I'm looking at a Multicomp OSW-8349 RGB LED, which has a Vf of 3V min / 4V max for the green and blue channels, and a Vf of 1.8V min to 2.8V max for the red channel, at 350mA. I'm planning on regulating 3.3V to drive the green and blue channels, but would like to avoid using another regulator just for the red one.

Is there a cheap way to cut the voltage for the red LED? I'm considering using the forward voltage drop of a diode to bring it down to 2.6V, but I'm not sure how well this would work, especially with the temperature varying between ~16°C and ~38°C.

  • 3
    \$\begingroup\$ You should be current limiting the LEDs, either with a constant current driver or a resistor. \$\endgroup\$
    – pjc50
    Commented Nov 21, 2012 at 14:05

2 Answers 2


Yes, use a 2Ω resistor in series.

  • \$\begingroup\$ Feel silly for not thinking of this now. Cheers. \$\endgroup\$
    – Polynomial
    Commented Nov 21, 2012 at 14:12
  • 2
    \$\begingroup\$ Shortest wrong answer ever? 2 ohm will drop the 3.3 V to 2.6 V, which may be too low for the LED to light properly. Same with the 3.3 V by the way: may be too low for the green and blue LEDs. \$\endgroup\$ Commented Aug 22, 2013 at 8:58
  • 2
    \$\begingroup\$ Least amount of research ever? "Vf of 1.8V min to 2.8V max for the red channel, at 350mA". 2.6V at 350mA is exactly what is desired according to the question and the datasheet. Same with the blue and green by the way, which operate with a Vf of 3V min to 4V max. To provide another correct short answer for your question: No. \$\endgroup\$
    – Samuel
    Commented Aug 23, 2013 at 16:37

LEDs need to be driven via current regulation rather than voltage regulation. Despite the G and B color channels being rated at 3 to 4 Volts, providing them with a 3.3 Volt regulated input without a current limiting resistor may to cause the LED to burn out.

One possible (but not practical in this case) solution is to use 3 resistors on the 3 driven legs of the LED (not just one resistor on a combined current return path). Calculate the resistance according to the current desired for each color, and the rated minimum forward voltage for each, e.g. :

R = (Vsupply - Vfwd) / I

Rr = (3.3 - 1.8) / 0.35 = 4.28 Ohms
Rg = Rb = (3.3 - 3.0) / 0.35 = 0.85 Ohms
(assuming 350 mA drive current, 3.3 Volt regulated input, no other voltage drops)

You probably will not want to look for 0.85 Ohm resistors for Green and Blue. Also the ~4.3 Ohm resistor for Red will need to be rated at 1 Watt, since the computed dissipation for it works out to 0.525 Watts. Hence this isn't a practical approach.

Please note that if these LED channels are to be switched or controlled, rather than permanently on, your switching solution (MOSFET, transistor, or something else) will introduce a voltage drop on each color's current path. This complicates the calculation further, especially when working with the 0.3 Volt headroom the proposed 3.3 Volt regulator will provide.

It would be better to use a higher voltage to drive the LED channels, dropping it down suitably with resistors, or preferably using current regulators such as SD42351 or one of the SuperTex LED driver products.

If using resistors, you need two of value Rg and one of value Rr calculated for Vsupply of 5 Volts, and wattage rating higher than the calculated power dissipation for each resistor ( = Vdrop * 350mA).


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