Capacitors sound like they would do pretty much what you want, although there are a couple of problems when used in this situation:
LED's, like all diodes have a voltage drop across them. That is to say that, for,say a white 'LED' its vdrop would be around 3.2-3.6V (depends on the current going through it). If you apply a voltage less tahn this across it, it will not light. This is because current only flows when the voltage is above this, and of course, no current = no power = no pretty lights. A resistor in series with an LED limits the current. If you apply 5V to the resistor/LED combination, the LED 'drops' its voltage, leaving the voltage across the resistor as 5V - LEDdrop, in this case 5v - 3.6V = 1.2V.
Using the wonderful formula: V=IR, rearranged to find out current we get: I = V/R. Where V is the voltage across the resistor (1.2v in this case) and R as the resistance.
For a resistor of 120ohms, I = V/R = 1.2/120 = 10mA. - pretty reasonable for an LED.
Now, if we place a capacitor across the LED and resistor, when we apply power the capacitor charges up, quite quickly, as we are essentially placing a voltage directly acrosds it with no series resistor. It starts at 0v across the cap...this means the LED will not light. The LED will only begin to light when the voltage across the capacitor is equal or greater than the LED voltage drop. Now it is like applying an increasing voltage across the resistor/LED.
Since the LED's drop is fixed, it is like slowly increasing the voltage across the resistor, from 0 (where the cap voltage is equal to the LED drop) to 1.2V (where the cap is fully charged and is equal to our 5v supply). Looking at the I=V/R...as R is fixed, increasing the voltage across the resistor, increases the current allowed to flow through it. - Its the current flwing through the LED which determines its brightness, and since the resistor and LED are in series, this current is the same for both.
So, when you turn on your flashlight, the LED won't light until the cap reaches the LED voltage drop. And then it will quickly go to full brightness. The only way to control how long it takes from being 'off' to 'fully on' would be the value of the capacitor used. Without yet another series resistor, the capacitor will undoubtedly charge very quickly...a lot less than a second.
The second problem, which may or may not be a problem, is down to how the capacitor voltage increases. When applying a constant voltage across a capacitor, its voltage rises non-linearly. That is to say, it doesn't 'ramp', it's voltage rises very quickly at first, then its increase gets slower and slower. google 'RC time constant' to see the charge curve.
Because as I mentioned before the LED will NOT light, until the cap voltage is equal to the LED voltage drop, this initial 'quick voltage rise' has no affect on our LED, so once the LED begins to light, the voltage rise on the cap is actually quite slow....so this 'may' be an advantage in this situation...since we do not want it to get from 'off' to 'on' within some stupidly small time-frame, like 3ms
Now, if we were to put the capacitor in parallel with the LED (not as before where we put it in parallel with the LED AND resistor) the same prinnciple applies. Except....this time the capacitor charges through the series resistor...limiting the current available to the cap...causing its voltage to rise much slower. However, this idea brings its own issues....if the caps voltage rises above the LED's forward voltage (voltage drop) then excessive current can flow from the cap, to the LED...causing it to meet its maker.
Bradass: I suspect the reason it 'sizzled' was because of the capacitors voltage rating...OR the series resistor being of incorrect value, perhaps even incorrect polarity of the capacitor used. Always pick a capacitor with a votlage rating much higher than your power supply is capable of providing. So, for a 12V supply, a capacitor with a voltage rating of 16V is the minimum, probably 25V just to be safe.
Sorry for along and detailed post..but, if you knew all this already, then it is I who have wasted their time...but I was hoping it would give you a basic understanding of whats going on
If you wish to know more, I'll psot some example schematics along with graphs and wonderfully boring maths.
Blueteeth