- 2078 Bhadra · 1+7 marks
Parallel counter; mod-6 synchronous counter (JK)
Short answer
A parallel (synchronous) counter applies the clock to all flip-flops simultaneously. A mod-6 synchronous up counter with 3 JK flip-flops has J2=Q1Q0, K2=Q0, J1=Q2'Q0, K1=Q0, J0=K0=1; it is self-starting.
Define parallel counter. Design a mod-6 synchronous up counter using JK flip flop.
Parallel counter
A parallel counter is another name for a synchronous counter: the clock is applied to all flip-flops in parallel, so they all change state at the same instant. Gates at the J-K (or T/D) inputs decide which flip-flops toggle.
Design: mod-6 synchronous up counter (JK)
Step 1: States. A mod-6 counter counts 0 to 5 (000 to 101) and returns to 000. , so 3 JK flip-flops () are used. Unused states 110 and 111 are don't cares.
Step 2: 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 |
|---|---|---|---|---|---|---|---|
| 000 | 001 | 0 | X | 0 | X | 1 | X |
| 001 | 010 | 0 | X | 1 | X | X | 1 |
| 010 | 011 | 0 | X | X | 0 | 1 | X |
| 011 | 100 | 1 | X | X | 1 | X | 1 |
| 100 | 101 | X | 0 | 0 | X | 1 | X |
| 101 | 000 | X | 1 | 0 | X | X | 1 |
Step 3: K-maps
| J1: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 0 | 1 | X | X |
| 1 | 0 | 0 | X | X |
| K1: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | X | X | 1 | 0 |
| 1 | X | X | X | X |
| J2: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 |
| 1 | X | X | X | X |
| K2: Q2 / Q1Q0 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | X | X | X | X |
| 1 | 0 | 1 | X | X |
since toggles on every clock.
Step 4: Equations
Step 5: 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 = Q2'.Q0 K1 = Q0
J2 = Q1.Q0 K2 = Q0
Two 2-input AND gates are needed.
Step 6: Unused states
| Unused state | Next state |
|---|---|
| 110 | 111 |
| 111 | 000 |
So the counter is self-starting.
Step 7: Timing diagram
CLK __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾
Q2 ________________|‾‾‾‾‾‾‾|_______
Q1 ________|‾‾‾‾‾‾‾|_______________
Q0 ____|‾‾‾|___|‾‾‾|___|‾‾‾|___|‾‾‾
Cnt 0 1 2 3 4 5 0 1
All outputs change together at the falling clock edge; gives .
✓ Answer checked by Claude Opus 5.5 (high effort)
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