Electronic Design

Low-Dropout Current Regulator Improves LED Driver Efficiency

With the proliferation of batterypowered designs, efficient power management becomes crucial in maximizing battery life. Although light-emitting diodes (LEDs) consume a large portion of the battery current in many applications, they’re necessary and cost-effective indicators.

In a device that only has room for three 1.5-V alkaline cells (4.5 V nominal) and has 5-V logic, there’s little choice but to power the logic with a stepup switching regulator. The circuit has to operate down to 2.7 V in order to extract most of the batteries’ available energy (0.9 V per cell). Boosting the battery voltage to 5 V, only to drop 3 V across a resistor to drive an LED, wastes most of the available power. Even an 85% efficient switching regulator can only deliver 34% of the available battery power to an LED this way. If a designer had the option to use 3.3-V logic, driving LEDs through resistors would be 52% efficient.

There’s a better way to power LEDs and control their brightness. A current regulator connected between the batteries and an LED could ensure constant brightness. But conventional current limiters drop too much voltage. To form a simple current limiter, a JFET or depletion-mode MOSFET and a resistor can be used. With any available MOSFET or JFET, however, the lowest guaranteed dropout voltage is 1.8 V and 1.5 V, respectively. Either one is too high for this application. The regulator can’t drop out above 0.7 V and drive a 2-V LED consistently.

In the standard two-transistor current source used in many ICs, the base voltage of Q1 plus the saturation voltage of Q2 determine the dropout voltage, which is about 0.8 V at 10 mA (Fig. 1). Although the dropout voltage is marginal, this circuit has a more serious flaw: Q2 requires base current to turn it on. Whenever the LED is off, R2’s current increases to well over 100 µA, constantly wasting power. If R2 is returned to the collector of Q2 instead of ground, the dropout voltage jumps to 1.5 V or higher, depending upon the value of R2.

Substituting a new low-voltage MOSFET for Q2 solves both problems (Fig. 2). The latest generation of low-voltage MOSFETs is designed for 12- to 25-V circuits. Many have a maximum gate threshold voltage of 1.5 V or less. The low RDS(ON) of Q2, which drops only 20 mV, eliminates Q2’s saturation voltage. Q2’s base current is replaced by the gate leakage of Q2 (less than 100 nA) plus the collector leakage of Q1 (less than 50 nA). Since Q1 conducts very little current through R2 to reach equilibrium, its base voltage is a remarkably low 0.53 V. On the bench, the dropout voltage measured a mere 0.55 V, which is 0.25 V less than the conventional current source.

While the LED is off, R2 is guaranteed to conduct less than 150 nA (2 nA typical). Also, when the LED is off, or even when the entire device is off, the LDO current regulator can remain connected to the battery without degrading battery life. The average efficiency of powering an LED is 56% across the life of the batteries. This increases battery life up to 65% over LEDs powered by the 5-V supply. Note that neither circuit is temperaturecompensated. Therefore, the current will drop about 10% for a 23°C increase in temperature. Out of curiosity, Q1 was replaced with an NTE160 germanium transistor. A dropout voltage of only 0.26 V was measured with a collector leakage of 300 nA (very high compared to silicon). This makes a unique blend of old and new technologies.

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