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

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Star Strider
Star Strider 2014년 8월 11일

0 개 추천

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.

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Hikaru
Hikaru 2014년 8월 11일

2 개 추천

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.

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Maroulator
Maroulator 2014년 8월 11일
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 2014년 8월 11일
@Maroulator — You can give him a vote (2 points).

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