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I don't understand how this circuit works. Its a clap switch circuit with transistors. I would appreciate if someone could explain how the current flows through the circuit and which transistor turns on or amplifies the current to turn on the LED at the end when the piezo detects sound. What i mean by this is the path of the current when the piezo detects sound.

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    \$\begingroup\$ Use LTSpice or something similar and simulate it. \$\endgroup\$
    – GER_Moki
    Commented Nov 9, 2014 at 18:56
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    \$\begingroup\$ Put some reference designators in your schematic so that we can talk about it. \$\endgroup\$
    – Dave Tweed
    Commented Nov 10, 2014 at 0:56
  • \$\begingroup\$ mkeith: But doesn't the current flow out from the emitter of the first transistor (BC547), so how does the second transistor(BC557) get the current to its base? \$\endgroup\$
    – Freakup2
    Commented Nov 17, 2014 at 1:40
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    \$\begingroup\$ Where does the emitter current come from? It's not just generated out of thin air. Follow the path, use KCL. \$\endgroup\$
    – Shredder
    Commented Feb 17, 2023 at 7:56
  • \$\begingroup\$ @Freakup2 That second transistor is a PNP type...If the first BJT is conduction, it is pulling base current from the PNP activating it. \$\endgroup\$
    – evildemonic
    Commented Jul 20, 2023 at 13:55

2 Answers 2

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I am not really sure of this but here is my take. This is an analog circuit. The piezo resonates in response to sound impulse, and the AC current goes through the base of the first transistor (BC547). Then the collector current (amplified by beta of transistor) goes through the base of the second transistor (BC557). Then there is ANOTHER beta and the collector current mostly goes trough the base of the last transistor (BC547). One more beta. So that is three betas. And the last transistor makes the white LED turn on. I imagine with all those betas, it is going rail-to-rail at the end, or would be without feedback

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    \$\begingroup\$ The second transistor seems to give negative feedback to the first, though. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 20, 2023 at 17:09
  • \$\begingroup\$ @user20574 - Maybe I am being obtuse, but the 2nd transistor / 3rd transistor looks to be POSITIVE feedback. \$\endgroup\$
    – Dwayne Reid
    Commented Aug 28 at 0:36
  • \$\begingroup\$ @DwayneReid As the first transistor turns on, the second transistor turns on, raising the first transistor's emitter voltage, decreasing its base-to-emitter voltage, turning it off. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Aug 28 at 15:57
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Just added a load capacitor of 10 uF on output.

Here is the behavior for an "pulsed" input of 100 mV (minimum).

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