Maintenance notice: These forum archives are read-only, and will be removed shortly. Please visit our forums at their new location,


edited July 2014 in General
i recently bought some mbi5026's and a little bag of common anode rgb leds from you guys.  i have finally gotten around to trying to use them, but have some concerns about the information on page 12 of the mbi5026 data sheet.  if i understand it correctly the supply voltage for the leds should be between .4 and 1.0 volts above the forward voltage of the leds.  that will be kind of tough with common anode leds.  will i have to select resistors to sit between the various cathode pins and the 5026 such that 0.030*r is about vled - vf + .4v for each color?

if this is the case, maybe i ought to go with surface mount leds with anode and cathode connections for each color and figure out how to come up with three different supply voltages.


  • edited July 2014
    I would definitely skip the resistors if possible-- and it certainly doesn't look to me like they're needed in the application that you're describing.

    The note on page 12 is based upon how it is designed to work given power dissipation considerations in the IC package. It is not a statement about the most voltage that is allowed there.

    But, let's see where that voltage value comes from, and what it really means. At the device maximum current of 90 mA per channel, with 1 V between LED voltage and power supply voltage per channel, the total power dissipation is P = I*V = .09 A * 1 V = 90 mW per channel. With 16 channels, that's 16 * 90 mW = 1.44 W.  That value is consistent with the maximum power dissipation for some of the smaller packages.  The 'GN (RoHS DIP) package that we use actually allows up to 2 W, so there's actually some additional safety headroom for us.

    If you wanted to stay below that level of power dissipation with only 30 mA per channel (the limit of those RGB LEDs), you would need to have less than 3 V headroom (on average) between the LED voltage and the power rail.

    Suppose that V_F is 1.8 V for the red LEDs and 3.6 V for green and blue, and that we have 6 red, 5 green, and 5 blue LEDs. Then with a 5 V power rail, the average drop is ((5 V - 3.6 V)*10 + (5 V - 1.8 V)*6)/16 = 2.075 V, and the total power dissipation is 16 * 2.075 V * 0.03 A = 966 mW.  That's about 2/3 of the power dissipation in the example given on page 12, and well within the safe operating conditions of the device.   If you wanted to run a bit cooler yet, pick a slightly lower system voltage, maybe 4 V or 4.5 V, which will make it so that the MBI5026 will run cooler yet.
  • ok.  thanks for the incredibly detailed reply.  i was thinking of connecting all of the leds of each color to one chip in case i wanted to try to tweak the color balance.  still 16*2.1*.03 is only about 1.4 watts.  or .92 watts if i run at 4v.  guess i'm good.

    thanks, again.
Sign In or Register to comment.