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LED reduction Resistor

Discussion in 'LEDs and Optoelectronics' started by t50ufo, Jan 19, 2013.

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  1. t50ufo

    t50ufo

    7
    0
    Jan 19, 2013
    Hi Group
    This is my first post here and I do hope you forgive the fact that I know very little about electronics!

    I am attempting to illuminate a red LED from a 12 V source. The intention is to illuminate the focuser (back end) of a telescope at night. I have sourced a fairly bright red LED and used an online LED resistor calculator to determine the value required. I hooked up the LED to resistor and connected to a 12vDC supply rated at 0.8A output. I noticed that after a few minutes the resistor was fairly warm. I understand that there will be some heat dissipation but I want to ask is: I am on the right track or should I be looking at something else?

    What I really want to achieve is to be able to use the LED powered from a 12v car battery as in the field this powers the telescope mount and other accessories.

    The spec for my components:
    LED: Fwd current 30mA. Fwd voltage 1.85v. Reverse voltage 5v. Luminosity 2300mcd
    Resistor: 390R 2 watt

    I also have a brighter 4500mcd LED with the same parameters as above. How would this effect the scenario if used?

    Any help would be much appreciated

    Thanks
     
  2. donkey

    donkey

    1,293
    56
    Feb 26, 2011
    ok I like these threads cos I can answer them.
    simply put you got it right (someone will add to this if the resistor value is wrong though) but the simple truth is you need LED resistor and power source (car battery or whatever)
    now the problem is that you are using a 12volt battery. this is about 6 timesthe volts needed so you WILL end up with some heat.
    you could downsize this and simply use 2 AA batteries (or D for longer life) and a different resistor (once again someone will probably tell you the right value) and this will give of alot less heat.
    also added to this isif you get any pics of something cool let me know I love space and all that happens up there.
    and last but not least have a read of this https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html
    its our guide to LEDs written by people in the know (the pic of the LED I found lol)
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,513
    2,651
    Nov 17, 2011
    The voltage drop across the resistor is 12V-1.85V=10.15V.
    The current is 10.15V/390 Ohm = 26mA.
    The LED is operated correctly.
    The power converted to heat within the resistor is 10.15V*26mA=264mW. The resistor will get warm, but at 2W rated power it will survive without problem.

    Since your other LED has the same electrical specs, the only difference will be more light output. No change on the electrical side.

    If you want to minimize the waste power, use a step down converter instead. But that is another story.
     
  4. t50ufo

    t50ufo

    7
    0
    Jan 19, 2013
    Thanks Donkey
    Hope you are enjoying the sunshine there - snowing here in Northern Ireland!
    Thanks for the info and link. Very useful.
    Regards
     
  5. t50ufo

    t50ufo

    7
    0
    Jan 19, 2013
    Hi Harald

    Very useful information. Is the current from the Car Battery a factor re the heat generated in the resistor or will the LED just draw the current it requires?

    I would be interested if you could expand on the step down converter option you mention - what is the minimum circuit required?

    Thank you for your kind reply.
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    11,513
    2,651
    Nov 17, 2011
    The LED would draw "infinite" current without the current limiting resistor. And be destroyed in a fraction of a second.

    As for the step-down converter, Google "led step down driver". You will find many applications using more or less sophisticated circuits. However, I wouldn't recommend a switched step-down converter as a beginner's project. If you can't resist, go for a kit like this one.
     
  7. t50ufo

    t50ufo

    7
    0
    Jan 19, 2013
    Thanks Harald
    A quick look on the web brought me to this LM317T Voltage regulator.

    [​IMG]

    With an input of 3 - 40VDC and output of 1.2 - 37V would this be suitable for illuminating my LED - from a 12V Car battery?

    LED Spec: LED: Fwd current 30mA. Fwd voltage typical 1.85v. (max 2.5v) Reverse voltage 5v. Luminosity 4500mcd

    On-line calculator shows me value for R1 240R and R2 180R resulting in voltage out of 2.19V which I hope should illuminate the LED to its best.

    So if I want to replace R2 with a potentiometer to vary the brightness - what value should I use? and what wattage should I be looking at for R1 & R2?

    Thanks :)
     
  8. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    LEDs are better driven by a current source than a voltage source. Look in the LM317 datasheet for the configuration of that device as a current source.

    Although, at 30ma, a resistor is fine.

    Bob
     
  9. t50ufo

    t50ufo

    7
    0
    Jan 19, 2013
    Thanks Bob

    Will my original setup with the resistor still be OK if I use a Car Battery? What with the different voltages at full charge etc. I assumed that the LM317 would be "kinder" to the LED and heat release from the resistor?

    Trev
     
  10. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    The heat is the same whether is it done with a resistor or an LM317. The LM317 can control the current more effectively, but, as I said before a resistor is just fine, people use them in even much higher current applications.

    Bob
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    With a single LED, there will be about 50% variance in current for battery voltages between 10V and 15V. This is not a large amount and you probably wouldn't notice it unless it was right next to another indicator which did not vary in brightness.
     
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