Two transistor LED flasher

Hello, I would be GREATLY indebted to whoever can help me understand how the devil this little circuit works. ITS BEEN DRIVING ME NUTS!!!!!!!!!!!!

C1 charges through R2 and R1 until the charge is high enough to turn on Q1. This causes Q2 to turn on. The LED is now connected to +Vcc and GND through Q2 and R3, thus lighting the LED.

After this happens, C1 discharges through the base-emitter of Q1. This causes Q1 to turn off, which then switches Q2 off. Once Q2 is switched off, the LED doesn't light up. Then the cycle repeats as the capacitor charges.

R2 can be adjusted to speed or reduce the charge rate of C1, this allows the user to adjust the flashing rate because C1 governs the oscillations(On and off fucntions) of the transistors Q1 & Q2.

D1 prohibits current from flowing to GND before the current goes through R2, C1 and R3.

R3 protects the LED and Q2 from excessive current when Q2 is switched on by Q1.

I hope this makes sense.

Thank you SOO very much johnson, it was crystal clear. One last question is, why is the diode there, couldn't you just have no connection there, and it would do the same thing?

zachtheterrible said:

Thank you SOO very much johnson, it was crystal clear. One last question is, why is the diode there, couldn't you just have no connection there, and it would do the same thing?

Click to expand...

The diode is there to prevent the first transistor blowing, as the circuit turns off the timing capacitor places too high a negative voltage on the base - this would blow the transistor. The diode reduces the negative voltage on the base to only 0.7V, which is well within the transistors specifications.

that makes a lot of sense, thanx!

I now understand how the two transistor led flasher works, but why won't this circuit here work? I would imagine that it would work just as well, but I put it on a simulation and played around with the values of components, but it didn't work.