Magnet coil help (turns, gauge, for coilgun)

[BobK]

Yea I already looked at the chart, I was just asking a hypothetical question.

My actual coil will be 100ft of 14gauge wire (not yet sure on how many turns that will be) So my resistance is .251 ohms

What is the difference using 40v @125amp batteries OR using 40v capacitors?

I know the capacitors pulse the amps really fast but when you have resistance in a wire the caps and batteries will both achieve same amps correct?

Another questions I have, I know @ high amps batteries have voltage drop. Does that mean when I try to draw 125amps from my 40v batteries that the voltage will really only be like 35v or even 30v?

Thanks guys

In order to get any efficiency at all, the current has to be limited by inductance and back EMF, not resistance.

The capacitors can be charged over a time much longer than the firing time and can then supply much more current than the battery could supply. If I can charge the capacitors for 10 seconds at X amps, I can discharge them for 1/10 of a second at X * 100 Amps.

Please educate me on the design of a coil gun, it must have multiple coils that are energized in series, no? Yet it sounds like you are talking about a single coil.

Bob

 

[supak111]

What do you mean limited by inductance? And I'm not trying to make it limited by resistance which is why I will be using 14gauge wire which can conduct more then my batteries can put out.

And I understand the Caps can put out more amp but my gun will not have time to charge caps so I have no choice but to just use batteries in series. Thats one thing I can NOT change in my design.

My coil gun will have multiple coils, I'm literally planning on making a 100 stage coilgun. Each coil will be the same (100ft of 14gauge wire, .25ohms) and all 100 will be connected to my 40v @125 amp car lithium batteries. Projectile 9.5mm 3.5g ball bearing.

QUESTION: I keep reading that: Amps x Turns is how strong the electromagnet is. How can that be? That would mean that 40v @125amps and 400v @125amps would produce a same electromagnet. One would just require thicker wire. That can't be right since one is using 10 times watts. Shouldn't the power of an electromagnet be defined by how many watts it uses?

ANYONE?

 

[(*steve*)]

QUESTION: I keep reading that: Amps x Turns is how strong the electromagnet is.

Yes, that's right (if you ignore the core).

How can that be?

Because the strength of the magnetic field is essentially defined that way.

Current causes the magnetic field, and every turn adds to the total magnetic field.

That would mean that 40v @125amps and 400v @125amps would produce a same electromagnet.

No. It would depend on the number of turns. If the number of turns were the same, then yes, they would have the same strength.

One would just require thicker wire.

well... ok, maybe. But just to create a suitable resistance so that the current is limited to the same value at a different voltage.

The required wire size can also be seen as that required to carry the current, and in this case it's both the same.

That can't be right since one is using 10 times watts. Shouldn't the power of an electromagnet be defined by how many watts it uses?

No, the strength of the magnetic field is determined by the current and number of turns.

The power dissipated is determined by the resistance and the current.

The two are only related in that current is a common factor.

 

[supak111]

So the using higher volts just makes the EM smaller and less efficient, but using lower volts, higher gauge makes the EM bigger and way more efficient??

I'm still comparing same 100 turn @125 amp. I'm just changing volts and gauge wire.

 

[(*steve*)]

No, for the same current and number of turns, the requirement for a higher voltage simply generates more heat.

OK, di/dt will also change, but in an electromagnet, I'm not sure that's important -- certainly not for DC.

 

[BobK]

If the current is limited by resistance, nearly all of the energy is going into heat, not into accelerating your projectile. The way an inductor works, is you place a voltage across it, and the current builds at a rate that depends on the voltage and the inductance. The equation is:

dI/dt = V / L

In a coil gun, each coil is receives a short pulse of current and they are energized in series, with timing that keeps the pull just ahead of the projectile. The fact that the pulse is short can let us limit the current via inductance, rather than resistance.

Let's say the inductance of the coil is 1mH. Then if you put 1V across it, it, the current in it will rise by 1A for each millisecond the voltage is present. with 100V it will rise by 100A for each millisecond. Note that the diameter of the wire does not come into this equation (as long as it is big enough to carry the ultimate current) So you want to use very big wire to keep the resistance low and therefore the loss to heat low. Note also, that unlike the resitance case, the voltage does matter now.

