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LED hallway lighting

Discussion in 'Electronic Basics' started by Ray, Apr 3, 2005.

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  1. Ray

    Ray Guest

    What I want to do is use 6 white LEDs that are rated at 3.5V forward v and
    max forward I at 30mA. I am going to use 4 1500mAh NiCad sub c batteries
    (because I have a few of them)in series to power up the LEDs in parallel. I
    know that I'll need about 50-75ohm resistance to limit the current to about
    25mA.

    I plan on using a 12V dc wall transformer to power the LEDs and charge the
    batteries when the power is on. What I want to happen is when the house
    power is on, the batteries get charged and then switch to trickle charge to
    maintain them and power the LEDs all the time. When the house power goes
    out, the batteries will power the LEDs.

    The system should work something like an exit sign from www.dual-lite.com

    What I'm looking for is a schematic for the exit signs from the link above
    or a simular type circuit.


    -Ray
     
  2. Those nicad C cells are 1.2V, not 1.5V. Thus, your resistors should be a
    bit smaller. However, 30mA is pretty high for most LEDs, so perhaps a
    larger resistor will let them live a bit longer.

    The other issue is that charging the batteries when they are in series
    is a potential problem, since if one of the cells fails to charge, the
    other cells will get toasted.

    Nicads like constant current at 0.1C, so a constant current charger of
    150mA is appropriate for these guys. However, you also need to notice
    when they get charged up, and turn off the current until they drop some
    amount.

    A constant current source is easy to make out of an LM317, a reference,
    and a comparator. You turn off the LM317 by pulling the common output to
    ground using the comparator output. The comparator needs enough
    hysteresis to not turn on until it gets down to 4.6, and turn off at
    4.8. A normal recharge circuit would wait until 4.4V, but for a system
    like this, seems like you would want to keep the batteries as high as
    possible without frying them with overcharge or trickle.

    12V
    | Constant Current Charger 150mA
    V With shutoff at 4.8V, Turnon at 4.6V
    -
    |
    o--. .------------------------------------.
    | | | |
    .-. | | _____ |
    1k| | | | | | ___ |
    | | '----------------)---------------|LM317|--|___|----. |
    '-' | |_____| 12R | |
    | | LM339 | ___ | |
    o----. | o-----|___|----o |
    | | | |\ | 10K | |
    .-. | '----|-\ | o--'
    10k| | | ___ | >----------o |
    | | o------|___|---o----|+/ | 4.8V ---
    '-' | 1k | |/ | -
    | | K | | |
    | R - TL431A | ___ | |
    o----^ '-----|___|--------' |
    | | 22k |
    .-. | A |
    11k| | | |
    | | | |
    '-' | |
    | | |
    -o----o------------------------------------------------'
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    You should also use a 12V relay, which is set up so that if the coil is
    powered, the LEDs are powered from a 5V regulator driven off of the 12V
    DC input; otherwise, they are powered from the batteries.

    When the power goes out, the relay switches, and the LEDs are powered by
    the battery. The LEDs need their own resistors, of course.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  3. Ray

    Ray Guest

    Thanks, that's exactly what I was looking for.

    30mA is the maximum current rating of these LEDs. They can take up to
    about 100mA and then the genie is released. They probably won't last the
    100,000 hours at 30mA. Your eye can't tell the difference in brightness
    when powering them with 20 - 30mA, so I will keep them down around 20mA or
    less.
     
  4. dB

    dB Guest

    Don't wire the l.e.ds in parallel and use a common R. Use a separate
    R for each l.e.d.
     
  5. Tom Biasi

    Tom Biasi Guest

    I agree.
     
  6. GotCoffee

    GotCoffee Guest

    What's the benefit of using seperate resistors as long as you keep the
    power thru the resistor below the watt rating?
     
  7. The problem is that LEDs vary as to the amount of current they require
    at a given voltage, and the ratio of currents is exponential in changes
    of voltage. Thus, even a tiny difference in devices can cause a big
    current difference, leading to one LED which is much brighter than the
    others. Since this difference gets bigger as the device gets hotter, the
    bright LED will tend to get brighter as it heats up. This can cause it
    to fail.

    The typical solution is to use separate resistors, because even a little
    bit of resistance will tend to compensate for this tendency.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
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