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Electromagnet with strength controlled by photoresistor

duke37

Jan 9, 2011
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The wattage limit is when the heat is distributed all over the device, this is achieved with 1A.
You should not pass more than 1A through any part of the wire or through the contact otherwise it might melt, it will certainly be unreliable.

The idea of the fet is that the potentiometer does not need to take significant current so can be smaller and cheaper. FETs are also very cheap, the biggest cost will be the heat sink but you can pop down to the nearest airfield and saw a chunk out of a plane main spar.:) I am lucky to be near a steel and aluminium fabricators and they let me look through the scrap bin.I got about twice the weight of aluminium than what the chap estimated.
 

hevans1944

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Jun 21, 2012
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@purj I notice that @duke37 has been giving you some pretty good advice. You need to be aware that rheostats have both a current limit as well as a power limit, as @duke37 alluded to in his post #18 and clarified further in his post #21. The 100 Ω, 100 W, rheostat that you cited with your link is actually a potentiometer because terminals are available at both ends, however it can be wired as a variable resistor, a rheostat, by using just the movable wiper contact terminal and one of the end terminals. It WILL NOT dissipate 100 W at all positions of the wiper when wired as a rheostat.

With one hundred volts applied to the end terminals, a current of I = E / R = 100 V / 100 Ω = 1 A will occur and the device will dissipate P = E I = (100 V) (1 A) = 100 W. So far so good. BUT this power is dissipated over the entire length of the resistance winding and every segment of that winding carries the same 1 A of current. This is the maximum current that you can pass through this rheostat (wired as a variable resistor with two terminals) for any setting of the wiper arm. So if, for example, you set it to one ohm, then you will only be able to draw one ampere through the rheostat and it will dissipate just one watt, not even close to 100 W, which would require 100 A at 100 V across the 1 ohm resistance. This amount of current would likely melt the resistance wire.

Your entire idea of controlling the electromagnet current with a rheostat is a non-starter. You may find it helpful during experiments to use a rheostat to determine the magnetic field requirements, but it is definitely not the way to run a railroad when you get to the final design stage. You need pulse-width modulation to control the average current delivered to your nine electromagnets from a hefty power supply. This is normally delivered by switching on and off (very rapidly) an FET, or again as @duke37 mentioned, you can operate a power MOSFET in its linear conduction mode by varying the gate voltage with a small potentiometer. That requires a huge, forced-air cooled, heat sink to remove the power dissipated in the MOSFET.

Your coils do not appear to have enough turns for the power supply you are using, and you are trying to make up for this by increasing the current until the magnetic field is strong enough to "blip" the ferrofluid. As @duke37 noted in post #5 and (finally) you noted in post #20, you can get the same magnetic field by increasing the number of turns while decreasing the current. Using the same gauge wire, any increase in the number of turns will increase the coil resistance in proportion, so the voltage to produce a given current proportionally increases. However, the current required to create the same magnetic field decreases, again in proportion to the increase in the number of turns, so you end up having a voltage-current product that is about the same as before, namely nearly one hundred watts.

So, based on your experiments so far, 6 A flowing through ??? turns produces a sufficient magnetic field. But what is ??? Nowhere in any of your posts do I see any mention of how many turns are on each electromagnet that actually worked with 6 A flowing through it from an overloaded 16 V power supply! Source of this factoid:
I took some measurements with my multi-meter while running on the power supply and have these specs:
  • Circuit runs at 16 volts
  • Circuit draws ~ 5 – 6 amps

From a casual inspection, it appears there might be 70 turns on each of two layers or 140 turns in all. Let's be conservative and say its 150 turns. That means your coil is producing a magnetic field from 840 ampere-turns. It is also dissipating P = E I = (16 V) (6A) = 96 W, about the same amount of power as an incandescent light bulb. So, yeah, sounds like you have a 100 watt coil. Learn to live with it. Maybe attach one end of each electromagnet to a forced-air-cooled heat sink since you will have a nice kilowatt heater running there with nine electromagnets.

I mentioned earlier in post #2 using magnetic flux guides to extend and direct your electromagnet poles. I don't personally know of any practical applications where this is done, but the principle is quite simple: soft iron attached to a pole of the electromagnet will concentrate the magnetic flux within the iron, allowing relatively little flux to "leak" around the iron into air. It would be even better if you wound your electromagnet around one end of the bundle of soft iron wires, thus avoiding the small air-gap between the electromagnet and the magnetic flux guide.

