Description
Source - http://www.bowdenshobbycircuits.info/
The LED flasher circuits below operate on a single 1.5 volt battery.
The circuit on the upper right uses the popular LM3909 LED flasher IC
and requires only a timing capacitor and LED.
The top left circuit, designed by Andre De-Guerin illustrates
using a 100uF capacitor to double the battery voltage to obtain 3 volts
for the LED. Two sections of a 74HC04 hex inverter are used as a
squarewave oscillator that establishes the flash rate while a third
section is used as a buffer that charges the capacitor in series with a
470 ohm resistor while the buffer output is at +1.5 volts. When the
buffer output switches to ground (zero volts) the charged capacitor is
placed in series with the LED and the battery which supplies enough
voltage to illuminate the LED. The LED current is approximately 3 mA, so
a high brightness LED is recommended.
In the other two circuits, the same voltage doubling principle is
used with the addition of a transistor to allow the capacitor to
discharge faster and supply a greater current (about 40 mA peak). A
larger capacitor (1000uF) in series with a 33 ohm resistor would
increase the flash duration to about 50mS. The discrete 3 transistor
circuit at the lower right would need a resistor (about 5K) in series
with the 1uF capacitor to widen the pulse width.
Circuit DiagramSource - http://www.bowdenshobbycircuits.info/
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