silvio2402, 1 year ago (edited 1 year ago) 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 = (Vs-Vf)/If = (5V-2.8V)/1A = 2.2Ω PLED = VfIf = 2.8V1A = 2.8W PResistor = VsIf-PLED = 5V1A-2.8W = 2.2W Let me know if the calculations are correct. Edit: Calculations
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 = (Vs-Vf)/If = (5V-2.8V)/1A = 2.2Ω
PLED = VfIf = 2.8V1A = 2.8W
PResistor = VsIf-PLED = 5V1A-2.8W = 2.2W
Let me know if the calculations are correct.
Edit: Calculations