Transistor

as A VOLTAGE REGULATOR

1Electronic Devices and

Circuits (2131006)

Index

1.Voltage Regulation

2.Zener follower

3.Two transistor regulator

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What is ”Voltage Regulation” ?

Besides being use in buffer circuits and

impedance , the emitter follower is widely

use in voltage regulators.

In conjunction with a zener diode ,the

emitter follower can produce regulated

output voltages with much larger output

currents.

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Zener Follower

In fig. shows a zener

follower, a circuit that

combines a zener

regulator and emitter

follower.

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How its works?

The zener voltage is the input to the base of the emitter follower.

The d.c output voltage of the emitter follower is

VOUT = VZ – VBE

This output voltage is fixed so that it is equal to the zener voltage minus the VBE drop of the transistor .

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If the supply voltage changes , the zener voltage remains approximately constant , and so

does the output voltage.

In other word the circuit acts like a voltage regulator because the output voltage is always

one VBE drop less than the zener voltage.

The zener follower has two advantages over an ordinary zener regulator: first zener diode

of in fig. has to produce a load current of only

IB=IOUT / βdc

Since this base current is much smaller than the output current , we can use a much

smaller zener diode .

For instance , if you are trying to supply several amperes to a load resistor , and ordinary

zener regulator requires a zener diode capable of handling several amperes .

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The second advantage of a zener follower is it low output impedance. In an ordinary zener regulator , the load resistor sees and output impedance approximately Rz, The

zener impedance.

But in the zener follower the output impedance is

Zout = r’e + Rz /βdc

In the fig(b) shows the equivalent output circuit. Because zout is usually very small

compared to RL , an emitter follower can hold the D.C. output voltage almost constant

because the source looks stiff .

In summary , the zener follower provides the regulation of a zener diode with the

increased current handaling capability of an emitter follower.

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Two-Transistor Regulator

In fig. shows two transistor

regulator

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The d.c. input voltage vin comes from a

unregulated power supply such as a bridge

rectifier with a capacitor input filter.

Vin has a peak-to-peak ripple of about 10

percent of the d.c. voltage.

The final output voltage Vout has almost no

ripple and is almost constant in value ,

even though the input voltage and load

current may vary over a large range

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How its works?

Any attempted changes output voltage produce an amplified feedback voltage that opposes the original change .

For instants , suppose the output voltage is increases. then , the voltage appearing at

the based of Q1 will increase .

Since Q1 and R2 form a CE amplifier , the collector voltage of Q1 will decrease

because of the voltage gain.

Since the collector voltage of Q1 has decreased the voltage of Q2 decreases .

Because Q2 is an emitter follower the output voltage will decrease.

In the other words we have negative feedback .

The original increase in output voltage produces an opposing decrease in output voltage.

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The overall effect is that the output voltage will increase only slightly , much less then it would without the negative feedback .

Conversely , if the output voltage tries to decrease less voltage appears at the Q1

base more voltage appears at the Q1 collector , and more voltage appears at the

Q2 emitter .

We have a returning voltage that opposes the original changes in the output voltage.

Therefore , the output voltage decrease only a little far less than without the negative feedback.

Because of the zener diode , the Q1 emitter voltage equals Vz. The Q1 base

voltage is one VBE drop higher.

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Therefore, The voltage across R4 is :

V4=VZ+VBE

With ohm’s law , the current through R4 is :

I4=(Vz+VBE)/R4 Since , This current flow through R3 in series with R4, The output voltage is:

Vout = I4(R3+R4)

After expanding;

Vout=((R3+R4)(VZ+VBE))/R4

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THANK YOU

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