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The number of flip-flops required to construct a binary modulo $N$ counter is __________
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binary modulo n counter means Asynchronous mod n up/down counter.
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Let say we have to design a mod-$8$ counter i.e $000$ to $111$. So we need $3$ bits to represent i.e $3$ FF.

For mod $N: 2^x = N$

$\implies x = \left \lceil (\log_2N) \right \rceil $
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@Shaik Masthan Sir.

According to me we can also construct a synchronous binary mod N counter. and it will have ceil(logn ) counters.

Counters can have any sequence of states.

And here binary mod N counters means that it follows binary sequence like (0-1-2-3-4..... upto 2^n-1 )and then this cycle repeats.

According to Morris Mano "A counter that follows binary sequence is called binary counters"

Am I correct?

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A simple mod N counter that counts from 0 to N-1 , so either you use a synchronous or asynchronous approach , you will be needing LogN FF . The only difference will be the amount of additional circuitry that will be needed in case of a synchronous counter.
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To construct a binary modulo $N$ counter, we need to store the current count in the counter. The number of bits needed to store a number in binary is equal to $log_2(N)$, where $N$ is the maximum number that can be stored.
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