Most readers are probably familiar with two-terminal current-limiting circuits, which have been around for many years. The circuit is very effective for applications where load current is relatively small and/or where the supply voltage isn’t too large. However, in the event of a fault, the power dissipated in the circuit’s “pass” transistor can easily become excessive.

In many applications, the cost, size, and weight of the components needed to handle such short-circuit power dissipation may be prohibitive. But by adding a few inexpensive components, you can adapt the circuit to provide both effective current limiting and protection for the pass transistor.

The shaded portion of Figure 1 contains a traditional two-terminal current-limiter circuit design. If these were the only circuit elements, the load current (I_{L}) and resistor R1 set Q1’s base-emitter voltage (V_{BE}) at V_{BE} ≈ I_{L}/R1. As long as V_{BE} is too small to bias Q1 ON, resistor R2 will bias transistor Q2 fully ON, and only the load resistance and the load voltage (VL) determine the load current.

If the load current increases to a point where V_{BE} is around 0.7V, though, Q1 starts to conduct and reduces Q2’s base-emitter voltage to a level that holds the load current roughly constant at a value given by: I_{CL} ≈ 0.7 V/R1.

This works fine as long as the load current remains within reasonable bounds. In the case of a short-circuit fault, however, IL can become large enough to stress the circuit. For example, if the circuit is designed to current limit at I_{CL} = 500 mA, and if the supply voltage, VS, could reach a maximum of 15 V, a short circuit across the load could dissipate almost 7.5 W in Q2. Not only must Q2 be able to handle this power with adequate margin, but additional heatsinking also could be required to keep its junction temperature at a safe level.

Adding the components outside the shaded area to the traditional design protects Q2 by cutting Q2 OFF if its power dissipation exceeds a predetermined level. In this circuit, the third transistor, Q3, remains OFF under normal conditions and has no effect on the current limiter. Yet in the event of a fault, either due to an excessive increase in supply voltage or by an abnormally low load resistance, Q3’s base-emitter voltage will become large enough to turn Q3 ON.

Turning Q3 ON clamps Q2’s base-emitter voltage almost to zero, so Q2 turns OFF and the load current drops to zero. The capacitor (C1) is in the circuit to provide a degree of filtering to prevent nuisance tripping. A value of around 10 nF to 100 nF should suit most applications. Note that in the absence of a load, the circuit draws no power at all from the supply.

The protection circuit operates on the fact that the voltage across the R3-R4 potential divider is equal to Q1’s V_{BE} plus Q2’s collector-emitter voltage. Under normal circumstances when the current limiter is active and holding the load current constant at I_{CL}, Q1’s V_{BE} is relatively constant at about 0.7 V and Q2’s collector-emitter voltage is relatively low. Thus, Q3 remains OFF.

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In the event of a fault, Q2’s collector-emitter voltage will rise and start to turn Q3 ON. As Q3 starts to turn ON it starts turning Q2 OFF, and the falling load voltage will result in an even larger voltage across the R3-R4 potential divider, driving Q3 ON even harder.

This positive feedback ensures that the protective function kicks in rapidly. The current limiter circuit will ultimately become latched OFF (I_{L} = 0) and remain that way until either the supply voltage cycles off-on or until the faulty load is disconnected.

Note that the power dissipated in Q2 is directly proportional to its collector-emitter voltage. Therefore, the voltage across the R3-R4 potential divider effectively tracks the power dissipated in Q2. The divider can then be sized to cut off Q2 during a fault when Q2’s power dissipation exceeds a safe value.

Assuming that the V_{BE} of Q1 roughly equals that of Q3, the following equation, which provides a first approximation of the required ratio in terms of the maximum safe power dissipation in Q2 (PD(max)), the base-emitter voltage (V_{BE}), and the current limit value (I_{CL}), allows determination of R3 and R4:

Consider an application where the supply voltage, VS, is normally 12 V and the circuit is to deliver a maximum current of 100 mA into a minimum normal load resistance of 80 Ω. Under these conditions, the maximum power dissipated in Q2 would be around 330 mW.

If Q2 is a 2N3906 transistor, which has a maximum power dissipation of 625 mW at 25ºC, set PD(max) = 500 mW to allow a safe margin. A value of 6.8 Ω for R1 would set I_{CL} to about 100 mA.

If Q1’s and Q3’s V_{BE} = 0.7 V, the equation produces a ratio of R4/R3 = 7.14. Standard resistor values for R4 = 6.8 kΩ and R3 = 1 kΩ are suitable choices and should result in Q2 cutting off when its power dissipation exceeds 470 mW.

The graph shows the results from a test circuit built using the resistor values quoted above, with the measured power dissipated in Q2 as a function of decreasing load resistance (*Fig. 2*) from 95 Ω to 45 Ω. All transistors were 2N3906 types, R2 was 1 kΩ, and the supply voltage was fixed at 12 V.

The black line shows how the current limiter behaved without the protection circuit (i.e., Q3, R3, and R4 omitted). Note how the power dissipated in Q2 continued to rise as the load resistance decreased, reaching dangerously high levels at 50 Ω. A short-circuited load would almost certainly destroy Q2.

The red line shows how the circuit performs with the protection circuit connected (Q3, R3, and R4 in circuit). Here, the circuit cuts off when the power in Q2 is approximately 420 mW (where the red line suddenly falls to zero), just slightly lower than the 470 mW predicted by the equation.