Calculation of Excitation potentialfrom I-V curve

[JamesJames]

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

 

[student1938]

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

 

[M98Ranger]

, your understanding is partially correct. The excitation potential can indeed be calculated by finding the difference between the peaks on the current-voltage (I-V) curve and taking the average. However, this method may not always be accurate as it relies on the assumption that the peaks represent the maximum current values. In reality, there may be other factors such as noise or fluctuations that can affect the peak values.

A more accurate way to calculate the excitation potential is by using the slope of the I-V curve. The excitation potential can be found at the point where the slope of the curve is steepest, also known as the maximum slope point. This point indicates the point of maximum energy transfer and can be used to determine the excitation potential.

Another method is by using the derivative of the I-V curve. The excitation potential can be found at the point where the derivative of the curve is zero. This method takes into account the entire curve rather than just the peaks, making it a more precise calculation.

It is important to note that the excitation potential is not a fixed value and can vary depending on the experimental setup and conditions. Therefore, it is recommended to use multiple methods and compare the results to get a more accurate estimation of the excitation potential.

I hope this helps clarify your understanding of how to calculate the excitation potential from an I-V curve. If you have any further questions, please do not hesitate to ask.