I want to build an electro-magnet for a HS physics class. I'm using a

6v dry cell battery for the power. I want to include a pot in my circuit to control the amount of power going to my magnet which will be about 6 layers of 100 turns each of 20 guage wire.

I bought a 1 meg pot and wired it in series in the circuit. It smoked.

Anyone have suggestions on good references on how to wire up a circuit like this to be able to use a potentiometer correctly?

the pot should control a device like a power transistor that is mounted to a heatsink.

the problem is that whatever power the pot is rated for iis the for the entire value of the pot... when you use the "wiper" to reduce the resistance the power handeling ability.

lets assume your 600 turns results in .6 ohms of resistance.

the 6V battery would then try to supply 10 amps of current. this will fry all but the largest of wirewound pots.

you might try putting a auto tail lamp in series to limit the current or add many more turns of wire to the electromagnet.

You're on the right track. The power rating of the pot you used probably wasn't high enough. There is a 98% chance that you bought it at Radio Shack, and the pots you can get there are rated maybe 1 watt. You can get pots with higher power ratings, but not at Radio Shack. The higher- power pots tend to be called "rheostats" or "power rheostats" or "wirewound rheostats". You might look at

formatting link

- they sell suitable parts, have no minimum order, and are also located in Texas, so even the cheap shipping should get to you relatively quickly.

If you have access to a multimeter, you might use it to determine how much current your magnet draws. If not, you can make an approximation based on the length of wire you used to build it and a "copper wire table", which will tell you the resistance in ohms of 1000 feet of a particular gauge of wire. Use Ohm's Law to figure out the current, since you know the resistance and voltage. Once you know the current, you can figure out the power rating you need on the pot.

You might go to the library and read the first eight pages of "The Art of Electronics" by Horowitz and Hill. This should give you a better idea about what you're trying to do.

Standard disclaimers apply; I don't get money from any of the companies or people mentioned.

Get a new pot - say 10 k, a 2n2222 transistor and a 2n3055 transistor. Connect the emitter of the 2n2222 to the base of the 2N3055. Connect the collectors together, and to the + side of the battery. Connect one side of the pot to + and the wiper of the pot to the base of the 2n2222 through a 1 K resistor. Connect the other side of the pot to battery -. Connect the emitter of the 2n3055 to one lead of the electromagnet. Connect the other lead of the electromagnet to the - of the battery. The

2n3055 must be mounted on a good heatsink.

The two transistors form what is called a Darlington Pair. You can see a diagram:

formatting link

that diagram for reference, but change the BFY51 transistor to a 2n2222, and change the TIP31C transistor to a 2n3055. The motor is replaced by the electromagnet, and you don't need the 1n4148 diode.

The starting point is what you are trying to do. You connect the electromagnet across the battery, current flows from the battery and through the coil. The current in the coil creates a magnetic field. Hey presto! An electromagnet.

You are quite right in that a resistor inserted in series with the coil will reduce the current and hence the strength of the electromagnet. But you need to work out what value of resistor you need and not just choose one at random.

To do that you need to know the current that is flowing without any resistor. Then, with a few simple sums you can work out what resistor would be needed to, say, halve the current, to quarter it and so on. When you know those values, you can choose a pot which is capable of replacing any individual resistor - just by turning its knob to the right place.

You can measure the current with an ammeter or with a multitester set correctly. As doing this incorrectly could damage the ammeter or tester, you need your teacher's help in doing this.

You also need to measure the voltage across the battery whilst the electromagnet is working.

The first simple sum is that the resistance in ohms of the electromagnet is the voltage from the battery divided by the current. Rc =Vb/Ib, where Rc is the resistance of the coil Vb is the battery voltage and Ib is the current that flows when the coil is placed across the battery. A typical value for Rc may be around, say, 1 ohm. Ib will be say 6 amperes and Vb will be 6 volt

Once you know Rc, if you want to halve the current, then the resistor you need, say R1 is going to be in series with Rc, so the total resistance across the battery is Rt= Rc+R1. We know the current that we want is going to be Ib/2, the voltage is still going to be Vb. Thus Rt= 2 Vb/Ib. However, as from above Rc= Vb/Ib, then Rt = 2Rc. But Rt = Rc+R1, thus R1 = Rc. So we need R1 to be about the same value resistance as of that of the coil - i.e. about 1 ohm.

Doing other sums similar to those above, for different currents, will give a range of values for the series resistor, from about Rc to about 10 Rc. So your 1M pot was about 100,000 times too high a value!

