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testing for 24vac

Discussion in 'Electronic Basics' started by George Economos, Jan 12, 2006.

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  1. I am currently writing software for an X10 capable sprinkler system.
    The system has the ability to control 8 zones. Each zone has a
    corresponding 24 vac output that connects to the manifold. When the
    voltage is present the zone turns on.

    I would like to be able to test that he sprinkler is turning on the
    zones that I request. I was thinking that I could use 8 diodes
    representing the 8 outputs. When voltage is detected on one of the
    outputs the corresponding diode lights up.

    Would someone please explain how I could accomplish this. I am open to
    other ideas as well that may be simpler (or maybe more elegant
    ?segmented LED?)

    -george economos

    BTW - Please keep in mind that I am a software engineer with limited
    electronics experience ;-)
     
  2. Noway2

    Noway2 Guest

    Probably the safest bet would be to use an opto isolator. These are
    simple devices, that work like a switch. The input to the opto on the
    24Vac side is an LED which gets turned on when the voltage is present.
    The output side is a transistor that gets turned on by the led. To
    make this work you would need a few simple components in addition to
    the opto. You would need to put a current limiting resistor in series
    with the 24Vac input and use a pullup resistor to your local power (I
    will assume +5v for this example) and then put a small cap to ground to
    filter out the glitches that occur when the AC input goes to zero.
    Basically the resistor and capacitor are in series between your power
    and ground with one side of the opto output connected between the two
    devices and the other leg at ground.. When no Vac is present, you will
    see your +5v at the component junction and when the Vac is present,
    this signal will go low (or pretty close to it - use a 1k or 10k
    pullup).
     
  3. Thanks for the reply. It seems what you have suggested would be a way
    for me to place your device between the controller and the manifold.
    Am I correct? (I hope otherwise I am more lost than I thought)

    My device will only be used during the software development phase. I
    don't actually have to pass the current on to the manifold. ( In this
    way I don't actually have to look out the window and see if the
    sprinklers are turning on and off during testing)

    I was really hoping to supply each of the 8 24vac-outputs to it's own
    LED? I'm not just sure how. For instance can I connect 24vac directly
    to a LED? To you wire the commons together?

    This is a real NEWBIE question!!

    thanks again for your patience,
    george economos
     
  4. You can connect LEDs to your 24 vac outputs with series diodes and series
    current limiting resistors and they will light up when the output is on.
     
  5. gecono

    gecono Guest

    So I would connect them in this order?

    24vac -> diode -> resistor -> LED
     
  6. Jasen Betts

    Jasen Betts Guest

    //
    --------[2.2K]-+-->|--+ LED
    | |
    24VAC +--|<--+ 1N914 (etc)
    |
    ----------------------+


    it may be easier to use 24v instrument lamps (they use them the
    instrument cluster of trucks)


    Bye.
    Jasen
     
  7. Noway2

    Noway2 Guest

    That should work, normally I think I would probably put the resistor in
    after the 24vac and before the first diode (there is nothing wrong with
    two diodes being directly in series. I don't have a "magic" reason for
    doing so, but I like to have a current limiting resistor between my
    source and my load for safety.

    The reason I suggested using an opto isolator (instead of a plain
    diode) is that you have the potential issue, pardon the pun, where you
    can't be sure if there is a voltage difference between the sprinker
    power and your power system. Using an opto isolator will give you a
    significant amount of electrical isolation, letting the two circuits
    operate independantly.

    The other pieces of information you need to know, is how to read a
    datasheet and how to apply ohms law. The diode (opto) and your LED
    will all have a voltage drop assocated with them typically on the order
    of 1 to 2 volts, at a given nominal operating current. As a rule of
    thumb for diodes and LEDs, I use 10ma, but you will need to see what
    the spec for your chosen parts call for. Once you have these figures,
    you subtract the voltage drops from 24 (from the 24Vac) and then divide
    this voltage by the desired current., which then gives you the nominal
    current limiting resistor value. For example, if each diode has a 2v
    drop, and you want 10ma you have 24 - 4 = 20 / 10ma = 2k ohm resistor.
     
