Introduction: Clap Switch ( 40 Claps in 5 Second)
Clap Switch has the ability to turn ON/OFF any electrical component by connecting the output of the circuit to a relay switch. Here we are going to make a clap switch with few components with very well explanations. Compared to all other clap switches, Here we use a single transistor and few resistor along with ic555 and ic4017 counter. since this circuit is accurate, I could switch ON/OFF the LED 8 times a second. The 8 times is only limited by my capacity of clapping in a second. This Instructables is mainly focussing on beginners
let's get started...
Step 1: Materials Required
4. PCB Universal Breadboard
6. Capacitors ( 10uf )
7. Power source 5v
9. Resistors ( 100 ohms, 1K )
10. Resistor pot (5k)
11. Electret microphone
Step 2: Circuit Diagram and Explanation
The entire circuit can be divided into three parts as shown in the figure
1. BC547 amplifier
2. IC555 monostable multivibrator
3. IC4017 counter
Let's get started
BC547 is a NPN transistor. Here 0.7v is needed at the base of the transistor to switch ON.That means if the base to emitter voltage is less than 0.7 V, no current will flow through the collector to emitter of the transistor and it will remain off.
When a voltage greater than 0.7V is applied to the base, current will flow from collector to emitter, switching on the transistor.
Here we use a voltage divider circuit consisting of a mic and a resistor. the input to the base is taken from between mic and resistor.
The resistor pot 5k is adjusted in such a way that the input at the base of the transistor is just below 0.7v ( it can be done by placing a LED at the collector of transistor)
when a sound detected by the mic, a voltage will be flowing through the mic which results in increasing the voltage at base more than 0.7v. which will switch ON the transistor and the collector will be connected to ground.Thus the voltage at collector will be zero.
since the clapping sound has a short time period, the collector voltage will be zero for only a few period of time.
thus it is used as a trigger input for ic555 monostable multivibrator
IC555 MONOSTABLE MULTIVIBRATOR
Refer the figure given above
in IC555 astable multivibrator, when a negative trigger is applied at the trigger input of ic555, The output will be high.But it will be HIGH only for a short time period depending upon the capacitor and resistor used in ic555
since we need it to be HIGH for an unlimited time, we used this output as a clock input for ic4017 counter
IC 4017 is a 16 pin out ic. which can count up to 10 states. IC 4017 mainly consists of two internal blocks
It has 3 inputs and 4 output. inputs are
1. clock input ------ connected to 14th pin of ic4017
2. reset---------------connected to 15th pin of ic4017
3. clock enable-----connected to 13th pin of ic4017
The output will be in the format of a 4bit binary code eg; 0000 or 1001 or 0111
The decoder converts the binary number to decimal number. Here, each output pin of ic4017 is referred to a decimal number. according to the input, the decoder decides which pin has to go HIGH.
Conversion of binary to decimal
0000 -----> 0
Pins of ic4017
1.) Pin 16
it is the positive power supply
2) pin 8
it is the ground
3) Pin 13
it is the Clock enabled pins to controls the clock.When it is “0” logic, the clock is enabled and the counter advances one count for each clock pulse. When “1” logic, the clock input is stopped, and the counter does nothing even when a clock pulse arrives.
4) Pin 14
It is the clock triggers one count.
5) Pin 15
It is the reset pin. Normally, it is “0”. When made “1”, the counter is reset to “0”.
6) Pins 1-7 and 9-11
They are the decoded output pins. The active count pin goes high and all others remain low
7) Pin 12
it is the Carry output, for the clock input of an additional counter or an external circuit that the count is complete.
IC4017 Circuit working
Here the pin 4 ( Q2 ) is connected to pin 15 ( reset ) and we took output from pin 3 (Q0)
1. Initial stage : Q0=1 Q1=0 Q2=0
2. First clock : Q0=0 Q1=1 Q2=0
3. second clock : Q0=0 Q1=0 Q2=1 ( temporary )
Since Q2 is connected to reset. The ic will reset and it will go to initial state. that is,
Q0=1 Q1=0 Q2=0
that's is for odd clock Q0 will be LOW
and for even clocks Q0 will be HIGH
here each clap act as a clock input
Hope you got how things are working :)
For any more information, please comment below
Step 3: Soldering and Circuit Completion
Step 4: Few Similar Project
Participated in the
LED Contest 2017