When S1 is closed, T2 no longer gets enough voltage to switch. Obviously it is because the capacitor is being discharged when S1 is opened, but what actually happens at the base of T2 when S1 is opened?
And when will T2 switch back again?
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\$\begingroup\$ Hint: when S1 is closed, what happens to T1 and R2? \$\endgroup\$– The PhotonApr 26, 2019 at 22:01
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1 Answer
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With S1 open and the circuit stabilized, T2 is on, LED2 is on, T1 is off, LED1 is off, and C1 has approx. 7 V across it.
When S1 is closed, T1 and LED1 go on, the right side of C1 goes below GND turning off T2 and LED2, and C1 starts discharging through R3. After less than one time constant, T2 starts to come on, turning on LED2. However, T1 stays on as long as S1 is closed. When S1 opens, T1 turns off. LED1 stays on but starts to dim as C1 is charged through it. Eventually LED1 fades off, based on some portion of the R2-C1 time constant.
Note - even with an exceptionally high gain part like the 548, 100K is pretty large for the base resistor. Consider reducing it to something like 22K to get more firm saturation when T2 is on.
