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The number of flip-flops required to construct a binary modulo $N$ counter is __________
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Let say we have to Design mod-8 counter i.e 000 to 111. so we need 3 bit to represent i.e 3 FF

for Mod N

2x = N

x = ceiling (log2N)

answered by Veteran (55.2k points)
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+3
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?
+1
to make mod8 ring counter, we will need 8 FFs.
0
this is for asynchronous counter