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

To solve this problem, we need to determine the maximum possible range of bit-count for an \( n \)-bit binary counter that uses \( n \) flip-flops.

1. Understanding Binary Counters:

- A binary counter is a digital circuit made of flip-flops that counts in binary. Each flip-flop represents one bit of the binary number, and the output of the counter increases by one for each clock pulse.

- An \( n \)-bit binary counter can represent numbers from 0 to a maximum value determined by the number of bits available. Specifically, with \( n \) bits, the counter can represent values from 0 up to \( 2^n - 1 \) because the range of values for an \( n \)-bit binary number is from \( 0 \) to \( 2^n - 1 \), inclusive.

2. Analyzing the Options:

- Option 1: "0 to \( 2^n \)" – This is incorrect because the maximum value is \( 2^n - 1 \), not \( 2^n \). The range includes 0 to \( 2^n - 1 \), not 0 to \( 2^n \).

- Option 2: "0 to \( 2^n - 1 \)" – This is correct. The maximum value for an \( n \)-bit binary counter is \( 2^n - 1 \), as this is the highest number that can be represented by \( n \) bits.

- Option 3: "0 to \( 2^{n + 1/2} \)" – This is incorrect. There is no fractional exponent when counting in binary. The range is always in whole powers of 2.

- Option 4: "0 to \( 2^n + 1 \)" – This is incorrect. The maximum value for an \( n \)-bit counter is \( 2^n - 1 \), not \( 2^n + 1 \).

Final Answer:

The maximum possible range of bit-count in an \( n \)-bit binary counter is 0 to \( 2^n - 1 \).