Question

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

• A. 3
• B. 4
• C. 5
• D. 10

Hint:

To design a MOD-N counter, always choose the smallest n such that \(2^n \geq N\).

Answer 1

The Correct Option is B

To design a MOD-10 counter, we need to first understand what this means. A MOD-10 counter is a counter that counts from 0 to 9 and then resets to 0, thus having a total of 10 states.

The number of flip-flops (\(n\)) required to design a counter can be determined using the formula for the maximum number of states (\(2^n\)), which should be equal to or greater than the desired number of states (\(N\)). In this case, \(N = 10\) for a MOD-10 counter.

Let's determine the smallest \(n\) such that:

\(2^n \geq 10\)

Calculating for different values of \(n\):

• \(n = 3\): \(2^3 = 8\) (not sufficient as 8 < 10)
• \(n = 4\): \(2^4 = 16\) (sufficient as 16 ≥ 10)

Therefore, 4 flip-flops are required to implement a MOD-10 counter. Each flip-flop contributes a binary digit to the counter, and with 4 flip-flops, we can represent states from 0000 (0) to 1111 (15), sufficient to cover the 10 states needed (0 to 9).

To summarize, the correct answer is that a MOD-10 counter requires 4 flip-flops.

Answer 2

To determine the number of flip-flops required to design a MOD-10 counter, we need to understand the concept of modulus in counters. A MOD-10 counter, also known as a decade counter, counts ten distinct states, from 0 to 9.

The number of flip-flops required to build a counter for a given modulus \(N\) is determined by the formula:

\(n \geq \lceil \log_{2} N \rceil\)

where \(n\) is the number of flip-flops and \(N\) is the modulus of the counter.

In this case, \(N = 10\). So, we need to find:

\(n \geq \lceil \log_{2} 10 \rceil\)

Calculating \(\log_{2} 10\):

\(\log_{2} 10 \approx 3.32\)

Taking the ceiling of 3.32, we get:

\(\lceil 3.32 \rceil = 4\)

Hence, a MOD-10 counter requires 4 flip-flops.

Every flip-flop can store one bit of information. With 4 flip-flops, this counter can have a maximum of \(2^{4} = 16\) states (0 to 15). However, a MOD-10 counter will reset to zero after reaching the 9th count.

Therefore, the correct answer is that 4 flip-flops are needed to design a MOD-10 counter.

Let's quickly rule out the other options:

• 3 flip-flops: Can provide only \(2^{3} = 8\) states, which is insufficient for a MOD-10 counter.
• 5 flip-flops: Can provide \(2^{5} = 32\) states, which is more than necessary and thus not optimized for a MOD-10 counter.
• 10 flip-flops: Would provide \(2^{10} = 1024\) states, which is significantly excessive and inefficient for a MOD-10 counter.

Thus, the answer is clearly 4 flip-flops.