PIR Light Switch(or Any AC Device) Without Microcontroller

1,038

28

3

Posted in TechnologyElectronics

Introduction: PIR Light Switch(or Any AC Device) Without Microcontroller

This is a simple circuit for activating a relay connected to an AC (or DC for that matter) Device like a bulb,

I am going to assume you know how to use a relay and basic electrical wiring (google is your friend)

The circuit is designed for the use with 12v DC supply but can be easily converted into 5v supply design by using a 5v relay and shorting out the Vin and Vout of the voltage regulator(7805), make sure to test out the design in a breadboard before moving on to more permanent set up like a PCB

Step 1: Materials Required

Testing

1.Breadboard

2.wires

3.Multimeter

4.Led

Final Assembly

PCB Clad/Perforated Board(This tutorial will be based on the PCB )

12v/5v Relay

Resistors (I used 1k,But make sure to keep a few standard values )

BC547 transistor

7805 voltage regulator(when using 12v supply)

Diode

Screw terminals,2 point,3 point

Female Header

PIR Sensor(Obviously)

Step 2: Schamatic

I have attached images of the schematic and the link to the schematic and PCB board file (EAGLE )

https://github.com/Xavier-John/Pir-Switch

Test the circuit on a breadboard with 5v dc instead of 12v with an led Instead of the relay once everything is ready

Step 3: PCB Etching and Soldering

U can Etch the PCB using the toner transfer method (u should check out other Instructables or google it for more info) and then solder it to the pcb and test it with 5v supply or 12 v supply (make sure to include a 7805v regulator)

Step 4: Testing and Calibration

Look up this link on calibrating the PIR sensor

https://learn.adafruit.com/pir-passive-infrared-pr...

After soldering and assembly get a light bulb and holder and move around in front of the sensor and see if it lights up, some calibration may be required on sensitivity and delay

And thats it

Make sure to mention any errors or improvements in the comments..

Thank you

Share

    Recommendations

    • Make it Move Contest

      Make it Move Contest
    • Woodworking Contest

      Woodworking Contest
    • Microcontroller Contest

      Microcontroller Contest
    user

    We have a be nice policy.
    Please be positive and constructive.

    Tips

    Questions

    3 Comments

    If you elect to go the +5 VDC route, why not just remove it and both capacitors to save parts count? Just a thought.

    2 replies

    Actually the capacitors are not required even for the 12v design , and completely unnecessary for the 5v design , I included it in the PCB design so if I need the capacitors in case the power supply doesn't output stable DC voltage.Thanks for ur comment by the way

    You had the right idea for C1 and C2. 7805 manufacturers recommend a small MLCC cap (usually 0.33uF) as close as possible to the input pin when the regulator is remote from the power supply filter capacitor. In your design that recommendation certainly applies. A small MLCC cap (usually 0.1uF) as close as possible to the output pin is recommended in all cases to prevent ringing and oscillation on the regulator output. I would definitely use both of those capacitors in your layout when a 7805 is installed. You were wise to include them.