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So, I have a 555 Timer in Monostable Mode, like below. I used a 47uF cap and a 47K resistor for the C and R. I press the trigger switch button and let go, and the output goes high for roughly 2.5 seconds, and then goes low. Just as would be expected.

Is there a way to make it so that when I press and hold down the trigger switch, but do not release it, that it outputs 2.5 seconds, and then goes low?

And if I let go of the trigger button and press it again, it starts over again?

enter image description here

Image from https://www.allaboutcircuits.com/tools/555-timer-monostable-circuit/

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    \$\begingroup\$ What you are looking for is something like the 74121. The 74121 includes a Schmitt-triggered input (which tolerates signals as slow as 1 V/s) and it completely ignores the input, once triggered, so the output pulse takes place regardless of whether or not you hold down the switch input. But they are hard to find, now. Kind of like hen's teeth. A Schmitt input monostable will likely require a capacitive input from the switch (edge triggered) or else two 555's. \$\endgroup\$ Commented Mar 23 at 13:20
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    \$\begingroup\$ Like millions of designers before you, you have discovered that the 555 is not a true monostable. The reason is that there is no positive feedback from the output to the inputs. It is this feedback that effectively disables the input once the circuit has been triggered. It is what makes the output completely independent of any input actions, including just holding the input in the trigger state, once it has been triggered. The solution, as in the answer, is to differentiate the input signal, converting a level change into a short pulse. \$\endgroup\$
    – AnalogKid
    Commented Mar 23 at 14:40

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Try something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

That probably should be close to what you want, I think, without using two 555s.

Here's a simulated output:

enter image description here

I used a couple of button presses (red) during the first output pulse. You can see that it doesn't re-trigger and just holds the correct output time (green.)

In the second period starting at \$5\:\text{s}\$ there is a long-press used that exceeds the output period. But as you can see the correct output time is still maintained.

It's just a simulation. It's not proof. And I didn't think too long about it. So I may have missed an important detail, too. But this is about where I might go with your circuit as a first step in the right direction.

(Note: I didn't actually use a clean switch in the simulation. I used a very bouncy switch that was set to bounce randomly for about \$25\:\text{ms}\$, which is probably even worse and longer than reality. So I did try to make this closer to a worst-case switch input.)

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    \$\begingroup\$ I just did the same on a breadboard and that does the trick. Thanks! \$\endgroup\$ Commented Mar 23 at 14:34
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    \$\begingroup\$ @ErikVincent Good to hear! Best wishes. \$\endgroup\$ Commented Mar 23 at 14:56

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