How to parallel wire 2 LED's, one 5W 6Vf 600ma (50% output=300mA) LED and one 3.2Vf 20mA 5mm

Discussion in 'General Electronics Chat' started by arjenvde, Aug 9, 2018.

  1. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    I need to power two LED's through 2 leads. The main 5W is connected through 20 ohm to 12 VDC to run at limited output. How can I get proper V and current from large LED terminals to run a tiny LED at limited power as well? Go to arjenvandeneerenbeemt on Facebook to see the art project for which I need the circuit. Thank you for your help! Arjen vd Eerenbeemt

    LV1aLR.jpg
     
    Last edited by a moderator: Aug 9, 2018
  2. ebp

    Well-Known Member

    Feb 8, 2018
    1,822
    632
    Treat the forward voltage of the high-power LED as if it were a supply voltage in order to calculated the resistor for the small LED. Put the resistor in series with the small LED and the resistor & small LED across the big LED. The current through the low-power LED is so small relative to the current through the high-power LED that it can be ignored in terms of its effect on the big LED's current.

    R = (6 - 3.2) V / 20 mA = 140 ohms

    If the high power LED goes open-circuit for any reason, the small LED will probably be destroyed, but presumably that isn't a serious issue.
     
    AliceLu likes this.
  3. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Thanks bro, that is what I was thinking too in terms of current load. Thank you for the 140 calculation. Greetings, Ar
     
  4. WBahn

    Moderator

    Mar 31, 2012
    22,986
    6,883
    What points do you have access to?

    +12 V ---- A ---- (20 Ω) ----- B ----- ( 5W ) ----- C ----- GND

    If you can connect to points A and B, then you can treat them as independent parallel circuits. To get 20 mA through the 3.2 V LED, you just need a (12 V - 3.2 V) / 20 mA = 440 Ω resistor (470 Ω is a standard size that should yield about 19 mA, though you might find 430 Ω as well giving a bit over 20 mA).

    If you can only connect between B and C, then use a 140 Ω resistor (150 Ω is the nearest standard, yielding about 19 mA). This will decrease the current in the main LED by about the same amount that is flowing in the small LED unless you lower the 20 Ω resistor to 18.75 Ω (18 Ω is a standard value, which might increase the big LED current to about 310 mA),

    If you can only connect between A and B, then again use a 150 Ω value. This time the big LED current will increase by about the same amount as is flowing in the 20 Ω resistor, so you could increase it (22 Ω being the next standard size).
     
  5. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Max current for large LED is 600mA, want to drive it at 40-50%. I have access to B/C for small LED. If small LED runs at half power I am happy. Key is to prolong longevity for both components as it is an art object retailing for 3K. Thank you for your input!
     
  6. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Got it working. Another problem: The 5W LED generates a fair amount of heat when running it in series with 40 ohms off 12 VDC.

    Led Calculator says 20 ohms to provide 300mA if I want to drive LED at half power.
    Is it correct to think half the max current cuts lumens output in half ?
    Or 40 ohms equates 25% output?
    Do I have to worry about thermal runaway if I run LED with 40 ohm resistor to drive the 5W power led off the 12 VDC supply?
    Wish I had an engineer buddy in my studio. LSR3wxLR.jpg
     
  7. ebp

    Well-Known Member

    Feb 8, 2018
    1,822
    632
    Light output at half current will be about "one stop" down (50% brightness), at quarter current about two stops down (25% brightness) - luminance is more or less proportional to current.

    300 mA at 6 V is 1.8 W total input. Without a spec sheet for the LED in question, I'd estimate about two thirds of that will be heat. 1.2 watts is enough to make a small thing quite hot. If you can stand to keep a finger on in, it is probably OK. LEDs in that power range, when operated at full nominal power, usually require some sort of heatsink. That can be easy if the LED is on a screw-mountable substrate, but difficult if it is intended to be soldered to foil on a circuit board.

    Can you post any more info (e.g. manufacturer's part number) on the LED, or a photo of it? That would help a lot with making suggestions for what you might use for a heatsink.

    Do you have easy access to aluminum oxide powder, as is sometimes used in glazes for ceramics? (it can be mixed with ordinary epoxy adhesive to make a pretty decent thermally-conductive adhesive that can be used to glue the LED to a heatsink - a gram would be more than enough)
     
  8. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Thank you for your input. I assumed the same lumens/current ratio you mentioned.

    I superglued the Chanzon 5W LED component onto a quarter sized starboard which I ground down to penny size for design specs. Then encase LED component onto starboard with superglue/baking soda glasslike hard mix. Works well.

    Will not grind starboard down again, too small. Thinking of attaching 1-2 thin disks to starboard to increase cooling surface. Starboard expansion suggestions are welcome.

    Is heat/lumens ration always 60/40 or does this vary with % vs max current/output? Have no interest in running LED's at more than 40-50% output as I need to keep heatsink tiny due to design. Quarter size or 25% bigger, no more.
     
  9. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Does anyone have a link to an article on the current/lumens/temp interactive dynamics of LED's? I am not clear on LED temp runaway dynamics. I will probably buy a IR thermometer gadget on Amazon (15 bucks only) to monitor LED temperature status and experiment with diff. current/ heatsink surfaces.
     
  10. arjenvde

    Thread Starter New Member

    Jun 26, 2018
    7
    0
    Hi Alice, no thank you.
     
Loading...