I have the following code which achieves what I need. However, when I tried to make it a bit more sophisticated and perform the substitution I note in the for statement below, I get no results. Cold someone please venture a guess as to why this is the case?
Io=2; K=1.38*(10^(-23)); q=1.602*(10^(-19)); id=-1:0.1:0.6;
for i=1:17 -----> substitute with for i=1:id(end)
T=75;
Vd(i)=((id(i)./Io)+1).*(exp(K*T)/q);
end

 Respuesta aceptada

Star Strider
Star Strider el 11 de Ag. de 2014

0 votos

Your id array has a maximum of 0.6. In your loop, i=1:0.6 would not execute because the value of the counter at the start, i=1 is already greater than the termination value, 0.6.

Más respuestas (1)

Hikaru
Hikaru el 11 de Ag. de 2014

2 votos

Instead of substituting with
i=1:id(end)
Use:
i=1:length(id)
You also might want to preallocate Vd since it changes size in every iteration.

2 comentarios

Maroulator
Maroulator el 11 de Ag. de 2014
Hi Hikaru,
Unfortunately this website only allows me to award one "Accept" and it looks like Star Strider was first, so my apologies for not awarding anything to you. Your answer was really helpful all the same and I wanted to note that.
Star Strider
Star Strider el 11 de Ag. de 2014
@Maroulator — You can give him a vote (2 points).

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