Question:medium

A ring counter (with 4 flip-flops) has how many unique states?

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Keep these direct state relations for $N$ flip-flops in mind: - Ring Counter states = $N$ - Johnson Counter states = $2N$ - Ripple Counter states = $2^N$
Updated On: Jul 4, 2026
  • $2$
  • $4$
  • $8$
  • $16$
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The Correct Option is B

Solution and Explanation

Understanding the Concept: A ring counter is a synchronous circular shift register where the true output ($Q$) of the final stage flip-flop is fed directly back into the synchronous input ($D$ or $J$) of the very first stage flip-flop. For an $N$-stage shift register configuration, the count sequence forms a circulating loop of binary patterns.

Step 1: State Space Comparison of Counters.

Let us analyze the capacities of various structures utilizing $N$ flip-flops:
Total Possible Arbitrary States: $2^N$
Standard Binary Counter (Asynchronous/Synchronous): $2^N$ unique states.
Johnson (Twisted Ring) Counter: $2N$ unique states.
Standard Ring Counter: Exactly $N$ unique states.

Step 2: Enumeration of States for $N=4$.

Given a 4-stage ring counter ($N = 4$), the circuit transitions linearly through a loop of exactly 4 unique states. If initialized to a standard one-hot encoded state (e.g., $1000$), the sequence behaves as follows: \[ \text{State 1: } 1000 \] \[ \text{State 2: } 0100 \] \[ \text{State 3: } 0010 \] \[ \text{State 4: } 0001 \] On the next clock pulse, the trailing $1$ recirculates back to the input, regenerating State 1 ($1000$). Thus, the number of distinct, valid operating states equals $N = 4$.
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