Explanation:

To understand why the minimum number of flip-flops required for a Modulus 15 counter is 4, let's first discuss what a Modulus counter is.

A Modulus counter is a type of counter that counts up to a specific value before resetting back to zero. In this case, we need a counter that counts up to 15 before resetting back to zero.

To determine the minimum number of flip-flops required for a Modulus counter, we can use the formula:

N = ceil(log2(M))

Where N is the number of flip-flops and M is the Modulus value.

Calculating the number of flip-flops:

In this case, the Modulus value is 15. Using the formula, we can calculate the number of flip-flops required as follows:

N = ceil(log2(15))
N = ceil(3.91)

Since the number of flip-flops cannot be a fraction, we need to round up to the nearest whole number. Therefore, N = 4.

Explanation of the formula:

The formula N = ceil(log2(M)) is derived from the fact that each flip-flop can store 2 different states (0 or 1). In a binary counter, the number of different states it can represent is equal to 2^N, where N is the number of flip-flops.

In a Modulus counter, the maximum count value is M. Therefore, the number of different states the counter needs to represent is M + 1 (including zero). In binary, this is represented by log2(M+1).

Since each flip-flop stores 2 different states, we divide log2(M+1) by 2 to determine the number of flip-flops required. Finally, we round up to the nearest whole number using the ceil function.

Conclusion:

In conclusion, the minimum number of flip-flops required for a Modulus 15 counter is 4. This is determined using the formula N = ceil(log2(M)), where M is the Modulus value.