If the load resistance R_{L} is reduced or load terminals are shorted accidentally, a very large load current will flow. It may destroy the pass transistor Q_{1}, diode or possibly some other component. Fuse protection will not prove adequate because the transistor may get damaged in a very small fraction of a second

To avoid this situation, a current limiting circuit is added to a series regulator, as illustrated in the figure.

The current limiting circuit consists of a transistor Q_{3} and a resistor R_{5} (approximately 1 ohm) connected between base and emitter terminals of transistor Q_{3}. With normal load current, transistor Q_{3} remains off because the voltage drop across resistor R_{5 }is small (less than about 0.7 V necessary for making the transistor Q_{3} on). Under this condition, the circuit works normally, as described above. With the excessive load current (exceeding 0.6/1, that is, 0.6 A or 600 mA) the voltage drop across R_{5} becomes large enough to turn transistor Q_{3} on. The collector current of transistor Q_{3} flows through R_{3} thereby decreasing the base voltage of transistor Q_{1.} This results in reduction of the conduction level of transistor Q_{1.} Thus further increase in load current is prevented.

Figure summarizes the current limiting. When load resistance R_{L} is infinite, the output voltage is regulated and has a value of V_{REQ}. The load current I_{L} is zero for this operating condition. When R_{L} decreases, the load current I_{L} increases upto the point where R_{L} becomes equal to R_{L(min)} At this minimum load resistance, I_{L} equals 600mA and V_{BE} equals 0.6V. Beyond this point, transistor Q_{3} turns on and the current limiting sets in. Further decrease in R_{L} produces decrease input voltage, and regulation is lost. When R_{L} is zero, the load current I_{L} is limited to a value between 600 m A and 700 m A. The load current with shorted-load terminals is symbolized as I_{SL}. When the load terminals are shorted in fig. 30.9, the voltage across resistor R_{5} is

**VBE = I _{SL} R_{5} or I_{SL} = V_{BE} / R_{5}**

where V_{BE} is typically between 0.6 and 0.7 V.

The minimum load resistance where regulation is lost can be estimated with the following equation

**R _{L(MIN)} = V_{REG} / I_{SL}**

The exact value of R_{L} _{(min)} will be slightly less or greater than this.

The simple current limiting circuit also has a drawback of large power dissipation across the series pass transistor. With a short across the load, almost all the input voltage appears across the pass transistor. So the pass transistor has to dissipate approximately

**P_{v} = (v_{in}-v_{BE})I_{SL}.**

where V_{BE} is the base-emitter voltage of Q_{3}, the current-limiting transistor.

**Foldback Current Limiting**

** **A problem with the simple current limiting circuit just discussed is that there is a large amount of power dissipation in series pass transistor Q_{1} while the regulator remains short-circuited. The foldback current limiting circuit is the solution of above problem.

The circuit of a transistor series voltage regulator with foldback current limiting facility is illustrated in the figure. In this circuit base of transistor Q_{3} is biased by a voltage divider network consisting of resistors R_{6} and R_{7}. The load current I_{L} flows through resistor R_{5}, causing a voltage drop of I_{l}R_{5}(approximately) across it. Thus a voltage of (IlR_{5} + V_{out}) acts across the voltage divider (R_{6} – R_{7}) network. The voltage applied to the base of transistor Q_{3} is equal to the voltage drop across resistor R_{7} and is given as

**V _{B3} = (R_{7} / R_{6} + R_{7}) (I_{L} R_{5} + V_{out})**

Emitter of transistor Q_{3} is connected to the positive terminal of V_{out}. Applying Kirchhoff’s voltage law to closed mesh of Q_{3} shown in the figure we have

**V _{out} + V_{BE3 }=_{ }V_{B3}**

**or V _{BE3} = V_{Ba} – V_{out} = K (I_{L}R_{5} _{+} V_{out}) – V_{out} = K I_{L} R_{5} + (K – 1) V_{out }where K = R_{7} / R_{6} + R_{7}**

Thus the magnitude of base drive of transistor Q_{1} is by this V_{be3}. Now if load resistance decreases, may be due to any reason, load current I_{L} will increase causing voltage drop Il,R_{5} to increase. This causes V_{B3} to increase and therefore V_{be3} to increase. This makes transistor Q_{3} on in a stronger way. The increased collector current I_{c3} of transistor Q_{3} flows through the resistor R_{3} thereby decreasing the base voltage of transistor Q_{1} This results in reduction of the conduction level of transistor Q_{1. }Thus further increase in load current is prevented.

From the equations above it is obvious that V_{BE} in this circuit is much more than that was in circuit illustrated in the above figure(only I_{L}R_{5}). It means that the increment in load current is limited by larger amount in circuit shown in figure.

Due to reduction in load resistance R_{L}, V_{BE3} increases to a level so that transistor Q_{3 }gets saturated. Now collector current I_{c3} becomes constant. Any further decrease in R_{L }will have no effect on I_{c3} . The corresponding load current is I_{Lmax}) and is given as

**I _{Lmax = }V_{BE3}/KR_{5} + (1-K) * V_{out }**

Beyond this point V_{BE3} also drops due to saturation. Therefore according to equations given above load current I_{L} begins decreasing with decrease in R_{L} from R_{L(min}), as illustrated in figure given above. When load resistance R_{L} is zero that is, when output terminals get shorted the output voltage V_{out} becomes equal to zero. Substituting V_{out} = 0 in the above equation we have

**I _{SL} = V_{BE3}/KR_{5}**

That is, shorted-load current I_{SL} is very much smaller than maximum load current I_{Lmax}) proving the foldback current limiting. The smaller I_{SL} limits power dissipation

in pass transistor Q_{1} preventing it from being damaged. This is the main advantage of this circuit.

**Transistor Current Regulator**

** **Transistor current regulator using a Zener diode and a PNP transistor is shown in the figure below. The main function of this regulator is to maintain a fixed current through the load despite changes in terminal voltage.

Suppose due to increase in output voltage V_{out} load current I_{L} increases. This causes an increase in collector current I_{c}, because I_{c} = I_{L}.

The increase in I_{c} causes an increase in I_{E} (because I_{E} = I_{c}) resulting in decrease in V_{EB} due to increased voltage drop across R_{E}. With decrease in V_{EB}, conduction level reduces and collector current decreases. Thus load current is maintained constant.

A similar logic applies in case of decrease in load current I_{L}.

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