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Thread: Electromagnet Control Wiring

  1. #1
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    Electromagnet Control Wiring

    Hello, I have a friend who owns a recycling company. He has an old electromagnet that he wants to hookup to his excavator to sort scrap. The previous owner had the magnet running off a generator on his excavator in a similar setup. The magnet has no markings on it, but he was told that it runs on 240V DC. The magnet has a 2 conductor 12awg SJO wired to it. This is a little outside my normal scope of work but I thought it would be a good chance to learn and also help out a friend. I am thinking I can use a relay to handle the control to the magnet but my question is how do I turn 240V AC into DC for the magnet? Half wave rectifier? Is there a good source to learn more about this?

  2. #2
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    Electromagnet Control Wiring

    crispysonofa,

    I'll try to give you a Scope of Work for this. First, I'll answer your last question (
    Is there a good source to learn more about this)? The best source is a Freshman College level (as in AC/DC Circuits 101) AC/DC Circuits Textbook. Try your favorite Book Seller or eBay. An experienced Textbook shouldn't cost very much and they last a long time.

    The next part goes back to you for some visual information. The Magnet Badge Plate is missing. That's a bummer. Measure the diameter of the Magnet and search eBay for a similar size "Scrap Magnet". Someone will have one listed that has a picture of a Badge Plate that is close to your size. I realize that your Magnet could be different but right now, we have nothing for information that we can verify. In general terms, the larger the Magnet, the more it will pick up.

    Since this is a "from scratch" startup; there are many things to consider. The Badge Plate information is a good place to start. The Standby Generator will need to be sized. Switching and Control Circuits will need to be drawn up. DC Power Contactor(s) will need to be selected and sourced. Getting the DC from the Generator / DC Conversion to the Magnet will need to be considered; as in Swing of the Boom and Angle changes of the Boom.

    The Control for the Magnet ON/OFF should be a Selector Type not Momentary. The Operator has other things to do and won't have enough Hands to keep a button held in.

    The Installation will be quite extensive and I can help you. It will probably need to go to PM if you want to continue.

    I hope that is enough to get you started.

    JimO

  3. #3
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    A simple bridge rectifier fed with 240VAC will give you 330V DC, but pulsating. Smoothing capacitors will take care of that, but raise the voltage a little more. What you really need is a DC power supply that allows you to adjust the voltage to what works for your magnet without burning it up. In addition, if your core becomes permanently magnetized, you may need to reverse the polarity to get the magnet to drop all of the smaller scrap, otherwise it just builds up until the whole thing becomes useless. Honestly it's just easier to buy a power supply designed for industrial magnets like this. They are out there.
    __________________________________________________ ____________________________
    Many people are shocked when they discover I am not a good electrician...

  4. #4
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    170930-1305 EDT

    crispysonofa:

    1. Measure the physical size of the magnet.

    2. Measure the room temperature resistance.

    As a guess of resistance I will assume 20 A at 240 V, or 12 ohms. So an ordinary DVM, a Fluke 27, could get an adequate resistance measurement. This voltage and current is about 5 kW.

    3. For the power supply use a bridge rectifier. This provides a full wave rectified DC output. From a 240 V sine wave this provides an average DC voltage of 240*0.636/0.707 into a resistive or inductive load.

    Diodes are cheap so still assuming 20 A use diodes rated 50 A or greater. Heat sinking will be required. Select 1000 V PIV rating, and with avalanche protection ( https://en.wikipedia.org/wiki/Avalanche_diode).

    4. Place a reverse biased diode in parallel with the magnet coil. This forces "inductive kick" current thru this parallel (shunt) diode and the magnet. The reverse voltage across the magnet at turn off does not exceed the diode forward voltage drop. This lengthens turn off time, but who cares in this application. Almost certainly less than 1 second.

    For this diode use the same rating as in the bridge rectifier.

    5. By using the shunt diode there will be very little switch arcing. Use a switch rated for the AC voltage and current.

    Provide information on DC resistance and magnet size.

    .

  5. #5
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    crispysonofa:

    Jraef's comment on residual flux may be an important consideration. If it is, then my shunt diode is probably a bad idea.

