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Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 11:19 AM

I do understand that in general the relay coil burden is discussed in VA.

However, lets say for a second that someone wants to know the current load caused by a relay coil at a certain voltage

In general it seems manufactures dont give a power factor of such a thing.

So how to determine the current?

Assuming the device is 100% ideal which i doubt

I noticed that in some spec sheets for hold in current i was told

9.5VA and then it also says 2.7 Watts

It seems that this ma be enough information to actually figure it out

So S = sqrt ( vars ^2 + Watts ^2) obviously

9.5 = sqrt ( vars ^2 + 2.7 ^2)

I solved for Vars and got 9.108

tan inverse (vars/watts) gives me an angle of 73 degrees.

This tells me the load is highly inductive of course which is to be expected given that its a relay COIL

cos(73) is about .28 which is pretty terrible but not to surprising of a PF

Given

VA = V*I*PF

im going to say I = 9.5/(120V * .28) using a 120 volt control

Meaning i actually is .28 amps also.

As far as i know the PF should not change from pick up to sealed VA

If you repeat this with the pick up VA of 75 you get 2.28 A

Roughly 10 times the sealed current. No surprise there.

Anyone see any flaws in this that im screwing up?

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#1

Re: Easy way to determine the current load of a relay coil

06/16/2017 11:53 AM

Measure it with an ammeter, or look it up on the data sheet. Sheesh.

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#2
In reply to #1

Re: Easy way to determine the current load of a relay coil

06/16/2017 12:03 PM

Not really feasible to measure 400 contactors that have not even been ordered yet to determine other upstream protections that have also not been ordered yet.

"Look it up on the data sheet"

The value is not given on any Data sheet of a contactor i have ever seen.

As i previously said its generally discussed in VA

"sheeesh"

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#3
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Re: Easy way to determine the current load of a relay coil

06/16/2017 12:08 PM

Why on earth would you ever wannabe a controller?

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#4

Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 1:14 PM

99% of the time when you are powering a relay, you are powering it from a transformer, and the transformer is rated in VA, not W. So by looking at the coil burden in VA, you don't NEED to know the power factor of the coil, it's irrelevant.

Selecting a Control Power Transformer

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#5
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 1:30 PM

just so you are aware, i posted that very same document in a post i made yesterday.

Im very aware of its existence.

The reason for the question is for sizing the upstream breaker.

say you have an 8 point digital output card.

The card is driving 8 motor starter relays.

The voltage feeding the card is 120 so the card is switching 120 AC.

The card 120 supply has an upstream breaker to protect it.

Yes a transformer is feeding the upstream breaker.

And you can size the transformer using the above mentioned document.

But last time i checked, circuit breakers are not rated in VA. They are in Amps

So if everything is in VA how are you going to determine the breaker size of the supply for the card?

You need to have atleast a guess at the current

a worst case perhaps if every single coil where to pick up at the same time.

and perhaps a nominal number which all coils are just sitting sealed.

The breaker size is then your discretion but you better know something more then VA. and if you dont you are not getting my stamp on it. Simple as that

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#16
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 6:00 PM

What are you attempting to protect with the breaker? A circuit breaker is really there to protect the conductors, it can't protect things like coils. So if you run #14AWG conductors, protect it with a 15A circuit breaker. Done. If that is feeding an I/O card that can handle 16A max, it will protect that too. Done.

I think you are over thinking this.

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#6

Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 1:32 PM

I think you are about right in your calculations! One gold brownie point awarded.

Why not just put a Styrofoam insulation box over the relay coil, and measure the temperature rise (using a small fan), and a calibration resistor temperature in, and temperature out, you turn on the cal resistor and see the change in temperature above the heat of the relay. Then ratio it all out, and you have it nailed to the nines.

Hint: Don't burn up the relay by totally not supplying air flow over it.

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#7
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 1:44 PM

that sounds pretty loosely unconventional...

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#8
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 2:05 PM

That was James unsubtle attempt to poke fun at me as usual

Please dont tell me you thought his answer was actually serious

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#10
In reply to #8

Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 2:07 PM

OH, I'm sorry, I was describing James....

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#11
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 2:35 PM

Actually, you could set it up strictly as a thermal calorimeter and gain some insight. Totally not necessary, when an ammeter and voltmeter would mostly do the same job.

