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Optical LED DIY Project question

  1. Jan 7, 2012 #1
    Hope I got the right section. First time poster! Hello all!

    I'm pretty much a Noob so forgive any ignorance.

    I'm running 31 Orange (2.2v) (24 ma) LED's on a parallel circuit. My power supply is 12v from the DVD power supply of a Xbox 360. I bought a 10 watt 15Ω clay type resistor per a suggestion of a calculator. The LED's are the correct Brightness I wished for but the resistor gets real hot I'd assume 50-60C°(too hot to touch). So logically I don't want to use it as it will damage other components with in the Xbox 360. And I don't want to use a heat sink just for the resistor. The question I have really is what resistor I'll need. I want to use just one and it being one that won't get hot to the touch but still provide the desired LED brightness.

    Thanks in advance! And sorry for being such a Noob all I know of electronics is trial and error along with google's help.

    P.S. I did two other projects before and didn't have an issue. I used blue led's and green leds before. I think they were both 32 led's and I ran in parallel but put a 470Ω resistor in between each positive end on the led. But that was way too much work.
  2. jcsd
  3. Jan 7, 2012 #2
    Your figures show that you would have just over 0.7A flowing through your single 15ohm resistor.
    This is a power dissipation in the resistor over 7Watts....it will get hot as you have found out.
    If you used separate resistors for each LED you would still need to dissipate more than 7watts but this would be over 31 resistors (of about 400 ohms each)
    Your problem comes from having a high supply voltage of 12V.
    This is probably not much help to you but I hope it gives you some idea of the problem
  4. Jan 7, 2012 #3
    Well my problem is i don't know anything about this stuff. I just do trial and error. I have a few different resistors laying around and none give the desired result as far as LED brightness. Some don't illuminate at all and others are too dim. I use a single resistor that gave me a somewhat desirable result and did not get hot but I prefer brighter the resistor I used is Gold, Brown, Orange, Orange. I'm not sure it's actual value. I tried a 1/4w 10 ohm resistor and it gives desired result but it also heats up and may blow or fail.

    I assume there has to be a happy medium for a single resistor I could use for this parallel set up I have. I just don't know what one so that's why I asked here. There seems to be many people much smarter than I on this forum.
  5. Jan 7, 2012 #4
    Sound Reactive LED Project

    I wanted to see if Someone can help me confirm if I have everything in order on this project. I previously installed this setup in Another Xbox 360 and it worked but the last two I tried it in it did not work. The switch is supposed to switch from constant LED on and sound reactive LED. For some reason the Sound reactive part seems only to work while playing an actual game (won't work in menu ect.). I'm not sure if it has to due with different audio voltages or what.

    I am able to get the constant on LED setting but when rotating the solder points at the switch "only" I'll get 2 different results for the two different switch modes.

    1. LED constant on works and no LED at all as well as no sound (as if something is interfering).
    2. LED constant on works and no LED at all sound works as well.

    FYI: I'm a complete Noob to electronic voltages and resistances. And this sort of thing in general. If you need any other pics or info to help you help me please let me know. Thanks in advance.

    The component is a TIP 31C

    Here are some pictures:
    R/L Audio = http://i.imgur.com/250Bi.jpg

    R/L Audio = http://i.imgur.com/fMQnI.jpg

    The Sound Reactive device itself = http://i.imgur.com/PjPLV.jpg

    The Sound Reactive device itself = http://i.imgur.com/7db2f.jpg

    Diagram = http://i.imgur.com/OTZY7.jpg
  6. Jan 7, 2012 #5


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    You could put the LEDs in series groups of four LEDs and give each four LEDs a resistor.

    The voltage across 4 LEDs would be 8.8 volts so each resistor would have 3.2 volts (12 - 8.8) across it and 24 mA flowing in it.

    So each resistor would be 3.2 volts / 0.024 amps or 133 ohms. You can buy 150 ohm resistors and this would be close enough. You would need 7 of them and one 220 ohm resistor if the last string only has 3 LEDs in it.

    The resistors would only be dissipating 70 mW so they would run cold.
  7. Jan 7, 2012 #6
    Thanks vk6kro.

