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1. Its a Shmidtt trigger
2. You have you Q1s and Q2s mixed up.
3. How it works, with Q1 off (Q1 Ie = 0). Q2 Vb = (1/1+1.5 )X12V = 4.8V. Q2 Ie = (4.8 - .6)/ 160 +330 = 4.2/490 = 8.5 mA. Therefore voltage at resistor tap = 330 X 8.5 mV = 2.8 V
So Q1 is sitting there with 2.8V on its emitter. When its base reaches 2.8 + .6 V it conducts ( 3.4V), this pulls the Q2 Vb down which switches off Q2, the current though it decreases, the voltage on its emitter falls, the tapping point falls so the emitter voltage on Q1 falls, this increases the current through Q1 which turns on fully.
The voltage across the 330 is 330/ (1.5k +330 ) X 12 = 2.16 so when Q1s base falls to this voltage you are back to square 1.
Frank
I think you didn't take voltage drop from collector to base of Q2 in the last line to calculate voltage across 330 ohms resistor. Why so?1.
The voltage across the 330 is 330/ (1.5k +330 ) X 12 = 2.16 so when Q1s base falls to this voltage you are back to square 1.
Frank