# How do I create a very long delay with Redstone?

The only things I've found so far are a cart method (which ceased working in 1.6), a dispenser method (utilizing the decay on dropped items, but this requires occasional feeding), and a binary counter (which won't work for a single input but rather a fluctuating one).

I want to have a circuit that when I apply an input signal, the output signal doesn't change for 2 to 5 minutes. When the input turns off, I again want a 2 to 5 minute delay.

I could accomplish this with (roughly) 240 to 600 redstone repeaters, but this is far in excess of being useful.

The only method I can think of, but would require a XOR gate (on the button-input and final output) to start and stop a clock attached to a binary counter. While I can make room for a device of this size, I was hoping for something simpler.

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 What do you want to attach it to? What other mechanisms/blocks are involved? What's your goal for the circuit? – Cyclops Aug 23 '11 at 22:33

Minecraftaddict's "Extreme Delay Redstone Timer" is the longest delay circuit I know of. (But I can see from the comments that you've already found it). However, directly using this circuit won't work with a light sensor output, since the output of a light sensor would be 10 minutes on and 10 minutes off.

The solution to this is to connect the output of your light sensor to a falling-edge and rising-edge monostable circuit that are OR'd together.

This is an image of a rising-edge trigger on the left, and a falling-edge trigger on the right. You would connect the A inputs together – to the output of your light trigger. When the A input is turned on, the Q output of the left circuit pulses. When the A output turns off, the Q output of the right circuit pulses. Because of this, you can connect the two Q output wires together (ORing them) to have an output that sends a signal whenever the light-sensor changes.

However, these circuits are designed to make a very short pulse. Because Minecraftaddict's delay circuit comes with a built-in pulse shortener, you will want to send a slightly longer signal to it. To do this, just add more repeaters to the edge triggers (add an extra repeater to when the current repeaters are). Leave the repeaters that should be on the first setting like that, but change the repeaters that should be on the second setting (according to the image) to the fourth setting.

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 Yep, that's exactly the video I had been looking for, and then found. :) – Draco18s Aug 23 '11 at 18:14

I would recommend using a slow automatic clock (See B') tied in to a chain of T flip flops, (if you have access to sticky pistons you can make them extremely compact, see the Z layouts).

In the above diagram A would take 2x long to cycle, B would take 4x long, c would take 8x long, D would take 16x long.

As you see it gets very long very fast, I doubt you would need more than 4 or 5 T flip flops to get a 2-5 min delay.

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 The TFF idea won't work, as I have a slow input signal (it's the output from a light sensor, so it toggles on every sunrise and off every sunset) and wanted to tie that to a door so the door would stay open for ~5 minutes. A TFF would require a clock-input, and the clock would have to be turned on and off based on the initial input signal. I found a solution though, digging through this one guy's youtube videos. youtube.com/watch?v=xsC2xoQi38I Extremely compact design that can be delayed exponentially through a linear increase in circuit size. – Draco18s Aug 23 '11 at 17:08 Sorry I did not understand the question. Glad you got help, you should post the details of your solution as an answer. There is nothing wrong with answering your own question on this site if you figured it out yourself. – Scott Chamberlain Aug 23 '11 at 17:23 I'd have posted my own solution, except I can't for about 5 more hours. :P Guy below, however, did have what I was looking for. – Draco18s Aug 23 '11 at 18:15

Use several clocks and an AND gate

a clock that takes 7 ticks and one that takes 8 ticks will out every 56 ticks, using only 5 repeaters instead of 15

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Minecart tracks can easily be extended to provide a very long delay, taking up (relatively) little space. The space they take up can easily be put anywhere you want too, which will avoid clogging up your circuit.

Wire up the output of this circuit to a flip flop, and have your input connected to a cart dispenser. When you hit toggle, a cart is released. 5 minutes later it will arrive and switch the flip flop. You can then return the carts to the dispenser.

Add more tracks as required

I'd recommend tight spirals of track to extend the time, which ensures they won't stray into a chunk that might get unloaded.

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You'd be surprised how much track you need for a 5 minute long loop! We've got a whole metro system that takes about 3 minutes for a complete loop and took several hundred track to build. (It's also going to get a track-switching upgrade here soon, as soon as I can work out the schematic with sticky pistons and assuming it's better than what we have now) – Draco18s Aug 23 '11 at 18:19
Couldn't you wire this to a counter? After 10 loops the counter fires, say? – Michael Aug 23 '11 at 19:28
Binary counter, yeah. Would work. – Draco18s Aug 23 '11 at 20:39
A mixture of methods is likely to be good. Make a binary counter out of minecarts! – fredley Aug 23 '11 at 20:41

for 5 minutes exactly you could have a dispenser drop an item onto a pressure plate, and the item would disappear after 5 minutes

I also have a video of what I use when I need long pulses on my youtube channel

You could modify this to use a system to activate the circuit several times sequentially to save space. I may do a video on that again some time.

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 +1 for a unique method of getting exactly 5 minutes – Fambida Sep 18 '11 at 20:26

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 Can you summarize what is in the video so that we know what it is even if the video is removed? – jeffreylin_ Nov 6 '12 at 21:02