Minimum flip-flop count for a modulus-5 (divide-by-5) counter: What is the least number of flip-flops required to implement a mod-5 counter?
Correct Answer: 3
Introduction / Context:Determining the minimum number of flip-flops helps size counters efficiently. Each flip-flop stores one bit, and the number of distinct states available is 2^n for n flip-flops. A mod-N counter must provide at least N distinct states, possibly discarding extras with gating when 2^n is not exactly N.
Given Data / Assumptions:
- Target modulus N = 5.
- Flip-flops are binary storage elements; state capacity is 2^n.
- It is acceptable to reset or skip unused states using combinational feedback.
Concept / Approach:Find the smallest n such that 2^n ≥ N. For N = 5: 2^2 = 4 (insufficient), 2^3 = 8 (sufficient). Therefore, at least three flip-flops are needed. The counter can be designed to count 0–4 and then reset to 0, effectively creating a mod-5 sequence within the 8-state space of three flip-flops.
Step-by-Step Solution:
Compute powers of two: 2^2 = 4 < 5; 2^3 = 8 ≥ 5.Choose n = 3 flip-flops to ensure enough states.Design gating so that upon reaching state 101 (decimal 5), the counter resets to 000.Result: a sequence of five states 000→001→010→011→100→000.Verification / Alternative check:Simulate a 3-bit synchronous counter with decode logic on state 101 to assert CLEAR. The output cycles through five states per period, confirming mod-5 behavior while using only three flip-flops.
Why Other Options Are Wrong:
- 5, 8, 10 flip-flops: Far exceed the minimum; 2^5, 2^8, and 2^10 states are unnecessary for mod-5.
- 2 flip-flops: Only 4 states available, which cannot realize a mod-5 sequence.
Common Pitfalls:Forgetting to add gating to truncate the 8-state space; confusing Johnson/ring counters with binary counters; neglecting propagation delays when forming synchronous reset logic (glitch avoidance may require synchronous clear or additional gating).
Final Answer:3