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}.

## 5 Comments

The current limiting circuit consists of a transistor Q3 and a resistor R5 (approximately 1 ohm) connected between base and emitter terminals of transistor Q3* It should be Q1. Hope other theory are fine.

Hi it is correct, the corrent sensing is between base and emitter of Q3 whose collector is connected to Q1 base to remove its bias once the current exceeds the set limit.

Kazkokia nesvankybe…

Thanks for such an informative article. but i’ve pointed out something that you should look ……. in fold back protection circuit discription, don’t you think that formula for maximum load current (Imax) is wrong? i think the term (1-K)*Vout should also be divided by K*R5 . Kindly reply asap!! Thanx again