Question

A counter with modulus 8 has:

• A. \(3\text{ flip-flops}\)
• B. \(4\text{ flip-flops}\)
• C. \(2\text{ flip-flops}\)
• D. \(8\text{ flip-flops}\)

Hint:

To find the required number of flip-flops for any modulus \( M \), take the ceiling logarithm base 2: \( m = \lceil \log_2(M) \rceil \). For example, \( \log_2(8) = 3 \).

Answer 1

The Correct Option is A

Concept: The modulus (or mod number) of a counter defines the total number of unique states the counter cycles through before returning to its initial starting state. A counter with a modulus of \( M \) counts exactly \( M \) distinct states. The mathematical constraint relating the number of flip-flops \( m \) to the maximum possible modulus is: \[ M \leq 2^m \] Where \( m \) is the total number of flip-flops. To design a counter for a specific modulus \( M \), the minimum number of flip-flops required is the smallest integer \( m \) that satisfies this inequality.

Step 1: Substitute the given modulus value into the inequality.
We are given a modulus \( M = 8 \). We need to determine the value of \( m \): \[ 8 \leq 2^m \]

Step 2: Evaluate powers of 2 to isolate \( m \).
Let's test consecutive integer values for \( m \):
• If \( m = 2 \): \( 2^2 = 4 \). Here, \( 8 \leq 4 \) is False. (A 2-bit counter can only track 4 unique states: 00, 01, 10, 11).
• If \( m = 3 \): \( 2^3 = 8 \). Here, \( 8 \leq 8 \) is True. Since \( m = 3 \) satisfies the boundary condition perfectly, a minimum of 3 flip-flops is required to construct a modulus-8 counter. The 8 distinct binary states range from ‘000‘ to ‘111‘ (0 to 7 in decimal).