Dc powered electromagnet

[aDogNameDude]

So I managed to make a metal rod into an electromagnet. I used speaker wire 18 gage. For power, I used a dc 24v 1.2A transformer. success it worked! After about 60 seconds the I could smell the transformer. I pulled out from the outlet and the transformer was smoking.

 

What should I do to prevent this? what is the best way to supply power and not have over heating?

[ewmon]

You could use a "load-limiting resistor", which is a resistor in the circuit whose purpose is to limit the load that the transformer must feed. Hmmm. Your coil drew too much current because of its low resistance. With this resistor in series (NOT parallel) with the coil, it will limit the amount of current that the transformer will attempt to provide.

 

Let's say for the sake of simplicity that your coil has a resistance of 1 ohm (although I'm sure it's probably much less than this). This coil would draw a current equal to the voltage divided by the resistance, 24v/1Ω, which would equal 24 amps. The transformer is rated at only 1.2 amps therefore you overloaded it, which is why it started smoking.

 

The nominal output voltage divided by the maximum current defines the lowest resistance the power supply can feed. It can used to be able to feed 24v/1.2A, which equals 20Ω. But don't use a load-limiting resistor of 20Ω because it will draw 1.2 amps from the transformer, which is right on the limit. To be on the safe side, use a load-limiting resistor that's larger than that (especially because you already smoked the transformer, which means it's probably not capable of putting out 24v or 1.2A anymore).

 

So, let's say 40Ω or even 60Ω. A 40Ω resistor will draw 0.6 amps (24v/40Ω) from the transformer. A 60Ω resistor will draw 0.4 amps (24v/60Ω) from the transformer.

 

Obviously, if you end up using another transformer, always use a load-limiting resistor with these experiments so that you don't smoke your transformer again. For example, if you use a transformer that's rated at 9 volts and up to 0.6 amps, the smallest load-limiting resistor would be 9V/0.6A, which equals 15Ω — but use a larger resistor. I recommend a safety factor of 2, so I would recommend a 30Ω resistor (15Ω × 2 = 30Ω).

Edited January 25, 2012 by ewmon

[Spyman]

So I managed to make a metal rod into an electromagnet. I used speaker wire 18 gage. For power, I used a dc 24v 1.2A transformer. success it worked! After about 60 seconds the I could smell the transformer. I pulled out from the outlet and the transformer was smoking.

 

What should I do to prevent this? what is the best way to supply power and not have over heating?

As ewmon is saying you need to choke the current to your electromagnet so that it gets inside the maximum rating of your power supply unit, I would use a 24 ohm resistor in series which would give a current of around 1 amps. But make sure you use a power resistor that can dissipate the waste heat, it will get very hot and needs to have a rating of at least 24 watt to not overheat.

[aDogNameDude]

please forgive my ignorence . never was great in the field of science.

 

So if i have a 24v 1.2A, then I also need connected to that a resistor to limit the overall current to the coils?

 

Are these two seperated divices? or can i find them together as one unit?

 

To better explain my overall goal. I'd like to be able to have control of voltage or current to the coils, allowing me control over the strength (flux) of the magnification.

 

Is this possible?

[ewmon]

please forgive my ignorence . never was great in the field of science.

 

So if i have a 24v 1.2A, then I also need connected to that a resistor to limit the overall current to the coils?

 

Are these two seperated divices? or can i find them together as one unit?

 

To better explain my overall goal. I'd like to be able to have control of voltage or current to the coils, allowing me control over the strength (flux) of the magnification.

 

Is this possible?

Yes, definitely. Variable resistors called "potentiometers" (say "poe-TEN-shee-AH-meh-terz") (also simply called "pots") exist, but I haven't seen them able to dissipate the heat yours will produce (although they may exist).

 

You might want to use separate resistors for each coil circuit you wish to use.

 

Or you could make one circuit that uses a resistance box that switches various resistances into the circuit as needed.

[aDogNameDude]

so to make sure that I am understanding this correctly I should be putting it a resistor on both ends of exposed copper wire? a resistor will allow me to keep both my coils and my power supply from overheating correct?

 

as for the potentiometer does it matter which end of the copper coil I catch it?

