I am new to working on 555 timer. I am running it in monostable mode. I am triggering it using a button on trigger pin. It works just fine. What I want is the LED connected to the out pin 3 to switch off when I press the trigger button again. If I don't press the button, LED will switch off after the specified time duration by the RC circuit. Is it possible? Thanks in advance.

 

It is only possible with a single button if you add two external chips. 74xx393 counter and a hex inverter. And one diode.

 

Same button or different button?

Same button.

 

Does it have to be a 555? I think it might be easier with a 4013.

 

Does it have to be a 555? I think it might be easier with a 4013.

please share circuit or something.

 

The two inverters should be two channels from one 74xx series hex inverter or a simple NPN inverter would probably work too.
The counter is cleared by a pulse from the capacitor (back to 0000) each time the 555 output goes low.
The first press of the button causes 2QA to go high (inverter triggers 555 by bringing its pin 2 low).
the second press turns on 2QB and "instantly" charges the 555's timing cap and brings the 555 output low (and resetting the 74393)

 

please share circuit or something.

4.7k/47k/100n is a switch debounce circuit - any switch debounce circuit will work.
Once triggered the circuit automatically resets after 0.7RC time period.
If you don't need to re-trigger it soon (within 3RC) after it has timed out then the diode can be omitted.
Use a HEF4013. They have schmitt triggers on the clock input, and CD4013 types don't. The schmitt trigger makes it much less sensitive to switch bounce.

 

View attachment 227549
4.7k/47k/100n is a switch debounce circuit - any switch debounce circuit will work.
Once triggered the circuit automatically resets after 0.7RC time period.
If you don't need to re-trigger it soon (within 3RC) after it has timed out then the diode can be omitted.
Use a HEF4013. They have schmitt triggers on the clock input, and CD4013 types don't. The schmitt trigger makes it much less sensitive to switch bounce.

it seems like R will be pulled high as soon as Q goes high. Once that is the state, R will stay high because the input will be disabled because R is high.

 

No. After the clock pulse, Q will go high. C will charge through R. After approx 0.7RC the voltage on C which is connected the the pin called "R" which is "Reset" (not to confuse it with the resistor), will exceed the threshold, the flip-flop will be reset and Q will go low and remain low until the next clock pulse. C will be discharged to zero through R (or through D if it is used)
If another clock pulse occurs before C has charged to the threshold, Q will go low immediately, and remain low until the next clock pulse. C will be discharged back to zero through R (or through D if it is used)

 

It starts to get complicated from anything besides a simple timed ON action.
If you want a minimalist solution then go with a single 8-pin MCU solution.

 

This might be possible with a short push...long push, using a single button.

Connect the button to the trigger as usual, then to the reset pin thru a RC delay network.

 

Below is the LTspice simulation of a modified version of Ian0's circuit in post #7.
Notice that a second PB input (green trace) before the timeout sets the output to zero.

Instead of an input debounce circuit I used a delayed feedback to the D input.
That avoids any possible risetime issues with the CD4013 clock input and the need for it to be a Schmitt-trigger.

 

Neat! No increase in number of components either.
I found that the 4013s without schmitt triggers didn't like the slow risetimes of the debounce circuits.
I note that you used a 5V supply. If you change to a 74HC74 the logic has to be reversed as the Set and Reset inputs are active-low.

 

Wally #12 - I was editing your *excellent* debounce circuit into the circuit in post #7, but you beat me to it.

ak

 

Don't forget the OP needs to drive a load.

 

Don't forget the OP needs to drive a load.

It's just an LED. Even a 4013 will manage that! (even at 5V)

 

Is it possible?

It is possible using a 555. Here's a version that requires 2 additional transistors.
Keep in mind it only works with brief pushes of the switch.

 

I found that the 4013s without schmitt triggers didn't like the slow risetimes of the debounce circuits.

Yes, as you can see from the CD4013 input clock maximum rise and fall time requirements, the input debounce circuit greatly exceeds those values:

 

Yes, as you can see from the CD4013 input clock maximum rise and fall time requirements, the input debounce circuit greatly exceeds those values:

View attachment 227582

Yet if you look at the HEF4013B it says
https://assets.nexperia.com/documents/data-sheet/HEF4013B.pdf
"The clock input’s Schmitt-trigger action makes the circuit highly tolerant of slower clock rise and fall times."
Why are all 4013 s not the same?

 

Why are all 4013 s not the same?

That's a good question.
Guess you have to know that the HEF prefix means the IC may be slightly different from the ones with a CD or other prefix.
Personally, I think they should have also modified the last part of the name to indicate it is different from a standard 4013.