## How To Make Simple Clap Switch

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How To Make Simple Clap Switch

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## Description

How To Make Simple Clap Switch

Introduction

In this project, we are going to make a simple clap switch.This will help us to search some objects in the dark, when you are watching TV or lying on bed etc.

Circuit

In simple, NE555 timer  was operating in 3 different modes

• Astable
• Monostable
• Bistable

For getting the output in Monostable mode,The values of the resistor R and the capacitor C decides how long the output pulse should stay high and time can be calculated using the following equation.

Time (secs) = 1.1 * R1 * C1

Where,

R – Resistance in ohms

So, when we give a low input to the trigger pin ( pin 2). We can get the high output in the output pin (pin 3). According to the values of the resistor and capacitor the width of the output pulse will be increased or decreased.

For eg.,

R = 4.5k ohms

C = 1000uf

Time (secs) = 1.1 X (4.5 X 1000) X (1000 X 10 ^ -6)

= 4.95 secs

Means, the output pulse will stay high for 4.95 seconds. After 4.95 the pulse will come back to low state.

Working

•  All we have to do is to provide a low signal to the 555s TRIGGER pin (pin 2) to switch on the light for a desired time.
• Due to this, the voltage across the collector-emitter of T1 is zero. So that no voltage across base-emitter of T2 that is the reason why the T2 is in off state by default.
• When T2 is turned off the voltage across collector-emitter of T2 was almost equal to the source voltage. As discussed before trigger requires low signal to on the 555.
• When the sound was detected by the mic the resistance of the mic will become high. Due to this, there is no current flow to the base of T1 which makes the T1 to turn OFF.
• Due to this, the voltage across the collector-emitter of T1 was high, this makes the T2 to turn ON.
• Once the T2 was in ON, the voltage across collector-emitter of T2 will be low. This LOW was sensed by the trigger pin (pin 2), which makes the 555 to produce the monostable output.
• When we power up the circuit, the current will flow from source to the (+ve) mic and leaves at (-ve). Due to the low resistance of the mic the sufficient amount of current will flow to the base of T1. That current helps the T1 to turn on.