You can obtain soft iron wires at a floral supply store. Florists use it to hold together floral arrangements and it is usually painted a dark green color to hide it among the leaves and stems. You squeeze a bundle of these wires into a diameter to fit the end of your electromagnet and then you are free to direct the other end wherever you need the magnetic pole to be. Visit your friendly local florist and see if they will sell you a handful of these wires so you can try it out. Make sure you explain to them why you need it, lest they think you are going into competition with them to make floral arrangements. Florist shops are literally a cut-throat (or cut-stems) business.

I got this idea from an ancient book, The Boy Electrician, by Alfred P. Morgan where floral wires were bundled to serve as the core for a high-voltage induction coil. Back in those days, parents didn't much care if their children "experimented" with dangerous voltages... well, mine didn't, thus leading me into a productive electronics career, albeit with a few shocking moments here and there.:rolleyes:
 

purj

Feb 14, 2016
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@purjSo, based on your experiments so far, 6 A flowing through ??? turns produces a sufficient magnetic field. But what is ??? Nowhere in any of your posts do I see any mention of how many turns are on each electromagnet that actually worked with 6 A flowing through it from an overloaded 16 V power supply! Source of this factoid:

@hevans1994, thanks for your response. Between yourself and @duke37, I'm starting to get it. Thank you both!

I wanted to fill in the blanks in your above quote. The number of turns in the winding in question is ~ 252. I arrive at this number (and more) through the following means
0.375" diameter core (0.03125 ft.)
Circumference is thus ~ 0.098 ft.
I estimate that I have ~ 252 turns (counting 6 turns per 1/8 inch, and 2.625" of turns, wrapped twice over the core)
so that's ~ 25 ft
As a double-check, I measured the resistance on the winding with my meter and after help I received here, was able to correctly ascertain that it's ~ 0.6 – 0.7 Ω, a figure very similar to that given by this wire resistance calculator for those specs. I love when things line up, so awesome!​

I get the sense that I'm shooting myself in the foot by asking too many questions at once :(. It's probably taxing to plow through my long posts, so I'll keep my writing brief in the future.

Thank you for the clarification on the rheostat wattage question. It makes much more sense than it did before, and you probably saved me from burning out a part.

Regarding my "come to jesus" moment in post #20, I was puzzled about the relationship between turns, amps, gauge and field strength. I did some more reading and was finally able to understand what @duke37 told me. I'm going to start experimenting with smaller wire and many more turns (to try to increase the copper mass to drop the amps while dramatically increasing the number of turns to increase the field strength). I just got back from the electronics store with some 26 and 28 AWG wire and I'll be making some new windings soon. I'm curious about how the heat will be affected. We shall see...

I also got some heavy-duty resistors (for example, a 10 Ω / 20 Watt one) so I'm going to stop abusing this power supply soon. BTW, I have only been activating these for a couple seconds at a time. I'm aware of the abuse I'm putting on this equipment -- I just didn't know how to stop until talking it out here. Thanks again!
 

hevans1944

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I get the sense that I'm shooting myself in the foot by asking too many questions at once :(. It's probably taxing to plow through my long posts, so I'll keep my writing brief in the future.
I wouldn't do that. You might have noticed that some of my posts are longish. It is better, IMHO, to provide too much information rather than to leave something out because you don't think it is important. If folks don't want to read a long post, well, they won't. Perhaps a few zillion electrons get terribly inconvenienced with longer posts, but that's their problem.
 

duke37

Jan 9, 2011
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You will need the same copper mass and will produce the same heat whether you use thin or thick wire.

Once you have determined the ampere.turns necessary you can start looking at PWM (pulse width modulation). This enables you to reduce the current without losing power in resistors. An inductance is necessary for this, your load is the inductance - ideal.
 

purj

Feb 14, 2016
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I wouldn't do that. You might have noticed that some of my posts are longish. It is better, IMHO, to provide too much information rather than to leave something out because you don't think it is important. If folks don't want to read a long post, well, they won't. Perhaps a few zillion electrons get terribly inconvenienced with longer posts, but that's their problem.

Ok then. Hey I had an eventful day today. Went to the electronics store, did some reading, and talked to a friend about series vs parallel circuits. Also, received an industrially produced magnet in the FedEx. All these things together gave me much to think about!

