- 2081 Bhadra · 2+6 marks
Mod-10 negative edge ripple counter with JK
Short answer
A sequential logic circuit output depends on present inputs and stored state (past history), requiring memory elements. A mod-10 negative-edge ripple counter uses 4 JK flip-flops in toggle mode (J=K=1), cascaded with Q outputs clocking next FFs, counting 0000–1001. It resets to 0000 on the 10th pulse via a NAND gate on Q3 and Q1 driving clear.
Define a sequential logic circuit. Explain the mod-10 negative edge-triggered ripple counter with its timing diagram. Use JK flip-flops in your design.
Sequential logic circuit
A sequential logic circuit is a circuit whose output depends on the present inputs and on the past history (stored state) of the circuit. It contains memory elements (flip-flops or latches) and a feedback path; flip-flops, registers and counters are examples.
Mod-10 negative-edge ripple counter
A mod-10 (decade) ripple counter has 10 states, 0000 to 1001. Since , four JK flip-flops are used.
Connections
- All J and K inputs are tied to logic 1, so every flip-flop toggles on its active (falling) clock edge.
- The external clock goes to FF0 only. is the clock of FF1, of FF2 and of FF3 (negative-edge triggering means a stage toggles when the previous Q falls from 1 to 0, which gives up counting).
- Count 10 = 1010 must not be held. It is the first state with , so (NAND) is fed to 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)
Q1 ---| |
Operation
- Initially all flip-flops are cleared: .
- Every falling clock edge toggles FF0. Whenever goes 1→0, FF1 toggles; whenever goes 1→0, FF2 toggles; whenever goes 1→0, FF3 toggles. The count rises 0, 1, 2, …, 9.
- On the 10th falling edge the outputs try to become 1010. At once the NAND output goes low and clears all flip-flops, so the counter returns to 0000. The false state 1010 exists only for a few nanoseconds.
| Clock pulse | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Q3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
| Q2 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
| Q1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| Q0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
Timing diagram
CLK __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q3 ________________________________|‾‾‾‾‾‾‾|___
Q2 ________________|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|___________
Q1 ________|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|___________
Q0 ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___
Cnt 0 1 2 3 4 5 6 7 8 9 0
Each output changes on the falling edge of the signal that clocks it. is high only for counts 8 and 9, and its frequency is . In a real circuit each stage adds a propagation delay , so the outputs ripple and a narrow spike appears on at the reset.
✓ Answer checked by Claude Opus 5.5 (high effort)
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