Consider a 4-bit Johnson counter with an initial value of 0000. The counting…
2015
Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is
- A.
0, 1, 3, 7, 15, 14, 12, 8, 0
- B.
0, 1, 3, 5, 7, 9, 11, 13, 15, 0
- C.
0, 2, 4, 6, 8, 10, 12, 14, 0
- D.
0, 8, 12, 14, 15, 7, 3, 1, 0
Attempted by 130 students.
Show answer & explanation
Correct answer: D
Key insight: In a Johnson (twisted-ring) counter the inverted output of the last flip-flop is fed back into the first flip-flop and the register shifts by one bit each clock. This produces a block of 1s that grows and then shrinks, giving 2n distinct states for an n-bit counter.
State transitions (binary -> decimal): 0000 (0) -> 1000 (8) -> 1100 (12) -> 1110 (14) -> 1111 (15) -> 0111 (7) -> 0011 (3) -> 0001 (1) -> 0000 (0)
Therefore the counting sequence is 0, 8, 12, 14, 15, 7, 3, 1, 0, which matches the provided correct option.
A video solution is available for this question — log in and enroll to watch it.