LEDs are semiconductors which convert electricity into light. Most LEDs have an operating voltage between 1.7 to slightly less than 4 volts, so when an LED is marked as a 12 volt LED consumers are often confused into thinking something must be special about these LEDs. In fact, these are nothing more than a normal LED with an appropriately sized resistor attached in series with it to supply it with the correct amount of current despite the extra voltage. In essence, the extra voltage is burned off, in the form of heat, through the resistor. To run a 12-volt LED with a 24-volt supply requires nothing more than a few calculations and an extra resistor.

Calculate the value of the resistor to be added. Since the LED module or bulb in question was originally designed to be driven by a 12-volt supply and now must be driven by 24 volts, subtract the old voltage from the new voltage. 24-12 = 12. The new power supply is 12 volts higher, and this is extra energy that the LED was not designed to handle. The simplest way is to burn it off in the form of heat through a resistor.

Calculate the value of the resistor to burn the excess energy. Divide 12 volts, which is the voltage difference, calculated in Step 1, by 50mA (or whatever current the LED in question uses). 12 volts / 50mA = 240 ohms. The LED module was already designed to be directly driven by a 12 volt supply, this is the reason it is not necessary to calculate the resistor value starting from the full 24 volts.

Determine the power requirements for the resistor. Power equals current squared times resistance. So in this case, P = 50mA^2 * 240 ohms or P = 0.05^2 * 240 ohms, which equals 0.6 watts. This is too high for a 1/2-watt resistor, which is likely to be found at an electronics supply store. Instead of special ordering a 1-watt resistor, simply connect two half-watt 120 ohm resistors in series. They will act as a single 240 -ohm, 1-watt resistor.