Capacitor question

Maybe this program can help your:

Making a rail gun Ken?

hahaha something like it :)

I do have a brief discharge from the cap to the coil. But the kind of repulsion i am getting seems that I have a 3v battery attached to it. I am getting the opposite effect that I am aiming for . 30uF does produce quit a spark

ken

I guess my questions would be, would getting a 300uF cap increase increase the electromagnet strenght ? MAybe a different of capacitor ?

ken

Perhaps most of the capacitor's energy is going into making sparks, instead of magnetic field.

How are you connecting the capacitor to the coil? Describe the coil dimensions, turns and wire gauge.

The spark comes from me putting a screw driver in there, Ithough tthe cap might be dead. Otherwise no sparks.

the last test I did was to charge the capacitor and immediatly put the positive and negative to the coil. The coil is gauge23, about 500 turns woth

10.6 ohms. I assumed if you connected a 30uF 127 v cap (charged) to a coil it would ( for a very brief time) repulse a magnet (assuming its north to north). I was thinking maybe I need a bigger cap 30uF is not enough? A dc cap or a flash cap would do a btter job? I was hoping to get the same result as if i would hook up a 120v battery (just not as long )

ken

What inside diameter?

Might repel or attract it, or even remagnetize it in the other direction. It will repel a closed loop of conductor.

Have you checked the capacitor voltage with a meter before doing the dump?

There is more energy in a larger cap, and the pulse will last longer.

The coil will interact with a magnet, in proportion to the instantaneous current. It will interact with a shorted loop of conductor, only when the current is changing.

2.2 cm outside is about 4.5 cm height 7cm

127v before dumping

load

12vto

to

get

It would be helpful if you told us what you are really doing and what the circuit looks like. The energy stored in the cap is 1/2*C*V^2. At best only half of that energy will be available in the coil. The rest will dissipate in the switch or spark, etc. What is the voltage the capacitor charges to? Have you measured it? Your coil is about 10 ohms. The RC time constant of 10 ohms and 30 uF is 300 us. If you are trying to get something mechanical to happen in 300 usec, you can just forget it. Mechanical interia will insure that little will be happen in that short time and it will all be over before anything apparent occurs. You need to determine how much force over how much time (impulse) is required to do what you want. You need to design the coil and discharge system to get that impulse. In otherwords, you have to do the math and understand the physics otherwise you are just screwing around which is ok but don't expect great results. I suspect your problem is insufficient energy storage and way too short of a time constant. Also, the coil will have inductace which will enter into the calculations. Good luck. Bob

Okay, by the formula, Energy (in watt seconds or joules)= 1/2 * C

*V^2 the capacitor has about 1/4 watt second in it. Not a big bang. But you would get almost all of that transferred to the coil if you dumped it with an SCR, instead of touching wires. Of course, if you could jack the cap up to 300 volts, the energy rises to 1.35 watt seconds. According to this inductance calculator: (with the wire gauge altered to give about 500 turns. You must have used vinyl insulated wire) your inductor should have about 2.8mHy of inductance, so the half period of the resonance with 30 uF is about 1 millisecond. If you switch the capacitor on to the inductor with no loss, that is how long the first pulse will be (excluding ringing, which you can force by having a diode in series). That is not a long time to be accelerating a magnet. More turns or more capacitance would extend the time.

coils are inductive. the energy will not get absorb fast enough for the size that your using most likely. that is just a guess, could be something much simpler. your chop rate maybe to fast.

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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

Ok so bigger coil. Need to buy more magnetic wire then :) Actually I was able to bring the voltage up to 280v, but the discharge is done no quickly I can barely feel a thing. Also Bob mentionned the time constant RC. ok If I added a 1000ohm resistor I,ll have a t=30.3mSec which is quit longer, but the resistance then prevented from getting much repulsion. Although the energy formula do not take into consideration the resistance.

K

Resistance is not the way to go, because it lowers the peak current. The inductance and capacitance forms a resonant circuit. When the charged capacitor is connected to the inductor, it produces a damped sinusoid. The frequency produced is roughly 1/(2*pi*sqrt(L*C). The first half cycle is what you want to deal with, and you can prevent there from being more than that, by adding a diode in series with the inductor. So increasing the capacitor by 4 times doubles the period of the dump pulse, and also quadruples the total energy in it for the same starting voltage. Quadrupling the inductance does nothing to increase the total energy, but doubles the time of the pulse.

putting the diode in parallel with the inductor would lengthen the pulse, in series just stops the oscilation (with the capacitor reverse charged - hope it's not a polarised type) after the first bump.

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Bye.
   Jasen

If you are charging this from the mains you could salvage a number of suitably rated rectifiers & reservoir caps from scrap equipment like TVs monitors & PC PSUs and make a Cockroft & Walton voltage multiplying latter, but be careful once you get into thousands of volts the charge can jump a surprisingly large gap to bite you and a few thousand volts on a bank of

220uF caps would be seriously lethal!!!!!

will try that

its Ac cap

use c=q/v

then find the value of charge ,use resistors to get required current