Hello Josh,
The simplest method i can think of right now is
to build a capacitance amplifier. This allows
you to use a relatively small cap and get large
delays like you are looking for.
One way to build the 'cap amp' is using an
NPN transistor (like maybe the 2N3055A)
and one electrolytic capacitor.
Connect the collector to v+ and the emitter
to the LED through it's dropping resistor
(for an LS at 9v this would be about 15 ohms)
The capacitor (+ lead) connects to the base
and the cap (- lead) to ground.
The size of the cap would be 1000uf, 10v.
To turn on, short the collector to base of
the 2N3055A. To turn off, remove the short.
If you dont mind adding another transistor
to create a 'darlington compound', you can
get away with a cap value of 10uf, 10v.
This second transistor (2N2222 or equiv) would
have it's collector connected to the 2N3055A
transistor's collector, and the emitter of the
2N2222 connected to the base of the 2N3055A.
The cap (+ lead) then connects to the base of
the 2N2222 instead.
To turn on, short the collector to base of
the 2N2222. To turn off, remove the short.
In both cases, the lower base current is supplied
by the cap. Since the base current is lower
then the emitter current, the cap discharges
much slower then a direct connection. This
gives the long delay effect.
If the circuit isnt turned on and off very often,
no heat sink will be required as the transistor
only dissipates heat during the relatively short
delay period (2 to 3 seconds).
total number of parts including the series resistor:
2 parts for the single transistor, 1000 uf circuit, and
3 parts for the darlington, 10uf circuit.
Just in case you find the circuit doesnt turn off all
the way when you remove the short, connect a 1Meg resistor
across the capacitor.
If you need a schematic let me know.
INRETECH:
Are you sure you got the F300 chip in dip
package??
Good luck with your LED circuits,
Al
