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My first coil gun... 100V battery or 100V capasitor?

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tangent1

Thu May 21 2009, 12:12PM

Registered Member #2130 Joined: Thu May 21 2009, 11:56AM
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Posts: 2

I am building my first coil gun.But I want to ask something.
If I have some way to connect 64AA batteries ,should I use it directly or connect it to a 100V capacitor?

Barry

Thu May 21 2009, 12:55PM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

Use the batteries to charge a capacitor. The battery's energy comes from a chemical reaction which is too slow to deliver its energy in just the few milliseconds needed. In contrast, a capacitor's energy comes from an electric field between two plates and is therefore available at very high speed.

Cheers, Barry
I'd like to meet the person who invented sex, and see what they're working on now.

Backyard Skunkworks

Thu May 21 2009, 05:21PM

Registered Member #1262 Joined: Fri Jan 25 2008, 05:22AM
Location: Maryland, USA
Posts: 451

Try to find a 100V capacitor that stores at least 10,000 microfarads if you want youg coilgun to have a decent amount of power, or parallel up several smaller capactiors.

And as Barry said you definatly will need to use a capacitor, the difference in peak current is up to three orders of magnitude.

hotcrazyfruit

Thu May 21 2009, 07:43PM

banned on 5/26/2009
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connect to a 100v capacitor, the capacitor will discharge in a quick pulse, where as batteries will supply a small amount of current. and lots of current is what makes a good coil gun, as well as a quick pulse.

Wessel

Thu May 21 2009, 11:51PM

Registered Member #2127 Joined: Wed May 20 2009, 03:35PM
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Betteries have this thing called internal resistance, due to the speed at which they produce free electrons. Basically, you add that resistance to whatever the resistance of your circuit is and that's total resistance. It's usually quite small and negligible. However, in a low-resistance circuit, it can be up double the circuit's resistance, hindering the ability to produce large currents.

tangent1

Fri May 22 2009, 01:25PM

Registered Member #2130 Joined: Thu May 21 2009, 11:56AM
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thanks all. Also,what is the expected time for charging the capasitor? And can I use a magnet as a projectile?

hotcrazyfruit

Fri May 22 2009, 08:02PM

banned on 5/26/2009
Registered Member #1877 Joined: Mon Dec 22 2008, 02:03AM
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tangent1 wrote ...

thanks all. Also,what is the expected time for charging the capasitor? And can I use a magnet as a projectile?

the charging time for a capacitor depends on the size of the capacitor (in microfarads) and the amount of current supplied by the charging supply. you may be able to find it on the hv-wiki.

Backyard Skunkworks

Fri May 22 2009, 10:21PM

Registered Member #1262 Joined: Fri Jan 25 2008, 05:22AM
Location: Maryland, USA
Posts: 451

tangent1 wrote ...

thanks all. Also,what is the expected time for charging the capasitor? And can I use a magnet as a projectile?

Using a magnet as a projectile tends to produce very odd results. Also, most magnets shatter on impact. Just use regular iron/steel.

Myke

Sat May 23 2009, 12:20AM

Registered Member #540 Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969

The high magnetic field that the magnet would experience might be strong enough to cause the poles to reverse. If not, it will probably demagnetize it.

Barry

Sat May 23 2009, 04:40AM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

tangent1 wrote ...

What is the expected time for charging the capacitor?

The charging time depends on the size of the capacitor, and the current from the charging source. If we assume a constant-current charge cycle, then:

t = C * V / I
where t = time in seconds, C = capacitance in farads, V = final voltage, I = current in amps

Some typical values might be something like a 1200 uF @ 250v capacitor charged from a neon sign transformer that limits current to 35 mA. In this fabricated hypothetical example:

t = (0.0012 farads)*(250 volts)/(0.035 amps) = 85 seconds

However, most charging systems are more like a "constant voltage source connected through a current-limiting resistor" than a constant-current source as above. These systems charge quickly at first, and more slowly as the voltage reaches its final value. Read about capacitor charge and discharge here.

For most simple charging systems, we simply calculate the RC time constant. And then assume it is fully charged after three of these RC time constants.

For example, suppose we have a 1200 uF @ 250v capacitor charged from a transformer + diode + 7K resistor. (I chose 7K for comparison because 7K limits the initial charging current to I = V/R = 35 mA.) The charging time t which is "3RC" is:

t = 3 * (7000 ohms)*(0.012 farad) = 252 seconds

So use your own values here, and you can accurately calculate your own charging time.

Cheers, Barry
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