Depending on the polarities involved (*), you might be able to add a suitable diode (or diodes, if required) in just the right location that would provide a path for the current generated as you've described.

* Obvious the first consideration is to ensure the added diode(s) aren't conducting when power is applied - shorting out the power supply.

If this is a large and high power system, then it would require some analysis to choose a sufficient diode.
 

24V DC, resistance through the magnets is ~2.4ohm when measured with a DMM.
They are powered by a small 24V, 120Watt control systems power supply.
There are 4 magnets at 8" x 4" x 2.5" thick.

All the specifications were done before I inherited this machine, and I have NO information on the magnets.

Right now, there is a 7700uF cap with a 8.2Ohm 3watt resister in series that are connected in a strange / useless manner, but I think the original thought behind them was to discharge the caps rapidly to kill any residual magnetism. (There were issues with residual magnetism being too much for the motor to pull open, but we since remedied that with a simple air cylinder to break them apart).

Thanks for the reply

Maybe a zener diode pair back to back, with high value voltage rating that only turns on when you pull the magnet away.  

What's the repetition rate? Is this effect happening only once in a long while, or (for example) many time per second? You'll need to know how much power you need to dissipate.

A diode in the right place should be able to steer the current and the IR loss in the coil would dissipate the power as heat.

By providing a current path you can keep the voltage down. If you don't provide a current path, that's when the voltage will build up and cause arcing.
 
 

I clamp for maybe 2 seconds out of an 6-7 second cycle time.

Figuring out the amount of power I need to dissipate will be the more challenging part. I'm trying to get some specifications on the magnets.

What kind of diode would I be looking for? Schottky? Or would that not be required for a 24v system? And how would I calculate the forward current for the diode?

Thanks for your help (and patience!)
 

I think that the diode won't work.

It will, of course, take down the arcing and the kick back voltage. But it will also slow down the release by several hundred milliseconds, perhaps a full second.

Also, if you had problems with residual magnetism, a diode will not be any good. You may need to accept the arcing - use a heavier contact - or find a working solution with capacitor/resistor that turns the circuit into a "demagnetizing device" by letting the current oscillate around zero with a decaying amplitude when you disconnect. That will effectively reduce residual magnetism.

Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.

What about a metal ring (with wire brush contact wiskers) around the magnet that has high power resistors to ground. As your magnet lifts up, brush wires contact this ring and dissipates energy resistively to ground. I'd guess the voltage developed in your magnet is proportional to the height it is lifted.  

Good point about the delay. Worth considering when determining the optimal solution.

Quote:

...to ground.

To ground? A local common point?
 

you might consider instead of de-energizing the main coil, placement of alternative flux paths that are separately saturable.

this allows you to hold and release your part without the large inductive kick of the main magnet  

Quote:

After removing charge from the magnet, I pull the magnet away from its workpiece, which causes the electromagnet to work as a motor and generate energy.

I don't really understand the problem statement.
What does it mean to remove charge from a magnet?
In what form is this energy manifested, and why do you consider it causing you a problem?  

=====================================
(2B)+(2B)'  ?

In response to electricpete

The core problem is arcing across the contacts of my relay.
I know this problem is caused by the inductance of the electro-magnet, but it seems to be complicated by the movement of the magnet so quickly after shutting it off.

To remove charge from a magnet = remove power from the electro-magnet

Form of energy = arcing across the relay contacts (well after the relay contacts are opened)

 

Quote:

What does it mean to remove charge from a magnet?

I (we) assumed it meant to remove (turn off) the power.

you are discribing an inductive kick that results from a drastic reduction in the inductance of the magnetic circuit, the energy originally stored has to go somewhere... d/dt[0.5*L*i^2]

you require a resistor across the coil limit the voltage kick

is the steady state dissapation is a problem then put a zener in series with the resistor

 

There are quite a few popular misunderstandings and assumptions here. Look at the basic facts first: 24V DC, ~2.4ohm, 120Watt power supply.

24 V and 2.4 ohm -> 10 A. That current is too much for the 120 W PSU (24*10 = 240 W). So, either voltage, resistance or both is wrong. Or the PSU is overloaded or working in current limit mode, or both. Find that out first.

