this post was submitted on 03 Jul 2023
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Hey, thanks for your comment. I looked at using a resistor in series with the LED, but if my calculations were correct I could only power the LED less than 3W and 2W would be wasted.
R = (V~s~-V~f~)/I~f~ = (5V-2.8V)/1A = 2.2Ω
P~LED~ = V~f~*I~f~ = 2.8V*1A = 2.8W
P~Resistor~ = V~s~*I~f~-P~LED~ = 5V*1A-2.8W = 2.2W
Let me know if the calculations are correct.
Edit: Calculations
Those calculations are correct.
Since the remote control signals are short and low duty cycle, you could use a capacitor to provide the peak current for the LED without going over the maximum current of the power supply.