Starter Coil Ratings

[mull982]

For a 480V contactor used in a size 2 motor starter the control voltage on the coil is 120V.

If the contactor is pulled in, what would the voltage on the coil have to drop below in order for the coil to dropout? I believe I saw somewhere that the contactor would stay pulled in up to 70% of its voltage which would be about 85V. I know the coil depends on the total VA, but with a voltage drop on the coil what voltage would the coil stay pulled in unitl?

Also is there a risk of damaging a coil if operated within 10% below its rating. If a 120V coil is operated continuously at 108V is there a chace of burning up the coils? Do the coils have a +/- 10% tolerance similar to motors?

What about if you power the coil with no plunger element attached on the contactor? In other words if you pull the coil itself off and power it what would happen?

 

[jim dungar]

I believe the standard is %85 pick up and 65% dropout.

Low voltage will cause heat and noise.

 

[mull982]

I believe the standard is %85 pick up and 65% dropout.

Low voltage will cause heat and noise.

I am currently in a debate with a co-worker about this subject.

He is stating that 90V on a 120V coil will cause the contactor to drop out. I said this was not true based on the 65% dropout specification that you stated.

He is also stating that running a 120V coil with 108V will cause the coil to burn up fairly quickly. I stated that the coil would probably pull about 10% more current but I dont think this would cause the coil to burnup. I was under the impression that you could operate the coil at +/- 10% of rating.

I was looking for further evidence or information to back up my statements.

 

[Jraef]

Keep in mind the context of what "Drop out" and "Pull In" voltage really means.

"Drop out voltage level" does not meant WILL drop out at 65% voltage, it means it WON'T drop out with voltage AS LOW AS 65%. So at 80% it will hold, at 70% t will hold, and at 65% it will hold. At 64% or 44% it might still hold and that's fine, as long as it didn't drop out at 66%. Some newer designs of IEC contactors are coming standard with electronic coils. They have a little SMPS built-in and can take any voltage from 23 to 267VAC. So if you apply 120V to that coil and it drops to 40V, it will not drop out.

The same holds for "Pull In". That just means that at 80% voltage, it WILL pull in. But it's just as acceptable if it pulls in at 50% or 30% voltage.

 

[mull982]

Is there any truth to the fact that operating these coils at 90% of rated voltage will burn up or damage the coil?

 

[gar]

090429-1304 EST

mull982:

Think about your question a little.

Is the coil for DC or AC operation? It makes a big difference.

In a DC coil application (solenoid, relay, contactor, or whatever) you are only concerned with a maximum voltage or current to the coil because this determines the power dissipation in the coil. Whether a plunger is present or not, or armature closes makes no difference because temperature rise is determined by the coil resistance and applied voltage. You will never burn out the coil of a DC solenoid valve if the applied voltage is less than the maximum allowed voltage and ambient temperature. This is true even if the bobbin gets stuck.

In an AC coil application it is quite different. See if you can reason why.

.

 

[mull982]

090429-1304 EST

mull982:

Think about your question a little.

Is the coil for DC or AC operation? It makes a big difference.

In a DC coil application (solenoid, relay, contactor, or whatever) you are only concerned with a maximum voltage or current to the coil because this determines the power dissipation in the coil. Whether a plunger is present or not, or armature closes makes no difference because temperature rise is determined by the coil resistance and applied voltage. You will never burn out the coil of a DC solenoid valve if the applied voltage is less than the maximum allowed voltage and ambient temperature. This is true even if the bobbin gets stuck..

My coil is an AC coil.

I understand here that the maximum power or heat dissipated in the coil will be a function of maximum current or voltage represented by either I^2R or V^2/R. So as long as the voltage is below the rated voltage the coil will never burn up.

In an AC coil application it is quite different. See if you can reason why.

.

The first difference that comes to mind with AC is the reactacne of the coil, however I cant seem to figure what the reactance would have to do with this. I could see if the voltage was kept the same and the frequency was decreasing that we would saturate the coil, but here the frequency is staying the same.

The only other thing that I can think of is that the coil is striving to keep a maximum power output to the load and with the voltage decreasing the current would increase. However I would think that this is the same case as a DC coil.

I cant seem to come up with any other logical answer to your question as to why an AC coil would differ.

Can you give me a hint???

 

[LarryFine]

Can you give me a hint???

The impedance depends on the core's (armature's) position.