There are no relay contacts in the world that can handle 10 A DC in an inductive load. For that, you need a DC rated contactor or breaker. So, that is what you need to fix first.

Arcing is quite normal and a DC contactor has blow magnets and chambers to take care of the arcs, where they are cooled down and extinguished.

The speculations about reduced inductance are not real. Of course, the inductance is influenced when iron drops, but my work with lifting magnets in steel works, ship yards and scrap yards has never shown that it could produce any ill effects. It is the breaking of current in the inductive load that causes the arc - not any dropping iron.

Find out what current you actually have (DC clamp), get a corresponding DC contactor and your problems will be over. Wear sun shades if you don't like the arcing.

The residual magnetism may still be a problem. If it is, then you need to do something about it. There are many different ways of handling that. The decaying oscillation (using a capacitor plus a low ohms resistor) is simple, but capacitor values may become prohibitingly high so a reverse and reduced current may be your best choice. The simplest way of doing it is to use another DC contactor and a series resistor that is adjusted to reduce remanent field to zero.

 

Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.

It would be very simple to refute or confirm the OP's assumption about the "motor" origin of the problem.

1) apply power
2) clip the system together
3) remove power
4) have lunch
5) suddenly pull off the clip and hit the air cylinder

This will separate in time (lunch) the normal and expected inductive kick-back from the supposed "motor" effect.



 

Good idea for an experiment.  It is hard to imagine residual magnetism causing arcing accross open (not opening) contacts.  

 

=====================================
(2B)+(2B)'  ?

I did the experiment mentioned. And it is not the movement of the magnets causing the issue, but solely the inductance of the magnets.

I was also incorrect on the power supply. It isn't 120 Watt, but 14A (336 Watt), which should be able to handle the magnets. I also don't have a DC clamp, so I can't get a measured value.

It sounds like a DC contactor is the best way to go.
Do they need to be rated high because it's an inductive load? Or would a 10amp rated DC contactor be sufficient?

As for the residual magnetism, like I said above, whoever it was that had originally setup the magnetic clamp had a 7700uF cap with a 8.2ohm resistor to help break the residual magnetism. I'll have to look into how it's wired and how it should be wired.

Thanks for all the hlep
Cheers


 

A 10 A DC contactor should work. But your margin is zero. OTOH, at 24 V, contactors usually can take more than at hundreds of volts.

7700 uF sounds like an electrolytic capacitor. They do not work well in oscillatory circuits where polarity changes. Also, an 8.2 ohm resistor is waqy too much. The damping coefficient will be too large and you will not get much of the necessary oscillation.  

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

I'd probably oversize the contactor anyways. Just wasn't sure if the rating takes into consideration the inductance, or if it needs to be factored in.

The 7700uF capacitor is actually 3 electrolytic capacitors in parallel. And I'm not sure the intention behind the setup. I'm guessing they were re-wired incorrectly at some point.
But from what I'm told by one of the mechanical engineers who has seen the project from the start, the intent behind them was to break residual magnetism by giving the magnets a short, reversed current.
I've attached a quick sketch of the circuit in the electromagnets on and off states.

Cheers


 

• http://files.engineering.com/getfile.aspx?folder=f0f4a92a-74c8-410e-b518-37

What you might want is a drop controller. Perhaps like one of these.

http://www.ecandm.net/products.aspx?category=LiftingMagnetControllers

Mike

10A, even inductive, is well within the practical limits for solid state switching with a nice big fat FET (or several in parallel if required). If designed correctly, it would last 'forever'.

 

That is something I would like to test next time, MJR2. Do you have any experience with them? Safety and such things?  

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

Drop controllers are used with lift magnets of all sizes. Particulary in rather automated systems. This particular brand I have no experieince with. Hubbell also has some.

Some controls actually drive the magnet negative to discharge/drop faster.

Safe? Never had what I thought of as safety issues.

They sure beat just pulling the powered lead off. Once is enough of that.

Mike

Soo right! Thanks

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

Thanks for the info.

Part of the reason I was hoping to stay away from a full blown magnet controller was due to the cost, and also the requirement. I'm only running 168 Watts (7Amps) @24Vdc. Most of the magnet controllers I've glanced at are rated for much higher.

