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The code I am using ''idx'' is giving me wrong Voltage value.

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MUHAMED SIDIBEH
MUHAMED SIDIBEH am 21 Feb. 2022
Kommentiert: Voss am 22 Feb. 2022
The voltage is suppose to drop as I increase the temperature, but the ''idx'' selects the highest Voltage value in the vector. Is there way to reduce the Vpmax as the temperature increases?
Io = 0.1*10^(-9);
IL = 9.5;
V = linspace(0,0.691,100);
q = 1.6*10^(-19);
k = 1.380649*10^(-23);
% Temperature at 45°C and Irradiance of 900 W/m^2
T = 273.15+45;
G =700;
Gr = 1000;
I = (IL*(G./Gr))-(Io.*(exp((q.*V)./(k.*T))-1));
n = 72;
Vp = V.*n;
P = (I.*Vp);
[Pmax,idx] = max(P);
Imax = I(idx);
Vpmax = Vp(idx);
Isc = max(I);
Voc = max(Vp);
figure(8)
plot(Vp,P,'r','linewidth',2);
xlabel('Voltage (V)','fontsize',15);
ylabel('Power (W)','fontsize',15);
title('PV Panel PV Curve at 45°C 900 W/m^2','fontsize',15)
grid minor
axis([0,55,0,400])
figure(9)
plot(Vp,I,'k','linewidth',2);
xlabel ('Voltage [V]','fontsize',15);
ylabel ('Current [A]','fontsize',15);
title('PV Panel IV Curve at 45 °C 900W/m^2','fontsize',15)
yline(Imax)
xline(Vpmax)
text(Vpmax,Imax,['MPP=',num2str(Pmax),'[W]'])
grid minor
axis([0,55,0,9])
Any help will be greatly appreciated
  1 Kommentar
Voss
Voss am 22 Feb. 2022
idx, as calculated, is the index where P (Power) is at its maximum, not Voltage.
[Pmax,idx] = max(P); % idx: index of maximum Power
Also, temperature is a scalar constant, so I'm not sure why you say Voltage is a function of Temperature in this code. Can you please clarify?

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