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Registered Member #90
Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301
Fortunately, arcing at turn-on is less of a problem. Some arcing is inevitable but it occurs at lower current. The turn-on time can be slower than even a slow IGBT.
With the relatively large inductive load of coilguns, it takes a short time for current to ramp up to high levels. This gives a small cushion of time (perhaps 50 to 500 usec depending on the LC time constant etc.) to establish a full-face contact area before the current begins ripping chunks out of the copper and dispersing them evenly around the room. I am expecting this extra time will allow switch closure rates substantially below 100 m/s.
<speculation type="wild">It sounds to me like a copper block makes a decent projectile for a good linear induction motor. Perhaps an induction coilgun can shoot the copper block into the bigger switch to fire the monster coilgun ... nah ... what switch would I use for the first linear induction motor? </speculation>
If this idea pans out, I'll be sure to provide a full write-up, and maybe even a video.
Barry Never use a long word where a diminutive one will suffice.
Registered Member #56
Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445
I wouldn't really worry about it, look at TDU's can crusher... There is very little contact in that switch (in fact his demo one uses 1/8" rods perpendicular to each other), and it is feeding a load of like 4 turns
Registered Member #358
Joined: Sat Apr 01 2006, 06:13AM
Location: UCSB
Posts: 28
If you know the shape of the contacts, you can calculating it by integrating up the infinitesimal resistance coming from the section with thickness dx (assuming the x = constant plane is an equipotential, maybe r = constant is more realistic). You can also solve Laplace's equation with the appropriate boundary conditions, which will figure out the equipotential surface deal, and give you the resistance, but that's not a fun way of doing it.
For example, lets calculate the resistance of 2 truncated pyramids contacting at their truncated tips. Since the pyramids are identical in this example, its just 2 times the resistance of a cone between the two points.
The resistivity of a pyramid is the integral of (d resistance/dx)*dx over the correct limits. The resistance of a slab with thickness dx is calculatd the same for a large thickness, so it is just (bulk resistivity)*dx/area.
That gives us the integral of (bulkresistivity)/(a*x)^2 dx, where a is the side length/distance from point and x is the distance from the point. When integrated out, you get -(bulk resistivity)/(a^2*x). lets plug in 1mm and 1cm for x, a = 1 and bulkresistivity = 1.678*10^-8 ohm*meters (copper). We get 1.5102*10^-5 ohms. If we plug in 1mm and infinity, we get an extra 1.678 micro ohms, while if we plug in 0 and 1mm, the result is infinite, so you know the resistance is dominated by the contact. <= that is all for one pyramid, multiply by 2 to get the resistance for the whole assembly
The downside of this approach is that you really have to know your shape. If you don't have enough pressure to really force the surfaces together, you don't know the real contact area. Garbage in garbage out...
The other approach is to just try it. Push two pieces of copper together, put a few amps through, and then put a mv meter on it, and just measure it with different amount of force on it.
Either way, more force = more better, and more softer = more better.
I would just go with IGBTs, but that's beside the point.
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[moved] how to estimate copper plate connection resistance?
</speculation>
