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Emitter-Follower Boosts Linear Regulator's Output Current


Contributing Author

June 26, 2000

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Automotive electronics often require memory for saving system information or for user inputs (the settings on a car stereo, for example). Flash memory can do the job, but flash is expensive. DRAM memory reduces the cost, but DRAM requires an always-on power supply. A standard linear regulator, such as the LM78L05, can generate the 5-V rail required for 5-V DRAM, but its supply current (as much as 5 mA) can discharge the battery over an extended period of inactivity.

On the other hand, linear regulators with low quiescent current may not have sufficient load-current capability to supply the DRAM during active read or write operations. The design solution in this situation is to add an emitter-follower output stage to a low-current, high-voltage, low-dropout (LDO) regulator (see the figure). Adding the transistor as shown, between the LDO output and load, increases the available output current without boosting the regulator’s low (15 µA) quiescent current.

Though not an issue for the application in question (regulating a 12-V battery down to 5 V), the dropout voltage (minimum-allowed VIN minus VOUT) for this circuit equals that of the regulator IC (350 mV maximum) plus the transistor’s base-emitter drop of approximately 0.7 V. The pass element in IC1 is a pnp transistor, whose saturation at low dropout voltage increases the quiescent current for that condition. To maintain low quiescent current, therefore, the regulator headroom should be kept well above 1 V.

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