I'm working on a small project where efficiency is very important (every mA counts). I need to drive two red 3528 LED's in series and only have a 9V DC source. I want the LED's to be only around half the brightness than what they would be if I drive them with the voltage/current that's in the spec sheet. If I take an adjustable switched DC converter and turn this up to the point where I like the brightness, would I still need a resistor? And would this be the most efficient way to do it? I was surprised to see that if I hook up the DC converter it already uses up 4mA even without load. That's already the same current as what the 2 LED's alone will be using. Just wondering if a 5V voltage regulator + resistor would use more current.
Electronic – Most efficient way to drive a low power LED at half brightness
dc/dc converterledresistorsvoltagevoltage-regulator
Best Answer
If you are talking about a fixed voltage converter, this will work until the temperature of the LEDs change. For example, because the LEDs are turned on and heating themselves up. As the temperature changes, the required forward voltage for a given current (and light output) changes. So this is generally a bad idea.
It is also possible to get (or design) a fixed current converter. This should work fine, but I'm not aware of any available off the shelf for 2 mA ouput.
Another option to consider is to run the LEDs at a higher current, but flash them at a low duty cycle and too fast for the eye to see.