Zener Diode Reverse Current

Question

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,

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.

Answer 1

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

2. 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.

Answer 2

The minimum specified value is 1mA where \$r_{dif}\$ aka "Rs or Zzr" rises at lower current from 10min to 50 min thus the drop of 4mA could be a drop around 125mV from a min of 11.7 for C parts.

The rated specified value is 5mA.
The maximum value of Iz depends on the power limit of Pmax=250mW @<=25'C with a temp rise of 0.33'C/mW of junction above the solder point. Thus Pmax must be derated for normal use by 30 to 50%.
Thus you may consider 10mA max but also consider C bin might be as bad as 25Ω with a rise of 5mA above rated Vz or 125mV above 12.7V max.

It also increases the voltage, Vz due to \$r_{dif}(Ω)\$ = 10 typ. to 25Ω max. @ 5mA which is the bulk resistance that is a major cause of the bin tolerances for A,B, & C.

Have you heard of bandgap ref diodes?

Rs will reduce as package power size increases in ALL diodes.

Answer 3

From your comment in response to Spehro Pefhany's excellent answer,

I am asking this question. Because my load is 100Ohms connected across the Zener. My maximum and minimum input voltage is 16V and 8V respectively (Load current range would be 80mA to 160mA). So, want to calculate what is the maximum and minimum series resistor values which I have to choose if I take this 12V Zener diode.

Let's take a look at the circuit you seem to have in mind.

^{simulate this circuit – Schematic created using CircuitLab}

As has been noted in comment, you CANNOT get 12 volts across a zener with 8 volts in.

For an input range of 8 to 12 volts, there is no value of RLIM which will produce an output voltage greater than 12 volts, so the zener will not turn on. This alone should tell you that this is the wrong approach to use.

Now, let's consider what happens at V1 equal to 16 volts.

Assuming the zener is regulating, there will be 4 volts across RLIM. The current through RLIM can be considered the sum of two separate currents, load and zener. Load current is 120 mA. How much zener current can you specify?

In your question you wrote,

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.

As Tony Stewart answered, 5 mA is neither. Maximum current is set by the power dissipation of the zener. In this case it's 250 mW, which implies a current of just about 20 mA (20 mA times 12 volts equals 240 mW - close enough).

So let's pick a maximum current of about 2/3 the rated max, or 15 mA.

Total current through RLIM will be 135 mA. By Ohm's Law, this gives a value for RLIM of (12/.135), or 88.88 ohms. Power dissipated in RLIM will be (4 x 4 / 88.88), or about .18 watts, so you could use a 1/4 watt resistor.

Is this a good idea? Not remotely. Remember that 20 mA upper limit on current? If you exceed that you will probably cook the zener. At 20 mA of zener current, total current is 140 mA, and RLIM equal (12/.14), or 85.17 ohms. 85.17 / 88.88 equals .96, which means that your resistor MUST be more precise than a 5% tolerance.

So, the short end to a long answer is that you don't want to do it this way. At best, you'll only get regulation over the upper half of your input range, and if you use a 5% tolerance on your limiting resistor you stand a decent chance of destroying your zener in the long run. If your load pulls 5% more current than you expected, you'll also kill the zener (or at least drive it over its nominal limits).

Short short answer - don't do it this way.