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Simple Vibration Detector Circuit

Last Updated on November 17, 2024 by Admin Leave a Comment

Regardless of whether it's the audio across the street, the kitty purring softly, or perhaps a bump on the entrance, the detector explained in this simple vibration detector circuit does not skip a thing. Anytime this detects a noise or vibration, it gives off an ear-piercing sound.

The circuit draws on the application of an 8 ohm loudspeaker as microphone/ loudspeaker.

As the impulses made by this mic are extremely modest, these are amplified in A1 plus rectified. The, ensuing DC signal will then be balanced with a reference voltage in A2.

Every time a disturbance or vibration is indexed from the mic, the voltage within the inverting input of A2 (pin 6), soars instantly to around 4 V and after that gently decays to 0V.

The fall period varies according to the time constant R6/C3. The voltage along at the non-inverting input of A2 (pin 5) is maintained constant at 0.7 V by means of R3/R4.

As soon as the input with pin 6 increases over 0.7 V, the output of A2 (pin 7) promptly changes to -4 V, which in turn causes the squarewave oscillator A3 to trigger.

The frequency (tone) of the oscillator are adjustable by means of preset potentiometer P1. The oscillator output (pin 8) is applied to amplifier stage T1 which in turn runs the loudspeaker.

The oscillator is going to carry on and function nevertheless, so that C3 charges progressively and may maintain your output at pin of A2 negative.

As this is not the purpose of the circuit, the incoming signal must be interrupted somewhere in the chain. To do this, an FET, T2, is used as a switch. As soon as the output of the comparator becomes negative, D3 conducts, T2 is cut/off and the incoming signal is interrupted.

When C3 has discharged to the extent that the voltage across it drops to below 0.7 V, the output of A2 (pin 7) becomes positive, D3 is cut off and T2 conducts.

This should, however, not happen too rapidly, otherwise there is the risk that a false alarm may be given.

Therefore, the gate (drive input) of T2 is connected to earth' via capacitors C2 and C8. The consequent delay. ensures that the circuit is not reactivated before half a second after the loudspeaker has gone quiet.

The earth potential is fixed by the voltage divider R9/R10 and impedance converter A4, which derive a symmetrical supply of  4.5 V from the 9 V battery.

When T1 conducts, the supply voltage will drop a little because a battery cannot deliver energy as well as a mains power supply. It can therefore happen that the output signal of A3 is superimposed on the supply voltage.

This undesired feed-back should be prevented by C5 and C6. If inspite of these capacitors difficulties are encountered, it may be beneficial to increase the values' of R5, C2 and C8 by trial and error. If that fails to improve matters, increase the value of capacitors for the proposed simple vibration detector circuit.

vibration alarm circuit

 

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About Admin

Hey friends, Thanks a bunch for stopping by this site! I am an engineer with a Bachelor of Engineering in Electronics and Telecommunication. One of my passions is gathering information from all sorts of electronics books and tutorials. I then take that information and compile it into a language that is super easy to understand. My goal is to make those complex electronics circuit concepts and technical terms much more accessible for all the new and budding electronics engineers out there. I can also design customized circuit diagrams as required by the users.
If you have any questions related to this field, please do not hesitate to drop a comment! I am always here and ready to help you out with any queries you might have. I cannot wait to hear from you!

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