Voltage drop in control circuit

[gar]

161202-2451 EST

To get closer to what could be happening with a larger contactor I increased the series resistamce with my #2 starter to 48.5 ohms compared to the DC coil resistance of 40 ohms, source voltage 123 V. Resistance a little higher and failure to pull-in occurs.

Simultaneously measured the resistance and coil voltages.

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Blue is coil voltage, red is resistance voltage and therefore current..

 

[drcampbell]

... You can put a 1000 ohm 20 w resistor in series with the contactor coil apply 120 V to the series combination, and measure the voltage across the contactor coil and get a rough estimate of impedance. Current is approximately 120/(1000+DC coil resistance).

Or you could measure the voltage across the resistor, in which case the current -- in the resistor and the coil, both -- is exactly E/R.

Once you know the current, you can measure the voltage on the coil and calculate the total AC coil impedance with Z=E/I.
(note that the voltage across the coil and the voltage across the resistor will add up to more than 120V because they're out of phase)

You'll want to make two measurements, one with the solenoid held open and one with the solenoid held closed. The AC impedance is radically higher when the solenoid is closed and the magnetic flux flows through a closed iron loop, than when it is open and there are large air gaps in the magnetic flux path.

 

[gar]

161204-1600 EST

drcampbell:

Actually making the two voltage measurements is a better way. Just that when I used the scope I would have had to move some wiring.

When I made my last scope measurement in post #21 I found my isolator plug and floated the scope so scope common could be common to both the resistor and contactor. This also meant that a simultaneous measurement of both items was possible.

In this last plot it is interesting to see that the contactor in the energized state does not look like a linear inductance. I don't have an explanation.

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[GoldDigger]

161204-1600 EST



In this last plot it is interesting to see that the contactor in the energized state does not look like a linear inductance. I don't have an explanation.

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I wonder whether with the magnetic circuit closed by the armature you are getting some saturation of the core?

Sent from my XT1585 using Tapatalk

 

[gar]

161205-1039 EST

GoldDigger:

My immediate reaction also because there is nothing but magnetic material and wire. But with what we know about current and saturating cores the wave shape does not correlate.

The integral of a sine wave is -cos, and therefore, still a sine wave. A +/- linear slope comes from the integral of a square wave. How would we get a square wave to an RL network in this situation?

Can the shading coil encompassing a portion of the magnetic circuit in some way cause this result? Don't know. If we replace the triangular waveform with a sine curve, then the result, a lagging current, is what is expected, and a very inductive circuit, about a 90 degree shift.

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