I was studying the function of this simple relaxation oscillator based on a single transistor, 2N4401. The schematic looks like this:

enter image description here

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.


2 Answers 2


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.

  • 1
    \$\begingroup\$ Thanks a lot for the explanation. \$\endgroup\$ Nov 7, 2019 at 17:34
  • 5
    \$\begingroup\$ It is interesting to note that most circuits that I have tried to simulate to utilize reverse breakdown of transistor B-E junction of a transistor do not model the breakdown. Thus such oscillator behavior of the shown circuit do not show oscillation. \$\endgroup\$ Nov 7, 2019 at 18:24
  • 2
    \$\begingroup\$ If testing multiple transistors with this, be aware that you cannot hot-swap the transistor, as the capacitor will kill the transistor if you do this \$\endgroup\$
    – Ferrybig
    Nov 8, 2019 at 9:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.