Determining ideality factor of diodes from IV curves

I'm currently doing a semiconductors lab and one of the tasks is to examine the IV characteristics of diodes to determine the values of the ideality factor, $$n_{id}$$, and the reverse bias saturation current, $$I_{0}$$.

For this, I used the Shockley equation

$$I = I_0 \rm{exp}\left[\frac{qV}{n_{id}k_{B}T}\right]$$

And then plotted the voltage, $$V$$ against natural log of current, $$\rm{ln}(I)$$

For a Germanium 1n34a diode we were expecting a straight line, since the pre-ln data was exponential in shape. Instead, the following was acquired.

Here, we fitted linear curves to three different regions of the graph, and determined that only the yellow region had an ideality factor of $$1 < n_{id} < 2$$. Thus my question is, is this an acceptable way to graphically deduce the value of $$n_{id}$$, and also there a specific reason why the Ge diodes's ln-curve was not as uniformly linear as the ($$\ln{I_{0}})$$ for the other measured semiconductors (1n4001 Silicon diode, GaAs LED, etc), which all had very clear straight lines.

Edit 1: Here is a plot of the original data before making it linear (ignore the exponential fit I haven't yet adjusted the fit parameters, and the y-axis should say Current (A) not (I)):

• You say the curve is for an LED, but the label on the graph says it's a Ge diode. Which is it? Commented Oct 15, 2022 at 16:39
• Just edited the post, it is supposed to be for a Ge 1N34A diode. Apologies, have been analysing data for a few semiconductors and mixed them up in the typing. Thanks. Commented Oct 15, 2022 at 16:44
• 1N34A is an ancient point-contact diode. I expect the contact resistance varies from device to device. Have you tried a model that includes series resistance? Commented Oct 15, 2022 at 16:46
• I have not, I'll look into that thanks. Commented Oct 15, 2022 at 16:52
• $$I = I_0 \rm{exp}\left[\frac{qV}{n_{id}k_{B}T}\right]$$ is not the Shockley equation. It should be $$I = I_0 \left(\rm{exp}\left[\frac{qV}{n_{id}k_{B}T}\right]-1\right)$$ Commented Feb 14 at 19:17

Your results surprise me and before you start thinking about linear fits you should consider the method of obtaining the data.

My first though is do you have a "dead" diode?

I note that you are taking current readings of $$e^{-20} \approx 2\,\rm nA$$ which is not unreasonable but how were the voltmeter and ammeter configured to take your readings and what are the specification of those two devices? However your practical arrangement might well be fine since you state that the data for other types of diode could be linearised.

I do not have the facility to investigate the $$iv$$ characters of a 1N34A Germanium diode so I found some data produced by somebody else in the form of a graph.

At the pixel level using a drawing package I analysed the graph and using Excel plotted $$v/\rm V$$ against $$\ln(i/\rm \mu A)$$ and that did give the expected linear relationship although not for the very small currents that you were measuring.

If you can you need to repeat your reading with the same diode as before and/or with another diode of the same type?

• I was taking measurements with a Keithley 2450 source meter which has a function to take sweep measurements across a defined voltage range (we chose 0 to 1V) and then it would measure the current accordingly. I have also edited the post to contain the original data which plots current against voltage (no log scales). Commented Oct 15, 2022 at 16:33