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Clap switch

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harrisonic

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Hello!! I just need help
i just wanna know how this circuit works?

especially the transistor..

Thnks for the help
 
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Looks like a microphone, an amp and a bistable switch.
Start by looking up microphone amp.
 
yea its a bistable switch..but i just wanna know how it will behave when the ac signal already enter from the mic to the 2nd and 3rd transistors?

tnks..
 
CLAP SWITCH
This circuit toggles the LEDs each time it detects a clap or tap or short whistle.
The second 10u is charged via the 5k6 and 33k and when a sound is detected, the negative excursion of the waveform takes the positive end of the 10u towards the 0v rail. The negative end of the 10u will actually go below 0v and this will pull the two 1N4148 diodes so the anode ends will have near to zero volts on them.
As the voltage drops, the transistor in the bi-stable circuit that is turned on, will have 0.6v on the base while the transistor that is turned off, will have zero volts on the base. As the anodes of the two signal diode are brought lower, the transistor that is turned on, will begin to turn off and the other transistor will begin to turn on via its 100u and 47k. As it begins to turn on, the transistor that was originally turned on will get less "turn-on" from its 100u and 47k and thus the two switch over very quickly. The collector of the third transistor can be taken to a buffer transistor to operate a relay or other device.
 
ahhh okie... but i just wanna ask what is the purpose of 1u F connected @ 2nd and 3rd transistor?

also the 33k connected @ diode?

tnks..
 
The 10u charges via the 5k6 and 33k when the first transistor is not receiving a signal. It will have about half-rail voltage across it.
When the transistor turns ON, the voltage on the collector will drop and this will take the positive end of the electro towards the 0v rail. The electro will have a small voltage on it and the negative end of the electro will drop below the 0v rail.
The second and third transistors will sit with one turned ON and the other turned OFF.
The voltage at the join of the two diodes will drop below 0v rail and it will turn OFF the transistor that is turned ON.
Now, both transistors are turned OFF. But the 100u's are charged to different voltages and the electro that was connected to the OFF-transistor has the least voltage across it. When both transistors are turned OFF, the electro with the lower voltage across it will start to charge with a higher current and it will turn on the base it is connected to.
This is how the transistors change states.
 
The circuit is too simple so a dog barking, a door closing or many other sounds will trigger it.
 
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