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Discussion in 'Electronic Design' started by pmoseley, Dec 3, 2007.

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  1. pmoseley

    pmoseley Guest

    I have a landscape lighting unit that can turn on whatever is wired
    into it for 2, 4, 6, or 8 hours from when the photo cell trigers the
    unit. My problem is that the unit needs a minimum load ot 20 watts to
    operate (max 300W). It puts out 12 or 14 volts. I want to use a 14v.
    relay to turn on a nouber of light which would far exceed the 300W
    max. If I can figure what the relay draws how do I figure what to add
    in parallel with the relay to add up to 20 watts to make the unit
    work. I guess it would be a resistor of some sort. But how do I
    figure the ohm rating and power requirement?

    TIA

    Peter Moseley
     
  2. Boris Mohar

    Boris Mohar Guest

    Light bulb
     
  3. Guest

    Could you just put 20 watts worth of lights in parallel with the relay
    to make the photo cell unit happy?

    M
     
  4. There's been an ugly rumor going around that Ohm's Law might be somewhat
    useful here.


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  5. pmoseley

    pmoseley Guest

    OK the way I understand it is as follows: 20W/V=1.429I. And using
    R=V/I (be gentle with me guys. It's been a long time since I've done
    this) R=14/1.429=9.8 Ohms. Radio Shack has a 10 Ohm, 10W wire-wound
    resistor. So, I guess two of these in parallel would do it. Right?

    TIA
    Peter Moseley
     
  6. Fred Bloggs

    Fred Bloggs Guest

    Sounds like you're using what is usually called an "electronic
    transformer" intended for low voltage lighting applications. That would
    then be a 12 or 24 volt output selection and not an oddball 14 volts.
    The minimum loading requirement of 20W seems to be a standard and is
    something they call a "demand circuit design." If your "unit" is in fact
    an "electronic transformer" of this type, then it is most likely a high
    frequency inverter type, producing a low voltage AC at several tens of
    thousands of kilohertz. This would make sense because high frequency
    operation makes for much smaller size (volume, "footprint," or what have
    you) which is what most people are after when they use the "electronic
    transformer." If your limited concentration span has lasted this far
    then you need to know that an ordinary low voltage relay will not
    operate at this of a high frequency, and this aside from the fact that a
    14V relay is rather rare.
     
  7. David Harmon

    David Harmon Guest

    On Mon, 3 Dec 2007 15:42:18 -0800 (PST) in sci.electronics.design,
    Just use 20 watts of your lighting, then put the rest on the relay.
    Don't waste the 20 watts.
     
  8. Guest


    Plus, if the original poster uses resistors, depending on how small
    (physically) they are, they can get pretty hot, dissipating 20 watts.
    That's a lot of heat, especially if placed in an enclosed space -
    could possibly start a fire.

    Just my 2 cents, and fwiw, I'm not an electronics/electrical expert by
    any stretch of the imagination

    M
     
  9. Active8

    Active8 Guest

    nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20
    W per, and goodby resistors. unless you need a space heater also,
    forget using resistors, and load it down with some automotive bulbs
    which handle 13.8 V or so nicely - if, in fact, that thing really has
    a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24
    auto bulbs are available since, IIRC military
    trucks use 2 batts in series for 24 V - can't remember why.

    mike
     
  10. Active8

    Active8 Guest

    nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20
    W per, and goodby resistors. unless you need a space heater also,
    forget using resistors, and load it down with some automotive bulbs
    which handle 13.8 V or so nicely - if, in fact, that thing really has
    a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24
    V auto bulbs are available since, IIRC military
    trucks use 2 batts in series for 24 V - can't remember why.

    mike
     
  11. pmoseley

    pmoseley Guest

    OK. Thanks Mike. How about a 5 ohm 50 watt resistor with the 12V
    output. That would pull 2.4 amps for 28.8 watts. I'm worried that a
    12V lamp would burn out. If the resistor doesn't put out too much
    heat I can just plug it in and forget it.

    peter
     
  12. Active8

    Active8 Guest

    Now you get more heat. At one time some people might not care, but
    with rising energy costs these days, you might put them in a box for a
    bun warmer ;) whatever works. Do it if there's nothing better staring
    you in the face.

    Mike
     
  13. ehsjr

    ehsjr Guest

    28.8 watts will make that resistor hot. Use 5 25 ohm
    25 watt resistors in parallel instead to spread the
    heat out over a much larger area.

    Ed
     
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