Electronic Design

LED Switching Driver Cuts Current Draw To 3 mA

This Idea For Design was originally published March 17, 1983.

A switching transistor and inductor combination, when substituted for a power-wasting series resistor, significantly reduces the current drain of a pilot light or a status-indicator LED. The efficiency of the circuit is somewhat in excess of 74% (vs. 22% for the resistor), dropping the typical 10- to 15-mA current draw from a 9-V supply to just 3 mA. The power savings, especially in battery-powered equipment, can be important.

When a 20- to 400-kHz clock signal, with a rectangular waveform, is fed to the input, the VMOS transistor toggles on and off. When it is on, energy is stored in the inductor. Rather than being wasted in a series dropping resistor, the energy held in the inductor lights the LED during the transistor's off state. The only waste is due to transistor switching and inductor losses, and the circuit is able to deliver an average of 10 mA to the LED for a total consumption of 3 mA

By adjusting the transistor's on time, and thus the energy stored in the inductor during each clock cycle, the 10-kΩ potentiometer acts as an LED power control. For systems without an appropriate clock signal, an ICM7555 CMOS timer, operated in the astable mode, supplies the needed signal at a cost of approximately 0.25 mA (see the figure).

Because many LEDs have a reverse breakdown rating of 5 V minimum and 15 to 25 V typical, caution must be exercised in selecting the LED, which must be able to withstand the supply voltage in the reverse direction. A suitable blocking diode, such as a 1N914, placed in series with the LED permits higher voltage operation. As with all switchers, care must also be taken to ensure that the inductor does not saturate and waste energy.

Update by John S. Rohrer
Betchel Nevada, P.O. Box 809, Los Alamos, NM 87544; [email protected]

Do you need to drive an LED efficiently from a voltage higher than 4 V? Here's an extrapolation of the original Idea For Design which does this even when the voltage varies. This improved circuit has come in handy in my circuit designs, and it should be useful to other readers as well. In addition to an LED, the circuit can power any similar current-driven device with a low "on" voltage, such as a laser diode or thermoelectric cooler.

An LED is a current-driven device that typically has a 2-V voltage drop. This suggests that when the supply voltage is much higher than 2 V, an inductor could convert the high voltage to the current for the LED. The basic concept behind the original circuit is shown in Figure A. As switch SW closes, current begins to build up in the inductor. SW is opened before inductor L1 saturates. When it opens, the current in L1 flows through the LED, turning it on. The current in L1 linearly ramps to zero because the LED holds the voltage constant at about 2 V. Therefore, the average current is half the peak value.

An improved circuit configuration allows the LED to be on during the charging of the inductor so that the light output will be more constant (Fig. B). For convenience, it's configured in the opposite polarity. If SW closes when the current in L1 is just exhausted, then the LED can almost always stay on.

At this point, adding a transistor switch (Q2) with current limit (R2, Q3), and a feedback transistor (Q1), yields a peak-current-limited LED driver (Fig. C). When Q2 turns on, the voltage V jumps to VS and the current through L1 starts to increase linearly. When the current reaches the limit, the rate of change of current in L1 causes V to swing negative, turning off Q1 and Q2. The energy stored in L1 then flows through the LED and D1. When it's exhausted, resistor RB starts the next cycle. A capacitor across the LED to smooth the light output doesn't affect the operation, because the LED voltage is almost constant and independent of the current.

With the component values shown, the circuit is set up for an LED that requires 35-mA average current. It may be scaled as desired and operates as follows: R2 sets the peak current through Q2 (as well as L1 and the LED). The peak current is twice the average current. For a peak current of 70 mA, R2 = 0.7 V/70 mA = 10 Ω. A peak base current for Q2 of 1 mA should be adequate, and R1 sets it. R1 ∪ (VS − 2 V)/1 mA. For VS = 15 V, then R1 ∪ 13 kΩ. RB supplies startup current for the circuit (Q1). That resistance is arbitrarily chosen to be about seven times R1, so that RB = 82 kΩ. L1 is 470 µH, arbitrarily. It keeps the operating frequency low, yet it's physically small.

When tested on the bench, the circuit gave the following results: The period was 14 µs, including the 0.5-µs delay between cycles. The average current through the LED was 33 mA, and the average current from the 15-V supply was 7 mA—a substantial power savings.

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