The minimum number of D flip-flops needed to design a mod-258 counter is

The correct answer is option 1

Concept:

Binary modulo N counter can count up to 0 to N – 1

Therefore, binary modulo 258 counters will count from 0 to 257 (order may vary)

Formula:

Minimum number of flip flop needed = ⌈log2 N⌉

Calculation:

An n-bit binary counter consists of n flip-flops and can count in binary from 0 to 2ⁿ – 1

Mod 258 counter has 258 states. We need to find the number of bits to represent max 257

\(\lceil{log_2n}\rceil = \lceil{log_2258}\rceil =9\) bits required

So the minimum number of D flip-flops needed is 9.