- 2068 Chaitra · 1+7 marks
Ripple counter; mod-10 ripple counter
Short answer
A ripple counter has flip-flops clocked by previous stage outputs, causing propagation delay. A mod-10 ripple counter uses four JK flip-flops in toggle mode. After count 9 (1001), the next pulse gives 1010 momentarily; a NAND gate on Q3 and Q1 resets all FFs to 0000, providing a divide-by-10 output.
Define ripple counter. Explain the operation of mod-10 ripple counter with timing diagram.
Ripple counter
A ripple counter (asynchronous counter) is a counter in which only the first flip-flop receives the external clock, and the output of each flip-flop acts as the clock of the next one. The change of state ripples through the stages one after another.
Mod-10 ripple counter
Construction
- 10 states (0–9) need 4 flip-flops, since .
- Four negative-edge JK flip-flops with (toggle mode). CLK → FF0, → FF1, → FF2, → FF3.
- At count 10 (1010), and are both 1 for the first time. A NAND gate of and drives the active-low CLEAR of all flip-flops, so the counter returns to 0000.
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
- Clear all flip-flops: 0000.
- Each falling edge of CLK toggles FF0. FF1 toggles every time falls 1→0; FF2 every time falls; FF3 every time falls. The outputs therefore count up in binary.
- After the 9th pulse the count is 1001. The 10th pulse makes fall, which toggles to 1, giving 1010 for a moment. The NAND output becomes 0 and clears all flip-flops to 0000. The next cycle begins.
| Pulse | Q3 Q2 Q1 Q0 | Count |
|---|---|---|
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| 2 | 0010 | 2 |
| 3 | 0011 | 3 |
| 4 | 0100 | 4 |
| 5 | 0101 | 5 |
| 6 | 0110 | 6 |
| 7 | 0111 | 7 |
| 8 | 1000 | 8 |
| 9 | 1001 | 9 |
| 10 | 1010 → 0000 | 0 |
Timing diagram
CLK __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q3 ________________________________|‾‾‾‾‾‾‾|___
Q2 ________________|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|___________
Q1 ________|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|___________
Q0 ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___
Cnt 0 1 2 3 4 5 6 7 8 9 0
- changes at every falling clock edge ().
- , , change at the falling edges of , , .
- is high during counts 8 and 9 and gives .
- In the real circuit each stage adds one propagation delay and a narrow spike appears on at the 10th pulse (state 1010 before reset).
✓ Answer checked by Claude Opus 5.5 (high effort)
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