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So my circuit is very simple: It's a lever, with an on or off state. I have a set of pistons with a small delay using repeaters, but I need these pistons to close immediately after the circuit goes off, despite being on a delay when they extend.

The delay of the circuit is a fully delayed repeater plus another repeater without extra delay (5 ticks?)

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    How much do you consider a "small delay"? Are we talking 1 second, 5 seconds, 10 seconds? Please edit your post to provide this information. Thanks! – ExpertCoder14 May 29 at 22:27
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    It would help to show an example or picture of what you have done already. Also adding what this is for will result in better answers. – Jason_ May 29 at 22:28
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    I don't understand the question. You put repeaters there and then you wonder why there is a delay? Of course there is, that's the main purpose of repeaters! – Fabian Röling May 30 at 10:04
  • @FabianRöling You completely missed the point of my question. I was trying to make a signal that has a delay when turning on, but immediately turns off. ExpertCoder14 answered my question bellow – Inevitable Entropy May 31 at 11:11
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An inverted pulse extender may be what you're looking for. An inverted pulse extender

Solution Notes

  • Adjust the repeater delay to gain the appropriate delay.
  • The redstone torches will cause a two-tick delay (0.2 secs) when the lever is turned on This delay is local to the inversions and cannot be removed without overcomplicating the circuit.
  • If you require an output that is normally on, simply remove the redstone torch next to the lamp and take your output from the block below it.

How it Works

  1. A pulse extender extends out the pulse for a specific length of time.
  2. When the input is on, the output turns on immediately.
  3. When the input is off, the output takes a little time to turn off.
  4. You need to invert the inputs and outputs, so that turning the lever ON will start the timer, and turning it OFF will reset it.
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