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Controlled Transistor Series Regulator With Overload and Sho

2014-12-13 02:07  
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If the load resistance RLis reduced or load terminals are shorted acciden-tally, a very large load current will flow. It may destroy the pass transistor Q1, 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.

Controlled Transistor Series Regulator With Short Circuit Protection

Controlled Transistor Series Regulator With Short Circuit Protection

The current limit-ing circuit consists of a transistor Q3and a re-sistor R5(approxi-mately 1 ohm) connected between base and emitter terminals of transistor Q3. With normal load current, transistor Q3remains off because the voltage drop across resistor R5is small (less than about 0.7 V necessary for making the transistor Q3on). 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 R5becomes large enough to turn transistor Q3on. The collector current of transistor Q3flows through R3thereby decreasing the base voltage of transistor Q1.This results in reduction of the conduction level of transistor Q1.Thus further increase in load current is prevented.

Figure? summarizes the current limiting. When load resistance RLis infinite, the output voltage is regulated and has a value of VREQ. The load current ILis zero for this operating condition. When RLdecreases, the load current ILincreases upto the point where RLbecomes equal to RL(min)At this minimum load resist-ance, ILequals 600mA and VBEequals 0.6V. Beyond this point, transistor Q3turns on and the current limiting sets in. Further decrease in RLproduces decrease input voltage, and regulation is lost. When RLis zero, the load current ILis limited to a value between 600 m A and 700 m A. The load current with shorted-load terminals is symbolized as ISL. When the load terminals are shorted in fig. 30.9, the volt-age across resistor R5is

VBE = ISLR5or ISL= VBE/ R5

where VBEis typically between 0.6 and 0.7 V.

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

RL(MIN)= VREG/ ISL

The exact value of RL(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

Pv= (vin-vBE)ISL.

where VBEis the base-emitter voltage of Q3, 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 Q1while the regulator remains short-circuited. The foldback current limiting circuit is the solution of above problem.

Foldback Current Limiting Circuit

Foldback Current Limiting Circuit

The circuit of a transistor series voltage regulator with foldback current limiting facility is illustrated in the figure. In this circuit base of transistor Q3is biased by a voltage divider network consisting of resistors R6and R7. The load current ILflows through resistor R5, causing a voltage drop of IlR5(approximately) across it. Thus a voltage of (IlR5Vout) acts across the voltage divider (R6– R7) network. The voltage applied to the base of transistor Q3is equal to the voltage drop across resistor R7and is given as

VB3= (R7/ R6R7) (ILR5Vout)

Emitter of transistor Q3is connected to the positive terminal of Vout. Applying Kirchhoff’s voltage law to closed mesh of Q3shown in the figure we have

VoutVBE3=VB3

or VBE3= VBa– Vout= K (ILR5Vout) – Vout= K ILR5(K – 1) Voutwhere K = R7/ R6R7

Thus the magnitude of base drive of transistor Q1is by this Vbe3. Now if load resistance decreases, may be due to any reason, load current ILwill increase causing voltage drop Il,R5to increase. This causes VB3to increase and therefore Vbe3to increase. This makes transistor Q3on in a stronger way. The increased collector current Ic3of transistor Q3flows through the resistor R3thereby decreasing the base voltage of transistor Q1This results in reduction of the conduction level of transistor Q1.Thus further increase in load current is prevented.

From the equations above it is obvious that VBEin this circuit is much more than that was in circuit illustrated in the above figure(only ILR5). It means that the increment in load current is limited by larger amount in circuit shown in figure.

Due to reduction in load resistance RL, VBE3increases to a level so that transistor Q3gets saturated. Now collector current Ic3becomes constant. Any further decrease in RLwill have no effect on Ic3. The corresponding load current is ILmax) and is given as

ILmax =VBE3/KR5(1-K) * Vout

foldback current limiting circuit graph

Beyond this point VBE3also drops due to satura-tion. Therefore according to equations given above load cur-rent ILbegins decreasing with decrease in RLfrom RL(min), as illustrated in figure given above. When load resistance RLis zero that is, when output terminals get shorted the output voltage Voutbecomes equal to zero. Substituting Vout= 0 in the above? equation we have

ISL= VBE3/KR5

That is, shorted-load current ISLis very much smaller than maximum load current ILmax) proving the foldback cur-rent limiting. The smaller ISLlimits power dissipation

in pass transistor Q1preventing it from being damaged. This is the main advantage of this circuit.

Transistor Current Regulator

Tran-sistor 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.

transistor current regulator

transistor current regulator

Suppose due to increase in output voltage Voutload current ILincreases. This causes an increase in collector current Ic, because Ic= IL.

The increase in Iccauses an increase in IE(because IE= Ic) resulting in decrease in VEBdue to increased voltage drop across RE. With decrease in VEB, conduction level reduces and collector current decreases. Thus load cur-rent is maintained constant.

transistor current regulator graph

transistor current regulator graph

A similar logic applies in case of decrease in load current IL.


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