Many RGB leds with a few resistors?

Discussion in 'The Projects Forum' started by jmh226, Jun 4, 2018.

  1. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Hi all,
    I am designing a "Plinko" game board, with an rgb led embedded into each of the clear acrylic pegs. My question is do I have to solder in 3 resistors per led for every peg, or can I run all of the red wires together, connect to a much larger resistor, and do the same for the red, and green, so that I only have 3 resistors total. My leds are common anode, red vf is 2.2, blue vf is 3.3 and green vf is 3.3. I am running from a 12v controller, so at 20 milliamps my resistor values are 490 ohms for red, and 435 ohms for blue and green. For a single led, I would need 1/4 watt resistors P = (12-2.2) * .02 = .196 Watts, so if I bundle all the leads and I am running 100 leds, could I use a 490 ohm, 20 Watt resistor instead? I think that the theory is sound (and it's a lot less work) What do you think? If this is a possibility, and I can only find 20 W resistors in 250 ohms, I could just run through them in series and the resistance adds, right? Thanks!
     
  2. Hymie

    Active Member

    Mar 30, 2018
    554
    131
    The problem you will encounter if you use a single dropper resistor for many LEDs, is that the LED with the lowest forward operating voltage will limit the voltage to all the other LEDs, resulting in a range of levels of illumination across the LEDs.

    You could experiment with a small number of LEDs to see if the illumination variation was likely to be acceptable for your application.

    Ebay if your friend for the power resistors you require.
     
  3. whitehaired novice

    Member

    Jul 15, 2017
    239
    28
    I do not know the game Plinko--how many LEDs might be on at the same time? Your calculation of needing a 20 watt resistor leads me to think all of them, all the time, but this seems unlikely in any game.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    If you bundle all of the LEDs in parallel (assuming you didn't have thermal runaway issues, which I'm not willing to rule out too easily), then the total current you would need for each color would be 100 * 20 mA = 2 A and the resistor value would need to be 4.9 Ω, not 490 Ω. Your calculation of 20 W is right, but you never want to run a resistor near it's maximum power dissipation. A common rule of thumb is to give it a factor of two, so you would need 40 W resistors.

    You have a number of issues with this approach. Let's consider some of them.

    1) If all of the LEDs are on at the same time, then you are looking at 6 A of current from a 12 V supply, or 72 W of power that has to get dissipated. Some of it will leave as light, but the vast majority of it will have to be given off as heat. Your resistors will dissipate about 54 W of it. Imagine putting a 60 W lightbulb in the location where the resistors are going to be. Would you want it there from a heat standpoint? A good fraction of the 12 W or so that the LEDs are absorbing is going to get dissipated as heat, as well. It's spread out across the LEDs, but if they are bunched together closely that's still a lot of heat to get rid of.

    2) Your approach means that all of the LEDs of a given color are either all on or all off. You can't control them individually (or in smaller groups). Is that what you need?

    3) If your LEDs are rated at 20 mA, you generally want to run them at well under that. A rule of thumb is to not exceed 85% of that and many people don't exceed 50%. But the human eye is quite insensitive to differences of a factor of two in light intensity. You might try your LEDs at 20 mA and at 10 mA and see if you can tell much of a difference. Don't put them side by side -- that's not fair. Instead, turn it on at one level, then turn it if and go away for a bit and come back and try it at the other level. Ask yourself if the difference between the two is important for your application. Running at 50% will not only make your LEDs last about forever, but it will cut your power dissipation in half.

    4) Running 100 LEDs in parallel is inviting big problems with thermal runaway. Each LED will have it's own voltage (so, near 2.2 V at 20 mA for the red LEDs, but all slightly different). So if they are all forced to have the same voltage by putting them in parallel, some will require more current than the others in order to sustain that voltage and others will require less. But the ones that require more will be dissipating more power and so they will heat up. But LEDs exhibit a negative temperature coefficient in terms of their voltage (for a given current), which means that the voltage needed to sustain the current running through the hot ones will drop and that will result in them hogging more current in order to keep the voltage the same. The other LEDs will then get less current. The aggravates the situation and the hot ones draw even more power and heat even more. This is known as "thermal runaway". With a few LEDs this often reaches an equilibrium point that works "well enough", but with 100 LEDs you can quickly end up with one or more LEDs operating at such a high current level that they fail in short order. But since they will usually fail as an open, that means that the original total current is spread out among fewer LEDs, so they all get brighter and the ones with the next lowest voltage drop will continue the process and burn out even quickly. You can end up with a cascade effect in which all of your LEDs are destroyed in a surprisingly short period of time.

