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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. 23=8≥62^3 = 8 \ge 6, so 3 JK flip-flops (Q2Q1Q0Q_2Q_1Q_0) 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 Q2Q1Q0Next Q2Q1Q0J2K2J1K1J0K0
0000010X0X1X
0010100X1XX1
0100110XX01X
0111001XX1X1
100101X00X1X
101000X10XX1

Step 3: K-maps

J1: Q2 / Q1Q000011110
001XX
100XX
K1: Q2 / Q1Q000011110
0XX10
1XXXX
J2: Q2 / Q1Q000011110
00010
1XXXX
K2: Q2 / Q1Q000011110
0XXXX
101XX

J0=K0=1J_0 = K_0 = 1 since Q0Q_0 toggles on every clock.

Step 4: Equations

J2=Q1Q0,K2=Q0J1=Q2‾Q0,K1=Q0J0=1,K0=1\begin{aligned} &J_2 = Q_1Q_0,\quad K_2 = Q_0 \\ &J_1 = \overline{Q_2}Q_0,\quad K_1 = Q_0 \\ &J_0 = 1,\quad K_0 = 1 \end{aligned}

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 stateNext state
110111
111000

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; Q2Q_2 gives fclk/6f_{clk}/6.

✓ Answer checked by Claude Opus 5.5 (high effort)

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