Q is equivalent to the integrated product of current and time I*t.

You can calculate dQ/dV by determining the slope of a I*t vs. V plot. I, t and V should be recorded in your data.

Q is equivalent to the integrated product of current and time I*t.

You can calculate dQ/dV by determining the slope of a I*t vs. V plot. I, t and V should be recorded in your data.

Thank you Sebastian.  I going to follow up on your suggestion and plot.  I know I am little bit rusty with my calculus, how do you relate the slope from I*t vs. V to dQ/dV?

I have attached the snapshot of recorded data.....there are two type of "time" data and seems the values are close to each each other...but not sure which one to use.  

One more thing...i have tried to plot I*t vs. V for the first cycle.  But the slope of I*t vs. V is not constant ....so how? 

As I*t = Q and the slope of (I*t) vs. V is d(I*t)/dV you will find that dQ/dV = d(I*t)/dV.

It seems that the columns test_time and step_time contain the same values so it does not matter which one you use.

Yes, you won't find a constant dQ/dV, it changes with V. It is therefore practical to plot dQ/dV vs. V. You can find an example in the attached paper:

http://www.nature.com/articles/srep03094

 Thank you again.  I am almost there.  But here is the problem.  

I hope I am right.  I did calculate, a) I*t   ..........then for d(I*t)= (I*t) @(t=t1)-(I*t) @(t=0) 

b) Similarly for V = V@(t=t1)-V@(t=0)

c) finally for dQ/dV = step a/step b............................

based on this calculation, the plot of dQ/dV was a mess.   

I would appreciate if you can shade some light on this.  Thank you.  

That procedure seems to be correct.

Could you attach a table (as .xlsx or .txt document or anything similar) of your data?

The table that you showed above contains no values for I, which makes it difficult for me to reproduce your calculation.

Please find in here attached Excel data sheet for the 1st cycle. I appreciate your time.  Thank you.    

By the way your file already provides two columns with charge and discharge capacity, so you don't need to calculate I*t.

I checked your data and yes, the plot is really kind of a mess, but the procedure is correct. The problem lies within your data. If you vary the difference quotient dQ/dV a bit for plotting, you may get a slightly better result in terms of presentability. I tried this for the discharge cycle, you can see the result in the attached table.

Thank you and I appreciate your time.  I looked up and looked better however is not presentable yet.  Do you think is there some issue with the way data was recorded in CC?

I wander if you have a template program for CC or CCCV?

My voltage window is 0.01 V (lower cut)-1.0  V (upper cut)

If possible, you can adjust the interval of data logging in your program. It seems to me that the time between two data points could be a bit longer. In this way you could at least get rid of all the zero values which are caused by consecutive V values that are identical.

In our lab we use EC Lab software which offers templates for CC/CV and many options of editing these methods.