    If instead of the diode you use a shunt consisting of a series resistor and capacitor, often called a snubber, then these values can be adjusted to produce a shorter dropout time, a lower residual flux level, and limit the peak inductive kick voltage.

    Without some sort of snubber the kick voltage will rise high enough to breakdown something. Usually this is the air gap between the switch contacts.

    A usual snubber will not have a capacitor large enough to raise the average rectified DC voltage much above its unfiltered average value.

    .

  6. #6
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    I will take a resistance reading on the magnet, also will provide dimensions. I appreciate the insight thank you!

  7. #7
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    Quote Originally Posted by gar View Post
    ... Place a reverse biased diode in parallel with the magnet coil. ... There will be very little switch arcing. Use a switch rated for the AC voltage and current. ...
    It's usually a bad idea to use a switch that's not rated for DC on a DC circuit.

    At the instant switch contacts open, and for several milliseconds more, the distance between them is small enough that the current can arc across the gap. This happens on all circuits, both AC & DC, and also non-inductive loads.

    In a DC circuit, the current continues to flow until the gap is wide enough to extinguish the arc.
    In an AC circuit, the current drops to zero twice per cycle and the arc goes out by itself.
    Because of this, a switch can be used at a much higher AC voltage/current than DC. This is why switches (and fuses & breakers) often carry two different ratings -- 125 VAC/32 VDC, for example -- or are rated for AC only.

    In this specific case -- rectified but unfiltered DC -- you might be able to get away with an AC-rated switch because unfiltered DC also drops to zero at the end of each half cycle, but an RC snubber might eliminate this advantage.

    Since this switch will be turned on & off hundreds of times per day, I recommend using a switch or contactor rated for considerably more current than is actually flowing.

    A snubber resistor can be used in series with a reverse-biased diode to speed up the turn-off time.
    (the inherent resistance of the magnet coil might be an adequate amount of resistance)

    The residual flux problem can be solved with a switch that reverses the polarity of the current through the coil, which the user can flip when it becomes a problem.

  8. #8
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    Quote Originally Posted by gar View Post
    170930-1305 EDT

    crispysonofa:

    ...
    ...
    5. By using the shunt diode there will be very little switch arcing. Use a switch rated for the AC voltage and current.
    You are assuming that the reason that DC rated switches are needed is entirely the inductive kick. But in fact, even with a resistive load the switch contacts will have more problems interrupting just the normal current flow and voltage because there is no mechanism other than that of the switch contacts to suppress the normal current arc. The current in an AC circuit will pass through zero 120 times per second, allowing the arc (actually the trail of ionized gases) to extinguish.


    Sent from my XT1585 using Tapatalk

  9. #9
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    drcampbell:

    In this application one puts the switch ahead of the bridge rectifier. Thus, the switch switches an AC current.

    However, with a reverse biased diode shunt across the inductor, if the switch was placed on the output side of the bridge, then after turn off the load voltage is only about -1.5 V. Thus, the maximum voltage across the switch is the instantaneous full wave rectified sine wave plus the approximately 1.5 V diode drop. The switch arc extinguishes quite quickly. But this is a bad idea because there is still some metal transfer between the contacts in one direction.

    When you have only an inductor and series resistor in a switched DC circuit, and the switch is opened, then the inductor will produce a voltage of whatever is necessary to maintain current flow (possibly many thousands of volts). This maintains the switch arc until the stored energy in the inductor is dissipated.

    .

  10. #10
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    I used the search term 'lifting magnet control' and found:

    http://www.hubbell-icd.com/icd/magc/magc4292.asp

    http://www.hubbell-icd.com/icd/magc/magc4291.asp

    http://www.walkermagnet.com/lifting-...trolmaster.htm

    https://www.hubbelldirect.com/Produc...et%20Controls/

    The latter appears to have schematics for some of the controllers.

    If the magnet is fed with 12 AWG conductors, then it probably is rated for pretty low current. I was visiting a shop doing some motor testing, and one of the other jobs that they were doing was re-winding lifting magnets. These magnets used flat copper strip as the conductor, perhaps 1/16" thick and 2" wide, coiled up into a spiral with thin 'paper' insulation (probably nomex).

    -Jon

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