I am building a coilgun and I have constructed a coil with several thousand turns, an ohm meter measures its resistance at 300 ohms, so I am having trouble powering it, I ran 240v through it and got nothing, it barely moved a paper clip. According to my calculations, I need to run 1000 volts at the minimum through the coil in order for it to work. From my research there are three ways to do that:

1)A really big transformer (such as a microwave transformer)

2)A Marx Generator

3)or with a Voltage Multiplier.

A Marx generator seems like it would be perfect for a coilgun, as it releases an extremely high voltage, high current pulse in a small amount of time, but it seems like a rather dangerous build. You are supposed to connect the voltage multiplier to the Marx Generator, but wouldn't it be easier to just to make more multiplier stages and remove the Marx Generator entirely? I've seen voltage multipliers go up to 15kv, which is more than enough.

  • \$\begingroup\$ 240 volts AC or DC? \$\endgroup\$
    – HL-SDK
    Feb 5 '14 at 18:53
  • \$\begingroup\$ Homebrew Marx generators have significant issues getting all the gaps to fire at the same time reliably. I am not sure you could get sufficient energy out of a multiplier like that. I'd use a step-up that charges a high voltage pulse capacitor array. Multistage coilguns really are the way to go also. \$\endgroup\$
    – HL-SDK
    Feb 5 '14 at 18:54
  • \$\begingroup\$ If it barely moved a paper clip at 240V then at a 1000V what can you really expect? Also what gauge wire did you use? \$\endgroup\$
    – Andy aka
    Feb 5 '14 at 19:11
  • 1
    \$\begingroup\$ @ChuckFulminata very true (maybe 4A) but we're still only talking 3 or 4 paper clips barely moving. What wire gauge did you use and maybe if you have a photo you can embed it in the question? \$\endgroup\$
    – Andy aka
    Feb 5 '14 at 19:29
  • 2
    \$\begingroup\$ AWG30 wire is not going to be very happy at anything much over 1A. Also, unless you took special care in the winding pattern, the insulation will break down at anything over a few hundred volts. I think you've created a nice paperweight, and you need a completely different coil (one with a lot less inductance, for one thing) for a coilgun. Think heavy-gauge wire, few turns, lots of current. \$\endgroup\$
    – Dave Tweed
    Feb 5 '14 at 20:19

Since you are making the coil yourself, you can wind it to various voltage versus current tradeoffs. It seems you used too many turns of too thin wire, which therefore has a high DC resistance and needs high voltage to generate the large magnetic field you want.

Use less turns of a thicker wire. That will require more current to generate the same magnetic field, but a lower voltage to cause the current.

You can maybe achieve this with the wire you have by breaking down the "several thousand turns" into several winding and connecting them in parallel. For example, let's say you have one winding of 3000 turns now, which a DC resistance of 300 Ω. 10 A would require 3 kV, and provide 30 kAmp-turns of magnetic field. If you make that into three windings of 1000 turns each, then each winding has 100 Ω DC resistance. Put in parallel, the combined coil has 33 Ω resistance. It would require 1 kV and draw 30 A to produce the same 30 kAmp-turns magnetic field. This is actually still unrealistic. You want to go lower resistance and higher current overall. I just used these numbers as example to show how to calculate equivalent magnetic field.

I would find the largest capacitors you can get your hands on, and design the coil for that. This means putting enough current thru the wire for a short time to heat it just below damage level. The time will be so short that there won't be opportunity to conduct the heat away. Therefore, figure all the energy goes into heating the copper. Figure the maximum copper temperature you are willing to tolerate, use the specific heat of copper to find the energy per unit length of wire, then figure how much of that wire you need to use to absorb all the energy of the capacitor.

  • \$\begingroup\$ Yes, I realize that my construction is sub-optimal, but the question was not "How can I get my coilgun to work?" that was simply background, I want to know if A) A Marx Generator will work in this situation B)If I can skip the Marx Generator and just use the voltage multiplier \$\endgroup\$
    – Devon M
    Feb 5 '14 at 19:40
  • 4
    \$\begingroup\$ @Chuck: "I'm on the side of the road with a flat tire. Which blade do I use on my pocket knife to take the lug nuts off?" \$\endgroup\$ Feb 5 '14 at 20:22
  • 1
    \$\begingroup\$ My favorite is "What's the better tool to drive a nail into wood, an old shoe, or a glass jar?" \$\endgroup\$
    – Phil Frost
    Feb 5 '14 at 20:57

A Marx generator won't work so well. Neither will a voltage multiplier, if you get it up to thousands of volts. A "high current", high voltage supply doesn't make that much current, by normal voltage standards. The reason is simple: power is the product of current and voltage:

$$ P = IE $$

If \$E=15k\mathrm V\$, then to make 1A would require:

$$ P = 1 \mathrm A \cdot 15000 \mathrm V = 15000\mathrm W $$

Not going to happen. You also have to take into account all the inefficiencies of the voltage converter. They can be substantial.

This is why you should do what Olin says, and build a coil more suited to your power supply. It will be more efficient overall, and given the high energy impulse you require, efficiency is key, if you intend on making a coil gun rather than a pile of molten copper.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.