LED Noob Q

Discussion in 'The Projects Forum' started by CMH70, Mar 12, 2010.

  1. CMH70

    Thread Starter New Member

    Mar 11, 2010
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    I want to make one of the pac lites mad with the 9V battery a resistor and switch and one LED. I have my 9V battery, a 3.2V 20ma 10mm Blue LED. now what i dont understand yet is 9v - 3.2v = 5.8v. And 5.8v / .02 = 290 ohm.
    now i got that but will my voltage still be above the 3.2 max for my LED?
    seems the most basic circuit got me confused.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Your calculations are correct.

    You should really measure the voltage on your 9v battery before you connect it up though. Most 9v batteries read about 8.6v with a 20mA load even when they are brand new. This is because most 9v batteries are made from six AAAA batteries; they look like miniature AAA batteries. They have a high internal resistance compared to other batteries.

    However, some 9v batteries are "industrial"; they have seven of those AAAA batteries in them, and they output closer to 10v when new.

    If you want to keep your LED's current very consistent over a fairly wide range of input voltages, you can use an LM317 regulator and a 62 Ohm resistor to get a constant 20mA output current.

    Connect the 62 Ohm resistor between the OUT and ADJ terminal of the LM317 regulator.
    Connect the IN terminal to your + supply.
    Connect the ADJ terminal to your LED's anode.
    Connect your LED's cathode (shorter lead) to the - supply.

    You can use the LM317 and the resistor with or without the LED, and a multimeter to check the battery voltage under load.

    In this configuration, the LM317 will drop a minimum of 3v across itself. However, you're using a battery that might range from 10v for a new industrial 9v cell down to perhaps 7.5v when it's dead for most intents and purposes. Once it's down that low, you won't get much current out of it anyway.
     
  3. Wendy

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    Mar 24, 2008
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  4. CMH70

    Thread Starter New Member

    Mar 11, 2010
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    cool, thanks guys.

    SgtWookie, didnt have a LM317 but did have a DE-SW033 worked perfectly tho rather bulky, but will suffice until i can get a 317. :) 9.7v in battery and 3.03 thru regulator.

    Bill, thanks for the link, that answered my Q on what the voltage . I was trying to get a measurement after the resistor without the LED attached.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Ack! :eek:
    The DE-SW033 is a switching regulator designed to replace a 78xx type for a 3.3v output. It will not work the same as an LM317 type regulator for a constant-current regulator. I'm afraid that your results measured with the DE-SW033 are not valid; I really have no idea WHAT you measured.

    Try measuring the battery without any load, and then with a simple 470 Ohm to 360 Ohm resistor, and report back the results. I'll show you how to calculate the internal battery resistance. That will give a much better idea of what the output under load will actually be.

    Don't hook it up yet - you haven't tested the battery voltage unloaded, and with a proper load, so that we can figure out what the internal resistance is.
     
  6. CMH70

    Thread Starter New Member

    Mar 11, 2010
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    ok i took the DE-SW033 out of the picture and will go try and buy a LM317 tomorrow. Set my beadboard up like the figure 1.1 on http://forum.allaboutcircuits.com/blog.php?bt=676 except for in inserted a push button switch so i could leave the battery attached.
    1. I tested ( no load from the ground to in between the 470 Ohm( yel vio bro) and the LED ( 3.2v 20ma rated) and result is 9.10v.
    2. pushed switch and test same points and result is 2.974v. I don't know how to test how much the ( ma ) is on my multimeter yet ( Fluke 179)
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    An LED sets its own voltage. The series resistor limits its current.
    A "3.2V" LED might be anywhere from 3.0V to 3.5V.
     
  8. gizmoman0

    New Member

    Jan 21, 2010
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    I just hooked up a Blue LED like that recently and found 20mA to provide a blinding amount of brightness to the LED. I was much happier with about 2 to 3 times the resistance of the calculated resistor.
     
  9. Audioguru

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    Dec 20, 2007
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    Some cheap LEDs are old dim ones in a sharply focussed case. Their beam is bright on axis but cannot be seen to one side. Better LEDs have a wide viewing angle and are bright over the entire angle.
     
  10. CMH70

    Thread Starter New Member

    Mar 11, 2010
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    No doubt, found out quickly not to look at the blue one head on. Tried experimenting with the resistance also, my White LED turned Green and died. Back to the study guides for me :)
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    What part of Texas are you?
     
  12. CMH70

    Thread Starter New Member

    Mar 11, 2010
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    Diboll, 100 miles North of Houston
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, you must be careful to not look directly at these newer super-bright LEDs, as you can actually cause permanent damage to your vision without knowing it.

    If I'm fiddling around with them, I'll use something to diffuse the light. A sheet of copier paper between my eyes and the LED works OK. So does an emptied & washed dairy products container, like the white tubs for cottage cheese and sour cream; just put them upside-down over the LEDs.
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    I have powered many LEDs. But I have never burned one out.
     
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