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I have a project where a PCB has been manufactured with an error, where a 3 lead BI color LED (red/blue) has a single resistor to limit the current for both colors. I can fix this by hacking apart the PCB but I would like to know what resistor value I could potentially use to satisfy both the RED and BLUE colors of the LED.

This is the datasheet for the LED: http://www.unique-leds.com/images/datasheets/CG/D3009R1B2SBDC.pdf

As per the datasheet the RED has a min/max forward voltage for 1.9/2.5 while the blue has a min/max of 2.9/3.5 If I were to take an average of the mins and maxes and then the average of that:

((1.9+2.9)/2) =  2.4v min
((2.5+3.5)/2) =  3.0v max
((2.4+3.0)/2) =  2.7v averages

and use this to calculate the resistance needed to satisfy both colors:

R = (5v-2.7v)/0.02a
R = 115ohms

Would this be the correct approach to solve my problem or is there a different way I should go about this. Is this even possible without negatively effecting one LED color or the other?

I can manage two separate resistors if I tear up the solder mask and hack it until it works but if I can use the already provided through hole it would make life easier.

Overall I am asking what is the best way to calculate the resistance that would satisfy two colors, RED and BLUE,in a single LED package with 3 leads using a single resistor.

Thank you for your time.

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According to your datasheet, the claimed brightness is the same at 20mA for either LED. So if you calculate a resistor to allow 20mA to flow through the BLUE LED then the current through the RED LED will be excessive and its life will be shortened.

Suggest you try it with a resistor calculated to allow 20mA for the RED LED (say 150 ohms) and see if the brightness is visually okay on the blue. It may be just fine. The current through the BLUE LED will then be about 15mA, which should be plenty bright. I suspect this is all you have to do.

If it isn't you could do something else that wouldn't involve hacking the board- you could parallel the RED LED with a resistor that would steal some of the current. Perhaps a small surface-mount resistor between the LED leads. So suppose the resistor you use is 100 ohms, then the current through the BLUE LED will be about 18mA, but the current through the RED LED would be more like 28mA. You could put a 220 ohm resistor in parallel with the RED LED to shunt away 10mA. But I doubt this is necessary- a 50% increase in brightness is really not that visually noticeable.

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    \$\begingroup\$ Thank you for the response. I will try running the blue LED with the resistor calculated for the red side and see if I like the brightness. If not I have a few options to fix the PCB or the LEDS in such a way to make it work, but it is a bit more work then there should be. Thanks again! \$\endgroup\$ – randy newfield Dec 1 '14 at 1:38
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A bi-color LED with three leads is just two LEDs sharing either a cathode or an anode. If you want only one resistor, then you can have only one color on at a time. You might do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

If you try to turn both on at the same time, then only the one with the lower forward voltage (red, in your case) will turn on.

If you exceed the maximum current for an LED, it overheats and is damaged. Less current just makes the LED less bright, but does not damage it. So, the thing to is calculate the resistor you might use for each one individually, then use the biggest resistor of all the options.

If the difference in brightness is unacceptable, or if you need the ability to have both colors on at once, you must have two resistors.

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  • \$\begingroup\$ Thank you for the response. I only need one LED lit at a given time as determined by a switch. As per the other answer I determined that I should use the maximum RED resistance and see if the blue brightness is acceptable. Thank you again kindly for your response. \$\endgroup\$ – randy newfield Dec 1 '14 at 1:41

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