 

4 my next test I plan on dropping my voltage to 12 volts. is there a specific amount of oil that you would recommend I use? is more always better?

 

thank you so much for all of your help you guys rock!

[doG]

Use a rheostat if you want to use a single load limiting device in the circuit, it will handle more power dissipation than a potentiometer. Alternatively you could limit the max load with a 25W ballast resistor of about 25Ω in series with a 0-200Ω potentiometer.

[ewmon]

so to make sure that I am understanding this correctly I should be putting it a resistor on both ends of exposed copper wire?

exposed copper wire — I assume you mean the bared ends of the wire used to make the coil.

a resistor on both ends — No. Think of the wiring as water pipes. (Actually, the English scientist Oliver Heaviside sarcastically called this hydraulic model, the "drain-pipe theory".) Resistors act like certain amounts of restriction in a piece of pipe. Coils are like water turbines, and the momentum of the turbine is the magnetic field. The high side (+) of the power supply is like the high pressure source in a water system. The low side (–) of the power supply is like a drain/sewer. So you have pressure coming out of the power supply, and going through a pipe that leads to a restriction and then the turbine, or vice versa. It doesn't matter their order, but you don't need two resistors unless one alone is too small. Resistances add when connected in series (in other words, resistors are additive in series).

 

a resistor will allow me to keep both my coils and my power supply from overheating correct?

You now know the equations to compute the current-limiting resistor needed to prevent you from blowing your power supply (even if the coil is missing from the circuit, and the resistor is connected to both power supply terminals). You know the resistor size that will protect the power supply because we know the power supply's current limit.

 

The coil is a different story because we don't know the coil's current limit. The current-limiting resistor does limit the amount of current flowing through the coil (actually, through everything in the circuit because it's all in series ... again, think water pipe). However, can the coil dissipate the power (I²R_coil) generated within it so as not to melt the wiring?

 

Here, several factors come into play.

1. How thick is the copper wiring? The thicker the wiring, the larger the cross-sectional area, the lower the resistance. Again, think water pipe.
   
2. How tightly/loosely is the coil wound? Loosely-wound coils might dissipate heat more readily due to the hot air within it flowing out and being replaced with cooler air.
   
3. What's the diameter (D) of the coil, and how many turns (N) does it have? These factors help determine the length of wire of which the coil is composed (NπD), which determines the resistance of the coil's wire, which helps determine the amount of heat it must dissipate.

as for the potentiometer does it matter which end of the copper coil I catch it?

No. That said, there's a danger which you must avoid — insufficient current-limiting resistance. You can set a pot's (or rheostat's) resistance to zero, which is as bad as not having any resistance in the circuit. So, you'll want to have a resistor in the circuit to ensure a minimum amount of resistance in the circuit.

 

4 my next test I plan on dropping my voltage to 12 volts. is there a specific amount of oil that you would recommend I use? is more always better?

 

thank you so much for all of your help you guys rock!

Oil?? What oil? If you mean "coil", we talk about coils in terns of: number of turns (N), diameter of core (D) around which the wire is wound, etc.

 

Last comment for the night. You don't want to be using your hands for trying to pull the power supply's cord/plug from the wall in an emergency. It could already be melting, or (though unlikely) the current might fuse the plug's lugs to the wall outlet. You could end up with live wires in your hands. Do you have a power strip with an on/off switch that you could use?

[Spyman]

please forgive my ignorence . never was great in the field of science.

 

So if i have a 24v 1.2A, then I also need connected to that a resistor to limit the overall current to the coils?

 

Are these two seperated divices? or can i find them together as one unit?

 

To better explain my overall goal. I'd like to be able to have control of voltage or current to the coils, allowing me control over the strength (flux) of the magnification.

 

Is this possible?

You can buy a regulated power supply where it's possible to control the output voltage and current to a specific value. It will hold the desired values nearly constant despite variations in the load.

[TonyMcC]

It's worth mentioning that the the strength of your magnet will be increased if you use more wire so that you can increase the turns around your iron core. This will also increase the resistance of your coil which will help in limiting current through it. But you still need to do your calculations and use a controlling/limiting device such as a rheostat or regulated power supply as mentioned in the above posts.

Edited January 26, 2012 by TonyMcC