I think I'll go the route of a DC contactor and look into using a reversed current for a short pulse.

Cheers

If you don't have a drain for the current the voltage spike will eventually cause a winding failure in your magnet. And eventually can be a rather short amount of cycles. We have them integral on 40 watt lift magnets.

Mike

If I understand the problem, you have an electromagnet that is powered by 24V DC.  With the power supply off/unattached/etc, if you move a magnet away from the coil, you get arcing somewhere in the system.

Sounds like you have reinvented the magneto!

Just put a large wattage zener diode across the coil terminals.  You put I would pick perhaps 50 volts.  The zener will clamp any reverse voltage greater than 50 volts, and any forward voltage greater than 0.7 V.  Problem solved.   

www.MaguffinMicrowave.com

Maguffin Microwave wireless design consulting

Problem with the zener is that the inductive kick is negative. So, to use the 50 V, you have to connect cathode to the negative terminal and anode to the positive terminal. Then, the zener will conduct heavily when you apply power. Not good at all. You could just as well use a diode, with the delay issues mentioned before.  Or a diode and a resistor.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

For relays, zener diode paired with a standard diode back to back is what Tyco recommends... see Figure 3 on page 2 here:
http://relays.te.com/appnotes/app_pdfs/13c3264.pdf

It provides the best compromise when we want to get the fastest reduction of current possible (fastest dropout) while clamping voltage to a specified value.

The requirement to get fast reduction of currrent is important for relays with time-sensitive outputs.  Also some have suggested slow operation of the relays can damage the output contacts if I understood correctly.

I don't think any of the concerns for fast deenergization apply to your application (??) so seems like diode with series resistor would do just as well.

There was also a lot of good discussion by contributors in this thread:
thread248-291225: dc coil surge supression
 

=====================================
(2B)+(2B)'  ?

"Problem with the zener is that the inductive kick is negative. So, to use the 50 V, you have to connect cathode to the negative terminal and anode to the positive terminal. Then, the zener will conduct heavily when you apply power. Not good at all. You could just as well use a diode, with the delay issues mentioned before.  Or a diode and a resistor. "

No, you connect the zener so the cathode is on the + terminal of the power supply, and the anode is on the - terminal of the power supply.  That way it does not conduct when the power supply is turned on.

When a inductive voltage kickback happens, it is thus limited to Either +50 Volts, or - 1 Volt.

www.MaguffinMicrowave.com

Maguffin Microwave wireless design consulting

The misunderstanding is that the kick-back is negative. Not positive. So your zener will act as a standard diode. Make a measurement to convince yourself.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

Gunnar is correct.

Connection using standard diode is shown in figure 2 of link above  (does not rely on any reverse breakdown).   Upon interruption the coil produces a voltage which (when seen by the rest of the circuit) will tend to keep the current flowing in the same direction as it had been.

The circuit using a zener is in figure 3 linked above, and provides best compromise of clamping the reverse voltage and rapid current reduction...but again rapid current reduction  may not be important to your application.

 

=====================================
(2B)+(2B)'  ?

Quote:

...coil .... tend to keep the current flowing in the same direction as it had been.

Perfectly correct (of course).


 

ElectricPete - thanks for your input and the link to the other thread on the forum. After reading through the other thread, I've come to the conclusion that a diode & zener would be the best option, as I do need rapid current reduction.

As I've never specified diodes before, what are the major characteristics to take into consideration?
(The parts will be purchased from Allied Electronics, and their online catalog shows ALL parameters, so it doesn't give me a good idea of what are the main characteristics to consider.)

As for spec's (*these have changed*) (I had all 4 magnets in parallel, but re-wired so each magnet has it's own relay contacts)
Electromagnet with 8.4ohm resistance @ 24v, therefore 2.85 Amp.
What size diode and zener would I need to take up the spike produced?

Note - it was also determined that the motion of the workpiece being moved away from the magnet was not causing the issue, as I initially suspected.

My apologies if these questions seem overly simple; I am outside my scope of study and have very much appreciated the input from everyone.

Cheers