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- Thread starter WindermereColvic
- Start date

If you're stuck pm me with the current in mA, but I might not be able to answer for a few days.

Cheers

The '1.2' is correct for an LED but you will already have a resistor to allow the lamp to work at 3.6 volts so the formula should be:

i.e. the charging voltage (say 14V) less 3.6, and divide by I (in mA) for a result in kOhms. So if it takes 30mA you get (14-3.6)/30 = 0.34 kOhms so use 330 Ohms.

The resistor will need to dump 0.33 Watts or greater.

Hope that clears it up.

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