ideality factor diode > 2 possible?

Question

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]:

Did you include a series resistance in your model? What kind of diode are you modelling (silicon pn junction, Shottky, LED, ...)? What parameters are you optimizing to fit your model? — The Photon, Feb 09 '16 at 23:32

Spehro Pefhany · Accepted Answer · 2016-02-10T17:40:48.923

2

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).

edited Feb 10 '16 at 17:40

answered Feb 10 '16 at 01:50

Spehro Pefhany

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Did you apply non-linear regression using excel? – Saavin Feb 10 '16 at 14:29
I used the solver function, in Excel 2003, but because it's so nonlinear I optimized one parameter at a time first. There's probably a better way (maybe linearize the cost function first) but this worked pretty well. – Spehro Pefhany Feb 10 '16 at 14:34
Can you help me improve my method so I can get a similar value to yours? – Saavin Feb 10 '16 at 16:39
Can I send you the file I used? – Saavin Feb 10 '16 at 16:40
I'm not sure you should be using a spreadsheet program for this anyway- maybe MATLAB or a free similar program such as Scilab. I just used it as an example. – Spehro Pefhany Feb 10 '16 at 16:51
Can you please tell me which function you used on matlab - that would really help me; did you use mdl = fitnlm(tbl,modelfun,beta0)? – Saavin Feb 10 '16 at 17:09
You can use the curve fitting toolbox and fit to an exponential. See above. You then have to back calculate the parameters given Vt. I did that too, just use the online help already. – Spehro Pefhany Feb 10 '16 at 17:33
Did you get 2.049871893 for n (ideality factor)? – Saavin Feb 10 '16 at 18:52
No, I told you it was well within [1..2]. Make sure you are using the correct temperature assumption - I've used Vt = 25 or 26mV which implies a temperature around 300K for Vt= kT/q. – Spehro Pefhany Feb 10 '16 at 18:53
Sir, I am instructed to use 38.68volt^-1 for (eV/kT)? – Saavin Feb 10 '16 at 19:01
That's fine, 25.8mV. – Spehro Pefhany Feb 10 '16 at 19:03

ideality factor diode > 2 possible?

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