on off touch switch
Single Touch ON-OFF Switch Using 555 And 4017
Introduction
Here I am Describing a Touch Switch Which Is operated by Only One Touch Plate. This Touch Switch Is Made using One 555 Timer IC And One Counter IC 4017. The Circuit Is Required Also Some Discrete Component Like Resistors, Transistors, Capacitors, and Diodes.
When Touch The Metal (Touch Plate) For A Moment (About 1 Sec), The Load Is Switched On. Now If You Touch Again The Same Touch Plate Metal, The Load Become Switched Off. So This Is a Very Interesting Circuit And Very Sensitive. At First, When Switch On The Circuit, The Load Which Is Connected Through this is OFF.
In This Project, I Used Only One Touch Plate Which Is a Small Piece Of Metal To Make The Contact Between Circuit And the Human Finger.
Must Watch The Video Of This Project At the Next Paragraph
Working
The 555 Timer is connected as Monostable Mode in The Circuit Which gives one-shot output For a fixed time duration, And 4017 Counter Ic is connected as Flip Flop. In this connection of 4017 Counter IC, Output At Pin 2 is Become High And Remains in the same state until the Clock Pulse At Pin 14 is not changed. 4017 is a 5-State Counter IC, Here Only One Output Of Pin 2 Is Used.
At First, When the Power Supply Is Turn On, Pin 2 Of the 555 Timer gets A Small Negative Signal through the Transistor Q2 As we know For an output of 555 it is necessary to Trigger Pin 2 By Negative Signal (supply). So, in This Case, The output Of 555 At Pin 3 gives a High signal Because Simply The Negative Signal is passed Through the Q2 at pin2. Therefore pin 14 of IC 4017 gets a Positive signal by Resistor R4 and Does not get the Low (Negative) signal Then the output Of IC 4017 Remains Low. When The signal At Pin 2 is Low Then The Transistor Q1 does not Turn On and the relay does Not activate so the Load is Switched Off.
When you Touch With the Finger At Touch Plate Then A Signal is By the Body Give on the base of the Transistor Q2, And the transistor Detects Through Transistor This Signal goes To pin 2 Of IC 555 And This IC is Triggered. When the Trigger occurs, Output Pin, 3 Of 555 Gives a Positive (High) Signal To pin no 14 Of 4017 And as the signal changes the output of 4017 At pin2 is changed ie LOW to HIGH. As the Output of IC 4017 is High Then The Transistor Q1 is ON And Relay is Activated, So the Connected Load is Switched ON.
Resistor R1 Provides a Positive Signal At Pin 2 For Returning The Previous State Of 555 Timer. A capacitor Is Used for Setting to provide a bit of duration To remain in One State. Timer 555 is returned in its original State But Once The State is changed Of 4017 It Does Not Return In the Previous Mode Until the Signal Is Repeated Again At pin 14. This Is Done By 555.
So When touching The Plate, The State Of IC 555 is Changed and On Every Touch Give Changed To an Alternate Signal Pulse and Then The IC 4017 also gets Changed Pulses at Pin 14, and On Each changed Signal Pulses The Transistor Q1 Turns ON and OFF. So The Connected Load is ON at one Touch And OFF at the Second Touch.
Resistor R4 is Necessary For The Reset the IC 4017 If You Not Connect The R4 Or IF You Connect Very High Resistance Value Then Each Time Of Starting The Power Supply, Load May Automatically On At Circuit Switched On.
• For Touch Plate, Use Any Small Metal Plate. Adjust The Preset For Satisfied Operation Of the Circuit.
• Capacitor C1 is Optional, You May Not Connect When I was Testing Without C1, No Any Notable Effect I Got.
Also, Read Fm Transmitter circuit
Demo Video Of This Project
Components
IC
555-1
4017-1
Capacitor
10uf 25v-1
Resistor
100k-2
56k-1
1k-1
Preset
500k-1
Transistor
BC548-2
Relay 9v or 12v -1
Diode
1N4007-1
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This is work bad cd4017 is very Hot
Connect circuit properly ag given diagram, and check carefully. It works well.
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Dear Author !
I wonder if it is possible to replace the CD4017 by a CD4022.
I try to find out the pin allocation “translation”.
Thanks by advance for your answer.
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