BCD synchronous counter using T flip-flops

A BCD (decade) synchronous counter counts 0000 to 1001 and returns to 0000 on the 10th clock. All T flip-flops share the clock. A T flip-flop toggles when $T = 1$ and holds when $T = 0$, so $T = Q \oplus Q^+$.

Truth (excitation) table

4 flip-flops are needed ($2^4 = 16 \ge 10$); states 1010–1111 are don't cares.

K-map simplification

Similarly $T_2$ is 1 for states 0011 and 0111, giving $T_2 = Q_1Q_0$, and $T_0 = 1$.

$$\begin{aligned} &T_3 = Q_3Q_0 + Q_2Q_1Q_0 \\ &T_2 = Q_1Q_0 \\ &T_1 = \overline{Q_3}Q_0 \\ &T_0 = 1 \end{aligned}$$

Circuit diagram

   +-----+       +-----+       +-----+       +-----+
T0>|T   Q|-Q0 T1>|T   Q|-Q1 T2>|T   Q|-Q2 T3>|T   Q|-Q3
   | FF0 |       | FF1 |       | FF2 |       | FF3 |
   |     |       |     |       |     |       |     |
   +--o--+       +--o--+       +--o--+       +--o--+
      |             |             |             |
CLK---+-------------+-------------+-------------+
T0 = 1           T1 = Q3'.Q0
T2 = Q1.Q0       T3 = Q3.Q0 + Q2.Q1.Q0

Timing waveform

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

• $Q_0$ toggles on every falling edge ($f_{clk}/2$).
• $Q_1$ toggles when $Q_0 = 1$, except at 9 where $\overline{Q_3} = 0$ blocks it.
• $Q_2$ toggles when $Q_1Q_0 = 1$ (at 3 → 4 and 7 → 8).
• $Q_3$ sets at 7 → 8 ($Q_2Q_1Q_0 = 1$) and resets at 9 → 0 ($Q_3Q_0 = 1$); its frequency is $f_{clk}/10$.

Unused states return to the count (1010→1011→0110, 1100→1101→0100, 1110→1111→0010), so the counter is self-starting.