Driving high-intensity LEDs in series ensures uniform brightness, but that approach typically requires a supply voltage greater than the total forward voltage across the LED string. A switch-mode boost converter can efficiently drive the LED string from inputs less than the LEDs’ forward voltage, but such circuits aren’t usually capable of regulating current through the LEDs as well. Adding some resistors and two garden-variety transistors allows the circuit to boost voltage and regulate current at the same time (see the figure).
Operating from 5 V, the standard boost converter (IC1, L1, D1, and all of the capacitors) applies a voltage across the string that’s sufficient to produce the LED current set by the value of current-sense resistor R4. The top of R4 could connect directly to the feedback terminal (pin 3), but R4 would then drop 1.5 V (the sense threshold internal to IC1). Consequently, it would dissipate excessive power equal to 1.5 V × ILED. R5 and R6 form a divider with Vin that introduces an offset, shifting the low-level feedback from R4 to a level usable at the FB terminal. Together, the R4-R6 values shown set the LED current at 88 mA.
When the SHDN terminal is pulled high, the switching regulator turns off. To ensure that the LEDs turn off completely during shutdown, the on/off signal also turns off a pnp transistor in series with the LEDs (Q1). Q2 and the low-battery comparator in IC1 form a protection circuit. Without it, the loss of feedback attending an open circuit in the LED string would allow excessive voltage to build on C3. If this voltage reaches the 15-V limit determined by R1 and R2, the open-drain switch at LBO turns off, allowing Q2 to turn the circuit off by pulling FB high.
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