# Volt-amps BS

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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. !

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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.

Tom

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VA rating = I*V !

I = VA rating/V = 15/24.

Too tricky ?

Graham

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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 current".

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.

Cheers Chris

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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?

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

Thaks....looks like about 625 milliamps

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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.

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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)

Thanks, Jon

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The shading ring delays the loss of current on each half cycle. It is usually a shorted turn of copper.

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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:

Paul

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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.

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On Sat, 31 Mar 2007 12:18:54 -0400, "Michael" Gave us:

For all intents here, VA equals Watts. The 24Volst is not a peak voltage, but an RMS voltage as most declared AC voltages are.

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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.

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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 higher.

Bye. Jasen

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

```--
Bye.
Jasen```
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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.

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Thank you for the feedback.

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