    In order to help you come at a good approach, we need to have more information about how the LEDs will be used. How many will be on at one time? What's the average amount of time each one will be on? What the maximum number that will be on at one time and for how long? What granularity do you need in controlling which one are on and which ones are off?
     
    -live wire- likes this.
  5. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Thanks so much to everyone for your thoughtful responses. My design documents are at work, so I will post a follow up with more details tomorrow!
     
  6. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Ok, so I've attached a picture of the plinko board schematic. The 10 triangular shapes on the sides and the 7 bars at the top are 3/4" clear obscured acrylic, and will be lit from the side for the triangles, and from the back for the bars with rgb led tape that I got here: (https://www.superbrightleds.com/mor...mds-per-ft-3-chip-rgb-smd-led-5050/1473/2892/) This will be controlled by an rgb led controller that I got here (https://www.superbrightleds.com/mor...olor-changing-modes-5-amps-channel/3050/6263/) The controller puts out 12v and can handle 5 amps per channel. My plan is to use the same controller to control a single rgb led that is embedded into the middle of each of the 58 (not 100) clear acrylic pegs. My power supply is 50W/12V (so not able to put out anywhere near the 15 amps that the controller could handle!). The bottom bars will be lighted clear only whenever the puck falls into a particular bay, but that is a whole separate project...

    How is it that a 16.5" loot led strip with 300 rgb leds only pulls 3.3 amps, but if I want to run 100 (actually only 58 though) individual rgb leds I am pulling 6 amps? Do I need much less powerful RGB leds? The ones that I got are here: https://www.mouser.com/datasheet/2/216/WP154A4SEJ3VBDZGC-CA-1145208.pdf

    If I transformed the 12v out of the controller down to 5v before connecting to the individual rgb leds that seems like it would eliminate a lot of my problems perhaps? But would the PWM from the controller still work, or would transforming the voltage screw that up?

    I basically want all of the leds on at the same time, (strips and singles) doing the same thing, and controlled by the controller. Is that just too much to ask :)

    Thanks everyone!!!
     
  7. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    I can't tell for sure, but it looks like that LED strip has the LEDs organized into 3 LEDs in series, with each group of three having it's own current limiting resistor. If each LED has 10 mA of current in it, that would then be a total of 3 A.

    But since your LEDs are all in parallel (a constraint imposed by having them all be common-anode), that not an option for you.

    Now, if you are willing to not have all of your anodes tied together, you could play the same game.
     
  8. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Is there a non-common led package in a non surface mount? I can't seem to find one... do you know if such a thing exists? A surface mount won't work for my application.
     
  9. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    I have no idea what a "non-common" LED package would be? LEDs come in a variety of packages, why do you only care that the one you use be "non-common"? What are the constraints you are trying to satisfy?
     
  10. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Being able to wire them in series for better efficiency, so basically what I'm looking for is three separate led chips (RGB) in a single 5mm package with no common pins.
     
  11. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    That's MUCH more useful.

    I don't know that you will find that in a 5mm package since six pins is getting awfully crowded. Sharp used to make a 6-pin DIP part that had independent RGB LEDs in it, but I think they stopped production quite some time ago.

    I would think that what you are trying to do would lend itself to surface mount since the acrylic posts should act as pretty good light pipes. Have you explored that possibility?
     
  12. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    I'm nervous about the very fine scale soldering with these chips...I'm fairly new to electronics and don't have very high quality tools at present, nor much practice at soldering. I can solder water pipes like nobody's business, but that's quite a bit different :) Also, how would I mount the light emitting surface of the surface mount led to the acrylic? For a 5mm package I was planning on drilling a hole into the acrylic and stuffing the led inside. Can I glue/silicone the light emitting surface of the led directly to the acrylic?
     
