Starter Coil Ratings

[broadgage]

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

No there is no truth in the idea that operating a coil at a reduced voltage will cause it to overheat.
If the coil is run at a reduced voltage, it will draw less current and not more, and will run cooler not hotter.
This is not simply my opinion, but an observed fact which may be confirmed by measurement.

An electric motor worked at reduced voltage is liable to draw more current and may overheat, but that is because it is turning at almost the same speed, and therefore supplying almost the same HP as at full voltage.
However a contactor coil is not the same.

In years gone by, large contactor coils in the UK and perhaps elswhere, were fitted with an "economy circuit".
This consisted of a small auxilary contact, that once the main contacts had pulled in, inserted an impedance in series with the coil.
This reduced the voltage on the coil, to save power and reduce the heat produced. (the impedance was, on AC supplies, a small choke which looked like a flourescent lamp choke, and probably was)

 

[gar]

090502-0820 EST

Larry:

A slight correction to your AC DC comment. If a relay is designed as an AC relay, then it has a shading coil, a slug of copper around a portion of the core to produce a phase shifted magnetic field in addition to the primary field. A relay of this type can be operated on either AC or DC but at different voltage levels. As an example KA5AY 120 V 60 Hz will pull in at about 37 V DC excitation. This relay did not show a large change in inductance relative to armature position. See the big difference in the A B contactor below.

A DC relay, no shading coil, can not be operated on AC because of the 120 Hz variation in magnetic force from 0 to maximum. Terrible noise. A bridge rectifier can be added to a DC relay to make it an AC relay, but now has lost the advantage of the AC relay in terms of pull in force.


broadgage:

No there is no truth in the idea that operating a coil at a reduced voltage will cause it to overheat.
If the coil is run at a reduced voltage, it will draw less current and not more, and will run cooler not hotter.
This is not simply my opinion, but an observed fact which may be confirmed by measurement.]

This statement is wrong for an AC relay using a shading coil, and of the structure of an AB motor starter. If your voltage is below the pull-in voltage, and the relay is not already pulled in, then because of the lower inductance of the magnetic circuit there may be a higher current thru the coil than at full voltage and the relay pulled in.

Here is some data on a 50 year old AB #2 709COD:
Dc resistance 41 ohms.
Following are 60 Hz measurements:
Voltage Current
..Volts......Amps
50 ........... 0.68
60 ........... 0.10 just pulled in
90 ........... 0.16
120 ........... 0.24
Note at 50 V the power dissipation in the coil is about 8 times that at 120 V. It will almost certainly burn out.

The coil inductance is 100 MH when the coil is de-energized, and 580 MH when I fully advance the solenoid plunger. These inductance measurements are at low voltage and 1 kHz. The corresponding calculated coil reactances are 37.7 and 218. If I calculate impedances from these values and the 41 ohms DC resistance and then use those values to calculate current I get higher current values than the measured values. May be a result of the 1 kHz measurement and the effect of the shading coil.

.

 

[mull982]

broadgage:

This statement is wrong for an AC relay using a shading coil, and of the structure of an AB motor starter. If your voltage is below the pull-in voltage, and the relay is not already pulled in, then because of the lower inductance of the magnetic circuit there may be a higher current thru the coil than at full voltage and the relay pulled in.

Here is some data on a 50 year old AB #2 709COD:
Dc resistance 41 ohms.
Following are 60 Hz measurements:
Voltage Current
..Volts......Amps
50 ........... 0.68
60 ........... 0.10 just pulled in
90 ........... 0.16
120 ........... 0.24
Note at 50 V the power dissipation in the coil is about 8 times that at 120 V. It will almost certainly burn out.

The coil inductance is 100 MH when the coil is de-energized, and 580 MH when I fully advance the solenoid plunger. These inductance measurements are at low voltage and 1 kHz. The corresponding calculated coil reactances are 37.7 and 218. If I calculate impedances from these values and the 41 ohms DC resistance and then use those values to calculate current I get higher current values than the measured values. May be a result of the 1 kHz measurement and the effect of the shading coil.

.

From these values I notice that at 50V the coil will have more power in terms of KVA than with the current value shown for 60V. However the coil does not pull in at 50V, and it not until 60V that it pulls in.

So even know there is more power in the coil at 50V, does it not pull in for some reason possibly related to a field excitation related to the voltage?

It would appear than the because the coil does not pull in the impedence never changes as a result of the armature position, and the coil continues to draw higher than rated current. But this is for a coil that is not yet pulled in.

It sounds like there is a general agreement that a coil already pulled in will operate at a lower voltae before it burns up, and this level should be right above the pull in and not anything more. It sounds like the reason is that there will be more current at a lower voltage although others have said otherwise.

 

[mull982]

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.

.

I'm guessing that when applying an AC voltage do a DC rated coil then the impedence of the DC coil will be high, and because the AC signal will now have reactane the total impedence of the coil will be too high and will not pull in with an ac voltage.

The opposite would be true with applying a DC voltage to an AC coil. An AC coil will have a low impedence, and a DC voltage because it has not reactance will not change this impedence value. This low impedence value with this DC voltage will cause the coil to burn up.

 

[gar]

090504-1140 EST

mull982:

If you apply an AC input to a coil with a magnetic core the magnetic force applied to another piece of magnetic material that does not have a permanent magnetic bias will vary from 0 to a maximum twice for every input cycle. There is a net average value of force with a large variation superimposed. In general this does not cause closure of a solenoid or relay at 60 Hz. It does cause lots of noise.

By adding a shading coil around a part of the core a phase shifted magnetic field is created in that part of the core. Thus, there are two magnetic fields out of phase with each other pulling on the armature. If the magnetic material surfaces are flat and contact flat, then the device is usually adequately quiet for satisfactory use.

Later I will try again to describe why the current and power dissipation can vary greatly as voltage is reduced.

.

 

[LarryFine]

Larry:

A slight correction to your AC DC comment.

Danke.

If I may return the favor:

Here is some data on a 50 year old AB #2 709COD:
Dc resistance 41 ohms.
Following are 60 Hz measurements:
Voltage Current
..Volts......Amps
50 ........... 0.68
60 ........... 0.10 just pulled in
90 ........... 0.16
120 ........... 0.24
Note at 50 V the power dissipation in the coil is about 8 times that at 120 V. It will almost certainly burn out.

My math tell me that 50 x 0.68 = 34va, and 120 x 0.24 = 28.8va.

 

[gar]

090504-1518 EST

Larry:

In the case of a solenoid or relay coil it is the current and the coil resistance that is the primary source of power dissipation. There are some other factors such as: eddy current, hysteresis, and shading coil that produce power dissipation.

Voltage applied to the coil in this application is not a good measure of power dissipation in the coil because of the very large change in inductance between an open armature and a closed one.

.