edited
14,871 views
27 votes
27 votes

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

6 Answers

Best answer
31 votes
31 votes

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

Answer is A.

edited by
39 votes
39 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-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.

18 votes
18 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).

6 votes
6 votes

In general, the minimum no of Flip-flop needed to count a number N = ceil( logN) base2.
So, Mod 258 counter needs = ceil(log 258)= 9 flipflops. It could be mod-258 up/down/random counter.

Although Asynchronous counters use T flip-flops for counting but can be designed by other flip-flops(j-k, D....) and combinational circuits using same no. of flip-flops(9) only.

Note- Only Asynchronous counters can count mod-N using only (log N) flip-flops. If we use any Synchronous counter is surely going to take more than (log N)flip-flops. There are many Synchronous counters which can be designed using a different type of flip-flop(J-K,T...) and combinational circuits which take lesser no. of flip-flops than ring and johnson counter to count mod N. So for Synchronous counters this no. could be (log N)<m<N.

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