IR LED array

Discussion in 'The Projects Forum' started by alexx, May 29, 2011.

  1. alexx

    Thread Starter New Member

    May 29, 2011
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    Hello!

    I am attempting to make night vision goggles, but I am confused about the IR LED component

    I've been reading many posts on this site for a few weeks and this is what ive gathered (Please correct me if it's wrong):

    At first I tried putting my friends 9 V battery to my 6 IR LEDs all in parallel with no resistors, and they lit up perfectly through my camera and illuminated well (across my room), but then later i put a new 9 V to a single LED and it fried. This doesnt make sense to me because in the parallel circuit wasnt 9 volts being applied to each LED? why didnt they all break? -- I came to the conclusion that my friends battery must have been pretty old making the voltage much less?

    After this happened i researched and now Im using a 9V battery (typical one from wal mart), and will connect 5 LEDs in series with a resistor, then repeat but attach it in parrallel so i have a total of 10 LEDs. It sounds like the common Vf for IR LEDs is about 1.5 V, so I assumed that my IR LEDs (which i took from remote controls) have a Vf of 1.5 V. Also, SgtWookie has said a number of times to be safe apply a max current of 25 mA, but i need these LEDs to be as bright as possible so is 100 mA okay? My understanding is that with a Vf of 1.5 V and 5 LEDs with 100 mA would give a resistor of 15 ohms:
    R = E/I
    R = [9 V - (1.5x5)]/0.1 A
    R= 15 Ω

    So if i have 9 V going to 5 remote control IR LEDs in series with a 15 Ω resistor, will I have bright IR LEDs and no malfunctions? also, will this make it so that there is 1.8 V (9/5) and 100 mA going to each LED ?


    A lot of that is probably wrong, so the important question is: how can I make these IR LED's as bright as possible?

    Thanks for your time!
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,795
    951
    In your first example with 6 LED's in parallel, the battery was a 9 volt one. The LED's were drawing a large current and 9 volt batteries do NOT like to deliver large currents. The 9 volt battery was 'loaded down'. Meaning its output voltage fell to something around 3 volts or less while trying to deliver that amount of current.

    With just one LED the current demand was lower so the battery put out a much higher voltage. The LED got hot and drew even more current and thermal runaway destroyed it in short order.

    The BEST THING you can do is get the data sheet on the LED's you have. Without the data sheet you are just guessing about them and you will never know if you are overdriving them, or only getting 10% of the output they are capable of.

    You make the LED's as bright as possible by KNOWING the maximum values found in their DATASHEET. Without that, like I said, you are just guessing, and so are we.
     
  3. nigelwright7557

    Senior Member

    May 10, 2008
    487
    71
    LEDs arent like bulbs you cant just apply a voltage.
    You need a resistor in series or a constant current source.
     
  4. #12

    Expert

    Nov 30, 2010
    16,346
    6,833
    The statement that the LEDs can use 25ma does not mean they can use 100 ma. You found that out the hard way, by letting the magic smoke out of one. The current listed on the datasheet is the maximum. To make the LEDs as bright as possible, give them the maximum amount of current they are designed for.

    You see, LEDs have a threshold voltage required to turn them on, and after that they are rather helpless at limiting their own current. For instance, with a 1.5 volt LED and a 9V battery, you have to use the excess 7.5 volts through a 300 ohm resistor to keep the current down to 25 ma. You can use this method to put several LEDs in series, like 5 LEDs on a 9 volt battery with the last 1.5 volts going through a 60 ohm resistor, again, to control the current at 25 ma. Try 6 LEDs with no resistor and you get smoke.

    Another point of interest is that a 9V battery is a bit weak on how much current it can supply. Theoretically, I could start my lawn mower 4 times with the power in a 9V battery, but a 9V battery is never going to produce the 12 amps required to run the starter motor. Keep your voltmeter handy and measure how much the voltage sags when you light up the LEDs. That information will do a world of good in teaching you how these things work.

    Good luck, and ask again if you get another question.
     
  5. alexx

    Thread Starter New Member

    May 29, 2011
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    0
    Thanks for the quick replys!

    Kermit: Oh okay that makes sense now! just out of curiosity, how long could that 'loaded down' parallel circuit run at 3 V with the extra current being pulled?
    And thats a good point about how if I dont know the max values, you cant really help me. unfortunatly I cant get the datasheet because as I said the IR LEDs are from random remote controls. So lets say we have 10 IR LEDs and the datasheet says 1.5 V @ 25 mA per LED, does this mean that if I go over 25 mA the LED will fry? I guess using 100 mA is too much, but could I put 40 mA through? making the needed resistor in the series circuit change from 15 ohms to 40 ohms I think

    Nigel: yes I'm going to use a resistor in series, but will the series i explained in my first post work?
     
  6. #12

    Expert

    Nov 30, 2010
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    Most LEDs the size for a remote control are in the 20 ma range. You can try 40 ma, but I'm betting on the smoke fairy at that amount of current.
     
  7. alexx

    Thread Starter New Member

    May 29, 2011
    29
    0
    thanks for the replys #12!
    #12: so your saying that I cant supply more mA to my LED then what it says on the LEDs datasheet because that is the max. So in order to get bright IR LEDs I need LEDs that can take a high maximum current?

    I dont have any of those, so alternativly, what happens if I supply 25 mA to the circuit but i use a higher voltage? will this result in brighter LEDs? by this i mean:
    12 V going to 5 IR LEDs in series with a 180 ohm resistor (making the nessessary 25 ohm current)
     
  8. #12

    Expert

    Nov 30, 2010
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    12 volts and 180 ohms puts you at exactly the same place as 9 volts with 60 ohms. 25 ma is 25 ma. The LEDs won't even know you have a 12 volt battery. They only care about "enough" voltage to get through the threshold and the current is the real determinant.

    If you want brighter LEDs or higher current LEDs, you simply have to buy them.
     
  9. alexx

    Thread Starter New Member

    May 29, 2011
    29
    0
    okay thanks for clearing that up!

    with this new info I think im going to try first putting 25 mA through one LED, and keep increasing that until it frys to find out my max current. does it matter how long I run the circuit for when testing? for instance say i get up to 30 mA and it appears to be working fine, is there a chance that after running for 2 minutes it will fry or am I good to up the current right away
     
  10. #12

    Expert

    Nov 30, 2010
    16,346
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    It's about lifespan versus overcurrent. At the rated current, LEDs go 100,000 hours or more. 10% less current might double their lifespan. 10% more current and they will die sooner. Double the current and you have a few seconds to live.
     
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