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The minimum number of Flip-Flops required to construct a binary Modulo $n$ counter is ________

  1. $n$
  2. $n-1$
  3. $2^n – 1$
  4. $\lceil \log_2 n \rceil$
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plz any one explain???
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n flip flops can be used to make modulo  2 counter.

So according to the question the answer must be  ceil(log2(n))

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The "MOD" or "MODULUS" of a counter is the number of unique states. The MOD of the n flip flop ring counter is n.

4 Comments

no, the options were wrong. See now.
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I think previously log n is not their right??
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yes..
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binary modulo n means the possible no.of states = n and mark them as 0,1,2,3,....n-1.

therefore you need to indicate n states with k flipflops

if n=2p then k=p

if n $\neq$ 2p then k= ⌈log2n⌉