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L.e.d. Help!

Discussion in 'General Electronics Discussion' started by olaf1axe, Aug 25, 2011.

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  1. olaf1axe


    Aug 25, 2011
    I want to create a simple DC circuit with 5 or 6 leds. How do I calculate total circuit value so I know total volts for batteries? How do I know how many volts each Led requires? How do I make sure that there is no drop of power between leds so they all shine just as bright? Help!!! I know it's easy but I can't figure it out. The first is brighter than the rest...
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Please read this. lots of your questions are answered there (or links given to calculators that will give you the answer).

    Presuming the LEDs are wired correctly in series, they will be sharing the same current -- don't worry about power dropping between them. If any power is dropped it will affect them all.

    The reasons they might be different brightnesses are:

    1) they are different LEDs with different efficiencies

    2) they are different LEDs with different beam widths

    3) one or more of the LEDs is damaged.
  3. olaf1axe


    Aug 25, 2011
    L.e.d. - Volts - Amps - Math!

    Thanks! Maybe you can help with a persistent doubt:

    - A 9 volt batt has approx 560 mAh. A regular led is rated at 2.5v and 20mAh. I set up six of these leds with the 9v and they work fine. How does the math work out!? I mean...

    - Are the 6 leds running on 1.5v each ( 9v div by 6)? Don't they need a min of 2.1v to run?
    - Most importantly what happens to the mAh? Is the current also divided evenly between resistances (in this case the leds)? Am I subjecting 20 mAh leds to over 90 mAh each!? Why don't they fry!?
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010

    mAh is a measure of capacity, or total energy. It really doesn't say much about current or time (other than giving an approximate relationship between them)

    Current is measured in mA (your LEDs probably have a Vf of 2.5V @ 20mA)

    If you read that link I gave to you, you will notice that Vf of a LED is dependant on voltage. at 20 mA (say) for these LEDs it may be 2.5V. However at 50uA it may only be 1.5V.

    Also, the voltage of the 9V battery may be anything up to 10.2V when new.

    If the LEDs are all in series, all of them will have the same current passing through them, and it will be the current that flows out of the battery. That current depends on the load.

    So yes, you divide the voltage across all LEDs, but the same current flows through each.

    Without measuring the voltage across a series resistor (or inserting an ammeter in series with the LEDs) I can't tell you what the current through the LEDs might be.
  5. TBennettcc


    Dec 4, 2010
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