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Look under the designation BH-321 (a 2-AA 3V snap holder); You can make a flasher that might appear to be an intrusion alarm to a would-be burgular. You need this part plus the BH-321, as well as a 9V snap, which you might get free if you cut off the top 3mm of a dead 9V battery. The cost is only $1.20, excluding the 2 AA's and the snap. BH-321 has simple instructions.
I built several 'moving light' displays with these over Xmas - and they were a huge hit!...
You can put standard LEDs in series with these, and the string will flash together. Simply allow 1.5V or so for each R,G or Y LED in series. These can even be 5mm ultrabrights!...
These LEDs' internal chip limits their 'on' current to 20mA @ 6V. Adding a LED in series barely changes this current, and you can flash both off a 6V supply. ... This means that with a 270 Ohm or so in series, four LEDs can flash (original + three added ones) off a 9V supply. ...
As to the poster regarding 12.5V operation ... I think this may be stretching it a bit. When the LED turns off, the entire supply voltage will be applied to the internal chip inside. ...I seem to remember these types of chips are specified to 9V.
A 6.2V zener in series would be preferable, with a 10K or so resistor across the LED (to bias the Zener) for 12V operation.
So in summary, use a maximum open-circuit voltage of 9V, and limit the 'on' current to 20 mA or so with resistors.
When a Flashing LED is fed from 12 vdc through a current limiting resistor, MEL is quite right when he states: "when the LED turns off, the entire suppy will be applied to the internal chip." A series zener diode is not the best solution, however, because any change in supply voltage will be seen in nearly it's entirety across the flasher. A better solution is
to feed the flasher through a current limiting resistor and to place the zener diode across the flasher to limit chip voltage when the LED turns off. The zener diode can also be used as shunt voltage regulator to minimize LED brightness change as a result of supply voltage changes.
Operation from a 12.5 vdc supply through a 330 ohm current limiting resistor resulted in an "on" voltage of 4.9 volts. A 7.6 volt drop across a 330 ohm resistor calculates out to an LED "on" current of 23 milliamperes.