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+16 votes

+21 votes

Best answer

+3

But how can D flip flop count 258 unless we convert to JK or T flip flop.As they have asked D flip flop so i think we should not assume that conversion is [email protected] sir ,please clear this.

+17 votes

Flip -Flop is a binary cell capable of storing 1-bit of information. so to store n-bit we need n Fli-Flip.

**n-bit Ring Counter can have n different output states.**

**n-bit Twisted ring counter(Johnson Ring Counter) can have 2n different output states.**

**n-bit ripple counter (mod-2 ^{n} counter) can have 2^{n }different output states.**

so i mean to say

A Ring Counter that consist n Flip-Flip will have n-states.

3-bit ripple counter is called as MOD-8 counter. So in general, an n-bit ripple counter is called as modulo-N counter. Where, MOD number = 2^{n}.

so MOD-258 counter needs 9 bit to store all states so ultimately 9 Flip-Flop.

Tell me if i went wrong somewhere.

0

Hi might be a silly query but still how did we make the assumption that its n bit ripple counter and not ring or twisted ring counter?

+11 votes

Following is the $mod-4$ counter using two D-flip flops:

Using the same technqiue we can build $mod-258$ counter using $nine$ D-Flip Flops.

**Hence, Answer is (A).**

0

@Manu Thakur can we say to count 2^n states, we need n flip flops(any flip flop) as we can design a combinational circuit which will provide input such that it will count those many number of states.

+1 vote

0 votes

Ok so let me explain - the first confusion is mod 258 counter - Should the counter count all 258 value or Should it count till values 258. and the Answer is it should count till 258 values just like mod 6 johnson counter counts 0-1-3-7-6-4 but it do not count 2 and 5 however it is called mod 6 counter and this requires ceil(log 6)=3bits=3 D-flipflop.

In the same manner mod 258 counter will take 9 D-flipflops.

In the same manner mod 258 counter will take 9 D-flipflops.

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