Johnson counter fundamentals In a 4-bit Johnson (twisted-ring) counter, how many distinct states (bit patterns) occur in one full sequence?

Difficulty: Easy

Correct Answer: 8

Explanation:


Introduction / Context:
Johnson counters are popular for generating multiple evenly spaced timing phases with minimal decoding. Knowing the number of states helps size the counter for a desired number of phases.


Given Data / Assumptions:

  • Counter type: 4-bit Johnson counter.
  • Feedback is inverted (twisted ring).


Concept / Approach:
An n-bit Johnson counter cycles through 2n unique states. For n = 4, that is 8 distinct patterns, doubling what a regular ring counter would produce with the same number of flip-flops.


Step-by-Step Solution:

Number of states = 2 * n = 2 * 4 = 8


Verification / Alternative check:
Enumerating the sequence shows the output walks through eight unique nonoverlapping patterns before repeating, used commonly to create 8-phase timing signals from four flip-flops.


Why Other Options Are Wrong:

  • 1 or 2: Far too small; even a fixed or toggle state machine exceeds these counts.
  • 4: This would be the count for a 4-stage ring (not twisted) counter, not a Johnson counter.


Common Pitfalls:

  • Confusing Johnson with simple ring counters.
  • Assuming 2^n states as in a binary counter; Johnson uses 2n states.


Final Answer:
8

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