Digital Electronics - Counters - Discussion
Discussion Forum : Counters - General Questions (Q.No. 1)
1.
How many flip-flops are required to make a MOD-32 binary counter?
Discussion:
26 comments Page 3 of 3.
Gopi said:
1 decade ago
Mod-32 means divide by 32 counter.
2^n=32. => n=5.
i.e 5 flip flops.
2^n=32. => n=5.
i.e 5 flip flops.
Ankita said:
1 decade ago
2^5=32.
Hence, it required 5 flip flops.
Hence, it required 5 flip flops.
Abdul wahab khokhar said:
1 decade ago
Simply it is:
2^5=32
That is required only 5 flip flop.
If there is mod-30 then also it requires 5 flip flop.
2^5=32
That is required only 5 flip flop.
If there is mod-30 then also it requires 5 flip flop.
T.Milhaan said:
1 decade ago
2^n>N>2^n-1
n-number of flip flops
N-number of states ie in mod32 N=32
n-number of flip flops
N-number of states ie in mod32 N=32
Nikita said:
1 decade ago
N=32
=> 2^(n-1)<=N<=2^n
if n=5......it satisfy the condition.
=>2^4<=N<=2^5
so.....n=5 i.e. 5 flip flops.
=> 2^(n-1)<=N<=2^n
if n=5......it satisfy the condition.
=>2^4<=N<=2^5
so.....n=5 i.e. 5 flip flops.
Ashwin said:
1 decade ago
Mod-32 means divide by 32 counter.
2^n=32. => n=5.
i.e. 5 flip flops.
2^n=32. => n=5.
i.e. 5 flip flops.
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