You will need a pot with a value of about 10 times Rc, able to carry the current of your coil, Ib. This will be quite big and expensive but your teacher may have one. Otherwise, you may be able to make one using resistance wire stretched between two screws on a piece of wood, using a crocodile clip as the slider on the wire. Again, your teacher should be able to help.

Of course, if this were being done for a product to sell, it wouldn't come with a piece of wood with wire and a crodile clip! Other posters have suggested a Darlington Pair to replace the resistor and these work very well.

What is even better is some electronics that switch the current to the coil on and off very quickly and adjust the power of the coil by varying the length of the short periods that the switch is on to those short times that it is off. So, if you want a very strong magnet, the switch is mostly on and if you want a very weak one, the switch is mostly off. Coils do strange and interesting things when the current to them is switched on and off like this, so this solution is going to need you to learn a lot more before you can use it.

These controllers can be very small and doesn't waste battery power. If you keep studying electricity and electronics, you will soon find that these circuits are indispensible and great fun to design and make work.

The Darlington pair would be a bit more sophisticated but as suggested you could calculate the value of a proper pot with the current drawn from the coil and the 6vdc supply battery ..... but you can't attempt to shut down the magnet with this pot aloneso place a blade switch in the circuit if you haven't., nevertheless, the transistors will definitely provide better control.

I think your electromagnet design could use a change as well, you didn't mention a core, are you using a Large Nail [similar metal object] or just winding upon winding? In My Opinion: A Core will give your E-Magnet more Reluctance and the Inductance from the coil will have a more desirous effect on your circuit, not to mention more stability if it has a core to work with.

Plus when the metal core magnetizes you can demagnetize it by reversing the polarity and show another cool electrical/magnetic effect ..... but even a small ferrite core will make your electromagnet better.

See what kind of componets your instuctor has available. Like a step down transformer for 6 or 12 volts or so and a variac. Most decent labs have these. Fuse the magnet for the current rating on the transformer secondary. If the fuse blows add more wire to the coil. make the core of tell strapping cut into lenghts about 6 inches long and spray them with varnish before stacking them together to make the coils core. This type of coil run off of AC will often pick up metal items otherwise considered non magnetic. I am a little short on a lot of explanation on purpose. Hopfully this will start a dialogue btween you and yor instuctor

You did it basically all right except for a couple small details - a limiter and not letting the equations to tell you what is wrong and spark plugs.

So first, use your numbers to cut thru the bull advice you'll be getting and see what really will work. What smokes potentiometers? Overpowering burns them up and smokes, or overvoltage from coil-current collapse that creates voltage and breaks down insulation and smokes (more likely for your problem).

Overcurrent first

1) If you had a 1 meg resistor rated at 1 watt and a 6 volt battery, and you smoked the resistor, how much power can you get into the 1 meg resistor to smoke it with 6 volts? From E=IR and P=IE

P=E^2/R = 36/1000000 = 36 microwatts = not enough to toast it by a LONG shot.

BUT IF you had the pot set so it was a milliohm rather than to a megohm, then how much power went thru it?

P=E^2/R =3.6 x 10/1 x 10^-6 = 36 million watts, but not for very long

2) SO what you need is a fixed resistor in series with the potentiometer in order to limit the current in the potentiometer to less than its power rating of 1 watt, just in case the operator should ever turn the pot to zero. (and it limits inrush)

P=E^2/R or R=E^2/P = 36 ohm resistor. And what watt rating does that one need?

P= E^2/R = 36/36 = 1 watt. (might seem obvious, but one always checks)

3) Now how much current will you get thru your loops?

E=IR gets you I=E/R = 6v/36ohm = 1/6 amp

You know that since it is all in series, that 1/6 amp that runs thru the resistor runs thru the wires.

Now you can go back to your physics book and figure out how strong the magnet is at 1/6 amp flowing, and compare it to one in the examples to get a feel for how strong it will be.

4) Overvoltage - VERY dangerous - knock you on your butt and kill you very dead shortly after the intense pain from cramps and suffocation, and leave your corpse cold from a big coil discharging thru your person.

You get that coil filled up with current running and then disconnect the circuit by opening a switch or by turning resitance down quickly, you can easily get enough voltage to smoke most potentiometers by punching thru their insulation - they get 20,000 volts in a car for the spark-plugs by opening the coil as current is flowing thru it. You need to slow down the filled-with-current-wanting-to-kill coil discharge just before you open the switch - like a fixed resistor about 100k or so (just guessing on the size-check it !!), to go from the off-switch coil side to the other side of the coil just before you shut off the switch, or you need a clipping device to limit voltage, or the like.

