IR Beam Alarm. Sensor Circuit.2

This is a highly sensitive broken beam alarm active both day and night. It uses invisible Infrared rays to inhibit the working of alarm generator. When the IR beam breaks, by a moving person, alarm will be activated. Ideal circuit as door protector.

1. IR LED is used to produce continuous IR rays. The IR rays are invisible to human eye because the IR LED emits IR rays in the 900 nanometer range which is beyond our spectral response. IR LED can handle up to 100 milli ampere current and when current increases, its output rage also increases.

Long lead is Anode ( + )

2.Photo transistor L14F1 is used as the IR Receiver. It is a Darlington NPN transistor with base exposed and present behind a glass lens. So light exactly focus on the base semiconductor and the photons in the light gives bias and it conducts. L14F1 also has a preamplifier stage to increase sensitivity. It cost around Rs.60.

See a Tab in the Rim. The pin close to it is Emitter. Its opposite pin is Collector and the middle one Base. Do not connect the base pin. Connect only Collector and Emitter.

3. IR LED and Phototransistor are fixed on the door frame face to face with exact line-o-sight. So that T1 will conduct. When T1 conducts base of T2 will be grounded and it remains off.

4. When the IR beam breaks by a crossing person, T1 urns off and T2 conducts .This activates the buzzer.

5. Adjust VR1 to set the base bias of T2. Adjust it till the buzzer stops in IR beam

6. Range of the circuit is 1 meter with proper alignment of IR LED and the Phototransistor.

7. Enclose both IR LED and Photo transistor in Black tubes to increase sensitivity and to avoid sunlight falling on Phototransistor

Trick to see Infrared

Just focus the mobile camera or digital camera on the active IR LED. A pink light will be available. It is infrared ray. We can confirm its working

4 thoughts on “IR Beam Alarm. Sensor Circuit.2

    1. Dear Venkatesh
      L14F1 Phototransistor has 3 pins. Collector, Base and Emitter. The base pin is rarely used in and is for feeding signals. You can leave it as un collected.

  1. One problem with the above circuit is that 10ohm resistor in series with the IR LED is going to limit the current to around 750ma! For an IR LED that can only handle a max of 100ma, that means POOF — blown LED. I would recommend a value of 78 ohms 2Watt, max. [Perhaps that’s a typo and a 100 ohm resistor was intended]

    An IR LED has a typical forward voltage of 1.5V so subtract that from 9V and that is the voltage across the current limiting resistor: 9-1.5 = 7.5V

    Then, divide that by 100ma to get the resistance value: 7.5V/100ma = 75ohms.

    At 10 ohms, the current through the LED is around: 7.5V/10 = 750ma. For a 10 ohm resistor to limit the current to 100ma, the IR LEDs forward voltage would need to be: 9V – 10ohm*100ma = 8Volts!!

    If you used 5 IR LEDs in series in place of the one IR LED, you might get away with this, but I wouldn’t recommend it.

    1. Thanks for your suggestions and detailed explanation about the LED series resistor. Actually it is 100 Ohms resistor, but the diagram, one zero is missing. I will correct the diagram and post again. I appreciate your comments, and these kind of comments are more useful because people get more ideas in circuit design. Keep this enthusiasm and write more about the circuits.

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