caractéristique courant-tension

[caran]

hi,

[caran]

hi,

in this program i want to find the value of the -(1/(di/dv)), and exactely at the point (0,max Current), it should be positive value and great, but i found it smalle and negative. did i forgot any thing in the program?

thank 's

[Zwired1]

Plot the output of the derivitive and you'll see it has a very large negative slope between index 0 and 1.  -1/slope will give you a small positive number.

Try wiring some initial conditions to the derivitive function.

[caran]

when i changed the method to (forward method), the first negative value is removed. 

[caran]

and i would like to ask you about the dt, because when i increase de dt value, the results of -(1/slope) is better,

can i change dt ?

thank you.

[Zwired1]

dt is the period of your samples and is equal to the reciprocal of the sampling frequency.   The sampling frequency in this case is determined by the "divide by 10" inside the FOR loop -- you are calculating 200 samples per the curve with an x-axis step size of 0.1 units to yield 20 seconds of data. 

Answer is "no", there is no justification to change the dt value as it corresponds the sampling frequency you set up earlier (assuming it is still set up as in post #12).

[caran]

i send you the main program to calculate the reciprocal value of the slope of I-V curve (for V=0), and he said that he had a good results (unlike me). 

what can you talking about this program and our previous program to calculate (1/slope).

thank you. 

[cody_a]

Caran,

It sounds like you are having trouble generating a points to be graphed.  What exactly is the issue that you are having?  And what kind of data are you trying to plot?

[caran]

i send you the VI program, it contain the data in the input, when you run the program, you will see the Current-Voltage curve

i want to extract the slope of the curve in each point and exactly in the point (I=I max, V=0) like in the picture, then calculate the vlaue of  (1/slope).

thank you.

[WenR]

Hello Caran

Maybe a workaround for this could be to divide the curve in several subsets of the waveform and calculate the slope for each subset. If you calculate the slope for every pair of points, you might have issues with the noise from your signal. The forum on the next link has more details about the implementation of this workaround.

http://forums.ni.com/t5/LabVIEW/How-do-I-find-slope-of-a-waveform/td-p/182207

WenR