This may be the simplest LED flasher circuit you can build, with the notable exclusion of LED's with integrated flashing circuits. This might be a good replacement for the LM3909 in some applications. Take a close look. Only the emitter and collector leads of the 2N2222 are connected. The base lead was cut off. The LED is from a string of Christmas lights and it has an integrated 100 Ohm resistor. This LED flasher occurred to me while reading about negative resistance in transistors. In this implementation, a common NPN transistor is used. In the circuit, a 1k resistor charged the 330 uf capacitor until the voltage became large enough to get the emitter-base junction to avelanche.
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If the resistor that charges the capacitor is too low in value (or if the power supply voltage is too high), the current through the transistor will not become low enough for the transistor to turn off. If the resistor that charges the capacitor is too high in value (or the power supply voltage is too low), the capacitor will not be able to charge to a high enough voltage to enable the transistor to turn on. This is because the transistor draws as small amount of current before switching on.
In the oscilloscope image, it can be seen that the peak voltage (yellow trace) was a little bit less than 9 volts. At this point transistor turned on quickly and partially discharged the 330 uf capacitor through the LED and the 100 Ohm current limiting resistor. The current wavform, which is the voltage drop across the 100 Ohm resistor, is shown in the blue trace on the scope image. Peak current was 26 milliamps, and the transistor continued to discharge the capacitor until conduction suddenly ceased at 6 milliamps (Many thanks to Luke in Australia for pointing out the correct current). After the transistor stopped conducting, the capacitor began charging again, thus starting a new cycle.
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