Zener diode and applications

The very steep portion in the reverse biased i-v characteristic curve is called the breakdown region.

In this region the voltage across the diode remains nearly constant while the current varies (i.e., small internal resistance). There are two physical mechanisms that can produce this behavior in the breakdown region. One is the Zener effect in which the large electric field in the depletion region causes electrons to be removed from the covalent bonds in the silicon. The second mechanism is the avalanche effect in which charges that are accelerated to high speeds due to the large electric field in the depletion region collide with atoms in the silicon lattice causing charges to be dislodged. In turn, these dislodged charges have sufficient energy to liberate additional electrons. In other words, this avalanche effect is a cascading, ionization process. Provided that the power dissipated in the diode is less than the maximum rated, the diode is not damaged when operating in the breakdown region. In fact, Zener diodes are designed to operate in this region. The circuit symbol for the Zener diode is

These diodes are usually operated in the reverse bias regime (i.e., breakdown region) so that IZ > 0 and VZ > 0. An enlargement of this bre akdown region is shown in text

The manufacturer specifies the –VZ0 and test current IZT. One can design Zeners with a wide range of voltages. The page below is from a Digikey catalog (www.digikey.com) and shows voltages from ranging from 3.6 V to 200 V, for example.

The rated VZ at the specified IZT is listed for these Zener diodes. The circled component, for example, has VZ = 8.2 V at IZT = 31 power is 1 W for this device. As the current deviates from the specified value IZT, the voltage VZ also changes, though perhaps only by a small amount. The change in voltage ÄVZ is related to the change in the current ÄIZ as

where rz is the incremental or dynamic resistance at the Q point and is usually a few Ohms to tens of Ohms. See the datasheet for the particular devic e you are working with. Because of the nearly linear relationships in the breakdown region, the reverse bias model of the Zener diode is

where vz=vz0+rzIz Applications of Zener Diodes
What are Zener diodes used for? Applications include: 1. Voltage overload protection. This circuit is from the NorCal 40A radio that is built in EE 322 Electronics II – Wireless Communication Electronics:

2. Voltage regulation. See the figure below. An example of such a regulator circuit will be considered next.

Example N6.1 (similar to text example 3.8). The Zener diode in the circuit below has the following characteristics: 6.8-V rating at 5 mA, rz = 20 Ù, and IZK = 0.2 mA. With

With these ratings

Note that the supply voltage can fluctuate by ± 1 V. Imagine this fluctuation is a random process rather than a time periodic variation. Determine the following quantities:
(b) Find VO with no load and V+ at the nominal value. The equivalent circuit for the reverse bias operation of the Zener diode is

From this circuit we calculate

Therefore,

(c) Find the change in VO resulting from a ± 1 V change in V+. Using the circuit above the V+ = 11 V:

Similarly, with V+=9 V:
.
The ratio of the change in output voltage to the change in the source voltage is called the line regulation of the regulator circuit. It’s often expressed in units of mV/V. For this example and no load attached,

(d) Find the change in VO resulting from connecting a load of RL = 2 kohm with a nominal V+ = 10 V.Assuming that the diode is operating in the breakdown region:

Is this a reasonable value? Calculate IS:

So, yes, this is a reasonable value because IL < IS , as it must. From (1), Ñ Vo = rzÑ VIz and since ÑVIz = -3.4 mA then

The ratio of the change in output voltage to the change in the load current ( Ñ Vo/ÑIL ) is called the load regulation of the regulator circuit. It’s often expressed in units of mV/mA. For this example,

(e) What is VO when RL = 0.5 kohm? Assume the diode is in breakdown. In this case,

Is this a reasonable value? No, because this value is greater than IS = 6.4 mA. Therefore, in this case the Zener diode is not operating in the breakdown region. Also, the diode can’t be forward biased. Consequently, we conclude the diode must be operating in the reverse bias region.

The equivalent circuit in this case is

From this circuit we calculate

This voltage is less than the breakdown voltage VZK, which is consistent with the reverse biased assumption. (f) Determine the minimum RL for which the diode still remains in breakdown for all V+. (We know from the results in parts (c) and (d) of this example that RL must lie between 500 Ù and 2 kÙ when 10 V + = V.) Referring to Fig. 3.21, at the “knee”Iz=Izk=0.2 mA and Vz=Vzk~VZ0=6.7

• If V+ = 9 V:

Therefore, IL = 4.6 mA-0.2 mA = 4.4 mA, so that

• If V+ = 11 V:

Therefore, IL = 8.6 mA-0.2 mA = 8.4 mA, so that

The smallest load resistance that can be attached to this circuit and have the diode remain in breakdown is RL = 1,522 Ù. The reason is that for any smaller value when V+ = 9 V results in the diode leaving breakdown and entering the reverse bias mode