    I knew of that method. The issue I have is I want to run it in parallel with just one resistor as it's easier to solder hat way and I prefer to do it that way. I am going to just use the resistor that I mentioned previously (Gold, Brown, Orange, Orange) at least for this one, not sure what the spec of the resistor is. now this resistor I'm using only provides about half of the total illumanation/brightness I'm looking for.

    Isn't there something I can use in the future with say 32 same LED's in parallel? That will not over heat? I know there are resistors that I can use since I have tried a few I just want to know which exact resistor I need that provides the most illumination/brightness.
  8. Jan 7, 2012 #7
    I would arrange the LEDs in a 5x6 array. 6 strings of 5 LEDs in series. That way the LEDs themselves are dissipating most of the power. Your resistor would then only have to drop about 1 volt and would generate a lot less heat. You could still just use one resistor. You seem to have the idea that changing your resistor will change the heat dissipation of the resistor. Increasing the value of your voltage dropping resistor may generate less heat by reducing the current flowing through it, but it may not supply your LEDs with enough voltage.

    As far as figuring out the correct voltage dropping resistor value for your LEDs I would figure out the current flowing in the circuit. If you have access to the datasheet you could just look up If for 2.2 volts. Then you would multiply that amount of current by 31. Then you would use Ohms law. V=IR means that the voltage drop (V) across the resistor is equal to its resistance (R) times the current (I) flowing through it. If your power supply is exactly 12 volts then you need to drop 9.8 volts. So IR = 9.8 and R = 9.8/I. If your current through the resistor were 1 amp you would need a 9.8 Ω resistor. For 500 mA you would need 19.6 Ω and so on. If your resistance value is a little bit too low then your LEDs may glow slightly brighter but will have a shorter life. If your resistance value is too low then you will just instantly burn out all of your LEDs. If your resistance value is a little too high then your LEDs will be less bright but will have a longer life. Ideally you would use the exact value of resistance that you need.

    As far as the power rating of your resistor, you need to calculate the power dissipation. That's easy once you know the current. It's just I^2 x R or the square of the current times the resistance. You can also figure out the power dissipation by measuring the voltage drop across the resistor. Then you would use V^2/R. Let's say your current were 500 mA and your resistor was 19.6 Ω. 0.5^2 x 19.6 = 4.9 watts. In this example you would need a 5 watt or higher rated resistor. It would be big. I think the cheapest ones are the axial cement kinds. For that low resistance you may also be able to find an affordable wirewound which has the advantage of being very precise. LEDs are not really the kind of thing for which you want to just pick a resistor you happen to have lying around the house.

    The bottom line is that in order to get the right resistor value you will need to know the current through your LEDs. LEDs are nonlinear devices. This means that the voltage across them which causes a particular amount of current to flow is not predictable. This is why you either need the datasheet or you have to measure the current yourself with a DMM. Note that this is just how I would do it. I'm sure there are people here who are more qualified to answer your question than I. They might have a different way.
  9. Jan 7, 2012 #8


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    It is poor practice to put LEDs in parallel because they usually don't share the current equally like that.

    To make sure they share the current equally, they are put in series and given one resistor which should drop about a quarter of the supply voltage.
    In this case, the resistor will drop about 3.2 volts if you put 4 LEDs in series with the resistor. So this is OK.

    It is also very inefficient to run single 2.2 volt LEDs from 12 volts. Most of the power is then lost in the resistor.

    If you just run one LED from the 12 volt supply, the resistor would be (12 volts - 2.2 volts) / 0.024 amps = 408 ohms at 0.235 watts. Multiply this by 31 for 31 LEDs.
    If you run 4 LEDs in series, the resistor needs to be (12 volts - (4 * 2.2 volts))/ 0.024 amps or 133 ohms at 0.077 watts.
    So, you can run 4 LEDs and waste 0.077 watts in the resistor or you can run 1 LED and waste 0.25 watts in the resistor.

