Using non-linear regression I have worked out the ideality factor to be greater than 2 (2.04987187490894) using spreadsheet software.
Is this realistic for a diode. Is the maximum ideality factor for a diode 2?
The data are as follows:
VOLTAGE := [0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]:
CURRENT(mA) := [0, 0, 0, 0, 0, 0, 0, 0.07, 0.92, 12.02, 158.29]:
Best Answer
No, the ideality factor should be between 1 and 2 inclusive. Of course with some random set of data it might not be representative of a real diode.
However, I don't think your optimization is very good. Below is the same exercise which I did myself. The ideality factor comes out well within the above range and Is at a quite reasonable 1.16pA. As you can see, the two curves are pretty much on top of each other.
Since this looks like a homework question, that should be enough of a hint.
Okay, using MATLAB, here is a similar fit (I used amperes for the Y axis). The value of Is came out a bit different at 1.243pA (I would trust this one more) and the ideality factor very similar to Excel. Again, as you can see, the data points are right on top of the fit. If I use the bisquare or robust algorithms, Is comes out very similar to the Excel version (1.17pA vs. 1.16pA).