First, the new electromagnet. It's made to run on 24 V, and its resistance measures at 87.7 Ω. I put it on my 16 V power supply and read the current, which was a nice, predictable 0.17 A. Its effect in the fluid is less than ideal, probably because of the electromagnet's short, fat shape. But honestly, the effect is pronounced enough, and it's a compromise I can live with. It can only get better when I give it the 24 volts it wants. I read here (Wikipedia) yesterday something that suggested that windings with a long core have magnetic fields which extend out from their ends. I'm going to assume that the converse is also true (i.e., that short squatty cores have more contained fields), because I'm not getting the spikes in the fluid with this store-bought magnet as I was with my own. However, it pulls less than 1/5 of an Amp, so that's of course a huge advantage.

< * the forum breaths a collective sigh of relief * >

Nevertheless, I'm undecided. I believe what you guys are suggesting is that if I'm going to build my own magnets, I start with the power supply requirements, field strength requirements, and build the parts to suit. Thinking about this, I started to wonder what it would take for me to make a magnet of my own that would give me what I see now in the one I like, but would play nice with a PSU the same way the store-bought one does. What if I had a 16 V / 4.5 A PSU like this one? What would I have to do to get 9 electromagnets working nicely on it? And what would I have to do with those electromagnets to get the same result (in the fluid) I'm already seeing?

Well the first thing is, each magnet would need to draw
(4.5 A) / (9 electromagnets) = 0.5 A
If I wired them in parallel, each would always be guaranteed the same voltage, and their amperage would simply be added up when evaluating against the PSU's current rating, correct? This is an important point, so if I'm wrong, please let me know.

My best electromagnet was drawing 5.5 A and had ~ 262 turns. The strength of the field in Ampere Turns was thus
(5.5 A) x (262 Turns) = 1441 AT
I'm not sure shooting for the same AT will give me the same result, but let's assume it would and run some calcs. If anyone has a comment on this assumption, I would be glad to hear it!

To get the same AT from 0.5 A, I would need
(1441 AT) / (0.5 A) = 2882 turns
Wow. But still...​

If I'm wrapping around a 0.375" diameter core, I get
(0.375 in) / (12 in / ft) x (pi) = 0.098 ft. per turn​

To get 2882 turns, that would be
(2882 turns) x (.098 ft) = 282 ft.​

282 ft is the length of wire for the turns on my given core size, but what gauge wire should I use? This only comes down to tailoring it to draw the current we used above. What resistance do I need to hit 0.5A on 16V?
(16 V) / (0.5 A) = 32 Ω
Looking it up in the calculator I see that
31 AWG gives me 37 Ω at this length (and that 30 AWG is only 29 Ω, so 31 AWG is the way to go)​

What would this magnet look like? According to Wikipedia's wire gauge table (matches against my measured results with the 24 AWG wire, so I trust it) 31 AWG gets 112 turns per inch. That's
(2882 turns) / (112 turns / inch) = 25.73 inches
If I made the core 6.5" long I would have to traverse it four times. And then I'd have to do that whole operation nine times (once for each magnet). Is it worth the effort? Or should I just settle for the store-bought mags and call it a day?

I see that there is 31 AWG wire available on eBay for a decent price... and I am tempted to favor the hand-wound approach because the exposed wire would look really good in a wood and plexi enclosure... and I'm pretty happy to know how to calculate all this stuff now!

+++

@duke37 I see you just posted info on this same subject. Can you look over these calcs and tell me if you see anything fishy? PWM on the Arduino is the next subject I'll take up, but I'd like to make sure my thinking here is straight before moving on. I'm going to bed now, though, and I'll check in the morning. 2 am is late enough for me! Thanks for your help!
 
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duke37

Jan 9, 2011
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If the magnets are in parallel, add the currents.
If the magnets are in series, add the voltages.
Nine magnets will need nine times the power however it is done.

I have not gone through your calculations but see nothing wrong with a quick glance.
Only you can say whether you wish to wind them yourself. You obviously do a neat job.
 

purj

Feb 14, 2016
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If the magnets are in series, add the voltages.

Just to clarify, if I wire the magnets from 16 V in parallel, when one shuts off, the rest will still get 16 V, correct? And the remaining ones would each only be using 8 watts because 32 Ω of resistance would be drawing only 0.5 A, correct? I want to isolate the voltage to each, and that's what parallel wiring does, right?

