I am trying to build a circuit which will have an IR sensor. The type of IR sensor is active high. I need to connect the output of the IR sensor to a circuit which will give low output when the sensor is not detecting any objects. If the sensor detects an object, the circuit should wait for a certain period of time, for example 6 seconds and the output should become high after the delay. Once the sensor stops detecting the object, the output should go low right away. Again if the sensor detects the object, same 6 seconds delay should be there and the output should become high. This cycle should repeat. How to build a circuit for this application using 555 timer?

The website of the sensor is given below. https://robocraze.com/products/eye-blink-sensor

The operating voltage of the circuit will be 5V.

Once the output signal goes from low to high after six seconds from the detecttion ofthe object(closing of the eye), it should stay high till the object is removed from the sensor(eye is opened again). Or else, once the output goes high, it should stay high for a longer period of time, preferably longer than 5 seconds.

The amplifier IC on my sensor is masked. So I couldn't determine whether it has open drain or open collector output.

The output signal will go to the positive terminal of a 5V buzzer if it is capable of driving it directly, or it will be connected to the base of the BC547 transistor via a 1KΩ resistor and the transistor will drive the buzzer.

I can build circuit for this application involving 555 timer or transistors. But I can't build the circuit using logic gates for the time being due to difficulty in components procurement.

I am sorry to not including the circuit diagram as I don't know how to create the circuit for this specific application. Thanks in advance.

  • \$\begingroup\$ Have a look at this question and the answer: electronics.stackexchange.com/questions/691321/… \$\endgroup\$
    – Saadat
    Dec 7, 2023 at 16:33
  • \$\begingroup\$ What do you want the circuit to do if the subject opens the eye before the 6 seconds has elapsed? Energize the buzzer immediately? Wait until the 6 seconds has passes and then buzz? Ignore that cycle and wait until the eye has blinked for the full 6 seconds? What if the subject blinks rapidly? Restart the timer with each blink, so that only long blinks are detected? \$\endgroup\$
    – AnalogKid
    Dec 7, 2023 at 18:54
  • \$\begingroup\$ The thing that you have told last is correct. When the subject keeps the eye open, the bizzer should NOT be energised and the timer should be in off state or idle state. When they close the eyes, the timer should start the delay timing and should wait till 6 seconds to pass. If 6 seconds are elapsed and still the subject don't open the eyes, the buzzer should be energized and should be on till they open eyes. If the subject opens the eyes before six seconds, the timer should reset and go to idle position again. So if the subject blinks rapidly, the timer should restart and stop continuously. \$\endgroup\$ Dec 7, 2023 at 19:44
  • \$\begingroup\$ I think the second schematic captures that. \$\endgroup\$
    – AnalogKid
    Dec 7, 2023 at 19:52

1 Answer 1


What you describe can be done with a 555, or two logic gates, or two transistors, but your question is missing some details.

What is the part number or website for the sensor?

What is the circuit operating voltage?

What does the output signal go to?

Does the output signal need to change from high to low very rapidly?

Does the sensor have an open-collector (or open-drain) output?

Please update your question with this information. Also, what is your skill set for assembling a small circuit? Which of the three options above do you prefer?

Here is a schematic from another thread that has the behavior you want. Now that I've seen your responses, I'll add a different schematic.

enter image description here

Here is a revised schematic based on the latest information.

enter image description here

When the signal from the sensor goes high, C1 begins charging up through R1. When the C1 voltage crosses 2/3 x Vcc (approx. 3.33V), the 555 output goes low. The bipolar (non-CMOS) 555 can sink 200 mA. Please post any data you have on the buzzer you want to use. D2 is included to suppress voltage spikes from an electromagnetic buzzer. C2 provides power supply decoupling for the 555.

C1 does not have to be rated for 50 V; that is the part in my design library. Anything 12 V or above is fine.

If your buzzer actually is a piezo beeper, you can delete D2.

When the sensor signal goes low, C1 is discharged through D1. This discharge current goes through the sensor output stage, which might limit the current to a relatively low value. If it takes too long to discharge C1, additional parts are needed to speed up the discharge time.

NOTE: The equation for the delay period t (in seconds):

t = 1.1 x R1 x C1

t = 1.1 x 270,000 x 0.000022 F (farads) = 6.5 seconds

  • \$\begingroup\$ Hello, I have updated my question. \$\endgroup\$ Dec 7, 2023 at 18:28

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