Once latched, that coil is effectively a "heater".

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#9

Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 2:06 PM

Replying to Forum Thread by

wannabecontroller: (Use Copy & Paste to quote the original text.)

I do understand that in general the relay coil burden is discussed in VA.

However, lets say for a second that someone wants to know the current load caused by a relay coil at a certain voltage

In general it seems manufactures dont give a power factor of such a thing.

So how to determine the current?

A volt-ampere (VA) is the unit used for the apparent power in an electrical circuit, equal to the product of root-mean-square (RMS) voltage and RMS current.[1]

https://en.wikipedia.org/wiki/Volt-ampere

If your voltage is 120V rms, divide burden by 120 to get rms current.

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#12
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 3:26 PM

please tell me this was an attempt at humor and you are not seriously trying to tell someone this is factual

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#15
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 4:29 PM

Mine was both, it is a fact you can determine power dissipation in a device that way. Heat is heat. It was humorous, in that I did not really expect you would go out and do it that way when there are easier, better ways to get the data.

I hope you got a good chuckle.

What is wrong is using rms values? You are talking a single phase, correct?

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#18
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Re: Easy Way to Determine the Current Load of a Relay Coil

06/19/2017 6:00 PM

Dear Wannabecontroller,

It is I expect, practice of contactor/relay makers to simply measure the coil voltage applied and the current drawn on the actual coil/armature when it is wedged open (maximum air gap and minimum inductance and maximum current) and when contacts are closed (minimum air gap and current). Hence the specification volts x amps = VA, with two numbers e.g. "inrush"/"pull-in" and sealed/closed. That means rms volts x rms amps for AC service. The mechanism only cares about the magnetic pull due to the ampere-turns.

Consider a contactor with two 100 turn windings - both energised at 120V rms operate the contacts and draw 1 amp rms each - that is 120VA x 2 . If you put the two windings in series you need 240 volts and still get 1 amp in each winding (same ampere-turns) and VA is 240 x 1 = 240VA.

So it is practical for the maker to make 110V, 120V, 230V and 240V coils and just give one VA figure.

The user just divides the nominal voltage into the given VA to find the current drawn -the manufacturer will have applied worst case tolerances of "gap" and resistance to give a VA figure which can be used safely.

If you have only been taught basic theory as a student (and of course worry about power factor) you may not be aware of such "common practice" in specifications or have not yet seen that VA rating is the biggest factor for size and cost of transformers and other supply equipment.

In the case of AC contactor coils, the dominant factors are the "pull-in" current drawn (the VA, when adding the nominal supply voltage) and necessary supply power transformer rating VA to feed them and that the power factor is low - these decide the necessary rating of the contact which switches the coil, also the (LV contactor/relay) supply transformer rating and regulation. An important factor in transfo rating is the number and VA of contactors which will be operated simultaneously - it is well worth sequencing operation to avoid heavy surges and volt drops and oversized transfo, which also affect primary supplies and motor starting.

If you are concerned about what it takes to switch a contactor coil, you should take more account of the saturation when a coil is energised at zero volt point in supply cycle. The safe viewpoint is that the peak current is limited only by the coil resistance - this is important for solid state switches and less important for relay/switch mechanical contacts.

I am sure you will find, if you research the point, that solid state switch/PLC makers are aware that a 1 amp rated AC switch/output will be expected to reliably control a "1 amp inrush specification" contactor coil and build-in suitable ruggedness, including voltage surge protection and have 40 years or more of experience that their design margins give reliability. Long before "solid state" the necessary size of a relay contact to reliably switch certain loads was also established by experience.

67model

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#13

Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 3:32 PM

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Abuse/Attack: This post was deleted because it was an attack on another user. Please review the Site FAQ and the Rules of Conduct.

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Re: Easy Way to Determine the Current Load of a Relay Coil

06/16/2017 3:38 PM

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Irrelevant This post was deleted because it is related to a deleted post and would otherwise be taken out of context.

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#17

Re: Easy Way to Determine the Current Load of a Relay Coil

06/19/2017 3:08 AM

A simple telephone call to the equipment manufacturer would yield the answer. Please do not be abusive or vulgar during the call.

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