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LEDs and batteries

  • 28-01-2005 01:01PM
    #1
    Registered Users, Registered Users 2 Posts: 3,209 ✭✭✭


    Been quite a while since LC physics, could anyone tell me which configuration would last longest please...

    A led powered by 9volt battery and series resistor
    or
    A led powered by 2 'D' size batteries.

    It is what it's.



Comments

  • Moderators, Science, Health & Environment Moderators Posts: 1,855 Mod ✭✭✭✭Michael Collins


    Ok first of all what type of LED are you using? Any normal LED will need a current limiting series resistor in either case.

    The length of time you'll get the LED ligthing for depends on the capacity of the battery, usually measured in mAh (milliamp hours), which differs with the make of the battery.

    You can work it out yourself, given that a normal LED will require around 20 mA to emit a decent amount of light, so take your millamp figure from above, divide it by 20 mA and you'll have the amount of hours your battery should last. (of course this figure will be ball park as the LED won't just suddenly go off at this time, but this should tell ya which is better...)


  • Moderators, Recreation & Hobbies Moderators, Science, Health & Environment Moderators, Technology & Internet Moderators Posts: 95,922 Mod ✭✭✭✭Capt'n Midnight


    2 D cells.

    If the resistor was too low then the LED would go pop on the 9V battery.
    most LED's apart from the Blue ones will run from two D cells which have a lot more mAh than a pp3 (~ 20 times)

    to calculate the resistor isn't easy though.
    the voltage accross the resistor is the difference between the battery voltage (1.6 to 1.1 - new alkaline to nearly discharged NiCSD) and the drop across the LED which depends on the colour 1.3v for InfraRed to 1.7 for red and then 2-2.1V for yellow/ green IIRC

    there was a way of cross wiring a FET so it would act as a constant current device ~ 20 mA - instead of a resistor - depending on type but you'd need about 5V (three D cells) to have this work.


  • Registered Users, Registered Users 2 Posts: 6,374 ✭✭✭Gone West


    definetly. think about it. your pumping in 9v into a little resistor. The resistor has to be little to protect the LED. eventualy, the resistor will fail. with the 2x1.5v D batteries, you will still prob need a resistor, but it wont be under the same load as the 9v one.


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