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Quick LED wiring question — Parallax Forums

Quick LED wiring question

bulkheadbulkhead Posts: 405
edited 2006-02-21 05:20 in General Discussion
Ok, I drew 2 diagrams. The circle is an LED, the rectangle is a resistor. The first circuit has 1 LED with a 100 ohm resistor. The second circuit has 5 LEDs. I don't want to get 5 100 ohm resistors, and put each one on each LED. Is there a way I can just hook up 1 resistor to cover all the LEDs? If so, how do I calculate the value of that 1 resistor?



Thanks.

Comments

  • gibbmangibbman Posts: 98
    edited 2006-02-20 12:07
    Hi bulkhead,
    A quick Google brought up these two resources
    http://led.linear1.org/1led.wiz·,
    which has a calculator (and schematic wizard) for both series and parallel setups,
    and here:
    http://www.kpsec.freeuk.com/components/led.htm
    which warns against wiring LEDs in parallel as a general design method.
    Hope it helps,
    Jim

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    In the end, it seems that it's all about getting the LEDs to blink....
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-02-20 15:05
    If you run LEDs in parallel, you'll come to a point where one will "current hog".· That doesn't happen when each LED has its own resistor or with LEDs in series.
  • bulkheadbulkhead Posts: 405
    edited 2006-02-20 18:02
    Thanks for the input. That LED calculator is very nice, but it doesn't allow you to just use 1 resistor for all the LEDs in parallel. If 4 more LEDs were added, would it mean you use 1/5 the value in the resistor, since there will be 5 times more current draw?

    I removed 16 LEDs from one of those LED flashlights. It has only 1 resistor thing. I call it a thing because it is a strange looking resistor, if it is one. It is blue, and has 5 bands instead of 3/4. Measuring the resistance, my MM read 1.6 ohms. The voltage is 4.5V. If my idea is right, then 1 LED needs a 1.6*16 ohm resistor, or 25.6. 4.5V/25.6 ohm is 176mA, which is huge for an LED. Is this right?
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-02-20 18:37
    Basically, the equation is:
    current = (Vsource - VLED) / R

    Once ON, the forward voltage across the LED (VLED) is·"constant" (at least, it does not vary widely.)··The remaining·voltage that is not across the LED, then, will be across the resistor.

    Assuming a 6V source (4 batteries):
    current = (6V - 4.5V) / 25
    ········= 1.5V / 25
    ······· =·60 mA

    Your cheapy flashlight uses one resistor because it costs less to do it that way, not because it's the right thing.· Obviously, if you have some "result" in mind and that's your sole consideration,·you can do this or that or whatever you like.

    Post Edited (PJ Allen) : 2/21/2006 3:18:49 AM GMT
  • StarManStarMan Posts: 306
    edited 2006-02-21 02:47
    I'm guessing that the LEDs in the flashlight are "binned" according to Vf, current draw, and light output.· I have a 21 LED flashlight with a single resistor and the LEDs are very uniform in brightness.

    But, still,·running multiple LEDs in parallel with a single resistor·is probably not the best approach.· Resistors·are one of the least expensive components you can buy.· I believe it is Jon Williams who says "they are cheap insurance".

    Chris I.
  • bulkheadbulkhead Posts: 405
    edited 2006-02-21 04:51
    Well, I was just thinking that if it were the same, using 1 resistor instead of 16 would be easier, since all 16 LEDs are the same type. I also thought it could save some space and make for a tighter configuration when soldering the 16 LEDs onto a board.
  • kjennejohnkjennejohn Posts: 171
    edited 2006-02-21 05:20
    You can buy LEDs with the resistors built in, according to driving voltage (5v and 12V versions, usually), but I can't say if that applies to super brights.

    kenjj
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