As white LEDs become more pervasive, there are ever-more options for driving them.
Edited by Leland Teschler
Linear constantcurrent drivers, for instance, offer low EMI, low cost, and use only a few components. Many designers favor them for low-current applications where power loss is a minor concern. If power dissipation is a problem, or if the application needs more current, you can parallel two or more drivers. Two drivers in parallel provide twice as much current and split power loss into two locations, which makes for easier heat dissipation.
High-side pass elements are preferred, especially in automotive uses, because they let you connect only a single current-limited wire (chassis return) to the load. To configure parallel drivers with high-side pass elements, however, the current-sense feedback circuitry must also be on the high side, and it must be able to withstand at least the voltage the LED load creates. Thus, linear drivers pose a dilemma if they have either low-side current-sense feedback or a requirement for limited common- mode voltage on the sense inputs.
Suppose we want to drive an LED string with 400 mA. Drivers U1 and U2 (MAX16803) are good for automotive use because they have a high input-voltage range, excellent thermal qualities, and the automotive temperature range. Their maximum output current, however, is 350 mA. Also, with maximum input voltages of 16 V, each IC would dissipate 3.6 W, far too much for one package to handle.
One solution is to use two drivers, with each handling half the current and generating half the heat. If the drivers are to share current equally, their current-sense resistors must be on the high (anode) side of the LEDs. That arrangement introduces another problem: Forward voltage for the three LEDs in series reaches almost 8 V, but the maximum rating for the currentsense (CS) pins is only 6 V. A matched transistor pair (Q1) solves this problem by translating the current-sense voltage (developed across R1) onto R3. Because the bases connect together, voltages at the emitters are nearly the same, and therefore R2 IR2 equals R1 IR1. High gain in the transistors means negligible base current, so IR3 is approximately equal to IR2. You can set the output current as follows:
R2 X V;Sense
R3 X R1
where VSense = the current-sense voltage between CS+ and CS- (203 mV). For best accuracy, choose R4 so the two transistor currents are about the same:
VLed - 0.7
where VLED = the nominal LED voltage at the desired load. This additional low-cost circuitry keeps the CS-pin voltages near ground, and lets the sense resistor (R1) be at any voltage up to the 45-V limit of U1.
The second driver must have an identical circuit. Tie the driver outputs together to get the full LED current. If you need more current or power-handling capability, you can add more drivers in parallel.
— Jim Christensen
Maxim Integrated Products Inc.