To get the maximum brightness and working life out of a high-power LED, it needs to be driven at the optimum specified current. Allowing the current to exceed the permitted value is to be avoided at all costs, since it will severely affect the life of the device. A power supply or a battery with a small current-limiting resistor is not really an ideal solution, since not only is energy wasted in heating the resistor, but, if a small value is chosen to minimise this wastage, small changes in the applied voltage will lead to large changes in the current that flows. It is well known that LEDs have a small dynamic resistance in the neighbourhood of their optimal operating point. We will therefore need more in the way of electronics than a simple series resistor to meet our requirements.
The most direct way to provide a highly constant current in the face of relatively small changes in supply voltage is to use a conventional regulated current source. Unfortunately this type of circuit unnecessarily wastes energy in its series transistor, which rather detracts from the charm of using a semiconductor-based circuit. The inefficiency can be mitigated by using a modern device such as a power MOSFET as the series component. Power loss is then limited to that in any current sense resistor that might be used and the dissipation in the relatively small ‘on’ resistance of the switching transistor. The circuit suggested here drives a commercially- available Luxeon LED using a BUZ71. The 5 W version of this LED draws 0.7 A.This means that 0.175 V is dropped across R9, making for a power dissipation of 122 mW. T1 has a typical resistance in the ‘on’ state of 85 mΩ. In the ideal scenario this means that about 60 mV is dropped across it, for a dissipation of at least 42 mW. The supply voltage therefore needs to be about 230 mV higher than the nominal voltageof the LED (6.85 V). To have something in reserve, 7.2 V, allowing 0.35 V for T1 and R9, is a good compromise. Serendipitously, a series of six NiCd or NiMH cells will give almost exactly this value under load!
A further happy coincidence is that an unregulated mains power supply with a 6 V transformer with bridge rectifier and smoothing capacitor will also give us almost exactly our target voltage when loaded. A 7.5 VA transformer is suitable, along with a 2200 μF/16 V electrolytic smoothing capacitor. Now to how it works. D1 acts as a reference, with a voltage of 2.5 V being dropped across it. IC1b, together with T1, form a current source, whose current can be set between 360 mA and 750 mA using P2. The otherwise unused opamp IC1a is connected to form an under-voltage cutout switch which prevents a connected battery from being discharged too deeply. The threshold point is set using P1. IC1a is configured as a comparator with a small hysteresis. If its output is high, IC1b is fooled into thinking that the current through R9 is too high, whereupon it switches off the LED. The same happens if R1 is not shorted by a switch. For the purposes of this circuit only opamps with input stages constructed using PNP transistors should be used. One last look at the energy budget: if six cells are used, the average voltage during discharge will be around 7.4 V. Subtract the nominal voltage of the LED and 0.55 V is left to be converted into heat. About 0.4 W will be dissipated by T1, which therefore will not require cooling. Efficiency is very good, at over 90 %.