Let's take your example of 125A. If the inductance is 1mH then we can apply 40V to it for 3.125 ms to get the current up to 125A. If we do this with wire that has very little Resistance , we can manage the heat by cutting off the voltage after this 3.125ms.

But even that is only a small part of the story of designing a good coil gun. The moving projectile becomes magnetic and is generating a voltage by moving through the coil. This is called back EMF. This subtracts from the voltage you are placing across to the coil, meaning you need a higher voltage to push the same current through.

This is what I mean by limiting the current via inductance and back EMF. If these are the dominant factors in limiting the current, rather than resistance, then you can transfer a much higher percentage of the power to the projectile and less to heat.

Ideally, the current would build up, limited by inductance until the projectile was in position to be accelerated. Then the back EMF would equal the applied voltage bringing the voltage to zero and all of the energy would be transferred to the projectile!

I don't know exactly what the ratio would be, but for an Ohm's law limited system, I am going to make a wild guess that you would giving no more than a few percent of the energy to the projectile.

Bob

 

[BobK]

By the way, how were you planning to control the timing of your 100 coils?

Bob

 

[CDRIVE]

If the current is limited by resistance, nearly all of the energy is going into heat, not into accelerating your projectile. The way an inductor works, is you place a voltage across it, and the current builds at a rate that depends on the voltage and the inductance. The equation is:

dI/dt = V / L

Bob

Coming from a guy that never gave coil gun theory much thought this was an enjoyable read. Since I have an affection for anything that launches projectiles I'm surprised that I've never investigated them. I doubt that I'd give up my Garand for one though. Come to think of it, I don't think I'd give up my pellet rifle either. I envision accuracy even worse than a smooth bore Match Lock.

Chris

 

[BobK]

I did actually give coil guns a little thought. Must be due to the congitive dissonace of my abhorance for firearms and my secret desire to off those damn squirrels that eat all of our bird seed. I think it would be a fun (and challenging) project.

Bob

 

[supak111]

Let's say the inductance of the coil is 1mH. Then if you put 1V across it, it, the current in it will rise by 1A for each millisecond the voltage is present. with 100V it will rise by 100A for each millisecond. Note that the diameter of the wire does not come into this equation (as long as it is big enough to carry the ultimate current) So you want to use very big wire to keep the resistance low and therefore the loss to heat low. Note also, that unlike the resitance case, the voltage does matter now.

Let's take your example of 125A. If the inductance is 1mH then we can apply 40V to it for 3.125 ms to get the current up to 125A. If we do this with wire that has very little Resistance , we can manage the heat by cutting off the voltage after this 3.125ms.
Bob

This helps me out a lot, so I need to know my coil inductance then.

How do I calculate coil inductance? Still planing on using 100ft of 14g wire, .25ohms.

Also is there any difference is wrapping a coil in so that its wider or taller (rows vs columns)? I want to make my coil only say 10 rows but many columns to keep my opto switch closer to the middle of the coil.

Each one of my coils will be fired by a opto/led light switch placed right before the coil. Opto switch will be ON for 9.5mm while the projectile is in it, switch will activate my IGBT's. So really POWER to each coil will be for 9.5mm or the time it takes the 9.5mm ball bearing to pass through it. First few coils will have a longer pulse, but as the projectile accelerates the opto switch will pulse shorter and shorter length of time for each subsequent coil.

This is where I need someone good with math to hopefully calculate how much speed I will have into the projectile for each 1 of my coils? Assuming:

- coil 100ft, 14awg, .25ohm, guessing I can get 250 turns from 100ft
- 9.5mm distance of acceleration per coil
- power: 40v@125amps
- projectile 3.5grams


PS. I find a coil inductance calculator so I understand it more now. I also have some idea about # of rows vs layers in a coil.

