Friday, March 29, 2024

Infrared Sensor Based Power Saver

--Fayaz Hassan He is a manager at Visakhapatnam Steel Plant, Visakhapatnam, Andhra Pradesh. His interest includes MCU projects, mechatronics and robotics

Quite often you forget to switch off the light or fan when going out of the room. The simple infrared sensor based power saver circuit presented here will automatically switch off electrical appliances like lights or fans as you vacate a room, after a predetermined time period. It will also switch on the light when you enter the room again. This will reduce unnecessary power consumption. Fig. 1 shows the pyroelectric/passive infrared (PIR) motion sensor used in the circuit.

Fig. 1: PIR motion-sensor module
Fig. 1: PIR motion-sensor module
Fig. 2: Circuit diagram of the PIR sensor based power saver
Fig. 2: Circuit diagram of the infrared sensor based power saver

Infrared sensor based power saver circuit

D7Z_PartsThe circuit diagram of the infrared sensor based power saver is shown in Fig. 2. It is built around bridge rectifier DB107 (BR1), PIR motion sensor connected across connector CON2, timer NE555 (IC1), two 1N4007 rectifier diodes (D1 and D2) and a few other components.

The circuit uses a PIR sensor, which detects the presence of people through change in the infrared radiation from the room when people enter or leave the room. The PIR sensor outputs around 3.3V high signal whenever it detects radiation change in front of it.

IC1, resistor R3, potmeter VR1 and capacitor C3 are used as a timer here to convert small time span of PIR signal to a long delay. Output of IC1 at pin 3 drives transistor T2, which, in turn, controls relay RL1. Electrical loads like lights or fans are controlled through this relay.

EF9_Test230V AC mains power is connected across connector CON1. It is stepped down to 9V through transformer X1, rectified by bridge rectifier BR1 and filtered by capacitor C1. Thus we get around 9V DC at test point TP1. This 9V DC voltage is used as power supply for the circuit.

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Circuit operation

When the circuit is first switched on, capacitor C3 charges through potmeter VR1 and resistor R3. During this time, voltage at pins 2 and 6 of IC1 is less than two-thirds of its supply voltage, and so output pin 3 goes high. This energises the relay through transistor T2, and the appliance is switched on.

When capacitor C3 charges above two-thirds of the supply voltage, IC1’s output pin 3 goes low and de-energises the relay and switches off the appliance after some delay that can be adjusted through potmeter VR1.

Whenever motion is detected by the PIR, its output pin goes high (around 3.3V) for a while depending on the setting on the PIR. The high signal from the PIR is fed to the base of transistor T1, which, in turn, discharges capacitor C3 through resistor R4. When the capacitor’s charge (voltage) reaches less than two-thirds of the power supply, output pin 3 of IC1 goes high again (initial stage) and load is switched on.

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When the load is switched off, LED2 glows. This indicates that the circuit is under power-save mode.

28 COMMENTS

  1. Hello sir….
    I am Divyanshu
    Sir i am trying to make a circuit of Infrared Sensor Based Power Saver.I connected all the components as per your circuit diagram but my circuit is not work. Then I check the voltage of TP1 . only 3.5 volts at TP1
    Sir you tell what is wrong in my circuit plzzz

    • Your problem is not clear whether the components are heated up excessively while measuring the voltage at TP1. If yes, there is a possible short circuit.
      First make sure that transformer output is about 9V. Remove IC1 and T1 from circuit and check the voltage at TP1 again, it should be around 9V as listed in the test point table. Also check voltage between pin 3 and 4 of BR1 which should be about 9V.
      If not, double check BR1 connections or replace it with a fresh one. The circuit is simple and working perfectly. It will work if all components are connected properly in the circuit as shown in Fig.2.

  2. I made this project.It is in running condition but i found one major fault in this circuit. 220k ohm resistor causes too much voltage drop . I removed potentiometer of 1M ohm and 220k ohm resistor and and i placed 2.2k ohm in it.

    • @ Muhammad farhan. This is the reply from author Fayaz Hassan. I felt happy to know that somebody tried the circuit and it is working fine.
      The main purpose of the 1M and 220K resistance is to slowly charge the capacitor C3 (1000uF), so that it increases time delay of the PIR which has limited triggering time. In case, the capacitor has more internal leakage than the charge input through the 1M and 220K, then the capacitor C3 will never get charged and the output of IC1 (555) will never change. so use new and good quality capacitor with atleast 25V rating. By changing the charging resistance, only relay hold time will vary else the circuit will function normally.

  3. Sir ,
    Can this project automatically switch ON or OFF light and fan .
    Can This project automatically work if two man or three man enter but one man out in the room??
    Pls reply this question .
    If it is not work then how solve this problem.
    Its my CLG project.

    • This is the reply from the author. Yes, it works. It does not count the persons. Actually, the PIR sensor triggers whenever a person is moved in front of it. (of course, within its range. refer datasheet for full details) if the person is not moved for more time, the relay will be switched OFF.
      It means the relay may be switched OFF if the person is sleeping (if no movement is observed for a long time). So the time delay is adjusted with 1M variable resistance. Use good quality and fresh capacitor C3 (1000uF) to have low leakage and high delay time. Also please read 2nd paragraph of Construction and Testing section.

  4. Sir,
    please give me a block diagram of infrared sensor based power saver without using microcontroller. In the circuit diagram we are not use the microcontroller so please give me some idea about block diagram.

  5. I have prepared this project for our office corridor. while i was trying to check the voltage at Test point 1 (TP-1), multimeter was showing descending volt up to 0 volts instead of 9 volts. the transformer which I have used was showing 10 volts. and the DB107 rectifier showing 13 volts at its terminal. te voltage after filter capacitor is -6.5 volt. for any short circuit probability, i have brake the further circuit but it still showing the same result. please help me.

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