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+14 votes
The number of flip-flops required to construct a binary modulo $N$ counter is __________
asked in Digital Logic by Veteran (52k points)
edited by | 1.5k views
binary modulo n counter means Asynchronous mod n up/down counter.

1 Answer

+22 votes
Best answer
Let say we have to Design mod-$8$ counter i.e $000$ to $111$. so we need $3$ bits to represent i.e $3$ $FF$

for Mod $N$

$2^x = N$

$x =$ ceiling $(log_2N)$
answered by Veteran (55.8k points)
edited by
I have a doubt: if it is possible to form mod 8 counter using 3 FF. Why do we use Ring Counter or Johnson counter?
to make mod8 ring counter, we will need 8 FFs.
this is for asynchronous counter
Counter Flip flops Counting
Ring N mod(N)
Johnson N mod(2N)
Asynchronous N mod(2^N)

I think here type of counter should be explicitly mentioned to find number of flip flops. @Shaik Masthan 

actually, the counting sequence is not in order for Ring or Johnson ===> we can't use them for requirement of the question !

the one and only choice is Go with asynchronous counter ===> ⌈log$_2$n ⌉

@Shaik Masthan for ring counter and johnson's counter is the output of left flipflop is MSB?

And also please mention which is the MSB for asynchronous UP and down counter.

in GATE exam, they will specify which is LSB and which is MSB

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