I was studying the function of this simple relaxation oscillator based on a single transistor, 2N4401. The schematic looks like this:
From 2N4401 datasheet, the collector-emitter breakdown voltage is 40V. My question is, how does then the above oscillator work? In the given schematic, the voltage on the +ve plat of 3300uF cap would never reach the required breakdown voltage so the transistor can fully conduct.
Collector-emitter breakdown voltage isn't the relevant parameter in this circuit.
Note that the B-E junction is reverse-biased, and the B-C junction is forward biased, so the relevant parameter is the B-E breakdown voltage, which is just 6.0 V.
When the B-E junction breaks down in reverse bias, charge carriers are injected into the base region, which allows the transistor to operate in a "reverse active" mode, passing currrent from the capacitor through the LED. This continues until the capacitor voltage drops too low to sustain this.
This mechanism creates a negative-resistance characteristic in the I-V curve of the transistor in this mode of operation, and it is this negative resistance that creates the relaxation oscillator. Without it, the circuit would just come to equilibrium without oscillating.
It is not the breakdown voltage that defines the behaviour of this circuit. In fact, the working of this relaxation oscillator is a bit of a mistery.
If you measure this circuit you will notice that the PNP transistor behaves like a Z-diode with negative resistance, i.e. the voltage decreases with increasing current. This negative resistance is one reason why this circuit oscillates.
