Motor Starter Voltage Drop

[mull982]

I believe that I am having a voltage drop issuue in a control circuit used to pull in a motor contactor. I have a control circuit that goes out a pretty long distance on a conveyor belt and then returns to an MCC to pull in the coil on a starter. After trying to run the motor I have found that the motor would run for a little bit, and the the coil on the starter would burn up/blow. When measuring the voltage across the coil (120V coil) during starting I am seeing about 110V which indicates that there is a slight voltage drop. Would this voltage drop be enough to damage the coil? I was told that with this voltage drop we would be pulling more current through the coil and that is what is causing it to blow. Is this correct?

I could understand why the voltage drop would cause more currnet to flow, but I cannot see it mathmatically. Rearranging ohm's law V=IR you would be able to calculate the current by I=V/R. But mathmatically if you decrease the voltage in the numerator of this equation with the "R" held constant it seems like your current value would be less. Is this an accurate representation of this scenario?

Thanks

 

[coulter]

...When measuring the voltage across the coil (120V coil) during starting I am seeing about 110V which indicates that there is a slight voltage drop. Would this voltage drop be enough to damage the coil? ...

I've never seen or heard of undervoltage burning up motor starter coils.

...I was told that with this voltage drop we would be pulling more current through the coil and that is what is causing it to blow. ...

No, It won't pull more current. The coil circuit looks like an inductor with a series resistance. Lower applied voltage = lower current.

Limiting the discussion to standard motor starter and coils, the only things I know of that will burn up the coil is the wrong voltage:
120V on 24V coils
DC to an AC coil

I'd start with some troubleshooting:
Has it ever worked as per design?
If it did, when did it change?
Are you really sure you are measuring the voltage at the coil?
Are you reallly sure the coil is rated for the applied voltage?

The only reason I mentioned the last two, is cause I been there at least once.

carl

edited in an attempt to fix my poor wording and fat fingered typing

 

[jim dungar]

If you do not have enough voltage (about 85% of nominal) to create a strong enough magnetic field to "pull in" the starter you will continue to draw very high currents.

The coil of the starter acts like the stator of a motor, lower voltage = more current drawn.

 

[mull982]

So is it safe to say that lower voltage results in more of a current draw? I dont see that happening with the simple V=IR equation. Do I have to add an inductance factor in that equation or is the circduit strictly resistance?

Thanks

 

[coulter]

Jim -

...The coil of the starter acts like the stator of a motor, lower voltage = more current drawn....

Isn't that true only if the contactor does not pull in?

...I have found that the motor would run for a little bit, and the the coil on the starter would burn up/blow. ...

My understanding is the contactor in question pulled in.

So, in the range where the contactor stayed pulled in, how would you characterize the V/I relationship?

carl

(Carl's bad pun of the day: "Well carl, V/I = Z")

 

[len149]

Motor Starter Voltage Drop

Check voltage at control transformer without conveyor running then check the voltage at the coil with starter pulled in, this will give you an idea of how bad of a drop there is. You can also check accross the wires going out to the stop Push button which should be causing the most drop since the holding contact shunts the start PB and pretty much eliminates any drop there. Make sure your contacts on the Stop PB are good. If, all else fails the voltage drop increases with load, so if this is a large size starter you could have the start pb bring in a small pilot relay mounted in the can which can then bring in the starter.

 

[coulter]

So is it safe to say that lower voltage results in more of a current draw? ...

Not in this context. Once the contactor pulls-in, the voltage can drop pretty far before it drops out.

Jim - If I am not understanding this correctly, I would welcome some science and references.

... Do I have to add an inductance factor in that equation or is the circduit strictly resistance?...

My understanding, for an AC coil, the current is mostly limited by the inductance. That is why, as Jim said, the current drops when the coil pulls in. The inductance goes up when the armature hits the coil pole piece.

That's why DC will burn up an AC rated coil. Inductance won't limit the current.

carl

 

[Jraef]

Low voltage will definitely increase the current in an AC coil, but I would not characterize 110V on a 120V coil as "low" with any significance. Any AC coil will have at least a +-10% tolerance and NEMA specs call for -85% to +110%. So a 120V rated coil on a NEMA design contactor should be good to 102V. A lot of IEC contactors meet that spec as well.

