Electrical – Zener Diode Reverse Current


I have this – Zener Diode – 12V Reverse Voltage

On the table 8, I can see the ranges for the 12V reverse voltage is between 11.4V and 12.7V for the "C" selection part. But this value if for 5mA of zener reverse current.

But, if we see the characteristics of zener diode, we see the below graph,

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We can see there is a minimum reverse zener current to enter the breakdown and there is a maximum zener current (above which, the device will get damaged).

But the datasheet table 8 , provides the value for a reverse zener current of 5mA only. But they didn't mention whether it is the minimum or the maximum zener current.

Question 1 :

I just want to know what is the minimum required zener current for the 12V "C" zener part so as to enter into the breakdown region? Where is it mentioned in the datasheet?

Question 2 :

On table 8, what does the 5th column, Reverse Current (uA) and VR indicate?

Please provide help.

Best Answer

  1. Behavior at low currents is not given in this particular datasheet, neither typical nor any guarantees. In part it depends on how you define "breakdown" (at what current). If you simulate a model of BZX84C12L from OnSemi in LTspice, you'll find this "typical" response modeled. Of course we don't know how accurate the model is, especially at the lower currents, and we don't know the variation from unit to unit. The voltage is only guaranteed in the datasheet at 5mA. I would expect at 10uA it would be behaving reasonably well as a regulator (note: this sharp breakdown behavior is highly dependent on the semiconductor mechanism used in the diode, and very low voltage models (< 5V) will not be nearly as well behaved).

5mA 12.2V

1mA 12.0V

100uA 11.7V

10uA 11.4V

1uA 11.1V

100nA 10.8V

10nA 10.5V

1nA 10.1V

  1. Table/column in the datasheet guarantees reverse current of < 100nA at 8V, The model suggests the actual current will typically be less than 1nA.

Edit: To address the particular problem as described somewhat in the comments:

To design a shunt regulator with a resistor and a zener (or TL431 or similar shunt regulator).

The series resistor has to supply all the load current, plus a small amount to keep the regulator working, at the minimum input voltage. Suppose you have an output voltage of 12V and a maximum load current of 100mA and a minimum load current of 0mA. Input voltage is 16V maximum and 14V minimum.

The resistor has to pass 100mA with 2V across it plus whatever the regulator needs. Let's assume that is zero. So the resistor has to be no higher than 20 ohms. Given say 5% tolerance and E24 values, maybe we'd use 18 ohms. Now we have at least 100mA for the load and minimum 5.8mA for the regulator. So we don't need to worry about minimum regulator current unless it's less than 6mA, just the tolerance takes care of it.

If the load now goes to 0mA the regulator has to eat all that current (106mA) so it dissipates 1.3W. That's a lot of power. But it gets worse.

Now, suppose we have 16V input, the resistor current is now (16-12)/18 = 222mA nominal. With our load drawing 100mA the shunt regulator has to eat 122mA, causing it to dissipate more than 1.4W. And if the load current goes to zero, then it has to eat the entire current, and dissipate about 2.7W of power, which would require a large device with a large heatsink.

You can see why series regulators are preferred for high load currents, and when the input voltage ranges above but also gets close to the output voltage.

I didn't address the quality of the regulation here, but the differential resistance of the zener you mentioned is similar to the series resistor we calculated at minimum input voltage. That means that (at minimum input voltage) a 0.1V change in input voltage will cause about 0.05V in output voltage, which is not very good regulation.