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LED circuit?

Discussion in 'Electronic Basics' started by REMUS, Oct 11, 2004.

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  1. REMUS

    REMUS Guest

    Hi i'm new to the group, and i'm not really to awesome when it comes to
    electronics, so I thought I would ask people who really know!

    Heres the situation: I play bass in a thrash/rock-punk band and I want to
    customise my bass a little. And my idea is to place red LED's under the
    scratch plate which I will raise up with washers/nut(s) to give a glow
    eminateing from underneath on dark gig's. Ideally I suppose 4 or 5 LED's on
    a 9v battery would be about as much as I could fit under there.

    I dont know how to make a circuit with a switch and LED's or where to buy
    them in the UK.

    A freind mentioned I could get a sort of fibre optic wire, that he has on
    his PC for cosmetic effect but I guess that runs off 12v?

    Cheers for any advice + sorry for being so clueless!
     
  2. Randy Day

    Randy Day Guest

    Here's a circuit for what you want; it's not complex,
    and any local electronic parts supply shop should have
    them:
    Resis
    Switch tor
    _/ ___ Led Led Led Led
    -o/ o--|___|-->|--->|-->|-->|-|
    | |
    --- + |
    - |
    | |
    -------------------------------|
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

    Parts List:
    1 - "Single Pole, Single Throw" switch
    1 - 9v battery and clip
    1 - Resistor, 470 ohm to 1000 ohm, depending on the type
    of LED you choose and the brightness desired
    4 - LED's of desired color

    As shown, everything just gets hooked up in a big loop; the
    only tricky part is making sure the LED's get hooked up
    'nose to tail' as it were.

    If you hook it all up and get the battery polarity wrong,
    don't worry; nothing will happen, nothing will blow up.
    Just reverse the battery and try again. If nothing lights
    in either direction, you probably got one or more LED's
    backwards; nothing can blow up as long as everything's
    in series, so just find the culprit(s) and turn them around.
    I don't know the current requirements of the fiber optic
    cable, but two 9v batteries in series run through a 12v
    regulator might work...

    Switch
    _/ .-----.
    |-o/ o-| |-------- +12v
    | | |
    9v --- | |
    - '-----'
    | |
    | |
    9v --- |
    - |
    | |
    -----------o----------- -12v
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

    12v regulator IC's are pretty cheap, so when you find a
    parts supplier, ask about them. Be sure to get a data sheet
    that shows how the pins get connected.
    Hey, everyone has to start somewhere!
     
  3. Richard

    Richard Guest

    There's a lot of stuff here:

    http://shopping.netsuite.com/coolight

    and here

    http://www.houseofrave.com/fiber_wire.html

    Some of it runs on as little as 2 AAs. or 9V.

    Google "Electroluminescent Wire" or El wire

    Seems like it's everywhere.

    Richard
     
  4. Buying is the easy part, most of the larger cities have a Maplin store,
    and you may also find some independents if you check the yellow pages.
    There are also several mail order stores for electronic parts in the UK,
    if you get a copy of elector magazine you should find their adds. For
    small orders however local stores are cheaper.

    LEDs work best with a current of 10..20 mA, the voltage required to
    drive that current depends on the colour of the LED (about 2 V for red,
    yellow and green and 3.5 V for blue and white).

    Thus assuming you want ot use red LEDs, you can have 4 of them in series
    from a 9 V battery, requiring a total of 8 V. The voltage of a new
    9V-batt is about 10 V, so the remaining 2 V needs to be fried in a
    resistor, the resistance is calculated from Ohm's law R = U/I = 2 V /
    0.02 A = 100 Ohms. As the battery empties, its voltage will drop and the
    LEDs will give less light.

    The power fried in that resistor is P = U * I = 2 V * 0.02 A = 40 mW, so
    a standard 125 mW resistor should handle the load easily without turning
    black.

    So your circuit will look about like this:

    /
    /
    +Batt o----o o-----/\/\/\--->|--->|--->|--->|-------o -Batt
    Switch 100 R 4 x LED red
    SPST 125 mW superbright

    Connection of the resistor and the switch may be in any orientation, but
    LEDs and the battery need to be connected in the right polarity.

    For blue LEDs, you can have only 2 in series, and the resistor needs to
    be increased to 150 Ohms. But you can put several such circuits in
    parallel:


    /
    / 150 R 2 x LED blue
    +Batt o----o o-----/\/\/\--->|--->|------o -Batt
    | |
    | |
    ---/\/\/\--->|--->|----
    | |

    Since blue LED appear not as bright as red it may become necessary to
    use several such branches.

    This of course is only the easiest version. As your knowledge and
    confidence with such projects grow, you could have the LEDs go on and
    off in the rythm of the music, or switch between different colours. You
    also may consider mains supply to avoid wasting batteries, but this is
    something you should do only when you understand the safety
    implications.
     
  5. Patrick

    Patrick Guest

    Hi im planning to do the same thing but im using it in my car. I plan
    to use a total of 6 blue led's. I just wanna know how many resistors
    should i use and it's rating. Is it possible to do it all in parallel?
    thanks very much
     
  6. Bill Bowden

    Bill Bowden Guest


    You can probably use 3 LEDs in series with a 150 ohm 1/4 watt
    resistor and another 3 LEDs and resistor for 6 total. Try this
    calculator to work out the exact values.

    http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm

    -Bill
     
  7. Patrick

    Patrick Guest

    Thanks! At the store though I found some color changing led's. It
    works well with 9v batteries but when I got home and tried it with 12v
    and a 750ohm resistor it just lit red and didnt change color. Tried
    again with lower voltage starting at 1.5 gradualy moving up without
    resistor... it changed color at 6v - and then burned! What am I doing
    wrong? It worked well with the 9v battery so why did it burn at 6v?
    Any ideas on how to get it to work?

    Thanks and good day
     
  8. I have no idea what circuit is inside the color changing LED. Does it
    come with a data sheet or any specs?

    When it was operating from the 9 volt battery, how do you know what
    voltage it was getting? 9 volt batteries sag under load quite easily,
    especially if they are not alkaline or fresh. Rechargeable 9 volt
    batteries put out something like 7 volts.
     
  9. Car probably means a 12 V lead acid battery, supplying about 14 V when
    full, blue LEDs need 3.5 V. So 3 of the LEDs in series would require
    10.5 V, the remaining 3.5 V would be fryed by a resistor of 3.5 V / 0.02
    A = 175 Ohms (180 Ohms is the next higher standard value) at 3.5 V *
    0.02 A = 70 mW.

    Since you want 6 LEDs, you need two such strings in parallel.
     
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