Determine material of Zener’s diode based on its characteristics

As you stated, there are 3 parameters that dictate the DC response of a diode. Those are the saturation current (IS), the emission coefficient (N), and the ohmic resistance (RS). I was able to fit the curve with a fairly high accuracy, so I'll document my model procedure.

The SPICE model for the diode closely matches the Schokley diode equation:

If = IS(e^(Vf/(N*Vt)) - 1)

where Vt = kT/q = 26mV at room temperature.

1. Get actual values from the graphs provided in the datasheet to use for comparison. The more points the better, and the more accurate the better. Below is a table that I estimated from the figure you provided:

   Vf  If (mA)
   1.3 0.001
   1.4 0.010
   1.5 0.080
   1.6 0.700
   1.7 5.000
   1.8 20.000
   1.9 40.000
   2.0 65.000
   2.1 80.000
   

2. Plug the values into Excel, and change the y-axis to a log scale. You should get a graph that looks identical to the original graph from the datasheet. Add another column for your graph, with If calculated from the forward voltage and the constants IS and N. We can use this configuration to iteratively find IS and N.

3. Solve for IS and N. We are trying to match the linear part of the graph (1.3 <= Vf <= 1.7). Adjusting IS will move the curve in the y-axis. Get the calculated graph to the same order of magnitude. The next step is to find the emission coefficient (N). N affects both the amplitude and the slope, so some adjustment of IS may be necessary to keep the curve in the same ballpark. Once the slopes match (the lines are parallel), trim IS so that the calculated data matches the datasheet values. I got IS = 1e-18, and N=1.8 for the diode you listed.

4. Identify RS. This is a bit tricky. RS is responsible for the curving of the current from 1.7V and above. Consider modeling the ohmic resistance as a resistor in series with the diode. As the current through the diode increases, the voltage drop across the ohmic resistance causes the forward diode voltage Vf to increase slower. At small currents, this effect is negligible.

The first thing to do is to get a ballpark estimate of RS to use in the more accurate solutions. You can calculate the effective value of RS from the datasheet values by back-calculating for Vf using the measured If. The voltage difference between the input value and the calculated Vf can be used with the forward current to generate a resistance. At the higher currents, this will be a good starting value.

To plot the diode current using RS, you need to first calculate the diode Vf given a voltage for the resistor-diode series combination. Wikipedia lists an iterative function - it converges easily if the resistor voltage drop is significant. This function was easy enough to set up in Excel. For Vf values below 1.8, I hard-coded the input value because the iterative function did not converge. Then take this Vf value to calculate the If of the ideal diode. I plotted this with the original datasheet graph.

Using trial and error, you should be able to get a RS value that gets pretty good overlap with the datasheet values. All that's left is to throw the model together in SPICE to verify your work.

Below is my diode model that I verified using HSPICE. The simulation data is almost a perfect overlay for the datasheet graph.

.model Dled_test D (IS=1a RS=3.3 N=1.8)

I used this article, which helped a lot with the diode spice parameters.

I cleaned up my spreadsheet, and tyblu has made it available for download here. Use at your own risk, results not guaranteed, etc... etc...

The behavior you describe reproduces in the simulated circuit exactly. If we configure the Zener to a 25 reverse voltage and have it in the circuit the right way, Q1 drops around 25V. If we reverse the Zener, we get approximately two junction drops, as you found empirically. So it looks like you have the right circuit, and that the Zener is marked wrong. The stripe on the diode should correspond to the cathode bar on the schematic.

A long-standing rumor is that NTE buys other manufacturers' semiconductors and stick their own name and their own NTE-specific part numbers on them, possibly after testing the parts to figure out what they are. Maybe your diode came from a batch that ended up as NTE because they were marked backwards.