    If you really want to run the LEDs in parallel, then the resistor should be (12 volts - 2.2 volts ) /( 31 * 0.024 amps) or 13.17 ohms, and it will dissipate 7.29 watts. So it should be a 10 watt resistor.
    BUT this is a bad way of doing it.

    The total current drawn from the supply would be 0.744 amps where it would be 0.12 amps if you put them in series strings.
    Last edited: Jan 7, 2012
  10. Jan 8, 2012 #9
    Thank you metiman and vk6kro for the education on running series. I'll probally do that next time. Now for this Project it would be too much work to start over since I already have them installed.

    I am assuming that the single resistor I used to get a less than desired effect for illumination is a 330 ohm 5% not sure on the watt. It doesn't over heat and works fine other than not being bright enough.

    I do not care about efficiency for this one but I will do the series method as you described for the next one. With that being said what resistor can I use that will make the LEd's brighter and also not overheat the resistor? 15ohm 10 watt? If so what type? I don't want the clay type because I have a 15 ohm 10 watt clay type and it gets way too hot. Is there a type that remains cooler in tempature?

    Will this one work? And will it stay cool in temp?:

  11. Jan 8, 2012 #10


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    If you really want to run the LEDs in parallel, then the resistor should be (12 volts - 2.2 volts ) /( 31 * 0.024 amps) or 13.17 ohms, and it will dissipate 7.29 watts. So it should be a 10 watt resistor.

    Any small resistor dissipating 7 watts will get hot.
    Ceramic types will get hot, but they are built to get hot and not be damaged as long as you don't exceed their power rating.

    The one you showed can be attached to a heat sink but it is in Hong Kong.
    Surely it is easier to wire up a few LEDs properly than to send to Hong Kong for a resistor.

    You gave a current rating of 24 mA for the LEDs. Check the data sheet and see if it can cope with any more current than this. More current will give a brighter LED.
  12. Jan 8, 2012 #11
  13. Jan 8, 2012 #12
    I would recommend this. It's difficult to say what the surface temperature of your resistor would be dissipating about 7.3 watts of power. The aluminum resistor would feel hotter to the touch than the cement one because aluminum is a better heat conductor, but if you wanted to use a heat sink the aluminum resistor would transfer the heat better to the heat sink.

    I was trying to analyze that circuit with a 330 Ω resistor in it and it got me all confused. Obviously you can no longer use V = IR when the resulting voltage drop would be more than the source voltage. A simple voltage divider analysis would work if we knew the resistance of the LED, but the resistance depends on the the actual current flowing through it.

    By putting a resistor of any value above 0 Ω in series with the LEDs you are also reducing the current in the circuit. I=V/Rtotal and Rtotal goes up with the resistor in the circuit. So it should not just be 31 x If. If we give the LEDs an equivalent resistance by assuming a 2.2 V drop across them this results in a 24 mA current and V/I = 2.2/.024 = 91.66 Ω. Putting 31 91.66 Ω resistors in parallel would result in 2.96 Ω. So for a 13 Ω resistor in series with the LEDs Rtotal ≈ 16 Ω. So the total current should be 12v/16Ω = 750 mA. If you ignored the resistance of the LEDs you would get 12v/13Ω=923 mA. If you ignore the resistance of the LEDs but assume that they are in fact dropping 2.2 volts (despite having zero resistance) then you'd get 9.8v/13Ω=754 mA.

    But what if you put a 330 Ω resistor in that circuit? IR = .75 x 330 = 247.5 volts. :smile: The total resistance of the circuit would be 330+3=333 Ω. The total current would be V/R = 12/333 = 36 mA. Each LED would only receive 1.16 mA of that current and because that is nowhere near If we don't really know the equivalent resistance of the LEDs anymore. If we had the datasheet and it actually gave a value for If=1.16 mA then we might be able to get the equivalent resistance, but the 1.16 mA wouldn't be correct anyway because the resistance would would be wrong. So the analysis gets kind of interesting with a 330 Ω resistor. I guess it's the same problem whenever you have a nonlinear element in the circuit. I'm going to look into this some more.
  14. Jan 8, 2012 #13
    Thanks metiman. I am lost when it comes to the gargin/termanalogy/formulas that you posted regarding the circuit it's all latin to me and I don't read/write or speak latin.

