Question

How many flip-flops are required to design a Mod-10 counter?

• A. 3
• B. 4
• C. 10
• D. 5

Hint:

To quickly find the number of flip-flops needed for a Mod-\(M\) counter, find the smallest integer \(n\) that satisfies \(2^n \ge M\). For a Mod-10 counter, \(2^4 = 16 \ge 10\), which gives \(n = 4\).

Answer 1

The Correct Option is B

Concept: The modulus (or "Mod") of a digital counter defines the total number of unique states the counter cycles through before resetting back to its initial state. For instance, a Mod-10 counter (also called a decade counter) sequences through exactly 10 distinct binary states (typically representing decimal values from 0 to 9) before looping back to 0. Each individual triggerable flip-flop is a bistable storage element capable of holding a single bit of binary information (representing either a 0 or a 1). When you connect \(n\) flip-flops together in a counter array, the system can generate up to \(2^n\) unique binary combinations (or states). To find the minimum number of flip-flops (\(n\)) needed to design a counter with a specific modulus (\(M\)), we use the following structural inequality constraint: \[ 2^{n-1} < M \le 2^n \] Let's evaluate this inequality for a Mod-10 counter, where \(M = 10\), by testing successive values for the integer variable \(n\):
• If we test \(n = 3\) flip-flops: The maximum number of states is \(2^3 = 8\). Checking the constraint: \(8 < 10\). This configuration cannot support a Mod-10 counter because it can only count through 8 unique states (0 to 7).
• If we test \(n = 4\) flip-flops: The maximum number of states is \(2^4 = 16\). Checking the constraint: \(10 \le 16\). This configuration can easily support a Mod-10 counter. The circuit will use 10 of the available states (0 through 9) and skip or reset past the remaining 6 states using logic gates. Therefore, a minimum of 4 flip-flops is required to build a Mod-10 counter.