GATE Overflow | Digital Logic | Test 1 | Question: 6

Question

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

• A. $n$
• B. $n-1$
• C. $2^n – 1$
• D. $\lceil \log_2 n \rceil$

Comments

plz any one explain???

Answer 1

n flip flops can be used to make modulo  2ⁿ  counter.

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

Answer 2

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

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..

Answer 3

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=2^p then k=p

if n $\neq$ 2^p then k= ⌈log₂n⌉