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 1 of 3.
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.
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.
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
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.
Ankita said:
1 decade ago
2^5=32.
Hence, it required 5 flip flops.
Hence, it required 5 flip flops.
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.
SAMPATH said:
1 decade ago
How can you people say that and represent no of flip flops ? CAN anybody explain it?
Neha said:
1 decade ago
5 flip flops will be required if we count from 0 to 31 for mod 32 we will need 6 flip flops.
D.rajeswari said:
1 decade ago
GIVEN MOD = 32.
We know for the no of clk pulses = 2^n.
where n = no clk pulses.
2^n = 32.
2^n = 2^5.
n = 5.
We know for the no of clk pulses = 2^n.
where n = no clk pulses.
2^n = 32.
2^n = 2^5.
n = 5.
Neelima said:
1 decade ago
N <= 2^n.
Where N is modulus of counter.
n is no.of flip flops.
So when n=5 condition is satisfied.
Where N is modulus of counter.
n is no.of flip flops.
So when n=5 condition is satisfied.
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