By the way - now that you have burnt up a potentiometer in front of you, I assume you have safety glasses whenever your big energy storage device (your coil) is on.

Your idea won't work - it will kill his experiment. His stated objective: "I want to include a pot in my circuit to control the amount of power going to my magnet"

The 36 ohm resistor you recommend will limit the current - as you said - to 1/6 amp. That reduces the amount of current control by at least 83 percent right from the start. Furthermore, a typical 6V lantern battery (which is probably what he is using) has an

11,000 maH capacity can can easily source 5 amps. Using 5 amps for the following: limiting the current to 1/6 amp (167 mA) could reduce the amount of control by a huge percentage - on the order of 97% (5000-167)/5000.

It gets worse.

A 1 degree rotation of the 1 meg potentiometer (it is vitrually impossible to achieve

*only* 1 degree of rotation of the pot) you are trying to protect with the 36 ohms will change the resistance by over 100 times that much. He will not be able to control the power demonstrably by rotating that pot - it will look more like an on off switch that works at only one spot. The pot will survive, but the experiment will fail.

Protecting the pot by current limiting (with a simple resistor) turns out to be a non-starter for the poster's need.

wrote [paraphrasing: if you use a 36 ohm resistor in series with a potentiometer, it won't work because the resistor removes pot control and the 1 meg pot is too big]

1) I think you have it all backwards - adding a resistor in series with a pot increases control of the parameter being controlled (current), not decreases it. See below. And see the numbers below.

2) The poster is a HS student making a coil. If he had some experence in electronics, he would not post about a 1 megohm pot and that coil. So you can either give him a fish, or you can teach him enough about fish that he won't glaze over, tell him to watch out for great whites, and let him discover, either in those equations which he can use or in testing, that the 1 meg won't work -- so he then uses the fish information to find the right size of pot for his application. Just like I do with young engineers. For some reason they remember it better and enjoy their job more when I tell them enough to let them discover it in short order rather than when I tell them the answer. I don't get to be God-for-the-moment doing that, but then anyone who ever was, doesn't want that job very long. And I don't get paid enough to be God, I get paid for being head engineer for the firm.

I will go >

I fail to see how a 36 ohm resistor in series with a 1,000,000 ohm resistor has an 83% effect on the 1,000,000 ohm resistor. Since the pot adds that much in 36/1000000 of a turn in a one turn pot (about 1/100th of a degree of rotation), it has 36/999,967 effect on control of the pot used. (And I definitely didn't say use a 1 megohm pot. My son built a similar coil a few years back, so I know what works.)

Your assumption is that he is pumping 5 amps thru the wires and the 36 ohms will interfere with the 5 amps he is controlling.

If he is using a 1,000,000 ohm resistor, how far does he have to turn the pot to pass by the 36 ohm mark - or to put it another way, the pot turns from infinte amp-let-thru to 1/6 amp-let-thru? 1/100th of a degree. He isn't going to lose control from a 36 ohm resistor in series with his pot, he is going to limit current to one watt max. He will rather lose control from using a too-large pot.

not worse, it gets obvious.

I agree with your techncial analysis here. In the reply, I was focusing on instruction about equations regarding the power, resistance, current, and voltage, protection from too many watts, and limiting the current in the coil to a safe level. If he got the idea about 36 ohms limiting it to 1/6 amp, he assuredly will get the idea that a million ohms will shut it down. I am going to give any student asking for help basic useful principles with which he can solve his problem, and I am not going to ever solve a problem for a student. That is just bad teaching and bad parenting. If the student does the numbers, he would see rather quickly that he has no current with a large resistance pot. If it fails again, and if he isn't just a skater trying to get foolish engineers to do his work for him, he should use the equations to figure it out in about five minutes. And go get the right size pot.

BTW, in that same vein, did you consider if the amount of amps left, at the voltage you would get interrupting your 5 amps in a coil of the size described, would be lethal?

The potentiometer got smoked by the *current* flowing through a small part of the restive film used to get 1 megohm.

The ignition coil in a car works by interrupting the current flow in the primary of the coil. The ignition coil is a step-up transformer that produces the high voltage to the spark plugs.

Nonsense!

Energy storage in a coil: J=1/2 L x I^2 You do the math. :-) Remember: L is in Henrys, and the short circuit capacity of the dry cell and/or the resistance of the coil limits the amps.

This is much too high a resistance! See below for why it smoked.

