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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 | 2.3k 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$.

answered by Active (4.2k points)
edited
+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.
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what about if it ask for JK flipflop?
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srestha same 9

+1
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.

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.

answered by Boss (12.2k points)
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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?
+2

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.

0

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.

Hence, Answer is (A).

answered by Boss (41.1k points)
+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
answered by Junior (785 points)
–1 vote
answered by Loyal (9.1k points)
0
If we want to buid with JK and  T ffs then they are also '9' or something different?

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