Comb vs seq; asynchronous mod-12 counter

Combinational vs sequential circuits

Asynchronous mod-12 counter

A mod-12 ripple counter counts 0 to 11 and returns to 0 on the 12th clock pulse. Four negative-edge JK flip-flops ($2^4 = 16 \ge 12$) are used with $J = K = 1$. CLK drives FF0, and $Q_0$, $Q_1$, $Q_2$ clock FF1, FF2, FF3.

The counter must be cleared at 12 = 1100. It is the first count with $Q_3 = Q_2 = 1$, so a 2-input NAND of $Q_3$ and $Q_2$ drives the active-low CLEAR of all flip-flops.

     J=K=1       J=K=1       J=K=1       J=K=1
    +-----+     +-----+     +-----+     +-----+
CLK-o FF0 |Q0--o FF1 |Q1--o FF2 |Q2--o FF3 |Q3
    +--o--+     +--o--+     +--o--+     +--o--+
      CLR         CLR         CLR         CLR

Q3  ---|      |
       | NAND |o--> CLR of all FFs (active low)
Q2  ---|      |

Operation: FF0 toggles on every falling clock edge; each later flip-flop toggles when the previous Q falls 1→0, so the count goes 0000, 0001, …, 1011. At the 12th pulse the counter momentarily reaches 1100, the NAND output goes low and all flip-flops clear to 0000.

Timing diagram

CLK  __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q3   ________________________________|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|___
Q2   ________________|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|___________________
Q1   ________|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|___
Q0   ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___
Cnt  0    1   2   3   4   5   6   7   8   9   10  11  0

$Q_3$ is high for counts 8 to 11, so $f_{Q_3} = f_{clk}/12$.