Ignore the BS advice given and use some common sense! Your electromagnet, assuming a 1/2" iron core, will use about 1.6" of wire per turn. At 600 turns this requires 80 feet of #20 wire. The resistance of #20 copper wire is about 10 ohms per 1000 feet, so your coil will have a resistance of about 0.8 ohms. At 6V this would draw

6/0.8 = 7.5 amps. Assume you want to reduce the current to 10% of this, or 0.75 amp. At 6V (E = IR) or R = E/I = 5/.75 = 6.66 ohms of which your coil makes up 0.8 ohms. A series resistance of 5.86 ohms will do this. As a practical matter, a variable rheostat of 10 ohms would do fine.

(Common sense would tell you that you would need to put 90% of the voltage into the series resistor. This would then be about 9 times the coil resistance, or about 7 ohms. Not exact, but much closer than 1 meg!) :-)

This would need to have a power rating high enough for the maximum wattage dissipated in it. A 10 watt rheostat will be adequate except when turned so that only the last few resistive turns are in the circuit. Although the watts will be much less than the rating, the current flowing will be almost that drawn by your magnet coil, and may exceed the current capacity of the rheostat. This is what "smoked" your pot. You may find it hard to find a suitable rheostat that can carry the 7.5A required.

An ungainly, but possible, solution would be to use ordinary 120V lamps for the series resistance. A 100W lamp has a cold resistance of about 9 ohms. By paralleling a number of bulbs is series with your magnet you could control the current to it. One lamp would cut the current to about 0.6 amps, and a second bulb in parallel would approximately double that. Not too elegant a solution, but do-able with common devices!

That's not correct. Take a simple example: A 10 volt dc source in series with 20 ohms. The 20 ohms is combined - a 5 fixed ohm resistor, a 5 ohm pot and a 10 ohm load. With the pot set to zero, 10V/15ohms (~.67) amps will flow. With the pot set to full resistance, 10V/20ohms (.5) amps will flow. The span of control is .67 - .5 or .17 amps.

Now remove the 5 ohm resistor. With the pot set to zero, 10V/10ohms (1) amp will flow. With the pot set to full resistance, 10V/15ohms (.67) amps will flow. The span of control is 1-.67 or .33 amps.

Adding the resistor is series with the pot limits the amount of control the pot can have over the current flow, as shown in the example above.

There is nothing backwards about that or the idea that a 1 meg pot in his circuit is too big.

The poster has been given a number of answers that will work. The addition of a 36 ohm resistor in series with the pot won't. It kills 5/6 (83%) of the controllable current, if you compute based on a 1 amp source. If the source is cabable of more current, the number is worse than 5/6. For example, a 5 amp source without a current limiting resistor as compared to the same source with your 36 ohm resistor would kill 4.833 amps. The ratio would be 4.833/5 or 96.66% of the controllable current. The lack of consideration of the size of the pot kills whatever is left over.

The approach (a pot with a current limiting resistance) is wrong in the first place, and the poster needs to be told that. Others recommended a big rheostat, coupled with the use of ohm's law to figure the powerrating needed. A power transistor control, a darlington pair, a pwm controller (without the use of the term pwm) were also recommendations that would work.

?? I never said anything about the 36 ohm resistor having an 83 % effect on the 1,000,000 resistor. I don't know where you got that or what you have in mind by it. You said, and I agree, the 36 ohm resistor limits the current though the pot to 1/6. That means that 5/6 or 83% of the current cannot be controlled by the pot.

I make no assumption of what "he is pumping" through the wires. I compared the assumed *available* current with no current limiting to the maximum current he could get with current limiting.

That fact (his pot is too big) needed to be pointed out to him, as well as the fact that it can't handle the current. But your statement "He isn't going to lose control from a 36 ohm resistor in series with his pot" is absolute nonsense. In fact, he loses a huge degree of control - 83 to 96 percent (and possibly higher) of the available current cannot be controlled due to the 36 ohm resistor.

Fair enough.

Cite cases where people were killed by an auto coil discharge (which would have a much greater "kick" than this magnet). Warnings about lethality of this coil sound like crying wolf.

if the span of control is larger using the same device, then by definition the control is lower. A potentiometer that has 1 amp/turn control has better control than a potentiometer which has 2 amps/turn. The amount of current controlled is inversely proportional to the amount of control

1) sorry, but based on personal experience with such devices across 35 years of engineering, from the 5MegaWatt 120KV ballistic missle detection radars to micropots to determining the wiping pressures and materials required on various interrupt switches on cranes and large and small machinery: a collapsing coil, being a storage device of energy, does not need a secondary coil to create high voltage when its flow is interupted and the field collapses - at least, turning off switches in those applications have not left the field sitting there on the coils for weeks waiting. They can and do arc like hell. If you have a better example of the collapsing of the magnetic field causing high voltage which might be commonly recognized by a high School Student, please let me know.

