Controlled Transistor Series Regulator With Overload and Short-Circuit Protection

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 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.

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. With normal load current, transistor Q₃ remains off because the voltage drop across resistor R₅ is small (less than about 0.7 V necessary for making the transistor Q₃ 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 R₅ becomes large enough to turn transistor Q₃ on. The collector current of transistor Q₃ flows through R₃ 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 Q₃ 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 R₅ is

VBE = I_SL R₅ or I_SL = V_BE / R₅

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 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 Q₁ 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 Q₃ is biased by a voltage divider network consisting of resistors R₆ and R₇. The load current I_L flows through resistor R₅, causing a voltage drop of I_lR₅(approximately) across it. Thus a voltage of (IlR₅ + V_out) acts across the voltage divider (R₆ – R₇) network. The voltage applied to the base of transistor Q₃ is equal to the voltage drop across resistor R₇ and is given as

V_B3 = (R₇ / R₆ + R₇) (I_L R₅ + V_out)

Emitter of transistor Q₃ is connected to the positive terminal of V_out. Applying Kirchhoff’s voltage law to closed mesh of Q₃ shown in the figure we have

V_out + V_BE3 = V_B3

or V_BE3 = V_Ba – V_out = K (I_LR_{5 +} V_out) – V_out = K I_L R₅ + (K – 1) V_out where K = R₇ / R₆ + R₇

Thus the magnitude of base drive of transistor Q₁ 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,R₅ to increase. This causes V_B3 to increase and therefore V_be3 to increase. This makes transistor Q₃ on in a stronger way. The increased collector current I_c3 of transistor Q₃ flows through the resistor R₃ thereby decreasing the base voltage of transistor Q₁ 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_LR₅). 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 Q₃ 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/KR₅ + (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/KR₅

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 Q₁ 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.