Modulus-10 (decade) counter — minimum hardware requirement For a MOD-10 (decade) counter that counts 0 through 9 and then recycles to 0, how many flip-flops are required at minimum (irrespective of synchronous or ripple style)?

Introduction / Context:A decade (MOD-10) counter is a sequential circuit that steps through ten distinct states (0 to 9) and then resets to 0. Determining the minimum number of flip-flops focuses on state capacity (2^n) rather than on whether the design is synchronous or ripple. This is a core idea in digital design when matching required modulus to available binary storage.

Given Data / Assumptions:

• Required modulus: 10 states.
• Binary storage uses n flip-flops giving up to 2^n states.
• Non-power-of-two modulus achieved via decode/reset logic.

Concept / Approach:

Find the smallest n such that 2^n ≥ 10. Then, add simple combinational decoding to reset the counter after reaching decimal 9 (binary 1001). The rest of the 2^n state space is unused and prevented by the reset action.

Step-by-Step Solution:

Verification / Alternative check:

Simulating a 4-bit counter with reset-on-10 shows the sequence 0000→…→1001 then reset to 0000, giving exactly 10 states per cycle.

Why Other Options Are Wrong:

10 flip-flops and 5 flip-flops greatly exceed the minimum for MOD-10.

2 flip-flops can represent only 4 states.

“Synchronous clocking” is a style, not a quantity of flip-flops; it does not answer the question.

Common Pitfalls:

Assuming a non-power-of-two modulus needs that many flip-flops; forgetting that decoding unused states yields the desired modulus with the minimal n.

Final Answer:

4 flip-flops