Capacitor question

Hi

I have a rated 525v 30uF Ac capacitor that I want to discharge into a load
(coil). I am using a battery as a source, using a chopper circuit, a 12vto
120 v transformer to a full bridge rectifer rated 1000v. everything seems to
be ok at this point. Althought I am not getting any discharge from the
capacitor. I let the capacitor charge before dumping it top the coil. I
guess I don't have enough charge to creat a great enough current. My
question is, how do I calculate the charge I need (in farads) to I could get
a discharge ?
tried fidling around with with v=dQ/dt=I R but I did not find anything I
could use
30uF was a a ball park figure, it does take a little while to charge so I
assumd it would be powerful enough

ken

 

Maybe this program can help your:
http://www.miscel.dk/MiscEl/miscelChargeCurve.html

 

Making a rail gun Ken?

 

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.

 

What inside diameter?
2.2 cm
outside is about 4.5 cm
height 7cm

127v before dumping

 

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:
http://www.oz.net/~coilgun/mark2/inductorsim.htm
(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.

 

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

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.

 

will try that

its Ac cap

 

use c=q/v

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