  8. Okay I am almost ready! One more question:

    Does it matter if I use a 1/4W or 1/2 W resisitor?

    thanks,
    george economos
     
  9. Rich Grise

    Rich Grise Guest

    You don't need optoisolators. At each output, I'm guessing you have
    two screw terminals or something, that you connect the two wires that
    go to the solenoid valve, right?

    If so, then take one ea. 2.2K resistor, one ea. LED, and one ea.
    1N4004-4007 (or any rectifier that's good for 50V or more) diode,
    and wire them in series- parallel, like this:


    +----|<---+
    24VAC 2K2 | LED | 24VAC Switched
    o-------/\/\/\------+ +---------o Return
    Switched | |
    +---->|---+
    DIODE

    where the two 'o's go in parallel with the solenoid coil. If all
    of the solenoids share a common return line, then you'll have to
    connect one end of each of these eight networks to it.

    Have Fun!
    Rich
     
  10. Rich Grise

    Rich Grise Guest

    No! Don't put them in series. Put the diode in antiparallel, to limit
    the reverse voltage on the LED:
    K
    +-diode-+
    24 vac -> resistor --+ +---24 V. ret.
    +--LED--+
    K

    K = "Kathode" ;-) (either way will work, as long as they're opposite.)

    Cheers!
    Rich
     
  11. The resistor will be dropping 22V. Current flow will be whatever you
    choose, e.g., 10mA. Since the diode is blocking half of the AC cycle, power
    dissipation is halved.

    So, power dissipation will be P = EI/2 = 22V * 10mA / 2 = 0.11W. You'll be
    fine with either a 1/4W or 1/2W resistor, assuming you choose 10mA of
    current (that is, a 2.2k resistor).

    Note that if you hook it up the way that Jasen suggests, with the diode in
    parallel with the LED but in the opposite direction, then current is flowing
    in both half-waves, so the power dissipation in the resistor doubles
    compared to the above calculation. The LED is still only turned on half the
    time, though, so its brightness does not change. For that reason, I think
    that Noway2's suggested circuit is preferable to Jasen's.
     
  12. Hey guys thank for all the help so far.

    Assuming I run the diode and reistor in SERIES, can I use the a
    1N914/1N4148 diode (for some reason or another I have some) with the
    2.2k resistor? I need to read up on what a diode actually does ;-)

    Hey another question what type of specifications should the LEDs have?
    I assume voltage would have to be around that after the resitor does
    it' s thing.

    Again thank you for all the patience. I find this all very
    interesting. Great group.

    thanks,
    george economos
     
  13. Rich Grise

    Rich Grise Guest


    No, because when the diode and LED in series are reverse-biased, there's
    no telling if the reverse leakage of the diode will allow the LED's
    reverse breakdown to be exceeded. There's no facility for them to "share"
    the reverse voltage, and if they did, the LED would pop.

    Use the diode in series if you need to, but put another diode in
    antiparallel with the LED, just to protect it.

    Good Luck!
    Rich
     
  14. The 1N4000 series rectifiers (standard for this type of application) will
    limit reverse leakage current to nanoamperes. Nothing will pop.
     
  15. Jasen Betts

    Jasen Betts Guest

    yes both types can handle the peak voltages (approx 32V) and currents
    (approx 15mA) involved
    bright enough to be visible if you're working in sunlight high intensity
    LEDs could be a good idea, otherwise any LED will do.

    Bye.
    Jasen
     
  16. Success! I managed to create a test circuit and hook it up to the
    sprinkler controller. I ended up wiring the LED and diode in parallel
    in opposite directions as suggested. I am able to turn the LED on and
    off using X10 commands from my pc. I would like to thank everyone for
    their help on this.

    george economos
     
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