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i am very new to electrical engineering and designing circuits and there is a particular project that i am working on that involves using a switch and a 555IC timer.

So basically my desired output logic is as follows (Note: I am using 2 different LED, green and red and an infrared sensor that acts as a switch)

1) Using an infrared switch, when a object is nearby, it closes the switch which starts the 555 timer. (Green LED lights up)

2) 555 timer counts down 5 seconds which triggers the 2nd 555 timer which then causes the red LED to light up.

3) Green LED turns off and 2nd 555 timer counts down 10 seconds and then RED LED switches off and the circuit resets. (where both green and red LEDs are off)

(For this circuit diagram please ignore Timer #1 and #2)

enter image description here

However, my circuit only works if the "start" is closed then opened immediately (i.e i close the switch and then open again). If i allow the switch to be permanently shorted, the green LED will be lighted and it does not trigger the 555 timer.

Hence, I am wondering if there is a certain kind of modification i can make to the infrared sensor that acts like a "button" that can close the switch and then open it immediately to trigger the 555 timer even if the object is still near the infrared sensor.

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    \$\begingroup\$ To be honest, IMO, using a small microcontroller instead of 555 hacking is both more relevant and more rewarding. \$\endgroup\$
    – crasic
    Nov 23, 2019 at 4:38

1 Answer 1

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I think you are asking what can be added to the switch input (actually your IR sensor) to make it act like a single pulse trigger and not a continuously held switch. To do this you can just use the same method you are already using between your existing 555's, use a capacitive input, (see below).

You may need to adjust the R1,C1,C2 values a bit to get the timing just right, and the original 33k pull-up might also need to be higher. Ultimately you could even include a buffer or one shot in-line with the first 555 to give you a better input pulse shape. If you look into 555 applications there should even be a circuit to use a 555 as a one-shot.

As another alternative you could use a tri-state buffer gate and enable it with an output of one of your 555's. That arrangement would temporarily disable the input signal and create a single pulse, (see second circuit).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ thanks for the answer! However i still dont quite understand how does the capacitive input act as a single pulse input. Wont the 555timer's trigger be permanently connected to ground and thus be switched on? \$\endgroup\$
    – Axois
    Nov 23, 2019 at 10:55
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    \$\begingroup\$ @Axois - In the first circuit closing SW1 quickly discharges C1, C2 then sees a negative going signal, this puts a low going spike on the 33K, that triggers the first 555. DC does not couple through a capacitor so the 555 input charges up again through the 33k, (even if SW1 is being held). When SW1 is released C1 recharges through R1. As long as R1 is not a low value the charge up should be slow and there would only be a very small positive pulse transferred to the 555. Note that if R1 were a low value another diode could be put across the 33k to prevent an over voltage condition. \$\endgroup\$
    – Nedd
    Nov 23, 2019 at 18:27
  • \$\begingroup\$ forgive me for asking this, but if i am not wrong, the 555 timer will be triggered when the voltage at the trigger pin is low. Hence by connecting a diode and grounding the trigger pin, wont the potential there be grounded and thus, it will be on? I am quite confused over the diode part of the circuit, or am i understanding something wrong here. \$\endgroup\$
    – Axois
    Nov 24, 2019 at 2:25
  • \$\begingroup\$ @Axois - The diode in the circuit has the anode on ground and cathode on the signal side, so no current passes in the normal state. The diode will only conduct if the input line dips below ground, which could briefly happen due to the capacitive coupling. Review the characteristics of a diode. \$\endgroup\$
    – Nedd
    Nov 24, 2019 at 4:01
  • \$\begingroup\$ @Axois - If my answer above is satisfactory you can tick the Accept (grey check mark) to label the original question as answered. \$\endgroup\$
    – Nedd
    Nov 24, 2019 at 4:14

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