This way I just have to plan the PSU so it has the capacity to push the current of all at once, but extra current will not flow through those which are left on when one switches off (because they're always given the same voltage). I believe this is how parallel works, but I wanted to be sure.


I have not gone through your calculations but see nothing wrong with a quick glance.

That's great. I just wanted to make sure that the principles I was applying seemed sound. Particularly the bits about parallel wiring, and keeping the Ampere Turns the same in order to preserve the strength of the magnetic field.


Only you can say whether you wish to wind them yourself.

This morning I've decided I'm going to go ahead and finish the project (the deadline is in two weeks) with the store-bought magnets. I'll probably try to wind some others as well, but that will be outside the scope of this iteration. Also, I dropped the number down to 7, as shown in the plans I'm working on. I love drawing plans:

160221-a-schematic.png

For installation, the unit will be suspended with metal rope (airplane wire). I'll have a light panel sitting underneath it and people will be able to wave their hands across its bottom surface to interact with the photoresistors and, of course, the fluid.
 
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purj

Feb 14, 2016
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I'm going to shift gears here and start talking about power management. I spent some time last night reading about MOSFETs and voltage regulators, and how to spec the parts. This is what I want to do
160221-b-schematic.png

Parts I need to source are the MOSFETs and the voltage regulator. I already have photoresistors, and the Arduino makes them pretty easy to use (although I haven't done it, someone else in my class has).

I spend some time last night learning about MOSFETs, and came up with this part:


In specifying the MOSFET, I'm looking at:
  • Drain Source Breakdown Voltage
    • 60 V
      • I'm thinking this is higher than 24 Volts, so my power is within its limit. Is this the right way to use this spec?
  • Drain-Source Resistance
    • 9 mOhms
      • I'm thinking this is the resistance the MOSFET will have when it's activated. Here I was just looking for a value negligible in comparison with what the other load (the coil) will have.
  • Gate-Source Threshold Voltage
    • 4 V
      • I'm assuming this is the voltage required to completely connect Source to Drain. Is that right?
      • I'll be hitting the Gate with 5 V, so I wanted to get something slightly below to make sure it's fully open.
  • Power Disspipation
    • 71 Watts
      • Assuming this is the limit on the product of the resistance and voltage in the circuit that connecting the Source and Drain will complete.
      • On 24 Volts, 87.7 Ω coils only pull 6.56 watts, so I'm within tolerance.
Am I looking at these correctly? Are there other criteria I should be paying attention to? There probably are...

I'll ask about the voltage regulator once I'm set on the MOSFET.

THANKS!
 
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hevans1944

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Actually, this is wrong. Resistors in parallel are not simply added together when determining the resistance of the entire circuit.
No, it wasn't wrong. Coils wired in parallel to a constant-voltage source will each draw their own particular current depending on the coil resistance. These currents are all independent of each other. Add or subtract coils and the total current varies, but not the current through each coil.

For the MOSFET you have selected, the gate-to-source threshold voltage is too high for use with a 5 V gate drive. The gate-to-source threshold voltage is the voltage required to just barely turn the MOSFET on. Your device is specified with 10 V gate-to-source drive to obtain full-on conduction. Instead look for a power MOSFET specified for TTL (5 V) gate switching levels, such as this one. Or visit this Google result page.

The commercial magnet you linked to is advertised for use as a "lifting magnet" which means the metal can surrounding the central pole serves as the other pole of the magnet. That will concentrate the magnetic field across the face of the lifting magnet between the central pole and the surrounding pole. When a ferrous flat surface is brought near this magnet, the magnetic field will be diverted into the ferrous surface to provide "lift" as that surface is attracted to the face of the magnet. This is probably not what you want!

Your hand-wound coils concentrate the magnetic flux in the center of the coil and then provide a very long "return path" for the magnetic flux in air and in the ferrofluid liquid. Every magnet, electromagnet or permanent, always has two magnetic poles, north-seeking and south-seeking. The magnetic flux always follows a closed loop from one pole to the other pole. If you explore a magnetic field with iron filings, or a small compass magnet, it appears that the magnetic flux forms lines of force from one pole to the other pole. These "lines of force" are not discrete lines, but it is convenient to think of them that way. They fill all of space between the two magnetic poles, including the space occupied by the core. Or maybe think of them as rubber bands that can be intercepted, concentrated or bent by nearby ferrous objects, but continuous nevertheless. And magnetic "lines of force" from a single pair of magnetic poles never cross. And there is no such thing as a magnetic mono-pole. Magnets of any kind always come with two and only two magnetic poles.

