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?
3
45
5
6
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 1 of 3.

Komal jadhav said:   2 years ago
How many flip flops are required to design the mod 29 counter? Please explain to me.

RISHI NIGAM said:   3 years ago
Yes, 2^n = 32 from where we get the answer is 5 is a generalized way of getting no of flip-flops and remember id mod is in between means not multiple of 2.

So we just higher 2 multiple for example 14 then we take 4 flipflops as we take count as 16.

Faisal yusuf said:   5 years ago
One flip flop store 1 bit 32 =100000 6 binary bit.

So, 6 flipflop use.

Piyush Kumar said:   6 years ago
Here, The minimum no of flip flops required for MOD 32 counter is 5.

But with 6 flip flop, we can also count 32 states. So I think n>=5. With 6 flip flop, we can represent 2^6 states, but if in this by some combinational circuits we can make it MOD 32 counter.
(1)

N.kiranmai said:   7 years ago
The formula is 2^p where p is the no of flip flops are used here we have to find the no of flip flops so 2^5 =32.

Daniel said:   8 years ago
What is the least number of jk flip flops it would take to create a mod 500 counter?

Usha kiran said:   9 years ago
In binary counter N = 2^n.
Where N = number of stats and n = number of flip-flop,
So 32 = 2^5,
i.e. 5 flip flops are required for mod-32 counter.

Paramesh said:   9 years ago
How many flip-flaps are required to construct Mod-10 counter?

Nikitha said:   9 years ago
How many flip-flops are required for the modulo-19 counter?

AKr-@2007 said:   9 years ago
32 means 10000 (5 digits); So 32 mod need 5 FFs.


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