cigarrete like fading led effect...cap value?

I'm trying to construct an electric cigarette..there very simple but VERY effective tobacco alternatives but I think I can do better Id like to replicate the fading LED tip (like a cigarette when you take a drag) but without destroying the ones I have I cant read the capacitor. If I made a small board with an LED (0603 package) and a 470 ohm resistor, what size capacitor would give me the effect. Hard to time my current one but it seems to be fully on in a fraction of a second and fade away in about 1.5 seconds. Any help Appreciated

EDIT::Li ion battery @ approx 3.6 VDC

About 7,000uF.

Maybe you need something like the attached circuit. D1 represents your LED. You can raise the current by lowering the value of R3.

Can you guys tell me the math involved with this? For future projects and for my information?

T = RC; The time it takes for a capacitor Cto charge to ~60% through resistor R
IC = BetaIB; relationship between transistor base current and collector current.
VBE ~= .7V, voltage drop from Base to Collector.
IE ~=IC; relationship between collector current and base current.

In this design, the capacitor charges through the 1M resistor with the time constant described above. Because the charging circuit has such high resistance ( 1 MEG ohm ) that signal is buffered by the two transistors. The current gain of the two trasistors is the Beta of a single transistor squared. The voltage at R3 is the capacitor voltage minus two VBE drops as described above. Current thru R3, and thus output current = V(R3)/R3. That's why decreasing R3 increases current thru the LED.