# Calculation of Excitation potentialfrom I-V curve

From a current v voltage graph, how does one calculate the excitation potential? I thought it was calculate the difference between peaks, take an average and this gives you the excitation potential? Is this right or is there a different way to calculate it?

James

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Plot distance along x axis versus peak number. calculate the slope using the method of least squares. Plot distance from 0th peak versus peak number. Calculate the slope as before. The ratio of the slopes gives the excitation potential.

There is another method using the current just before and just after to locate the peak precisely. IT depends on whether you can accurately measure the current which is hard.

Perhaps someone else can describe another method. It is WRONG TO do WHAT YOU ARE SAYING as I have detailed below:

Say you measure minima position V1,V2,V3,V4,V5,V6

Then you calculate differences:

a1=v2-v1, a2=v3-v2, ... a5=v6-v5.

And then you calculate "average difference"

a= (a1+a2+a3+a4+a5)/5

But a=(a1+a2+a3+a4+a5)/5 = (v2-v1+v3-v2+v4-v3+v5-v4+v6-v5)/6=(v6-v1)/5 !

so you lost all the information about v2,v3,v4,v5 in your procedure, because you cancel them out. It makes no sense ! a1, a2 a3 etc. are not independent measurements and you are not allowed to take an average of them like that ! What you really measure is v1,v2,v3 etc. and you have to use all your information.

student