Let's say your "inverter" works just like a battery backed up UPS, and that it has a pretty wide margin on its storage voltage, going from 14V down to 7V before it taps out.
That means your main capacitor in case of the low voltage option can only drop 7V.
The voltage drop across a capacitor is an integral over time of the current signal taken from it, which becomes a very nice fiddly bit of magic concerning systems of differential equations once you add in a 10W constant load and a variable converter efficiency, so I'm going to roughly ball-park it, because I'm lazy and it's evening and I have a million things to do.
(( Ref: Wikipedia page jumped to the spot where the voltage current relation of a capacitor is given ))
How will I do this?
- Coursely take 80% efficiency for a converter going from 12VDC to 120VAC (which may be seriously overestimating it in a DIY scenario, to be honest).
- Estimate the current draw to be constant, calculated at a capacitor voltage of 9.5V, rather than the exact average, which voltage I drew from my large hat of "that'll probably do". If you want to do other use cases, you can take the average, since a constant current assumption will be off any ways.
- Simplify the integral for constant current, which then becomes a simple linear equation: V = (I*t) / C.
So, the current from the capacitor will be:
I = (10W / 0.8 [=efficiency]) / 9.5V =~ 1.32A
Which then can be put into the simplified linear equation for the assumption of constant current (be aware, this is a very broad and lazy assumption):
V =~ (1.32A * t) / C
Let's say you want only ten seconds of power, with the known voltage drop of 7V across the capacitor, that becomes:
7V ~= (1.32A * 10s) / C
Which becomes:
C =~ 13.2As / 7V =~ 1.88F
Let's quickly do that for 120VDC as well:
Same assumptions, but the voltage range will be 80V to 120V, probably, so a drop of 40V is allowable, estimating the constance of current at the 90V point:
I = 10W / 0.8 / 90V =~ 139mA
with t=10s:
40V =~ 1.39As / C
C =~ 1.39As / 40V =~ 35mF --> Charged up to 120V = very, very lethal.
So, you see, I've already used a lot of assumptions about all the stuff you're not giving us about your project, and how you will personally be able to complete the electronics and it's still a lot of calculation, even though I made a very bad and broad assumption of constant current
The final choice will depend on fixing all the parameters and some will intertwine. There's no solution to that and that's what makes electronics design a difficult field.
This is just your very first, very broad ball park. But to be honest, re: "very, very lethal", if you are asking this question I don't really think you should be considering anything above 30VDC to store energy the likes of this.
Just adding a cap seems easier to me, (rechargeable battery will need recharge circuit, and pass-through diode [kind of like a voltage source-selector] at the very least, no?) but the question remains whether that will be enough to hold you over.
You've gotta do the math though to see how much juice you need. Current will have to be measured from the leds + any leakage / quiescent through the step-down system. Voltage max might be 9 V but more than likely is 9VRMS if you're turning the crank steadily. Voltage min is min input to the step-down circuit.
http://www.circuits.dk/calculator_capacitor_discharge.htm
Here's a good time to make you aware that capacitors store a lot of energy. If you short them, especially large ones, you're in for a bit of a surprise. I think anything over 500uF starts to garner my attention, and you're going to need a lot more than that. You'll need to consider a discharge circuit to make sure you don't hurt yourself or blow stuff up with the super cap (no joke), but I believe if the light circuit is constantly connected / on you should be ok (anyone else care to pitch in on that?).
edit1: and on second thought (for the very same reasons described above) just do the battery. If you don't know what you're doing things can get problematic quickly. Search for battery power source. There's plenty of info around on this sight. Main points. Separate with diodes. Make sure your step-down output is HIGHER than the battery output (but that your led's can still run off battery output). If the step-down is putting out 3.3 (probably?) then use 2 nimh (2.4v). Done.
Battery in parallel with main power source
edit2: Okay so maybe a 9V cap isn't going to kill you..., but it's going to to put on a show if it shorts.
Guidelines for determining shock hazard of capacitors
Best Answer
NO, there will be no side effects.
The current will be shared between the capacitor and battery for both charge and discharge. In a solar panel usage configuration as you suggest, the current from the panel will be limited and the voltage will track the battery charge/discharge characteristics.
It is however very inefficient use of the Supercaps when you simply parallel them with a battery.
Consider that a 12V battery may have a fully charged state of 13.4V and a fully discharged state 10.5V. Putting a large supercap in parallel with the battery does not change the terminal characteristics. You still would have low voltage trips at 10.5V, and still classify as fully charged at 13.4V.
The charge stored in a capacitor is: W = 1/2 * C * V^2
For a capacitor in parallel with a 12V battery the total charge in the capacitor would be:
W = 1/2 * 88 * 13.4^2 ---> 7900 Joules
But since the lowest voltage is the fully discharged level of the battery you can only access a portion of the stored energy:
W =!/2 *88 * (13.4 - 10.5)^2 ---> 370 Joules
What a waste of an energy storage capability!
Does the supercap improve the battery system, absolutely it does. Large pulse current can be provided by the supercap so it can help where the output loads are DC-DC convertors (which tend to have high current spikes) especially where the battery is close to discharged (and has a higher impedance).