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I have 3.3V switching power supply. I want to power 3W RGB LED.

Forward voltages are: 2.2~2.8V(Red), 3.0~3.8V(Green), 3.0~3.8V(Blue) Current: ~350mA

Should I use current limiting resistors? As I understand, for Red I need to drop 3.3V to under 2.8V and the resistor is necessary, but how to be with Green and Blue? Can I connect them directly?

Please explain.

Thanks.

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  • \$\begingroup\$ Do you have a datasheet for the LED? \$\endgroup\$ – CHendrix Jan 25 '17 at 14:59
  • \$\begingroup\$ You should always use some form of current limiting, whether a constant current source or resistor. \$\endgroup\$ – Colin Jan 25 '17 at 14:59
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    \$\begingroup\$ Green and blue cannot be lit. Possible if Vf is exactly 3.3 or lesser but you never know. The variation of forward voltage and current is too high. You may have to pump up the voltage using boost regulators and then have constant current circuit or simple resistors limiting current. \$\endgroup\$ – Umar Jan 25 '17 at 15:12
  • \$\begingroup\$ @CHendrix These are cheap chinese LEDs, something like aliexpress.com/item/… \$\endgroup\$ – Roman Simonyan Jan 25 '17 at 15:14
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    \$\begingroup\$ Look at en.wikipedia.org/wiki/Light-emitting_diode#/media/… once your powersupply voltage is greater than the LED's forward voltage current rises very quickly. Matching your power supply exactly to the LED's Vf is a dodgy game as the Vf will change with temperature. \$\endgroup\$ – Colin Jan 25 '17 at 15:21
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You can't get there from here.

3.3V is not enough the reliably light the green and blue parts. and the red part with only between 0.5 and 1.1V spare may give variable performance if you use a 3.1 ohm resistor.

look instead for switched-mode LED driver that's compatible with your supply and LED.

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I think it won't be a problem, for diodes have a particular current value for each voltage, thus for 350mA rated current, rated voltage would be the point on its V-T characteristics plot.

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    \$\begingroup\$ There will be a problem, once the diode exceeds it's drop voltage it ligths up proportional to the intensity of the current, if no resistor is used then the current will be limited by the power supply's internal resistance implying any small peak of voltage will turn into a huge amount of increase of current burning the LED. \$\endgroup\$ – Marcelo Espinoza Vargas Feb 25 '17 at 19:34

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