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T4 is turned on by R9 (although that might be a bit marginal as it relies on a very good gain on T4)
The microphone signal is amplified by T1 and T2.
When there is a big enough signal on the collector of T2, the voltage comes down, causing the other end of C3 to come down as well. That turns off T4.
The collector of T4 goes up, turning on the Triac, and T3.
T3 holds the left end of C3 low, holding T4 off until current through R9 has brought the voltage at the right end of C3 up to 0.7 V, where T4 turns on, the Triac turns off, T3 turns off and the circuit goes back to how it was.
It should take several minutes for C3 to be charged though R9.
Apart from C3 being the wrong way round, a BC548B has a gain of 220 to 475 (Data sheet here) so the 11 μA through R9 will only be amplified to about 2 - 5 mA, so the voltage on the collector of T4 won't ever come down enough to turn off the Triac or T3. You should probably use two transistors instead of T4 to improve the gain.
When T3 turns on, C3 will take the base voltage of T4 very negative, past the base-emitter breakdown voltage. While that probably doesn't do any harm, half the capacitors charge will be lost quickly, so the capacitor has to be larger than it would otherwise be.
It is a horrible circuit.
At rest, transistors T1 and T2 are cutoff since they are not biased enough to work as an amplifier.
A very loud sound will turn on T1 which causes its output to go low which cuts off T2 even more than it was so its output does nothing.
good day driver300..can i ask some follow up qeustion?
what do you mean that c3 is in the wrong way around?
does it value is small enough? i tried it but in not exaclty 2mins that the bulb
will turn on. You mean half of the charge of c3 will go directly to the ground?
why is it sir that T3 is put up that manner? is it because when it saturates even if you introduce
a new ac signal like clap it will not add up to the time that the bulb will light??
thank you for you answers and reponse sir.. i appreciate it a lot. GODBLESS
C3 is shown as a polarised capacitor, probably an electrolytic capacitor. The +ve end of it should never be at a lower voltage than the negative end.
In the circuit, the base of T4 can never be higher than 0.7 V, because the emitter is connected to ground. So the right hand side of C3 is always below 0.7 V. The left hand side of C3 is at 12 V when T2 and T3 are turned off. That means that the voltage between the two is 11.3 V, with the left hand side at a higher voltage. That is likely to damage C3.
T3 is there to hold the light on until C3 discharges. If you don't have T3, the light would come on when there is noise, but would immediately go back off again when the noise finished.
If you put a diode, such as a 1N4148, in series with the base of T4, with the cathode connected to the base of T4, and the anode connected to C3 and R9, you will get a longer time. The diode will prevent base-emitter breakdown in T4.
You cannot get accurate times with a capacitor - resistor circuit like this. If you want 2 minutes, a circuit like this might give between 1.5 minutes and 2.5 minutes.
thank you for the answers diver.
so t3 is there to hold the bulb at stage until c3 is
finished discharging? right? i have a question.
where does the Ib of t4 came from? where in infact 1M is connected
to its base junction? its a large resistance for a currnt to pass through. does c3. helps in producing it?
in DC analysis all transistors are off .right? now in AC , the ac signal saturates it and c3 discharges and t4 is saturated as well as t3?right? if i will change the value of R9 and make it lower than 1M , where c3 is charge. does it make the time of charging short? and is it enough the current in r9 can produce Ib to saturate t4?
thanks for your response...
The 1M base resistor for transistor T4 has 9.4V across it so its current is 9.4uA. A BC548 transistor could have a current gain as low as 110 or as high as 800 so its collector current is 1mA to 7.5mA.