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

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?

[federico]

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

Answer 1

[Star Strider]

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.

.

[federico]

so https://it.mathworks.com/matlabcentral/answers/2014371-polynomial-fit-for-a-schottky-diode-and-evaluation-of-its-characteristics-ideality-factor-barrier#comment_2865296 this straight line that they use is actually not as intended and should result more like the one you are suggesting?

or am i forgetting something?

i feel like the article has some catch that i can't figure out.

[Star Strider]

My plot is essentially their ‘Fig.3’. A straight line (their ‘Fig.4’) fit would be a lilnear fit to those data, and that actually works reasonably well.

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

381.8467 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')
title('Nonlinear Fit')
text(0.5, 2, sprintf('$I = %.2f (e^{\\frac{V}{%.2f}}-1)$',B), 'Interpreter','latex', 'FontSize',14)

BL = [xdata(Lv) ones(size(xdata(Lv)))] \ ydata(Lv)

BL = 2×1

10.0648 -0.1544

figure
plot(xdata(Lv), ydata(Lv), '.')
hold on
plot(xv, [xv(:) ones(size(xv(:)))] * BL, '-r')
hold off
grid
xlabel('V')
ylabel('I')
title('Linear Fit')
text(0.5, 2, sprintf('$I = %.2f V %+.2f$',BL), 'Interpreter','latex', 'FontSize',14)

.

[federico]

tried on matlab and it gives me this error for every temperature, do you know why?

with the run option in here it does not

[federico]

solved, i had to run by section, and not the full script, thanks

[Star Strider]

As always, my pleasure!

That error had to do with th file names. That is not something I could solve without more information.

[federico]

while i was watchin the whole thesis i came across a doubt about the linear fit we discussed and found out that this linear fit is not of Ln(i) vs V, since values between 1 and 0 should be negative.

i modified it in order to go from V==0 to 1.5V values for all temperatures but when i try to modify it's axis i can't seem to figure out what i'm doin wrong, do u mind if i ask you to help me find a solution?

also the model for custom eq to fit this value is: a*(exp(x*b)-c)

[Star Strider]

I do not understand what you want to do. I specifically do not know what to do about ‘values between 1 and 0 should be negative’ since they do not appear to be.

This uses all the data, and an appropriately-modified ‘objfcn’ —

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

format long
xdata = T1.V;
ydata = T1.I;
% Lv = xdata > 0;                                                           % Estimate Using All The Data
%  y=a*(exp(x/b)-c)
objfcn = @(b,x) b(1).*(exp(x./b(2))-b(3));
% B = fminsearch(@(b) norm(ydata(Lv) - objfcn(b,xdata(Lv))), rand(3,1))       % Use 'fminsearch' To Estimate The Parameters
B = fminsearch(@(b) norm(ydata - objfcn(b,xdata)), rand(3,1))       % Use 'fminsearch' To Estimate The Parameters

B = 3×1

1.778669403038519 0.560973868192308 0.192130984472511

% xv = linspace(min(xdata(Lv)), max(xdata(Lv)));
xv = linspace(min(xdata), max(xdata));

figure
% semilogy(xdata(Lv), ydata(Lv), '.')
plot(xdata, ydata,'.')
hold on
plot(xv, objfcn(B, xv), '-r')
hold off
grid
xlabel('V')
ylabel('I')
title('Nonlinear Fit')
text(0.5, 2, sprintf('$I = %.2f (e^{\\frac{V}{%.2f}}-%.2f)$',B), 'Interpreter','latex', 'FontSize',14)

% BL = [xdata(Lv) ones(size(xdata(Lv)))] \ ydata(Lv)
BL = [xdata ones(size(xdata))] \ ydata

BL = 2×1

1.288111372561517 3.029151300350628

figure
% plot(xdata(Lv), ydata(Lv), '.')
plot(xdata, ydata, '.')
hold on
plot(xv, [xv(:) ones(size(xv(:)))] * BL, '-r')
hold off
grid
xlabel('V')
ylabel('I')
title('Linear Fit')
text(0.5, 2, sprintf('$I = %.2f V %+.2f$',BL), 'Interpreter','latex', 'FontSize',14)

.

[federico]

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]

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)

.

Answer 2

[Sulaymon Eshkabilov]

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

[federico]

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]

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