If the coil is burning up, are you sure there is nothing that is interfering with the pole faces of the armature coming cleanly together? Because any obstruction will interfere with the shading coil being able to function, so the armature will vibrate at twice the line frequency and the coil current will remain high. At 120Hz, sometimes you can't hear the contactor buzz. Or the shading coil itself is damaged; same difference.

Another remote possibility is that you have the control circuit run in the same conduit as the power leads (separate but adjacent in plastic conduit will do it too). With enough distance, the control wires my be picking up an induced voltage which is raising the coil terminal voltage beyond it's limit. When you tested the control voltage, was the motor running, or were you doing it with the disconnect open? It could also be next to the power leads of a different motor, in which case you wouldn't have noticed the extra voltage if that other motor was not running.

 

[coulter]

Low voltage will definitely increase the current in an AC coil, ....

Let me make sure I understand what you are saying. for this case and this limited case only:

120VAC contactor coil.
The contactor is pulled in.
Measure the voltage across the coil and it is 10% low, 108V.

You are saying the coil will draw more current than if the voltage is at 120V?

carl

 

[mull982]

What is the typical current rating on a size 5 coil?

 

[don_resqcapt19]

Jeff,

With enough distance, the control wires my be picking up an induced voltage which is raising the coil terminal voltage beyond it's limit.

How common it that? How long do the runs have to be? We always run the control wires with the motor leads as long as the motor leads are #4 and smaller and have not seen this problem. Most of the runs would be less than 500'.
Don

 

[coulter]

...How common it that? How long do the runs have to be? ...

I've never seen an over voltage, however, one time I had one that wouldn't drop out. 3wire S/S station about 1000' out, 120V AC controls. Control wires were run with a lighting circuit. Lighting ckt was 3#8 feeding two each - 100W incandesents. As I recall (20+ years ago) it was a small pump, maybe a size 0 or size 1.

Stop button out at the end would not drop out the contactor. Went to 120VDC controls and it worked. That was when I still knew all there was to know about this stuff.

carl

 

[mull982]

I have been noticing that the coil has been blowing only when we try to stop the motor, thus drop out the contactor. I have read that when the electromagnetic field in the coil collapses, that it produces a voltage transient in order to dissipate some of the stored energy in the magnetic field. I read that this voltage transient can become very large, and this is why surge suppressors are used accross the starter coil. Could I be having a problem with the surge supression not working correctly and the coil being damaged because of a high voltage transient? There is a surge suppressor across the coil, but is there a chance that something is happening here?

Just for my own knowledge can someone explain technicall why this voltage transient would occur when the magnetic fied drops out or collapses. Does anyone have any resources as to where I could find learning information for electromagnetic fields in general?

Thanks for all the help.

 

[frenchelectrican]

the other possiblty is the control wire size with that long distance i am not sure if the OP have ran #16 ga wire in that if so it might be pretty close to the limit but if ran #14 gauge wire it should work fine for that distance.

so i really cant say why if it was about 500 feet [ 157 metre ] if that the case the #14 will work due the 120 volt coil have very short peak during start up but when get in holding mode it draw very little current [ i cant say the spec size yet unless the op give us little more info on that ]

someone say about the coil curent on size 5 starter well ,, it depending on which manufacter it used i cant say for now because i dont want to do the " WAG " on this one because there are few factors it will fall in is voltage of coil and if this is ac or dc it will make the differnce there.


thanks,

Merci , Marc

[* WAG = wild A** guess ]

 

[rattus]

The inductance of a solenoid is greatly reduced if the plunger is left out. If the plunger is only patly pulled in, excess current may result. One might see this on contactors, but I doubt that you would see it on ordinary relays.

This happens because the solenoid plunger provides the iron core of the coil.

 

[jim dungar]

A coil and armature are not a simple resistive circuit so it requires the more complex formula V=IZ and taken into consideration that Z is not linear.

