## Zener voltage regulator Operation with example

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Zener diode as voltage regulator

From the VI characteristics of zener diode it is clear that the zener diode in reverse breakdown region has very small variation in voltage across the zener diode for large variations in the current through the diode. Hence it can be used for regulator applications for providing constant output voltage independent of load and supply voltage variations. A simple shunt voltage regulator designed using a Zener diode is shown in the figure

Shunt voltage regulator

The load is in shunt with the load resistance hence the name shunt voltage regulator. Assume for the entire range of load and supply voltage variations the zener diode remains in reverse breakdown region. The condition to ensure the zener diode is in reverse breakdown region is that the current flowing though it should be greater than the knee current IZK.

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Model for Zener diode

The VI characteristics of zener diode in breakdown region are almost linear with a slope of 1/rZ. Hence the zener diode can be represented with a model containing a constant voltage source with a zener resistance rZ. The model of Zener diode is shown in the figure below

It is described analytically as follows VZ=VZO+IZ*rZ it applies for I>IZK.

Analysis of shunt voltage regulator

Lets analyse the circuit for load variations when there is change in load, the load current which is given by IL = VZ/RL (neglecting the voltage variation across the zener diode) remains constant. But the total current through R changes due to variation in Vin and is given by (Vin–VZ)/R. Hence the difference of total current and load current should flow through Zener diode which is approximately ∆Vin/R. Assume that this variation is ∆IZ, corresponding to this variation in zener current the variation in Zener voltage is given as ∆VZ = rZ*IZ which will be order of milli volts. Hence the output voltage is regulated.

The voltage regulation for supply voltage variations is quantified by a parameter called line regulation. It is defined as ratio of output voltage variation to supply voltage variation. It is specified in the units of mV/V.

Line regulation = ∆VO/∆Vin

Similarly whenever there is a variation in load, the load current which is VZ/RL varies. But the total current through R is almost constant and is given by (Vin–VZ)/R. Hence the difference of total current and load current should flow through Zener diode. Assume that this variation is ∆IZ, corresponding to this variation in zener current the variation in Zener voltage is given as ∆VZ = rZ*IZ which will be order of milli volts. The maximum value of load resistance is limited by power dissipation rating of Zener diode while the minimum value of load allowed is determined  by the knee current of Zener diode.

The voltage regulation for load variations is quantified by a parameter called line regulation. It is defined as ratio of output voltage variation to supply voltage variation. It is specified in units of mV/A.

Example: Consider a shunt voltage regulator with a 6 Volt zener diode at a zener current of 6 mA and rZ = 10 ohms, knee current IZK = 0.2 mA, VZK = 5.8 volts,R=500 ohms. The supply voltage is 10 Volts with a tolerance of +or- 1 volt.

a) Find no load output with Vs at its nominal value

b) Find line regulation

c) Find load regulation when load draws a current of 1 mA

e) Find minimum value of load resistance for which the zener diode still will be in reverse breakdown region

Solution: The parameter of VZO for rated current of 6 mA is given as

VZO = VZ-IZ* rZ = 6-0.06 = 5.94 volts.

a) With no load IZ = (VS-VZO)/(R+ rZ)=(10-6)/(500+10)= 7.84 Amps

Hence no load output = VZO+ IZ* rZ = 6.02 volts

b) For a change of 1 Volt in supply voltage change in output voltage = ∆VO * rZ/( R+ rZ)

∆VO = 1*10/(510)=19.6 milli volts

Line regulation = ∆VO/∆VS = 19.6 milli volts/volt

c) When load draws a current of 1 mA the zener current decreases by 1 mA. Hence the corresponding change in Zener output  ∆VO = rZ*∆IZ =10*(-10-3)=-10 milli volts.

Load regulation = ∆VO/∆IL = -10 milli volts/milli Ampere

This calculation is based on the assumption that he change in current is negligible. A more accurate value is derived from the model

d) When a load of 2 kilo ohms is connected, the load current will be approximately VZ/RL.

IL = VZ/RL = (6/2)*10-3 = 3 milli Amps

Thus the change in load current is approximately ∆IZ and the corresponding change in output voltage

∆VO = rZ*∆IZ = 10*(-3*10-3)=-30 milli volts.

e) The Zener diode will be breakdown region if the IZ = IZK and VZ = VZK = 5.8 volts. At this point the worst case current through R is I = (VS,min-VZ)/R = (9-5.8)/500 = 3.2/500 = 6.4 milli Amperes and thus the load current is I-IZK = 6.4-0.2 = 6.2 milli amperes. The corresponding value of RL is

RL,min = VZ/IL = (5.8/6.2)*103 = 935.5 ohms