LED

Discussion in 'The Projects Forum' started by whycanot, Jul 25, 2011.

  1. whycanot

    Thread Starter Member

    May 22, 2011
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    0
    I connected 1 Infrared LED Transmitter with a 1.5V battery
    It lighted up, but after a few minutes, it overheat n spoil
    I want to know how much resistance should connect in series between the Infrared LED and the 1.5V battery
    Help please.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It would help a great deal to know what the particular diode's specifications are.

    However, a Vf of 1.2v with 20mA current would be pretty typical.

    You calculate the limiting resistor as:
    Rlimit >= (Vsupply - Vf_LED) / Desired_Current
    and then use a resistor that is >= the result.

    A table of standard resistance values is here: http://www.logwell.com/tech/components/resistor_values.html
    Bookmark/favorite that page.
    E12 and E24 values are pretty commonly available E48 and higher are more expensive and usually need to be ordered.

    If you can't find a value close enough, you can always use pairs of resistors in series or parallel to get much closer.
    Here is a resistor calculator: http://www.qsl.net/in3otd/parallr.html
    Very handy.

    OK, let's say you have 1.5v as your supply, your IR LED needs 20mA current and will have a Vf of 1.2v.
    Rlimit >= ( 1.5v - 1.2v) / 20mA = 0.3 / 0.02 = 15 Ohms.
    Looking at the table of standard resistance values, you can see that 15 Ohms (shown as 150 Ohms; this is a decade table where you can multiply or divide the shown values by powers of ten) is a standard value of resistance.

    As your battery runs down, your LED will become more dim.
    If you want the LED to stay on longer, you could use two batteries in parallel.
    If you wanted the LED to maintain it's brightness for a longer period of time, you could use two batteries in series, and a different resistor because your source voltage increased.

    Two batteries in series would be 1.5v+1.5v=3v.
    Then you would need to calculate:
    Rlimit >= ( 3v - 1.2v) / 20mA = 1.8 / 0.02 = 90 Ohms.
    Looking back at our standard values of resistors, we can see that (910/10) = 91 Ohms is a standard E24 value of resistance; very close to 90 Ohms.
    But what if we didn't have a 91 Ohm resistor? What values could we use?

    Going to that resistor calculator and asking for a 90 Ohm resistor from E12 resistors results in:
    Code ( (Unknown Language)):
    1. 68  +   22  =   90      (0 %)
    2. 180 ||  180 =   90      (0 %)
    3. 82  +   8.2 =   90.2        (0.222 %)
    4. 150 ||  220 =   89.189      (-0.901 %)
    5. 100 ||  820 =   89.13       (-0.966 %)
    6. 56  +   33  =   89      (-1.111 %)
    7. 120 ||  390 =   91.765      (1.961 %)
    8. 47  +   47  =   94      (4.444 %)
    9. 82  +   0   =   82      (-8.889 %)
    The "nn || nn" means the resistors are wired in parallel, "nn + nn" means wired in series.
    Almost all of them would be a reasonable choice except for the very last entry; too much current.
     
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  3. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    The usual way is to find out the specified voltage drop of the device (e.g., read the specifications sheet) and its operating current. This may be given as a range or a maximum value may be given; use the highest value. Subtract the voltage drop from 1.5 volts; this is the voltage that must be dropped across the resistor -- call it V0. Calculate the value of the needed resistor from V0/i, where i is the rated current of the device. Use volts and amperes for units and the answer will be in ohms.

    The best way is to find out the rated current of the device, then use a constant current DC power supply to set this current through the device. Read the voltage drop across the device and proceed as above.

    If you answer that you don't have the spec sheet, then that's kinda like using a medical drug without reading the label... :p
     
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  4. whycanot

    Thread Starter Member

    May 22, 2011
    54
    0
    the attached photo is my IR LED~
    i dun have it's spec~
     
  5. whycanot

    Thread Starter Member

    May 22, 2011
    54
    0
    so you means that connect 1.5V with IRLED and 150ohms together in series then can work dy? the IR LED is shown on my previous reply~
     
  6. RRITESH KAKKAR

    Senior Member

    Jun 29, 2010
    2,831
    89
    IR led require a switching pulse ( may be 38khz which is good for its app.) and connetcing directly to supply will d o nothing..!!


    now i think you can connect the resistance in series with it..

    If you want to see is Glowing use a digital camera like Mobile,etc.... not with naked eyes..!!
     
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  7. wayneh

    Expert

    Sep 9, 2010
    12,090
    3,027
    It won't hurt to try. If it does not work, you'll know because the LED will not light. You can tell this by viewing it with your cellphone's camera, or maybe most digital cameras. You can test the camera with any working remote control you have - you should see a bit of light from the emitter LED if your camera can "see" IR.

    150Ω may be too large a value, but will protect your LED. Only go to smaller values (120, 100, etc.) as you determine that the current is not too high. Be aware that a fresh battery will deliver more current than a used one. You want to choose a resistor that will protect your LED even with a fresh battery.
     
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  8. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    You can't tell the specs just by looking at it. You need to look where you bought it. Just from what you are saying I'm willing to bet you didn't have any resistor on it, this is almost always a mistake. LEDs require resistors. The only time this is not true is if you are using a very small watch battery, and that will work only because they have an effective resistance inside.

    Use the 150Ω Wookie suggested.

    LEDs, 555s, Flashers, and Light Chasers
     
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  9. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    IF your device is a typical led with a Vf of 1.2V and requires 20mA then you would use a 15 ohm (not 150 ohm) resistor in series.
    The formula is (Supply voltage - led VF)/current = required resistance
    So (1.5-1.2)/.020 = 15 ohms
     
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  10. wayneh

    Expert

    Sep 9, 2010
    12,090
    3,027
    That sounds better, and is what Wookie recommended. The OP missed his comment on the table and (Bill and) I didn't catch it.
     
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  11. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I posted 15 Ohms as the value of the limiting resistor.
    150 ohms got confused in there, because that is what is shown in the standard decade resistance table, and I don't think that our OP understands that the table represents the values in each decade, not absolute values.

    It depends upon the OP's application. They may be using it for IR illumination, not for signalling.
     
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  12. RRITESH KAKKAR

    Senior Member

    Jun 29, 2010
    2,831
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    What is IR illumination...?
     
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  13. nerdegutta

    Moderator

    Dec 15, 2009
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  14. whycanot

    Thread Starter Member

    May 22, 2011
    54
    0
    thanks for the help ... really appreciate it .... thanks guys !
     
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