Most coil surge suppressors contain MOVs which will eventually fail open and therefore no longer protect the coil.

For long control circuits a "capacitor" can be created by the paralled conductors. This capacitor can be strong enough (especially in water filled conduits) to prevent the coil from dropping out or to create high discharge spikes when the circuit is opened. The exact effect depends on the location of the control power source, the starter coil, and the "stopping" control contact as well as the size of the coil and the applied voltage. Almost every starter manufacturer has some type of application note concerning "long distance control".

 

[coulter]

A coil and armature are not a simple resistive circuit so it requires the more complex formula V=IZ and taken into consideration that Z is not linear. ...

Jim - Let me ask again.

Considering only the limited case where:
120VAC contactor coil.
The contactor is pulled in.
Measure the voltage across the coil and it is 10% low, 108V.

You are saying the coil will draw more current than if the voltage is at 120V?

If I am in error in my thinking, I would welcome some physics, or references to clear me up.

carl

 

[jim dungar]

Jim - Let me ask again.

Considering only the limited case where:
120VAC contactor coil.
The contactor is pulled in.
Measure the voltage across the coil and it is 10% low, 108V.

You are saying the coil will draw more current than if the voltage is at 120V?

If I am in error in my thinking, I would welcome some physics, or references to clear me up.

carl

I am saying that the coil requires a certain amount of VA to create the magnetic flux that does the work of holding the armature assemby in a sealed position. The physics are similar to that of a single phase motor trying to drive a fixed amount of output horsepower.

 

[coulter]

I am saying that the coil requires a certain amount of VA to create the magnetic flux that does the work of holding the armature assemby in a sealed position. The physics are similar to that of a single phase motor trying to drive a fixed amount of output horsepower.

I'm not seeing that - here's why:

Once the armature is pulled in, and the pole pieces are touching, no further work (physics definition of work) is being done. So the analogy to a motor doesn't fit - no energy output.

Second difficulty is with a change in inductance caused by a change in the voltage:
As I see it, once the pole pieces are touching, what would change the inductance. More coil voltage would just piush the pole pieces together harder. No motion, no work, just more pressure. Conversly, less voltage = less pressure. As long as the pole pieces are still touching, what would change the inductance?

carl

 

[winnie]

I am saying that the coil requires a certain amount of VA to create the magnetic flux that does the work of holding the armature assembly in a sealed position. The physics are similar to that of a single phase motor trying to drive a fixed amount of output horsepower.

The assumption of constant magnetic flux is not correct. As the supply voltage goes down, the current through the coil will go down, and the flux will reduce. The _force_ holding the armature assembly sealed will go down, and eventually the armature will release.

Once the armature releases, the inductance will go down, and at any given voltage the current will go up. If the supply voltage is low in just the wrong way wrong, then higher then acceptable current will flow but the armature will never pull in, causing the coil to overheat. But if the armature pulls in, then lower voltage will mean lower current flow and lower dissipation in the coil.

Just for my own knowledge can someone explain technical why this voltage transient would occur when the magnetic fied drops out or collapses. Does anyone have any resources as to where I could find learning information for electromagnetic fields in general?

Thanks for all the help.

Take a look online for a discussion of the concept of 'self inductance'. If you want to take the long (but fun detailed) way around to it, I recommend the following: http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm . This is a freshman Electricity and Magnetism course, available for free if you just want to watch through it. (No one will check your homework, and if you want the text, you'll have to buy it, but the lectures and class notes are free for the watching.)

In a nutshell: 1) when you have a conductor in a _changing_ magnetic field, a voltage will be induced in that conductor. This is used to make generators function; the 'changing' magnetic field is created by the rotation of the rotor. 2) A conductor carrying current _creates_ a magnetic field. 3) When you interrupt the current flowing in a conductor, you are trying to change the magnetic field created by that current flow; now you have a conductor in a changing magnetic field. See step 1. 4) The voltage induced by self inductance when you try to interrupt current flow will be in a direction to try to maintain that current flow; you can think of self inductance as electrical inertial mass.

-Jon