 

[gar]

090429-1652 EST

mull982:

Take a coil of wire with an air core and measure its inductance, record the value. Insert a straight ferro-magnetic (iron) rod in the center of the coil, record the inductance. Next place the center leg of an E section of an E I core thru the coil center, again measure the inductance. Last place the I part of the E I core against the E section, and measure inductance.

You will find there was a progressive increase in the inductance and therefore at a fixed frequency the reactance also progressively increased.

When the relay armature gap is large the inductance is low, and when the armature is closed the inductance is higher. Thus, there is an advantage to an AC coil over a DC coil in an otherwise identical mechanical structure. Can you tell me why?

As an example I took an unusual relay, a Sigma Instruments 4F 10000 S-Sil high sensitivity unit, and measured open and closed inductance. The values were 85 and 140 Henrys for the two states. A more ordinary relay will show a greater ratio.

This relay requires 14 V and 1.4 MA to close (19.6 milliwatts) and opens at 7.5 V. From this you can see it is a 10,000 ohm coil and is wound with very fine wire.

.

 

[mull982]

090429-1652 EST

mull982:

Take a coil of wire with an air core and measure its inductance, record the value. Insert a straight ferro-magnetic (iron) rod in the center of the coil, record the inductance. Next place the center leg of an E section of an E I core thru the coil center, again measure the inductance. Last place the I part of the E I core against the E section, and measure inductance.

You will find there was a progressive increase in the inductance and therefore at a fixed frequency the reactance also progressively increased.

So with a DC coil the resistance is a fixed value no matter what, and no matter if there is an armature presnet or not. Because this is a fixed impedence the power is only dependent on the max current or voltage.

With an AC coil because reactance is involved the total impedence of the coil is dependent upon the inductance of the coil based on Xl = jwL. When an armature or any other core is present in the coil then the inducatnce goes up and in turn the impedence goes up as well. Becasue the impedene of the coil is variable, the total power dissipated in the coil depends on not only voltage but the inductance of the coil controled by the presence of an armature.

When the relay armature gap is large the inductance is low, and when the armature is closed the inductance is higher. Thus, there is an advantage to an AC coil over a DC coil in an otherwise identical mechanical structure. Can you tell me why?

The advantage that sticks out in my head would be based on the fact of the coil size. I would think that since a DC coil impedence is strictly depended on just the resistance of the coil, then the coil needs to be large enough and have the correct amount of copper to allow heat dissipation for a given value.

With an AC coil, since the impedence is based on inductance, the coil can be made smaller and designed to dissipate less heat with the idea that when an armature is present the impedence will be high and thus limit the current. This may explain why when the coil is removed from the armature the impedence is low and the power dissipated is more than the coil can withstand. Based on this, my answer to you question is that an AC coil can be smaller in size and less expensive.



I dont see where a lower voltage plays into any of this as far as damageing the coil. I would think that based on V^2/R the power dissipated in a coil would be lower for any given impedence when the voltage was lowered.

 

[LarryFine]

. . . my answer to you question is that an AC coil can be smaller in size and less expensive.

More, the coil can have a greater pull upon initial energization, then the coil current drops as the armature engages, allowing a lower maintain power.

In other words, the AC contactor's coil power varies with the need for the magnetic field. A DC coil continually uses the power required for pull-in.

 

[mull982]

More, the coil can have a greater pull upon initial energization, then the coil current drops as the armature engages, allowing a lower maintain power.

In other words, the AC contactor's coil power varies with the need for the magnetic field. A DC coil continually uses the power required for pull-in.

So it would be true then that with and AC coil if it is pulled off the armature the impedence will remain low, and the current will remain high and thus burn up the coil?

What about operating coil at 90% voltage, I cant seem to see this causing damage?

As an example I took an unusual relay, a Sigma Instruments 4F 10000 S-Sil high sensitivity unit, and measured open and closed inductance. The values were 85 and 140 Henrys for the two states. A more ordinary relay will show a greater ratio.
.

How do you measure the inductance? With a meter or do you calculate it?

This relay requires 14 V and 1.4 MA to close (19.6 milliwatts) and opens at 7.5 V. From this you can see it is a 10,000 ohm coil and is wound with very fine wire.

.

Was this an AC or DC coil? If AC then this 10,000ohms would be the total impedence value including the resistive and reactive portions?