    I do appreciate your help with this and your time analyzing my problem.

    Now the leads broke off the original resistor I was using ( which i assume it 330 ohm via the color bands).

    I'm going to radio shack to buy another. Do you know what watt I should get?

    P.S. In the future I'll just do a 4-5 LED series as it will draw less power through the xbox and determining resistors will be easier as there are calculators with actully dummy proof diagrams.

    P.P.S. However if there were a way to use just one resistor in a prallel to get the desired brightness that didn't over heat I'd do it.
  15. Jan 8, 2012 #14
    Haha. That analysis wasn't for your benefit. It was for vk6kro. You are right about your resistor color codes. It is in fact 330 Ω with a 5% tolerance. Unless vk6kro is way off on his current estimate 330 Ω is WAY too high for your circuit. I am amazed that your LEDs are lighting up at all. My analysis indicated that your LEDs should only get about 1 mA of current each with a 330 Ω resistor. Assuming that my analysis is even remotely close then the power dissipation for a 330 Ω resistor should be something like 430 mW. So one 1/2 watt or two 1/4 watt resistors would be enough. If you want to use two 1/4 watt resistors in parallel to equal one 1/2 watt resistor you would use 660 Ω resistors.

    If you want the correct resistor value to light up your LEDs properly you will need something like a 13 Ω 10 W resistor just as vk6kro says. So you want to buy something from Radio Shack? Well I think they do sell 1/2 watt resistors in packs of 5. You could buy 3 of those packs and put those 15 resistors in parallel. That would give you a total power rating of 7.5 watts which would just get you over the wire. If you want those 15 resistors in parallel to give you around 13 Ω then the value of the resistors should be around 195 Ω each. Rat shack won't have 195 Ω, but they will probably have 220 ohm. 15 of those in parallel would be the equivalent of a 14.67 Ω 7.5 W resistor. That should result in a current of around 679 mA and a voltage drop across the resistor of 9.96 volts. That would leave the LEDs with only 2 volts. So they will be a little less bright than they could be, but they should be much brighter than they are now.
  16. Jan 8, 2012 #15
    Ok I'll buy them then. The 220 ohm x 15. any chance you could point me to a diagram as to where to put each resistor?

    I'm going to have to really break down and learn these calculations. Like how to calculate voltage drop = 9.96 volts, 15 resistors in parallel give 13 Ω, 679 mA,195 Ω, ect.
  17. Jan 8, 2012 #16
    You put them in parallel. Like if you take them out of their packages and then just twist all the ends together and solder them. Let an 'I' be a resistor then: IIIIIIIIIIIIIII. The bodies of the resistors would literally be parallel to each other.
  18. Jan 8, 2012 #17
    So you're saying I can just solder all 15 220 ohm 1/2 watt 5% resistors like the five that are in this pic and attach the opposite resistor end to 12v?

  19. Jan 8, 2012 #18
    Yes. Exactly like that pic except you'll have 3 times as many resistors. That guy did a terrible soldering job though. Never rely on solder bridges. Twist the ends of all the resistors together with a pair of needle nose pliers (or whatever) before soldering them.
  20. Jan 8, 2012 #19


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    There seems to be some confusion about how to calculate the resistors for LEDs.

    The resistance of the LEDs doesn't matter as long as you know the voltage for a given current.

    So, you can read off the voltage for the current you are designing for and then calculate the voltage drop across the resistor by subtracting this from the supply voltage. Then you divide the voltage across the resistor by the current through it to get the resistance.

    If you did use a 330 ohm resistor, even if it was put straight across the 12 volt supply, it could not deliver more than 36 mA. (12 volts / 330 ohms = 36 mA)

    If this was spread among 31 LEDs the maximum possible current would be about 1.1 mA

    As we don't have a data sheet for this type of LED, we have to guess what the LED voltage will be at such a low current. It will be something less than 2.2 volts and you could guess something like 1.8 volts.

    So the actual current would be about 1 mA per LED.
  21. Jan 8, 2012 #20
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