as to your other comments, you can see from the equations provided that you lack enough information to but provide speculation and guesses. Provide numbers, please. See responses below for specifics

DEH, PE

Nonsense? Indeed. That is an odd comment from someone who it looks suspiciously like one who has never designed switches or coils and who has not done the numbers. You do not even know the Henrys of the coil.

so back to Fields 101 - what happens to the energy contained in a current carrying coil when the current is interrupted? Do those flowing charges just go poof and magically disappear without manifestation? Or do they stop and sit still and the magnetic field magically disappears without manifestation? Or does is it true that entropy takes over and they seek to neutralize?

You must admit that your approach requiring a second coil to get a coil discharge kind of blows the core theories of electromagnetism, doesn't it?

sure will - if can you just give me the dimensions so I can figure the Henrys. For the six layer coil - BTW, do you suppose the student who posted the question used a brooks coil so we can dispense with all those ln parameters?

tell you what - you give me the Henrys and I'll tell you if the arc voltage and amps from the collapsing coil exceeds skin punch-thru levels and at the same provides the requisite milliamps to kill. And I can check to see if we get enough voltage to break down the 1000 volt on a good day radio shack pot insulation and thus let the field collapse as current thru the breakdown hole.

You can use this eqqaution for his 6 turn coil - L = 2.4×10-6? (mean radius of coil) (Number of turns)2((0.5+ceff/12)ln(8/ceff) -0.85+0.2ceff) henrys,

where ceff = (outer radius-inner radius)^2 /(2 x length x [mean radius] ^2 )

Is it your firm's practice to design circuits for coils without knowing the radius and thickness of the coil?

Remember:

Yes, using the 20 gauge and his 6 volt, he gets about an ohm resistance per

100 feet.

And what is this "short circuit capacity" of the 6 volt lantern dry cell?

that should read A potentiometer that has 2 amps/turn control has better control than a potentiometer which has 1 amp/turn. I.e., the amount controlled per degree is smaller and thus finer and thus the two turn pot has more control over the parameter. Each increment is smaller -control is better -when the range is smaller.

Interesting! I worked on the design and testing of the 5MegaWatt S band magnetron used in that radar. I am amply aware of the dangers of high voltage!!! The 120KV was obtained from a hydrogen thyratron with a step up transformer. I designed the voltage divider that monitored the HV. The primary voltage was, (only!), 10KV as I remember, but at several amps.

Why then to all automobile ignition systems use a "step-up" ignition coil?

I don't know of any way opening the circuit for a 6V battery to the coil described can cause a "lethal voltage" to appear anywhere. You don't even need to open the circuit to "smoke" the pot. Just turn it from a high value resistance to the low end. At some point the current through the resistive element will be high enough to burn it up. (With smoke!) No voltage over the 6V battery required.

Plug in the numbers yourself. You'll see it's not possible to store more than a few Joules in such a coil at 6V! To accurately calculate the inductance of the coil, you'd need knowledge of the core size and material, which your formula ignores.

Where are you registered as a Professional Engineer? I assume that's what you mean by "PE".

Yours apparently were not the same ones I worked on, since mine were L-band, and they used 14 foot klystrons (BMEWS). Over 100,000 volts and a very decently amped beam, unless it burped up to half again as much. The circuit breakers alone were out of science fiction: those flat iron plates, used to create magnetic fields to contain and move clouds of ionized gas into the weird ceramic chutes during a trip, always were an impressive engineering concept.

I also worked on the old TPS-22 Air Search with its 5 MW L-band magnetron. That had a thryatron (one is now a lamp) that triggered the PFN. The pulse bandwidth on the 22 was somewhat different from the BMEWS klystrons. :-)

the core size factors out in the equations to leave the outer-inner radius and mean radius, and the equations assume an air core As i undertsand the core effects, the other cores affect density and they create a frequency response of sorts in that you can lose all inductance at certain frequencies - (and I am not talking about skin effect)

A LM317 1.5 amp voltage regulator would work nicely, Just about any Pot over 5K will work properly LM [ 317 ] | | |___________Vin | |_____________Vout | ____________To Vin | | |-->Pot |___________To Ground

Your ground is common to both Vin and Vout. You should put a 1K load resistor to Vout and Ground, this helps stabalise the output voltage.

The above circuit is only good for DC voltage, from 1.5 - appr 37 V Absolute 1.5 Amp with a full heat sink.

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.