 

[CDRIVE]

I did actually give coil guns a little thought. Must be due to the congitive dissonace of my abhorance for firearms and my secret desire to off those damn squirrels that eat all of our bird seed. I think it would be a fun (and challenging) project.

Bob

Google "Squirrel Launcher". Now that's funny and they don't get hurt...usually!

Chris

 

[(*steve*)]

Without capacitors the best you'll get is a linear induction motor.

But sure it will be marginally safer.

 

[supak111]

Ok so lets say that I use a single 40volt 34000uf capacitor so 27joules of energy per each stage, I would in that case need have 2,700joules after 100 stages, correct?

Thats if the cap on each stage discharges all of its energy. Does anyone know how long it would take this cap to discharge or charge?

 

[(*steve*)]

The time taken to discharge a capacitor depends on its charge, its internal resistance (ESR essentially) and the load it is discharging through.

Depending on the capacitor, you might be able to discharge it single digit milliseconds without a lot of effort. The problem you'll have is the energy stored in the coil if the projectile doesn't absorb it all (which it won't). This residual energy will try to recharge the capacitor backwards. This will either damage the capacitor, or lead to a series of oscillations that won't be helpful.

The amount of energy you transfer to the projectile will depend on how good your timing is. As the projectile accelerates, you'll need to discharge the capacitor through the coils faster and faster because the projectile will spend less and less time in the coil. And you need to detect when to discharge the capacitor so that you accelerate rather than brake the projectile, or completely miss doing anything to it at all.

I have no idea what efficiency you'll get, but I expect that it will fall as the speed of the projectile increases.

 

[BobK]

100 stages is probably not practical, I would start with 1 stage and work up to four or five before even thinking about 100 stages. As Steve said, timing these is not going to be trivial. Note that even with 1 stage the timing is important. If the current hasn't gone to zero when the projectile gets half way through the coil, it is going to slow it down.

According to Wikipedia, a Russian Scientist achieved 90% efficiency, but home-built amateur devices are typically 1 to a few percent. If you could achieve 10% efficiency you woiuld be ahead of the pack.

Bob

 

[supak111]

I've read about backwards voltage, anti-parallel diode seems to fix/help that. My timing shouldn't be a problem either since I'm using a optical switch that will turn IGBTs ON about 10mm before coil, and OFF about 1/4 of the coil. There shouldn't be any voltage in the coil after the first 1/4 of the coil so def nothing there after the crucial halfway point.

I know a 100 stage isn't practical but thats what my gun is going to end up being so I just wanted everyone to understand why I'm only using 1 capacitor on each stage and not 10 or 20 caps per stage. I guess what I'm trying to make everyone understand is that for my new design its easier, much cheaper, and more practical to use more stages with fewer caps per stage, then fewer stages and more caps on those. I also think this more stage design might be more efficient. And yeah If I get 10-20% efficiency I would be happy but I believe I can do better with this design hopefully.

And Steve said "Without capacitors the best you'll get is a linear induction motor." That's ok with me since my understanding is that linear motor is way more efficient and theoretically it has no projectile speed limit?

 

[supak111]

Hey guys I just got my opto switch that will be activating my coil, but I have a problem... The output wire from the switch has power when the opto switch IS NOT block, and has NO power when the switch IS blocked. I need it to be opposite and have power on the output when the switch is blocked by my ball bearing projectile.

I know I can use a mechanical relay to fix this but they are slow and usually only up to 50amps so that wont work.

I wanna use a SSR but they don't seem to have 5 plugs like mechanical relays so I don't see how I can get this to work?

I basically need to turn ON my IGBTs with no signal, and have them OFF when there is a voltage signal from this opto switch I just got.

 

[BobK]

Google "inverter"

Bob

 

[supak111]

DC to AC inverters come up. I don't see how that would help? Are you referring to something different BobK?

PS: actually just found "positive to negative inverter" on ebay, any idea how fast these are? Don't those just use a mechanical relay which is slow, inefficient, and usually less then 50 amps?

 

[CDRIVE]

Try BJT Inverter, TTL Inverter, CMOS Inverter and Logic Inverter.

Chris