Looking for DIY projects....

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  • ziggy53ziggy53 Super Moderators Posts: 24,127 moderator
    edited July 1, 2005
    ziggy53
    Moderator of the Cameras and Accessories forums
  • XO-StudiosXO-Studios Registered Users Posts: 457 Major grins
    edited July 1, 2005
    Stan wrote:
    Hi Tim,


    I took my inspiration from here

    And bought my stuff from here in Germany, I am sure it can be sourced elsewhere but this was much cheaper than the high street

    I think I bought This because of the wider degree of luminosity but I think the brighter one at 22000mcd is brobably better. The description thread shows the diagram here using 62ohms 1/8th watt resistors headscratch.gifheadscratch.gifne_nau.gif so I did and it turned out to be 6volts total.

    Oh and I bought little clip cups to hold the leds I glued them in place.

    Good luck

    Stan
    Ok here is your challenge.

    The LED's you substituted are 3.6 V, 20 mA
    That is what they need to ignite.

    With a 6V battery that would need (6-3.6= 2.4 V to get a 2.4 volt drop at 20 mA you need a U=IXR 2.4=.020AXR R=120 ohms) or with a 9V battery you would need (9-3.6=5.4V 5.4 drop at 20 mA => 270 Ohm).

    The error snuck in your replacement, most LED's are 5 V and 30 mA, in which

    6V battery => 33 ohms
    9V battery => 133 ohms

    or for lower intensity LED's 20 mA

    6V battery => 50 ohms
    9V battery => 200 ohms

    For more information on Ohms law
    http://www.google.com/search?q=ohms+law

    FWIW, YMMV,

    XO
    You can't depend on your eyes when your imagination is out of focus.
    Mark Twain


    Some times I get lucky and when that happens I show the results here: http://www.xo-studios.com
  • Tim KirkwoodTim Kirkwood Registered Users Posts: 900 Major grins
    edited July 1, 2005
    XO,


    thanks for the help.


    Tim
    www.KirkwoodPhotography.com

    Speak with sweet words, for you never know when you may have to eat them....
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