# Volt-amps BS

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

1. ### MichaelGuest

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.

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 BiasiGuest

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

3. ### EeyoreGuest

VA rating = I*V !

I = VA rating/V = 15/24.

Too tricky ?

Graham

4. ### ChrisGuest

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

5. ### MichaelGuest

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. ### MichaelGuest

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

7. ### Homer J SimpsonGuest

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. ### Jon SlaughterGuest

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

9. ### Homer J SimpsonGuest

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

10. ### Paul E. SchoenGuest

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

Paul

11. ### John PopelishGuest

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. ### MassiveProngGuest

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. ### John PopelishGuest

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. ### jasenGuest

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

15. ### jasenGuest

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

16. ### MichaelGuest

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. ### John PopelishGuest

Thank you for the feedback.