- 2081 Baisakh · 2+6 marks
Sync vs async; 3-bit synchronous down counter
Short answer
Synchronous counters have all flip-flops share a common clock, changing simultaneously with minimal delay; asynchronous counters ripple with accumulated propagation delays. A 3-bit synchronous down counter using JK flip-flops has J0=K0=1, J1=K1=Q0', J2=K2=Q1'Q0', and counts from 7 down to 0.
Differentiate between synchronous and asynchronous counter. Design 3-bit synchronous down counter using JK flip-flops.
Synchronous vs asynchronous counter
| Point | Synchronous counter | Asynchronous (ripple) counter |
|---|---|---|
| Clock | Common clock to all flip-flops | Only FF0 gets the clock; each next FF is clocked by the previous output |
| Change of state | All flip-flops change together | Flip-flops change one after another (ripple) |
| Delay | One FF delay + gate delay | Adds up: |
| Speed | Fast, high clock frequency | Slow, frequency falls as grows |
| Circuit | Needs extra gating logic | Simple, few or no extra gates |
| Decoding glitches | None (no false states) | Momentary false states give glitches |
| Example ICs | 74160/74163, 74190 | 7490, 7493 |
Design of 3-bit synchronous down counter (JK)
A 3-bit down counter counts 7, 6, 5, …, 0 and then returns to 7. Three JK flip-flops () share one clock.
Excitation table (JK: 0→0: 0X, 0→1: 1X, 1→0: X1, 1→1: X0)
| Present Q2Q1Q0 | Next Q2Q1Q0 | J2 | K2 | J1 | K1 | J0 | K0 |
|---|---|---|---|---|---|---|---|
| 111 | 110 | X | 0 | X | 0 | X | 1 |
| 110 | 101 | X | 0 | X | 1 | 1 | X |
| 101 | 100 | X | 0 | 0 | X | X | 1 |
| 100 | 011 | X | 1 | 1 | X | 1 | X |
| 011 | 010 | 0 | X | X | 0 | X | 1 |
| 010 | 001 | 0 | X | X | 1 | 1 | X |
| 001 | 000 | 0 | X | 0 | X | X | 1 |
| 000 | 111 | 1 | X | 1 | X | 1 | X |
K-maps
| J1: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 1 | 0 | X | X |
| 1 | 1 | 0 | X | X |
| J2: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 1 | 0 | 0 | 0 |
| 1 | X | X | X | X |
and are 1 in every row (Q0 toggles every clock), so . and have the same map, and so do and .
Equations
In words, a stage toggles when all lower bits are 0 (the "borrow" condition), just as an up counter stage toggles when all lower bits are 1.
Circuit diagram
+-----+ +-----+ +-----+
J0>|J Q|-Q0 J1>|J Q|-Q1 J2>|J Q|-Q2
| FF0 | | FF1 | | FF2 |
K0>|K | K1>|K | K2>|K |
+--o--+ +--o--+ +--o--+
| | |
CLK---+-------------+-------------+
J0 = K0 = 1
J1 = K1 = Q0'
J2 = K2 = Q1'.Q0' (one 2-input AND gate)
Timing diagram
CLK __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q2 ____|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|_______________
Q1 ____|‾‾‾‾‾‾‾|_______|‾‾‾‾‾‾‾|_______
Q0 ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾|___
Cnt 0 7 6 5 4 3 2 1 0
All outputs change together on the falling clock edge, so no ripple delay or false state appears.
✓ Answer checked by Claude Opus 5.5 (high effort)
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