For best results, I think you should wind your own coils and use 30 or 31 AWG magnet wire, whichever size you can get quickly. You are going to pulse-width modulate the current, so fitting the wire into the space available is your major concern. Either wire gauge should work fine. And figure out some way to remove the heat if you are going to operate the coils continuously. Maybe blow air across them, or mount them to a non-ferrous (aluminum or copper) heat sink/ heat spreader.

Nice drawings.
 

purj

Feb 14, 2016
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No, it wasn't wrong. Coils wired in parallel to a constant-voltage source will each draw their own particular current depending on the coil resistance. These currents are all independent of each other. Add or subtract coils and the total current varies, but not the current through each coil.

@hevans1944 you caught me! I tried to delete that msg after I reconfirmed that I had been correct. I was kinda panicked for a sec...

For the MOSFET you have selected, the gate-to-source threshold voltage is too high for use with a 5 V gate drive. The gate-to-source threshold voltage is the voltage required to just barely turn the MOSFET on. Your device is specified with 10 V gate-to-source drive to obtain full-on conduction. Instead look for a power MOSFET specified for TTL (5 V) gate switching levels, such as this one. Or visit this Google result page.

Thanks for this info. I'm actually reading about MOSFETs right now and I can see that my assumptions were totally incorrect. I've updated the diagram to show that I don't know what I'm doing in that area :cool: and I'll circle back.

The commercial magnet you linked to is advertised for use as a "lifting magnet" which means the metal can surrounding the central pole serves as the other pole of the magnet. That will concentrate the magnetic field across the face of the lifting magnet between the central pole and the surrounding pole. When a ferrous flat surface is brought near this magnet, the magnetic field will be diverted into the ferrous surface to provide "lift" as that surface is attracted to the face of the magnet. This is probably not what you want!

I did test the magnet under the fluid and it made an effect that I can live with. It created a bulge in the fluid above it. But I think we're on the same page, as the one reservation I have about this magnet is that it doesn't give any spikes in the fluid like I often see when looking at ferrofluid demos. Your explanation makes it clear as to why. The hand-wound ones look a lot more like this:

FerrofluidHVoff.jpg

... but with more, smaller spikes.

My result (albeit with only one bulge, since I've only tested one electromagnet so far) looks more like this:
Halbach_Ferrofluid_Weak.jpg


As far as the effect is concerned, I think this is fine. It might even be preferable, since what I'm going for is a kind of shadow of whatever's under the unit. With only 7 magnets, it won't really be a shadow, more of a proof of concept.

Yet still, I like the way you think. I'm gonna get some wire and have a backup plan. It will arrive mid-week, so I can settle on a final design next weekend, and work on mastering the PWM and stuff this week.

Nice drawings.

Thanks!
 

hevans1944

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Jun 21, 2012
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I'm going to shift gears here and start talking about power management. I spent some time last night reading about MOSFETs and voltage regulators, and how to spec the parts. This is what I want to do
160221-b-schematic.png

Parts I need to source are the MOSFETs and the voltage regulator. I already have photoresistors, and the Arduino makes them pretty easy to use (although I haven't done it, someone else in my class has).

I spend some time last night learning about MOSFETs, and came up with this part:


In specifying the MOSFET, I'm looking at:
  • Drain Source Breakdown Voltage
    • 60 V
      • I'm thinking this is higher than 24 Volts, so my power is within its limit. Is this the right way to use this spec?
  • Drain-Source Resistance
    • 9 mOhms
      • I'm thinking this is the resistance the MOSFET will have when it's activated. Here I was just looking for a value negligible in comparison with what the other load (the coil) will have.
  • Gate-Source Threshold Voltage
    • 4 V
      • I'm assuming this is the voltage required to completely connect Source to Drain. Is that right?
      • I'll be hitting the Gate with 5 V, so I wanted to get something slightly below to make sure it's fully open.
  • Power Disspipation
    • 71 Watts
      • Assuming this is the limit on the product of the resistance and voltage in the circuit that connecting the Source and Drain will complete.
      • On 24 Volts, 87.7 Ω coils only pull 6.56 watts, so I'm within tolerance.
Am I looking at these correctly? Are there other criteria I should be paying attention to? There probably are...

