I am still a student and so this question might sound too easy to be asked here, for which I apologize.

I want a circuit that consists of a single LED. When this LED recieves a 0V potential difference, it should turn off and when it recieves a 5V input, it should turn off. Although this sounds like a NOT gate, but there is a complication. I want the circuit to work like this. At the beginning, when no input (0V) is provided, the LED should remain OFF. Then a timed 5V input is provided for some time, for which the LED should remain OFF. After that when the input is 0V, the LED should turn ON.

I have searched the Internet for any solution but in vain. Please help me with this problem if you know any possible way to achieve the above mentioned work. Thanks for the attention.

Figure 1. Timing diagram added by @Transistor for the OP to edit. 'T' is the minimum required on-time.

               _________               ___               ________
Input:  ______|       >T|_____________| <T|_____________|      >T|_______
                         _____________                            _______   
Output: ________________|             |__________________________|
  • Output turns on if input pulse has been longer than T.
  • Output turns off on rising edge of input.
  • 1
    \$\begingroup\$ According to your question, the LED never turns on. Then you contradict this later on. BE CLEAR. \$\endgroup\$
    – Andy aka
    May 17, 2020 at 9:59
  • 1
    \$\begingroup\$ Draw a timing diagram. \$\endgroup\$
    – Transistor
    May 17, 2020 at 10:05
  • \$\begingroup\$ You need to present your requirement in an unambigious manner. A diagram is usually an effective way to do this. Q1: It SOUNDS as if your turn on condition is a low to high transition. Yes? Q2: It is not stated how long the LED should stay on and/or Q3: what turns it off again or ... Q4 Is there any limitation on the high time before a high/low transition enables the LED? | As stated: AND gate, inverter, diode, capacitor. Diode to cap. Cap to ANDIN1. Input to ANDIN2 via an inverter. AND drives LED \$\endgroup\$
    – Russell McMahon
    May 17, 2020 at 10:20
  • \$\begingroup\$ I've added a timing diagram for you to verify or edit. Your timing diagram should show what happens in all possible input cases so I've included a "working" case where the input pulse is long enough and a "non-working" case where the input pulse isn't long enough. I've also shown that the output turns off on every rising edge of the input pulse although I don't know if this is what you want. Please edit to verify or correct. \$\endgroup\$
    – Transistor
    May 17, 2020 at 11:44
  • \$\begingroup\$ "When this LED recieves a 0V potential difference, it should turn off and when it recieves a 5V input, it should turn off. " - This statement doesn't make any sense. \$\endgroup\$
    – Sadat Rafi
    May 17, 2020 at 20:08

2 Answers 2


What you have asked for is impossible. Because you said that you need to turn off the LED at the rising edge of the input pulse (when the pulse width is less than T).

First, you have to measure the pulse width. Then you have to decide what to do. So you can't change the state of LED before measuring the width.

If you want to turn off the LED after the input pulse is provided, then it is possible.

The easiest approach for you will be to use an Arduino. In your code start a counter at the rising edge of your input pulse and if the counter value exceeds a specific range then turn on the LED and stop the counter. And if the falling edge comes before your desired limit, then turn off the LED.

And if you want to go in a hard way then take a capacitor and a resistor. Charge the capacitor through your input pulse. With time the voltage across the capacitor will increase. Use analogue comparators to find out when the voltage reaches the desired level. But you will need an oscilloscope and the process will be tough. You will need a D-Flip flop to hold a state.


That sounds like a latched circuit, the latch trips when the input goes high, supplying power to the common of another relay powered when the input goes high, the led is on the normally closed contact

To begin everything is off, next a positive is supplied, both relays change to there normally open contacts, the first relay is a latch, the second just opens the contact to the LED, once the input goes low, the latch remains on, and the second relay then connects the LED to that latched supply

  • \$\begingroup\$ This is not a simple digital circuit. The input is pulse width, not 1/0. \$\endgroup\$ May 17, 2020 at 20:35
  • \$\begingroup\$ This answer was from well before he updated his question, at that point no mention of minumum pulse width was included, so this appeared to be a value answer \$\endgroup\$
    – Reroute
    May 17, 2020 at 20:39

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