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In most circumstances (unless specifically stated by an IC specification), an LED must have a resistor in series with it to prevent it destroying itself.*
For voltages between 6 to 12v, a 470 ohm resistor will be fine but to get maximum brightness, without damaging it, you should work out the correct value.
This will depend on
- The forward voltage (VF) of the LED
- The forward current (IF) of the LED
- The voltage Vin.
To work out the resistor do the following:
- Subtract the forward voltage (usually 2 to 4v) from the Vin,
- Divide the result by the LED current.
So if Vin =12v, VF = 2v and IF = 30mA then R1 would be
= (Vin - VF) / IF
= 12v - 2v) / 0.03 = 333 ohms, or 330 ohm resistor
*You can match the LED voltage exactly thus eliminating the need for a series resistor, but this is a risky practise keeping the voltage exactly right, especially if driving a display of LEDs. The outcome can be catastrophic.
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