = [xdata(Lv) ones(size(xdata(Lv)))] \ log(ydata(Lv)) % Regression: log(y) = m*x + b
BL = 2×1
2.5868 0.0258
figure
plot(xdata(Lv), log(ydata(Lv)), '.')
hold on
plot(xv, [xv(:) ones(size(xv(:)))] * BL, '-r')
hold off
grid
xlabel('V')
ylabel('log(I)')
title('Linear Fit')
text(0.5, 0.5, sprintf('$log(I) = %.2f V %+.2f$',BL), 'Interpreter','latex', 'FontSize',14)
.

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更多回答（1 个）

Sulaymon Eshkabilov
Ran in:
If understood your question correctly, here is how it can be simulated:
D=load('ST_n_293K.txt') ;
xdata=D(:,1); % v (riga, colonna)
ydata=D(:,2); % i (riga, colonna)
%plot dati
semilogy(xdata,ydata,'ro');
title('fit I vs V a 293K')
xlabel('V [V]');
ylabel('I [mA]');
IDX = find(ydata>0);
V = xdata(IDX);
I = ydata(IDX);
ft = fittype('a*log(1+exp(b*x))-c')
ft =
General model: ft(a,b,c,x) = a*log(1+exp(b*x))-c
[Fmodel,Fmodel_quality] = fit(V,I,ft)
Warning: Start point not provided, choosing random start point.
Warning: Negative data ignored
Fmodel =
General model: Fmodel(x) = a*log(1+exp(b*x))-c Coefficients (with 95% confidence bounds): a = -186.6 (-573.7, 200.4) b = -0.1105 (-0.3471, 0.126) c = -129.2 (-397.6, 139.2)
Fmodel_quality = struct with fields:
sse: 0.1074 rsquare: 0.9993 dfe: 10 adjrsquare: 0.9991 rmse: 0.1036
hold on
I_estimated = Fmodel(V);
plot(V, I_estimated,'k', 'linewidth', 2)
legend('Data','FitModel')
grid on
hold off
[V, I_estimated]
ans = 13×2
0 -0.1595 0.0900 0.7664 0.1800 1.6878 0.2700 2.6046 0.3600 3.5167 0.4500 4.4242 0.5500 5.4271 0.6400 6.3249 0.7300 7.2181 0.8200 8.1066
  2 个评论
federico
federico 2023-8-29
not quite right, i did the same thing just by using the fitting app on matlab using exclusion rule and excluding x<0.
the problem here is based on the document i linked, how to find the paramenters such as n, Φ and i0 from this graph since i don't get it at all.
fig.3 and 4 are, supposedly the same thing zoomed in, in which the researcher are using the least squared fitted option to evaluate those parameters from the plot.
federico
federico 2023-8-29
the article says " i0 the reverse saturation current can be extracted by extrapolating the straight line on lnI to intercept the axis at 0 voltage"
so for the firts method they found that i0, if i'm not mistaken, is around 3,748*10^-7A but i can't seem to understand how to found it in my plot

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