  13. DNA Robotics

    Member

    Jun 13, 2014
    397
    146
    -live wire- likes this.
  14. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    I ordered some smd leds and some 5mm leds from mouser, and will let you know how I make out once I've trialed a few :)
     
  15. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Ok - I got some smd leds from mouser, and they are a no go - way too big a pain to solder 58 of them together. I'm going with these: https://www.mouser.com/datasheet/2/216/WP154A4SEJ3VBDZGC-CA-1145208.pdf which, at 20mA, match with my led strips as far as color and intensity. When calculating resistance with ohm's law and a 12v source at 20mA: for red I get 12-2.2/.02 = 490 ohms, and for green/blue I get 12-3.3/.02 = 435 ohms. The closest resistors I found to these values are 493ohm and 432ohm on mouser. Are these sufficient, or is it good practice to bump them higher than the calculated value? Also the power dissipation is calculated at 0.196 watts for the 490 ohm resistor, and 0.174 watts for the 432 ohm resistor. Can I use 1/4 watt resistors or is 78% of rated capacity (0.196/0.250) too close to the rated power? Thanks for your help!!!
     
  16. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    If you need 490 Ω for an application like this, then go with standard E24 (or even E12) values. If you don't know what that means, Google E24 resistor values and you'll soon know.

    This is a none critical application and you could get by with 20% tolerance resistors (which you can't even find any more). In the 10% set (E12), the neighboring values are 470 Ω and 560 Ω. Today you would almost certainly be buying these as 5%, if not 1%, values because today's production tolerances are so much better. Either value will likely do. Since you seem to be intent on running these up near their absolute max ratings, I'd recommend going with the higher value. So 560 Ω for the read and 470 Ω for the green and blue.

    Running 200 mW in a 250 mW resistor is probably okay, provided it has good free-air cooling. But if you are putting lots of these things in fairly small space or one without good air flow, then this is probably not a good idea. So either use a 500 mW resistor or two 250 mW resistors either in series or parallel. If in series, make each about half the target value and if in parallel make each about twice the target value.

    Regardless of the size of the resistor, you probably need to address cooling because of the total power dissipation you are talking about (something like 72 W, if I recall).
     
  17. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    I am confused about running these LEDs at 20mA and that being near their top rating. I am newish to electronics, and reading product datasheets is still a bit confusing for me. I think that the max ratings are 30mA for red and blue, and 25mA for green, is this correct?(https://www.mouser.com/datasheet/2/216/WP154A4SEJ3VBDZGC-CA-1145208.pdf) I was using 20mA as a safe middle of the road since nearly every parameter is specified at a test condition of 20mA. I am happy with the light output at 20mA, but if it is not a good idea to run them this high, I will reconsider. As far as cooling, the resistors and LEDs are spread out about 5" apart from each other in all directions, and they will be more or less open to the air with a piece of tape covering them, but that's it. I can bump up to 500mW without too much trouble though :) Turns out I'm running 58 LEDs, not the 100 I thought about before, so the total maximum power out when all three channels are on is 60mA*12v per LED = 0.72W * 58LEDs = 41.76W. Thanks for all of your help :)
     
  18. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    You are probably okay at 20 mA. But do this test. Try them at 10 mA and 15 mA and see if you would be satisfied with that amount of light output. You might be surprised how little difference there is. If all the resistors have 5" of clearance, that is quite possibly enough for free air cooling. Try it and see. If things are getting too warm, a small fan will probably take care of it.
     
    DNA Robotics likes this.
  19. jmh226

    Thread Starter Member

    Nov 7, 2015
    34
    1
    Wow - really there is really not much of a discernible difference between 20mA and 10mA. I'll just go with the 10mA then - no heating worries, and it will keep me further from hitting the edge of my 5A power supply :) Thanks for all of your help - I'll be sure to post pictures when I finally have this thing put together :)
     
    ebeowulf17 likes this.
  20. WBahn

    Moderator

    Mar 31, 2012
    23,157
    6,972
    Yep, that's the point I was trying to make way back in Post #4. Humans are used to seeing light intensities that span several orders of magnitude and, as a consequence, we have a more-or-less logarithmic response to it, which means that we perceive a factor of two difference as only a minor change. We can leverage that insensitivity to significantly reduce the power and cost requirements for many types of circuits.

    This past solar eclipse really drove that home for me. Where we are at saw a drop of 91% in the light intensity. Yet when the eclipse happened few of us were able to discern any significant difference in the perceived light level, although all of us noticed a very easily discernible change in the perceived temperature.

    Looking forward to the pictures.
     
Loading...