No... A voltage divider won't work!With a high-power LED (1W or more)

the best solution is a special constant-current LED power supply. (You can build one or buy one.)

If you connect anything to a voltage divider that requires significant current, the resistance of whatever you hook-up, messes-up your calculated voltage. Voltage dividers work fine for low current "signals", but not as power supplies. LEDs make the situation

*worse* since the resistance of an LED is not constant.

LEDs are not powered (properly) by a constant

*voltage*. You need to supply a constant (or approximately constant)

*current* with enough voltage

*available* to turn-on the LED.

You can create an approximately-constant current source, by using a series resistor (the voltage gets divided a lot like a regular voltage divider). You subtract the rated LED voltage from your supply voltage to find the voltage across the resistor. Then you use

Ohm's Law to calculate the required resistance from the required current and the voltage across the resistor. (In a series circuit, the voltage is divided among the series components but the same current flows through all components.)

For best results, the supply voltage should be twice the LED voltage, so the voltage across the resistor is at least equal to the voltage across the LED. The more voltage you have across the resistor, the closer you are to a constant current source.

A series resistor works fine for regular low-power LEDs.

For high-power LEDs, you need a high-power resistor. Power is calculated as Voltage x Current. So, with 12V across the resistor, it will have to dissipate the same (approximately) 3W as each LED element.

It's inefficient, the resistors get hot, and this is the main reason for using a proper constant-current switching power supply.