How many flip-flops are required to design a MOD-10 counter?
Show Hint
- 3
- 4
- 5
- 10
The Correct Option is B
Approach Solution - 1
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.
Approach Solution -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.
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