# Relationship between Zener voltage and current?

What's the relationship between the current flowing through the Zener Diode? I simulated the following for various values of R, and the voltage drops quite a bit with small currents. The 1N750 has a Vz of 4.7V.  The only information related to it on the datasheet I could find was the following, however I'm not sure what curve I'm meant to be looking at: So then, I simulated the circuit with a current source instead of R1:

• 10mA of current gave a voltage of 4.66V
• 0.45mA of current gave a voltage of 4.13V

This suggests it is the righter curve. However, then I simulated the circuit at 0.01mA (the bottom of the graphs). This gives a voltage of 2.65V, which suggests the left curve. Which is it (if either)?

## 2 Answers

The zener voltage rating is right at the top edge of your graph (the 20mA point). If you look on page 2 of the data sheeet it tells you the test current. So, looking at your graph and ignoring the black line at the very left (because they don't seem to be specifying a zener voltage at that rating!), the 2nd trace is the 3V3 zener, 3rd trace is 3.6V, 4th is 3V9 etc..

Zener diodes are notoriously slack in their operating knee at low voltage ratings and this improves a lot when the voltage rises above 5V.

For the 4.7V zener you fed with 10mA, the voltage is about right and if you looked at the graph this confirms that. However, there is a discrepancy when testing at 0.45mA but this could easily be due to the model used not being a perfect representation of the data sheet. It looks like 0.45mA would produce about 3V8 according to the data sheet.

Stick to the data sheet for most things but be prepared to do real tests on a batch if operating points significantly below test currents are required. It's nice if the simulator agrees but at this level of detail it doesn't surprise or worry me.

Adding more basic information to what Andy Aka already said: reverse biased Zener diodes keep the voltage across their terminals approximately constant only if the reverse current is high enough to drive them fully in the breakdown region of their characteristics. If the reverse current is too low, they are either cut-off or just barely entering breakdown (i.e. their operating point is on the knee of the characteristics).

You can draw a plot of the reverse voltage vs. reverse current using this arrangement in LTspice: You get the following results: You can see the "knee" starting about at 1mA and ending at about 2mA. Of course this doesn't seem too sharp because of the logarithmic scale on the horizontal axis. Using linear scale you will get this: 