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The minimum number of $\text{D}$ flip-flops needed to design a mod-258 counter is

1. 9
2. 8
3. 512
4. 258
edited | 3.2k views

Mod $258$ counter has $258$ states. We need to find no. of bits to represent $257$ at max. $2^n \geq$258$\implies n \geq 9$.

edited
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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.
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srestha same 9

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Doubt why we need conversion? why D flip flop can't count.if we use some combinatory ckt before giving input to D Flip flop then it is possible to count 0 to 2^n-1.
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Is this count same for a synchronous and asynchronous counter?

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-2n counter) can have 2different 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 = 2n.

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

Tell me if i went wrong somewhere.

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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?
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Because that gives us the minimum no of flip flops i,e ceil(log2(258)) =9, while Ring counter needs 258 FFs and Johnson counter needs 128 FFs.

But D FF cannot be used in case of Ripple Counter?

By converting each D FF to J-K FF, we can.

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Because that gives us the minimum no of flip flops i,e ceil(log2(258)) =9,

But if the question is like

How many D flip-flops are needed to design a mod-258 counter ? Then also we get the same answer.

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.

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

2^n ≥ 258
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.
–1 vote
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If we want to buid with JK and  T ffs then they are also '9' or something different?

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