Exponential Increase instead of Linear

2 Ansichten (letzte 30 Tage)
Gautam Khanna
Gautam Khanna am 24 Sep. 2020
Kommentiert: Looky am 25 Sep. 2020
The plot between V and I should be as shown, but it produces an exponential increase of I with V
q = 1.6e-19;
E = [-0.5:0.001:0.5]; % [eV]
V = [0:0.01:0.5]; % [V]
mu = 0; % [eV]
kB = 1.38e-23; % [J/K]
kB = kB/q; % [eV/K]
h = 6.626e-34/(44/7); % [J.s]
h = h/q; % [eV.s]
T = 300; % [K]
g1 = 0.2; % [eV], gamma1
g2 = 0.2; % [eV], gamma2
g = 0.1; % [eV], g1*g2/g1+g2
dE = 0.001; % [eV], delta E
D = zeros(1,length(E));
mu1 = zeros(1,length(V));
mu2 = zeros(1,length(V));
f1 = zeros(1,length(V));
f2 = zeros(1,length(V));
f = zeros(1,length(V));
F = zeros(1,length(V));
I = zeros(1,length(V));
for j = 1:length(V)
for i = 1:length(E)
if E(i)>= -0.1
D(i)=0.5;
else
D(i)=0;
end
end
mu1(j) = mu + ((q*V(j))/2);
mu1(j) = mu1(j)/q; % [eV]
mu2(j) = mu - ((q*V(j))/2);
mu2(j) = mu2(j)/q; % [eV]
f1(j) = 1 ./ (1+exp((E(i)-mu1(j))/(kB*T)));
f2(j) = 1 ./ (1+exp((E(i)-mu2(j))/(kB*T)));
f(j) = f1(j)-f2(j);
F(j) = D(i)*g*f(j)*dE;
I(j) = ((2*q)/h)*sum(F(j));
end
plot(V,I,'b-','linewidth',2); hold on;
set(gca,'fontsize',20);
xlabel('V');
ylabel('I');
xlim([0 0.5]);
set(gca,'xtick',[0:0.1:0.5]);
  1 Kommentar
Looky
Looky am 25 Sep. 2020
Check your inner for loop. It computes the D values for index i 1:length(E) but only the i = length(E) is used afterwards. Also this loop does not contain the index j and E is nether changed, so why is it nested anyway? You could compute it beforehand?
Look at f1(j) = 1 ./ (1+exp((E(i)-mu1(j))/(kB*T))); Here the E(i) will always be the same value ( E(length(E))? is this intended?

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by