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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
0
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
+5
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?
+3
to make mod8 ring counter, we will need 8 FFs.
0
this is for asynchronous counter
0
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 

+1
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 ⌉
0

@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.

+1
in GATE exam, they will specify which is LSB and which is MSB

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