4-bit asynchronous down counter

A 4-bit asynchronous down counter counts 15, 14, 13, …, 0 and then back to 15, decreasing by one on every clock pulse. Only the first flip-flop gets the external clock.

Principle

In a down count a flip-flop must toggle when the next lower bit changes from 0 to 1 (a "borrow"). With negative-edge triggered flip-flops, a 0→1 change of $Q$ is a 1→0 change of $\overline{Q}$. So each stage is clocked from the $\overline{Q}$ of the previous stage.

Circuit

Four negative-edge JK flip-flops, all with $J = K = 1$ (toggle mode). CLK → FF0; $\overline{Q_0}$ → FF1; $\overline{Q_1}$ → FF2; $\overline{Q_2}$ → FF3. The count is read from $Q_3Q_2Q_1Q_0$.

     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
    +-----+     +-----+     +-----+     +-----+

Working

1. Start with 0000. On the first falling edge FF0 toggles: $Q_0$ goes 0→1, so $\overline{Q_0}$ falls and FF1 toggles; $Q_1$ goes 0→1, so FF2 toggles; and then FF3 toggles. The count becomes 1111 (15).
2. Next edge: $Q_0$ goes 1→0, $\overline{Q_0}$ rises, so FF1 does not change: 1110 (14).
3. Next edge: $Q_0$ 0→1 toggles FF1 ($Q_1$ 1→0), which does not clock FF2: 1101 (13).
4. The count keeps falling by one per pulse: 12, 11, …, 1, 0, and then 15 again.

and continuing 0111 (7), 0110 (6), …, 0000 (0).

Timing diagram (falling-edge clock)

CLK  __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q3   ____|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|_______________
Q2   ____|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|_______________|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
Q1   ____|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|_______
Q0   ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___
Cnt  0    15  14  13  12  11  10  9   8   7   6   5   4

Each $Q$ changes when the $Q$ of the stage before it goes high (its $\overline{Q}$ falls). In practice each stage adds a delay $t_{pd}$, so the change from 0000 to 1111 takes $4\,t_{pd}$. The circuit also divides frequency: $Q_0 = f/2$, $Q_1 = f/4$, $Q_2 = f/8$, $Q_3 = f/16$.