I have a lithium-ion battery of 4.2 V maximum voltage. I want to power a 1 W LED at 3 V. What size resistor do I need?
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\$\begingroup\$ What's the volt drop of the LED? What do you mean "at 3V"? Do you mean you want to bring the battery voltage down to a 3V power rail for the LED? If so, why? What is the minimum voltage of the battery? Or are you only looking at using the battery a maximum voltage? Are you sure you want to do a 1W LED via a resistor? It would make a lot more sense to use a constant current source to power the LED to improve efficency. \$\endgroup\$– PuffafishAug 8, 2022 at 9:45
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\$\begingroup\$ That kinds of current source are already exist because they are widely used in flashlights. \$\endgroup\$– user263983Aug 8, 2022 at 14:14
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1 Answer
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Given that the battery voltage will vary between 4.2 V and 3.2 V during its discharge, a simple resistor will give you a very variable brightness. What you need is a current source.
You have a choice of a linear and a switch-mode current source. At this voltage drop, there is little to choose between them on efficiency averaged over the range of battery voltage.
If you do want to stick with a resistor, then to get 1 W at a nominal LED voltage of 3 V, you would require 330 mA to be passing through a resistor dropping 4.2-3 = 1.2 V. R = 1.2/0.33 = 3.6 Ω, which is a standard value in the E24 series. Once the battery is down to the 'typical' 3.6 V, you would get 166 mA or about 0.5 watts, and at 3.2 V battery voltage (unwise to go lower), you would have only 55 mA or 160 mW.
