Question

What is the maximum possible range of bit-count specifically in an n-bit binary counter consisting of 'n' number of flip-flops?

• A. \( \text{0 to } 2^n \)
• B. \( \text{0 to } 2^n - 1 \)
• C. \( \text{0 to } 2^{n+1/2} \)
• D. \( \text{0 to } 2^n + 1 \)

Hint:

For any digital system with \( n \) bits (or \( n \) flip-flops), there are \( 2^n \) unique possible states. If counting starts from 0, the numeric range spans from \( 0 \) to \( 2^n - 1 \). This is key to understanding registers, memory, and counter designs.

Answer 1

The Correct Option is B

An \( n \)-bit binary counter consists of \( n \) flip-flops, where each flip-flop can store one bit (either 0 or 1).
The total number of unique states it can represent is \( 2^n \). Since counting typically starts from 0, the range of values is: \[ \text{Minimum count} = 0, \quad \text{Maximum count} = 2^n - 1 \] Examples:

• For \( n = 1 \): \( 2^1 = 2 \) states → 0 to \( 2^1 - 1 = 1 \)
• For \( n = 2 \): \( 2^2 = 4 \) states → 0 to \( 2^2 - 1 = 3 \)
• For \( n = 3 \): \( 2^3 = 8 \) states → 0 to \( 2^3 - 1 = 7 \)

Therefore, the correct and general range is: \[ \boxed{0 \text{ to } 2^n - 1} \]