 

[gar]

090501-0720 EST

mull982:

In an AC coil application --- An armature held open or removed might cause a coil to burn up if normal voltage is applied. If 1/10 normal voltage is applied it probably would not burn up.

I think in general AC solenoids will burn up if the armature does not pull in. In particular solenoid valves.

As you lower the voltage to an AC coil relay or solenoid and go just below the point of pull in and maintain this voltage you will probably burn out the coil. Not so of a unit designed for DC operation.


I measured the inductance with a LRC bridge, a General-Radio 1650-A.


This was a high sensitivity DC relay with very small motion of the armature and thus contacts. I used this particular relay because it was small, open, convenient, and I did not need to look far for an example. The 10,000 ohms was DC resistance and thus very fine wire and maybe over 10,000 turns.

.

 

[mull982]

090501-0720 EST



I think in general AC solenoids will burn up if the armature does not pull in. In particular solenoid valves.

I understand now that this is because the impedence will be kept low, and alot of heat will be dissipated across the coil.

As you lower the voltage to an AC coil relay or solenoid and go just below the point of pull in and maintain this voltage you will probably burn out the coil.

.

What would cause this?

And if you are saying that this voltage would be all the way down near the point of pull-in then we shouldn't see this problem at 90%

 

[gar]

090501-0947 EST

mull982:

The equivalent series resistance of the coil is not very much different whether there is or is not a ferromagnetic core in the coil. With the core present, because of losses in the core, the equivalent series resistance of the coil increases a little.

If the inductance is low then for a fixed input AC voltage the current will be high. Insert a core into this same coil and the impedance goes up and the current down and there is less power loss in the coil. This results from the resistance of the coil being the same and there is less current at the same input excitation voltage.

What would cause this?

And if you are saying that this voltage would be all the way down near the point of pull-in then we shouldn't see this problem at 90%
Today 07:53 AM

Low voltage for some reason. Under normal conditions you should not have a low voltage. But under experimental conditions to understand the circuit or device theory a Variac can be used to lower the voltage.

If the voltage is just below the pull in point and you apply that voltage, then the device will not close and burn out will probably occur. In a sense this is what happens with stuck or gummy solenoid valves. Here the problem is the required force has gone up and there is a higher pull in voltage.

.

 

[mull982]

Low voltage for some reason. Under normal conditions you should not have a low voltage. But under experimental conditions to understand the circuit or device theory a Variac can be used to lower the voltage.

.

Say there was a large voltage drop on the feeder circuit feeding whatever is providing the control power for the coil?

My collegue was telling me a story that something happended with a transormer tap and only 108V was supplied to a dozen or so starter coils and burned them out within 20min. I found this hard to believe

 

[gar]

090501-1352 EST

mull982:

If the voltage at the starters only dropped to 105, and they failed at this voltage, then I would suggest the starters were poorly designed.

If I were designing the starters I might design for pull in at maybe 85 V, and drop out would be whatever it turned out to be. This would certainly allow me to normally operate down to 95 V with a reasonable safety margin. This may not be the criteria used by the manufacturers.

.

 

[mull982]

What would happen if you put an AC voltage on a DC coil? I'm guessing that no damage would occur, only that you would constantly be using the maximum power on the coil weather the armature was pulled in or not, as Larry mentioned earlier.

Would the impedence of the DC coil be too low for an AC voltage and thus maybe be damaged due to greater current, and the impedence not changing due to inductance?

 

[gar]

090501-1724 EST

mull982:

If you compare two identical relays of the same voltage rating. one being an AC unit and the other a DC unit, then the coil resistance will be much higher on the DC unit. Applying 60 Hz AC power to the DC relay will probably cause 120 Hz noise and no closure of the relay and probably no burn out.

.

 

[LarryFine]

What would happen if you put an AC voltage on a DC coil? I'm guessing that no damage would occur, only that you would constantly be using the maximum power on the coil weather the armature was pulled in or not, as Larry mentioned earlier.

Not exactly. A coil designed (rated, really) for DC will behave on AC as would any AC-rated coil with the same impedance characteristics, and vice versa.

Except for any physical differences, there is really no such thing as an AC or DC coil, they're merely constructed to operate at a design impedance for a target voltage.

There is an AC voltage at which a DC coil will operate satisfactorily, and vice versa. If needed, one could experiment if one couldn't obtain such info from the maker.