Timed pulses with flip-flops

[Shibe]

Hopefully this merits a thread.
Sometimes, you'll want to create a pulse of delay S (in seconds), and most people approach this by placing down 2S delay blocks in a loop (as a delay block creates a 0.5 second delay).

However, you can save space by realising that the addition of a flip-flop as the output to a repeating pulse (like 2 delay blocks cross-linked) will double the total delay: The flip flop will toggle on each cycle of the delay circuit. If you add another flip-flop, you will double the delay again, as this flip flop will toggle only after the previous one has, itself, cycled.

You can therefore achieve any delay with minimal blocks by taking S and halving it until you're left with a non-integer value or 2, we'll this number D. The number of times you halved S to reach D will be called F.
Create the loop of delay blocks using exactly D delay blocks. Then, attach a chain of F flip flops as output. The last flip-flop in the chain will take exactly S seconds to pulse.

In this image, the final flip-flop will take 8 seconds to pulse: the delay cycle takes 1s, which is doubled three times.

In this way, you can also see that the delay of a circuit like this will equal D*(1+F/2)

 

[Forgedrake]

and now I'm gonna go try to figure stuff to jam this idea in

 

[plusnine]

very handy condensation advice. nice work.

 

[ruiner333]

Would you mind clarifying your instructions for odd numbers? If I wanted 7, what exactly would be the setup? If I follow the instructions exactly. Halving 7 = 3.5(non integer) = D and you divided once = F. 3.5 delay blocks and 1 flip flop?

 

[Jaaskinal]

Would you mind clarifying your instructions for odd numbers? If I wanted 7, what exactly would be the setup? If I follow the instructions exactly. Halving 7 = 3.5(non integer) = D and you divided once = F. 3.5 delay blocks and 1 flip flop?

You kindof don't. The domain of the function is limited to counting numbers, and the range is 2^x.

 

[ruiner333]

Well, I imagine you could still use it to reduce the number of blocks. I haven't tested it but I assume it should work to use the flip flops to get to the nearest power of two <= your target value and then chain a few delay blocks after that to hit the desired value.

 

[Adonis0]

Would you mind clarifying your instructions for odd numbers? If I wanted 7, what exactly would be the setup? If I follow the instructions exactly. Halving 7 = 3.5(non integer) = D and you divided once = F. 3.5 delay blocks and 1 flip flop?

well, that'd be 8 blocks in total instead of the 14 to do purely with delays, so it still saves space, it's just more efficient on saving space for 2^x numbers.

Well, I imagine you could still use it to reduce the number of blocks. I haven't tested it but I assume it should work to use the flip flops to get to the nearest power of two <= your target value and then chain a few delay blocks after that to hit the desired value.

For this instance, that'd result in 4 blocks to get to 4s (Delay-Delay-Flip-Flip (or Delay-Flip-Flip-Flip)), and then another 6 delays for the 3s on top, resulting in 10 blocks in total.
I believe you'd want to test out for each method, but this particular method I believe would be more useful for numbers slightly above 2^x instead of slightly below, but it still saves some space.