Use a series resistor, calculate it using this formula:
R= (V supply - V LED)/LED current
If your led has a forward voltage of 3.6 V, your power supply is 12 V and the nominal current of the LED is 500mA (which is fairly high, found only on modern high light output LEDs), then it should be:
R= (12-3.6)/0.500 = 16.8 ohms
You can use a broad range of voltages and resistor values, only make sure you don't exced the nominal LED current.
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It really depends on which LED you select. The average LED only uses 20 mA not the half amp that was suggested previously.
There are several suppliers who provide LED replacement bulbs for existing sockets. Sometimes, however, the bulb is too big to fit the fixture.
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A great troubleshooting tip...."When you eliminate the impossible, whatever remains, however improbable, must be the truth." Sir Arthur Conan Doyle
The least efficient but simplest method is to just provide a series resistance to handle the voltage drop from 12V to the diode voltage at the desired current. One of the most efficient approaches would be to design a switch mode power supply that outputs a constant current into the LED. A SMPS can be designed to handle a wide input voltage range so a failing 12V battery will produce the amount of light. With circuit miniaturization a SMPS can be incorporated into the base of many light bulbs.
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