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Volt-amps BS

Discussion in 'Electronic Basics' started by Michael, Mar 31, 2007.

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  1. Michael

    Michael Guest

    Some of these industrial ratings want to make me pull my hair
    out.....horsepower....volt/amps.....I've never bothered to commit any of it
    to memory.

    Could some please tell me...

    If a relay has a 24vac coil and a "hold VA" of 15. what does that translate
    to in current draw at 24 volts? Who knows what they're talking about.....I'm
    sure the relay probably kicks in around 18 volts or so. I'm just trying to
    get an idea of the transformer I'll need.

    Thanks. !
  2. Tom Biasi

    Tom Biasi Guest

    The Hold VA is the power being consumed to keep the relay operated.
    When you say the relay "kicks in " at around 18 volts you are referring to
    pull-in voltage.
    If you decrease the voltage until the relay drops out that is called the
    (take a guess) drop-out voltage.
    If your relay is using 15 VA at 24 volts the current can be calculated;
    P=VA, A=P/V.

  3. Eeyore

    Eeyore Guest

    VA rating = I*V !

    I = VA rating/V = 15/24.

    Too tricky ?

  4. Chris

    Chris Guest

    Hi, Michael. AC relays and solenoids are constructed differently than
    their DC counterparts. If you'd like, try to run a 12VDC coil relay
    with 12VAC. You'll just create a buzzer. AC relays have shading
    rings to counter this effect.

    An AC relay or solenoid draws a heck of a lot more current when it's
    pulling in (sometimes called "pull current") than when the plunger or
    relay actuator is in place. This lower current is called the "holding

    You should use common sense when sizing transformers for AC relay or
    solenoid loads. Since the pull-in only takes milliseconds, you can
    usually get away with sizing the transformer for the devices' holding
    current. I like to make sure the transformer is oversized (by, say,
    50%) just in case.

    Hope this has been of help.

  5. Michael

    Michael Guest

    I know exactly what hold in current is.

    That wasn't what I was asking.

    I'm simply asking someone to do the math for me as VA
    gets past me.

    What kind of actual current at 24vac does the "15 Hold VA" refer to?
  6. Michael

    Michael Guest

    OK thanks Graham.....I see you spelled it out for me.

    Thaks....looks like about 625 milliamps
  7. You need 24 VAC to pull it in. The 15 VA hold rating is probably constant
    over all voltage models. It still doesn't tell you the drop out voltage.
  8. Can you explain a little about how the AC relays work a bit? I'm confused
    because it would seem that only DC would work unless there was some sort of
    rectification inside. What are these shading rings and stuff? (I tried
    searching but really can only find info on "DC" relays)

  9. The shading ring delays the loss of current on each half cycle. It is
    usually a shorted turn of copper. ring
  10. There are some AC relays with built-in rectifiers to a DC coil, but they
    are rare.

    For AC devices with capacitance or inductance, VA does not equal W (watts).
    However, for calculating current draw, you can use VA/V to get A. The
    transformer needs to be sized for the current, not the power. When an AC
    relay or solenoid operates, the "clapper" or plunger moves to a position
    where the inductance increases, so the current drops.

    The hold VA is determined by the current drawn at rated voltage with the
    relay pulled in. The relay will drop out at a lower voltage, something like
    70% of operating voltage, which would be about 1/2 the hold VA.

    I found a good explanation for shading coils and shaded pole motors in the
    following link: ring

  11. If a relay was pulled in by a simple AC coil, there would be
    two moments every cycle (when the current passed through
    zero) that there would be zero pull in force applied to the
    armature, and the whole thing would buzz, badly.

    This is fixed by slotting the end of the pole piece that
    attracts the armature, and surrounding one half or so of the
    pole piece with a thick copper shorting ring. You can think
    of this ring and the part of the pole piece passing through
    it as a phase delay mechanism. When the flux level is
    rising in the other part of the core, the ring circulates
    current in the direction that bucks that flux to slow its
    rate of rise. But when the flux level is falling in the
    rest of the core, the ring circulates current in the
    direction that delays the drop of flux in the surrounded
    part. The net effect is that the armature is attracted by
    two poles excited by 2 phases of current, with the peak of
    one roughly at the zero of the other. So the attraction
    force is smoothed out, much like the way a two piston engine
    smooths the torque, compared to a 1 piston engine.

    The shorting ring does generate considerable heat (from the
    large circulating current), so AC relays are often less
    energy efficient than similarly constructed DC relays.
  12. MassiveProng

    MassiveProng Guest

    For all intents here, VA equals Watts. The 24Volst is not a peak
    voltage, but an RMS voltage as most declared AC voltages are.
  13. Actually, since relay coils generally have a low power
    factor, watts and VA are not at all the same. The relay
    coil may have a power dissipation of only 2 or 3 watts while
    having a VA of 15. The simple definition of VA is RMS volts
    times RMS amperes. This product tells you nothing about
    watts except that absolute value of watts is limited to
    something equal to or less than the VA. Watts can be
    positive or negative (power can be arriving or leaving), but
    VA are always positive, since RMS voltages and currents are
    always positive values.
  14. jasen

    jasen Guest

    15 VA with the plunger in and 24vac (at rated frequency) 15/24 A
    about 625mA. If the plunger doesn't go all the way in the current will be

  15. jasen

    jasen Guest

    VA is Volts times Amps just like in algebra, it's that simple.
  16. Michael

    Michael Guest

    Just wanted to say something....

    I want to thank John Popelish for his answers and "mini-tutorials" on so
    many topics that have come up on the electronics newsgroups.

    I've learned quite a bit from you.

    You always manage to make sense of it.
  17. Thank you for the feedback.
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