I'll ask about the voltage regulator once I'm set on the MOSFET.

THANKS!
Your N-channel MOSFETs need a positive voltage on their drains. Reverse the polarity on the 24 V DC supply and reverse the diodes across the coils. Ground the negative terminal of the 24 V DC supply.

Don't power the 3-terminal 5 V regulator from the 24 V DC power supply. It will drop 19 V across the 5 V regulator and waste too much power. Use a separate power supply and provide only about 8 V to 9 V to the regulator input. Provide a heat sink for the regulator.

The power dissipated in the MOSFETs should be almost zero if they are either fully on with virtually no voltage drop from drain to source, or fully off with no current from drain to source. In either instance, the power is either dissipated in the coil or no power at all is dissipated.

Your 24 V 4.5 A power supply should easily provide enough current for nine of the commercial coils or nine hand-wound coils with 0.5 A per coil in the case of hand-wound coils, or about 0.27 A per coil for the commercial coils.

Good luck finishing this project in less than two weeks.
 

purj

Feb 14, 2016
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Don't power the 3-terminal 5 V regulator from the 24 V DC power supply. It will drop 19 V across the 5 V regulator and waste too much power. Use a separate power supply and provide only about 8 V to 9 V to the regulator input. Provide a heat sink for the regulator.

I think I'll just get a proper supply for the Arduino then. Thanks!

Your N-channel MOSFETs need a positive voltage on their drains. Reverse the polarity on the 24 V DC supply and reverse the diodes across the coils. Ground the negative terminal of the 24 V DC supply.

Like this?
160221-c-schematic.png



Good luck finishing this project in less than two weeks.

You'd be amazed how quick we get through 'em. Once I know what to do, this fabrication is actually going to be pretty simple (if I don't do the hand-wound mags, that is). But I suppose that's the hurdle, knowing what to do, then doing it right, so who knows?

Thanks for all these pointers!!
 

purj

Feb 14, 2016
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reverse the diodes across the coils

One other thing, can you give me some clues about how to spec these diodes? I haven't delved into it learning them yet, but it might be good if I have something to keep in mind as I read.
 

duke37

Jan 9, 2011
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The common diodes are 1N4001 to 1N4007. These are 1A diodes an any one in the series will do. I got a lot of 1N4007 since they can stand 1000V and will do for any project I have.
 

purj

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The common diodes are 1N4001 to 1N4007. These are 1A diodes an any one in the series will do. I got a lot of 1N4007 since they can stand 1000V and will do for any project I have.

Sweet, the components selection I recently bought just happens to have some 1N4001s. Thanks!
 

Alec_t

Jul 7, 2015
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Good luck with the project. BTW, those FETs should smell sweet if they're CHANEL :).
 

hevans1944

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Yet still, I like the way you think. I'm gonna get some wire and have a backup plan. It will arrive mid-week, so I can settle on a final design next weekend, and work on mastering the PWM and stuff this week.
If you can find some threaded bolts of appropriate diameter, perhaps made of mild steel (iron would be better) you can heat them with a blow torch to incandescent red temperature and then allow them to slowly cool. This will "temper" and soften the steel, possibly making it more permeable to the magnetic flux. It would certainly work with low-carbon iron bolts, but you have to worry about the iron oxidizing when it is heated because that oxidation will form a scale that is difficult to remove. Or you could visit the florist and get some soft iron wire. With threaded bolts, you can use a cardboard or plastic sleeve as a coil form and screw the exposed end of the bolt into the heat sink. With a bundle of soft iron wires as the core it becomes a little more difficult to attach the bundle to a heat sink, but you sound like a pretty ingenuous person.:D
 

hevans1944

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One other thing, can you give me some clues about how to spec these diodes? I haven't delved into it learning them yet, but it might be good if I have something to keep in mind as I read.
The diodes need a peak-inverse-voltage (PIV) somewhat greater than the power supply voltage applied to the coils. When the MOSFET turns off, whatever the current is in the coil at that moment will be transferred to the diode, so the peak current capacity of the diode needs to be at least as large as the coil current. 1N400x series diodes work well for this, even the lower PIV-rated 1N4001 diodes.
 

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