LCR meter

[Metal john]

We have many 5,000 volt vacuum breakers & 480 volt motor controllers (AC 3-phase) with 120 volt AC coils. Occasionally an 120 volt coil will become defective (burn up) and be replaced. The coils are not subjected to under or or over voltage and are installed according to manufacture specifications.
Can a LCR meter with the added ESR feature detect if the coil meets manufacturer spec's? As the coil ages & being in service for a long period of time, can this meter measure the inductance and be compared to the manufacturer data sheet?
I have used a LCR meter to troubleshoot a 480 volt 3 phase 30 hp motor finding an ohm imbalance. My Fluke 87 did not detect any issues nor did megging the motor but my LCR meter did because it can measure ultra low resistance. The motor was replaced and service was restored without any issues.
If a coil is subjected to heat, moisture, etc., can a LCR meter be of value?
Thank you

 

[gar]

160826-0844 EDT

I don't know.

Why do coils fail. Coils are made of conductors with a coating of insulation. Electric current thru most conductors produces heat. Heat raises the wire temperature, and in turn the insulation temperature. The lifetime of an insulator is inversely proportional to its temperature. The smaller a coil is made to do a specific job the higher is its internal temperature when doing that job. From an engineering perspective a balance is made between size (cost) and expected life. Also note that lifetime is a random variable dependent on many variables.

The temperature distribution within a coil at constant excitation and ambient conditions will vary from some lower value near the outside surface to some inner point called the maximum hot spot. I believe the maximum hot spot is often considered to be about 10 C (18 F.) higher than the average temperature rise and average temperature rise is considerably above the minimum surface temperature. Thus, the most likely failure point is the maximum hot spot point. This failure could initially cause some shorted turns, but not yet an open circuit of the coil viewed from the coil terminals. This is what you are trying to detect.

Shorted turns will cause a DC resistance reduction. How many turns are shorted will determine the percent resistance change. A few turns and the change is small. Likely this could be less than the manufacturing tolerance on coil resistance. If you knew the actual coil resistance when at a stable room temperature condition when the coil was new and compared this with a later time and the same conditions, then with a high accuracy DC ohmmeter you could detect shorted. For high accuracy low resistance measurements you need to do a 4 terminal test. For 1% accuracy your contact resistance needs to be substantially less than 1% of the resistance to be measured. An old AB #2 motor starter with a 120 V coil has a room temperature DC resistance of about 40 ohms. My Fluke 27 with a good pair of leads reads 0.1 to 0.2 ohms. It is a two terminal resistance measuring device. With a good banana plug short it reads 0 ohms. Resolution is 0.1 ohm.

On the Fluke 27 my best accuracy on a 1 ohm measurement is about 20%. With two Fluke 27 instruments used to make a 4 terminal resistance measurement I can approach 0.1% accuracy on a 1 ohm resistance.

Can an impedance test more easily detect a shorted turn condition than a DC resistance test? I don't know. With my 1650-A bridge the #2 coil resistance is 39.6 ohms. The inductance is 0.108 H at a Q = 8 at 1 kHz. The inductive reactance is about 40 ohms at 60 Hz.

Haven't proofread anything above. When I return later I will comment on the open and closed inductance of a solenoid or AC relay coil.

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[gar]

160826-1246 EDT

Continuing:

My previous post has a least one missing word and some comments may not be real clear.

The motor starter is an AC coil type.

Measurements on the #2 motor starter coil are:

1. Coil DC resistance = 39.6 ohms, round this to 40 ohms, and assume this is also the 60 Hz resistance.

2. Assume that magnetic core losses do not alter the apparent AC coil resistance whether the core (or armature) is in its de-energized or energized position.

3. Inductance is measured at 1 kHz, assume value is valid at 60 Hz.

4. Armature in de-energized position --- L = 0.108 H, Q = 8. Calculated impedance at 60 Hz --- Z = ( ( 377*0.108)^2 + (40)^2 ) )^0.5 = 57 ohms.
At 120 V the calculated coil I = 120/57 = 2.11 A. Coil power dissipation 40*(2.11)^2 = 178 W. This is huge and if the armature does not move quickly, then coil burns out quickly.

5. Armature in energized position ------ L = 0.456 H, Q = 3. Calculated impedance at 60 Hz --- Z = ( ( 377*0.456)^2 + (40)^2 ) )^0.5 = 177 ohms.
At 120 V the calculated coil I = 120/177= 0.678 A. Coil power dissipation 40*(0.68)^2 = 18 W. This is large and implies that hand holding the armature (plunger) did not fully seat the armature.

6. Actual measurement of energized coil current at 125 V was 0.23 A. Calculated coil power dissipation for this current is (0.23)^2 * 40 = 2.1 W. The physical sizeof this coil can probably handle this continuous power formany years.

The point of all the above is that coil burn out is not likely if the armature seats properly, and quickly if the armature does not properly seat. If you have a burned out coil look for possible problems with armature orplunger motion.

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