You should use balancing resistors across the capacitors, assuming it's a simple series connection of electrolytic capacitors.
The appropriate resistor to use across each capacitor is
R = \$ \frac {n V_M - V_b}{0.0015C V_b}\$
R is in M ohms
n is the number of capacitors
C in uF is the capacitance of each capacitor
\$V_M\$ is the voltage allowed on each capacitor
\$V_b\$ is the total voltage across the string
If we allow 400V across each cap and total voltage will not exceed 15kV
then we have a balancing resistor value of 555K ohms.
Each one needs a power rating suitable for the highest possible voltage,
so
\$P_R = \frac {V_M^2}{R} \$ = 0.29W for this example
The balancing resistors will suck 0.54mA in this example, which is 8.1W total,
a significant power loss, so you may not be able to get full voltage from your coil.
Reference: Cornell-Dublier Aluminum Electrolytic Capacitor Application Guide
Note: Some of the information below is anecdotal, since my measurements use a Canon 1D Mark III dSLR, rather than specific light-sensing instrumentation
I've just started work on a strobe softbox for my own product photography studio use, especially motion capture (multiple exposure of moving parts in a single frame): I will be using RGB LED strips to get color control.
Couple of things to note:
- If you have 12 Volt or 24 Volt LED strips, those already incorporate the current limiting resistors in each cut-able piece, so you do not need current limiting - Just provide the rated voltage.
- A 3 millisecond strobe every 500 ms is trivial to achieve with most LED power supplies and LED strips: I have been experimenting with a 40-Watt LED power supply, using MOSFETs to switch power to the R, G and B lines, and I am able to achieve strobe duration reliably down to 2 strobes in 1/8000 second (0.125 milliseconds).
My results get muddy below around 30 microseconds per strobe. My camera does not support a shorter shutter time than 1/8000, so I cannot capture more precise data.
Lessons learnt:
- A 24 Watt power supply used on a 24 Watt LED strip leads to spurious pulses during the time my shutter is open, but a supply with at least double the LED strip's rated power works fine.
- Minimum clean strobe duration varies widely between LED strips of different manufacturers, and slightly between different LED strips from a single manufacturer.
- Pulse edges get softer due to capacitance along the conductors, as the length of the LED strip increases, so connecting lots of the "cuttable" cut pieces in parallel works better.
- The green intensity needs to be reduced a lot compared to the other two, the red a bit lower than the blue, else a color cast appears.
- The green needs to have duration reduced by about 0.25-0.5 microseconds compared to the other two colors, which may be due to secondary phosphor afterglow: See this answer for a detailed discussion on the "long tail" subject.
The last two points are not applicable if white LEDs are used. However, white LEDs will have a slight color cast when using very short duration (say less than 0.1 millisecond, exact limit depends on specific LED used) strobes, due to the yellow-emitting secondary phosphor continuing to glow after the blue / UV junction emission stops at each pulse. At 3 millisecond strobes this color cast will not be discernible.
Recommendation:
Experiment with the specific LED strip you intend to use, to determine the point at which any shorter strobe duration causes a color cast. The figures should be far shorter than the 3ms specified in the question.
Best Answer
Based on Brian Drummond's comment:
The maximum energy your capacitors will be able to store is:
E = 1/2 C V^2
Where C is the capacitance, V is the voltage the capacitors are charged to, and E is the energy stored. Using this, if all of the capacitors are charged to their maximum voltage, you will have an energy of:
(1/2)*2200*12*400^2 = 0.0011 * 12 * 160000 =.0132 * 160000 = 2112 Joules
Assuming an approximate weight of .989 oz, or mass of about 28g, and perfect energy transfer(impossible), the velocity necessary to have the same amount of kinetic energy as stored electrical potential is:
E_k = 1/2 m v^2
where E_k is kinetic energy, m is mass, and v is the velocity. so
2112 = (1/2)(.028kg)v^2
150857 = v^2
= 388 m/s or 867 mph.
Now, given the highly imperfect nature of home building your own rail gun, I would say you need to be very lucky to get even half of that energy transferred to your nail, the case would instead be
1050 = (1/2)(.028kg)v^2
75000 = v^2
= 273 m/s or 610 mph.
Even that figure is very optimistic, and ignores all energy lost to friction, air resistance (which is no where near negligible at these speeds), and even energy that wasn't able to get transferred to the nail before it leaves the barrel.
One more note on Brian Drummond's comment:
The capacitors will store the same amount of energy in series or in parallel (assuming the voltage across every individual capacitor is kept the same), however, current will rise to a more useful level much faster in the series configuration (since the coil of the gun is just a long inductor) which means more energy has the opportunity to be transferred before the nail leaves the barrel.
These are very rough estimates that assume perfect conditions. Better estimate could be gotten if we knew the exact inductance of your coil, the exact mass of the nails you plan on firing, etc. But that would require some calculus that I don't want to do.
Be safe, I don't want to hear about anyone getting killed by their own rail gun on the news.
Also, its possible my math is off (please correct me if it is), its been a while since I've really done classical physics.
EDIT:
I hadn't done much research on the subject, I simply used the formulas I knew to get approximations. After looking into the subject a little more you probably mean a coil gun instead of a rail gun. They use the same principles to work, but the construction of a rail gun is much more complex, requiring the use of a sliding armature and more advanced mechanics. A coil gun is simply an object accelerated through a tube by a magnetic field, from a coil around the tube.
On efficiency http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA389605 this document discusses the efficiency of rail guns, (lets ignore the fact that it is questioning if the efficiency is a useful metric), and states that typical efficiency is around 40% (100% minus 5% residual inductive, 20% lost in the armature, and 35% lost to friction [what they dubbed "rail resistive"]) I would hazard a guess that your best efficiency before any air resistance or other unknowns will be somewhere from 20-30%. Not great efficiency, but still enough to get a nail moving pretty fast.