high current pulse?

Wed Aug 05, 2009 6:13 am   

I'm trying to create a permanent magnet. I have a steel rod. I wrap AWG
6 around the rod twice (so I got a two layer coil) and directly connect
it to a 12V DC battery for like a tractor or snowmobile (235 cold
cranking amps).

Very little happens. A max of 5A comes out and I can't even make an
electromagnet.

Can anyone tell me what I can do to get an extremely high amperage
pulse? Are there any power supplies that will generate a high amp
current? I'm thinking like 80 amps at 12V.

Wed Aug 05, 2009 8:59 am   

"Ken Williams" <kenw232_at_yahoo.com> wrote in message
news:gq9em.65625$cf6.56534_at_newsfe16.iad...

Wire sufficient FGCs in parallel to give you 1 Farad.

Charge to 1kV

Discharge through 10 turn 6" diameter coil made
from 1/4" copper rod.

Important .... just before you discharge, phone for the ambulance
because the ignorance that you display is certain to kill you
during the process.

(Wasn't his advice given here recently?)

Thu Aug 06, 2009 2:31 pm   

On Wed, 05 Aug 2009 02:13:00 -0400, Ken Williams <kenw232_at_yahoo.com>
wrote:


---
Well, let's see...

#6 AWG wire has a diameter of about 0.167" and a resistance of about
0.395 ohms per 1000', so if we assume that you've got a 12" long steel
rod with a diameter of 1", then you'll be able to get:

12" length
n = ------------------ = 71.86 ~ 71 turns
0.167" diameter
on it.

The steel rod has a circumference of:

C = pi D = 3.14* 1" = 3.14 inches,

so the length of the first layer will be

L1 = C n = 3.14" * 71 turns = 222.94" ~ 223"

Since the wire has a diameter of 0.167", the second layer will be wound
on the diameter of the rod plus twice the diameter of the wire:

D2 = 1" + 0.167" + 0.167" = 1.334"

and its length will be:

L2 = pi D2 n = 3.14 * 1.334" = 297.4" ~ 297"

So, the total length of the winding is:

L = L1 + L2 = 223" + 297" = 520"


Since the wire has a resistance of 0.395 ohms per 1000 feet, that's
0.000395 ohms per foot, and the resistance of the winding would be:

0.000395R
R = ----------- * 43.33 = 0.017 ohms.
1'

Assuming your battery's internal resistance is about equal to the
winding's resistance means that you'll have 12V across 0.034 ohms,


and the current into the coil will be:

E 12V
I = --- = -------- ~ 353 amperes
R 0.034R

Since you're only getting 5A into the coil I'd suggest that your battery
isn't fully charged, that it's bad, or that whatever's between the
battery and the coil is limiting the current.

BTW, if you do get 353A into the coil, it'll be dissipating:

P = IE = 353A * 6V = 2118 watts

which is about 50 watts per foot, so it'll get very hot very quickly.

The battery will also be dissipating the same amount of power, so you'll
need to be very, very careful when you're experimenting with it and if
you ever get it working right.

JF