15 W \$\times\$ 5 s = 75 J. If you want to store that in a capacitor you'll need a capacitance of
\$ C = \dfrac{2 \times 75 J}{(10 V)^2} = 1.5 F \$
There are supercaps with even higher capacitances, but types which can supply a 1.5 A current are between expensive and Damn Expensive™.
A battery will be a better choice. For instance a LiPo followed by a boost converter, to have a nicely regulated 10 V, even at varying battery voltage. Use another boost converter to charge the battery between load bursts; I would shut-down the charger while you're transmitting.
If the 3.3 V can supply 100 mA that's 330 mW. Suppose the two boost converters have an efficiency of 85 % then to get the required 75 J output you'll need 105 J input. At 330 mW that will take 315 seconds, that's 5 minutes and 15 seconds. So you can transmit a 5 second burst once every 5 minutes and 20 seconds. If you need a higher frequency you'll have to find another power source.
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I almost forgot: the battery has a less than 100 % efficiency too. You can't get every joule you put into it out again. So in practice the charging time will have to be longer than the 5 minutes.
I would use two LiPo cells in series. Then you'll have 7.4 V nominal, 8.4 V fully charged. That's relevant, because the closer the voltage to the 10 V the less current will be required from the battery. If you draw 1.5 A at 10 V a 3.7 V cell would have to supply 4.8 A if we take again the 85 % for the booster. Two cells in series will only have to supply 2.4 A.
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Dave's ultracap seems to be a good alternative, but he should mention that the 3.3 V to 2.5 V regulator also has to be a switching regulator, unless you're satisfied with a 75 % efficiency there. If we use eBay as a price reference you can buy a set of two 240 mAh LiPo's for 4.5 dollar, free shipping. That means they only will be discharged for 1 % during the 5 s burst. So while 14 dollar for the cap isn't bad, and definitely something to remember, the LiPo's are only a third of that.
It's also worth noting that stepping down from 3.3 V to 2.5 V and then up to 10 V is less efficient than up from 3.3 V to 7.4 V, and then to 10 V. That's because a switcher's efficiency is related to both the in-out voltage difference and their ratio.
Wow, where to start...
If you blind yourself from the arc or electrocute yourself, it's your own fault. These sorts of do-it-yourself circuits can produce LETHAL amounts of energy and are easily FATAL.
Now to your questions:
I don't know what this exactly means, but I think it means to wind 5 turns with two separate coils and connect the two middle ones together.
Correct. What you're describing is two windings of 5 turns with the end of the first winding connected to the start of the second winding (technical speak for 'the middle ones').
I used fairly thick magnet wire from a radio shack roll of three thicknesses (i used the thickest). (tell me if i need thicker).
"Fairly thick" is completely relative and not helpful. The 5+5 turn windings are used to source energy to the arc that's formed by the open HV terminals. It's difficult to predict just how much current can flow since (I believe) this sort of self-oscillating, non-controlled design is going to be dominated by parasitic elements and hard-to-control elements like transformer coupling, the resistance of the windings, the layout of the switching devices with respect to the transformer, etc. - so, use the thickest magnet wire that you can fit on the core.
I am planning on winding one myself due to price, so what size toroid core should I buy, is that same magnet wire reasonable for 10 amps or do I have to buy larger, aprox. how many winds do I need to get close enough for the circuit to work!
You should do a complete inductor design. The number of turns on the toroid depends on the core's inductance factor (\$A_L\$) which of course depends on the exact toroid you're going to be using. There's no magic solution here. As for wire, I'd guesstimate 18AWG magnet wire or thicker to minimize DC losses. Go for a toroid that has room for more turns that you calculate, so that you can more easily add more turns if you find you need more inductance.
Third, I have a bunch (like 30) aerovox capacitors. The schematic calls for 6 1μf 270 volt capacitors to make a large bank but I looked and they can get quite pricy especialy when buying 6 of them so I am wondering if these would work.
The idea is to use multiple capacitors to divide up the current, so these in parallel should work. The inductor and capacitor values define the operating frequency (or so a few websites say) so try and keep the same capacitance value as the original schematic as a starting point.
Next, is the flyback itself suitable for a ZVS driver? And is it possible I don't even have to wind my own primary? (maybe it has something like 5+5 turns already built in)
You tell us. It's your transformer, after all. Seriously, "flyback transformer" is a broad term that covers many more devices than those found in CRTs. And I wouldn't trust any windings other than the multi-turn high-voltage one (that's the reason you're recycling a CRT transformer and not building your own transformer, right?)
My main concern is winding the primary of the flyback CORRECTLY and EFFICIENTLY and the capacitor bank and the inductor.
This sort of homebrew work doesn't lead itself to immediate efficiency. You probably won't hit the sweet spot the first few times, especially if you don't have any power electronics knowledge.
Best Answer
Q = C * V Charge stored in Coulombs is Q Rated capacitance in Farads is C Voltage across the plates in Volts is V
Divide both sides by a period of time and interpret V as the change in Voltage across the plates over that period of time and... Q / t = I = C * V / t Current is I in Amps if the is time in seconds.
More precisely, and since things vary continuously in the real world, I = C * dV/dt
The time derivative of Voltage is dV/dt.
Rearrange these with the usual algebra to answer your question.
C = I * t / V = 7A * 4s / (13V - 7V)... as a coarse estimate.