Can't fit a curve for an exponential function, anyone knows why?

**slow_rider** · 12-11-2012, 06:49 AM

Hello,

I mapped out a relation between voltage and resistance on a device and need to fit a curve to this graph, however Excel can not do it although it seems like a pretty standard exp. function to me. Any help will be greatly appreciated. I am also interested in linearising the data later on, I don't know if this this helped at this stage or not. Thanks! File is attached.

**MrShorty** · 12-11-2012, 11:55 AM

What do you mean by "Excel cannot do it"? Does it crash the program or do you mean that it returns an equation that does not give a good fit of the data?

If by "standard exp function" you mean something of the form y=ae^(bx), then I think the reason it is failing to provide a good fit (assuming it is returning some kind of equation) is that a standard exp function is a straight line on a semi-log plot. If you convert your y-axis to a log scale, you will see that your data are not linear on a semi-log plot. Again, assuming it is returning an equation, I am guessing that a standard exponential is not a suitable equation for the data.

1st thing I'd probably suggest is not really Excel related. Obviously we are not looking at ohm's law or anything like it in this case. I would suggest you go back to your electrical schematic and analyze it for clues to the nature of the curve that would give a good fit. For example, perhaps R is one resistor in parallel with other resistors. From ohm's law and the formula for the overall resistance of resistors in parallel, we might hypothesize and equation of the form E=f(1/R). Beyond that, I'm not real familiar with electrical theory, but I would expect that somewhere in a good circuit analysis should be clues to indicate what equation would give a good fit for these data.

**slow_rider** · 12-11-2012, 01:03 PM

A more accurate description of the situation would be that Excel will fit a curve, but very poorly as the coefficient of determination is below 0.8 and that's using a high order polynomial function which is not a possibility in my case since I need a monotonous function.

I took your approach as well in the electronics field, these figures are LDR resistance vs. voltage applied to an LED shining directly on it.
The voltage applied to the LED is proportional to the current flowing through it and we could say it is linear.
The light intensity of the LED is proportional to the current flowing through it and we could say it is linear.
So we could say that the light intensity is proportional to the voltage applied and is behaving like a squared function.

I took sqrt() of R but the curve ends up looking about the same and I still can't get a good fit which is not Polynomial.

**MrShorty** · 12-11-2012, 04:41 PM

I'm not sure what to recommend, as I am unfamiliar with LDR's. These people http://www.scribd.com/doc/30226030/L...dent-Resistors suggest that a "typical" LDR's resistance is related to intensity by something like R=A/L where L is light intensity and A is a constant. If L is directly related to E, then maybe a function of the form R=sum(Ai/V^i) might be suitable.

You might search further in the literature on LDR's and see what function types others have used to correlate the resistance of an LDR.