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Another question regarding mA

Discussion in 'Electronic Basics' started by AB, Mar 2, 2007.

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  1. AB

    AB Guest

    As I mentioned in previous posts, I have this alarm panel that uses
    12VDC to power a set of output contacts for an optional sounder
    device. The only requirement is that you don't exceed 120 mA current
    draw @12VDC for these contacts. Thanks to the help I've gotten here, I
    was able to measure the draw from all of my sirens, as well as some
    small strobe lights. Now here's my question. My lowest draw siren
    draws 50mA. That leaves me another 70mA free on that circuit. I have a
    low draw strobe assembly (strobe light w/ little circuit embedded
    under it) that I'd like to add to the siren, but it draws. 140mA @
    12VDC...too much still.

    Is there anyway I can modify the strobe so that it will draw only 70mA
    rather than 140? Maybe with a resistor or something? Before you say
    it, I know that there is the possibility that the strobe won't fire or
    be pretty dim...but...I'd like to try it anyway. Thanks again for the
    great advice I've gotten here.
    Bob G.
     
  2. It would be easier, and probably better, to use the alarm output to
    operate a relay, and use the relay contacts to apply power to the
    sirens and strobe.



    --
    Peter Bennett, VE7CEI
    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
    GPS and NMEA info: http://vancouver-webpages.com/peter
    Vancouver Power Squadron: http://vancouver.powersquadron.ca
     
  3. You could change the charging/firing resistor to slow the strobe down to
    less than half speed.
     
  4. ehsjr

    ehsjr Guest


    Assuming the power supply in the alarm is capable
    of supplying the current the strobe needs, use that
    for the following circuit:

    +12 ----+----Strobe----+
    | |
    o |
    S1 / /c
    o---[470R]---| BC337
    \e
    |
    Gnd -------------------+

    It's about 20 cents worth of parts - 1 resistor and
    1 transistor. When the alarm contact (S1) closes,
    the transistor conducts and the strobe operates.

    If the alarm contact is normally closed instead of
    normally open, use this circuit:

    +12 ----+----Strobe----+
    | |
    [470R] |
    | /c
    o------------| BC337
    S1 | \e
    o |
    | |
    | |
    | |
    Gnd ----+--------------+

    If you have NPN transistors on hand, give one
    of them a try - almost any NPN will work fine.

    If your alarm power supply cannot provide the
    current your strobe needs, use a separate 12V
    DC wall wart supply.

    Ed
     
  5. Marra

    Marra Guest

    You need a constant current source to achieve this.
    Just google it.

    www.ckp-railways.talktalk.net/pcbcad21.htm
     
  6. John Fields

    John Fields Guest

    ---

    The strobe resistance:


    E 12V
    Rs = --- = -------- ~ 85.7 ohms
    I 0.14A


    The total resistance required to drop 12V at 70mA:


    E 12V
    Rt = --- = ------ ~ 171.4 ohms
    I 0.07


    The resistance you need to add, in series, between the strobe and
    the supply in order to limit the current to 70 mA:


    Rx = Rt - Rs = 171.4R - 85.7R = 85.7 ohms


    The closest standard 5% resistor is 82 ohms, so it'll drop the
    current to:

    E 12V
    I = --------- = ------------- = 0.072A
    Rs + Rx 85.7R + 82R


    And will dissipate:


    P = I²R = 0.072A² * (85.7R + 82R) = 0.869 watts,


    so I'd use at least a 2 watt resistor in order to keep it from
    getting _real_ hot.
     
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