Ripple Counters — Natural (Maximum) States for 4 Bits How many natural states (untrimmed combinations) exist in a 4-bit ripple counter?

Introduction / Context:
Ripple counters advance through all binary combinations defined by their width unless additional gating trims the sequence. The number of natural states equals the maximum count before wrapping back to zero. This determines both the modulus and the frequency division ratio.

Given Data / Assumptions:

• Counter width: 4 flip-flops (Q3..Q0).
• No external reset logic to skip states.
• Binary counting (not Gray or BCD).

Concept / Approach:
For N bits, the number of natural states is 2^N. Thus a 4-bit counter cycles through 16 distinct states (0000 to 1111) and divides the input frequency by 16 at the MSB output. If a lesser modulus is required (e.g., MOD-10), extra logic is used to asynchronously or synchronously reset at the desired terminal count.

Step-by-Step Solution:
Compute 2^N for N = 4 → 16.Enumerate conceptually: 0000 … 1111 (16 total patterns).Wrap behavior: next count after 1111 returns to 0000.Therefore, the 4-bit ripple counter has 16 natural states.

Verification / Alternative check:
Simulation or truth-table expansion verifies that 16 unique states occur before repetition. The MSB toggles at fin/16, confirming the divide ratio.

Why Other Options Are Wrong:

• 4 or 8: Correspond to 2-bit or 3-bit counters, not 4-bit.
• 32: Would require 5 bits.

Common Pitfalls:
Confusing “natural states” with “desired modulus”; trimming requires extra logic beyond the natural 2^N count.

Final Answer:
16