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 2 of 3.

Niraj sharma said:   9 years ago
32 mod means till 5 digit number.

So 5 flip flop is required.

Rahul Sharma said:   10 years ago
MOD 32 means that counter can justify 32 states.

Now, these 32 states can be represented using 5 bits (2^n = 32 => n = 5).

Manisha said:   10 years ago
What is full form of MOD?

Krishna Kumar said:   10 years ago
N = 2^n, for binary counter.

Taking log where N = No. of state.

N = No. of flip flop.

N is equal to logarithm of N to the base 2.

Hence n = 3.32 logN to the base of 10.

N = 3.32*1.5051 = 4.997.

Thus n = 5 flip flop.

Mallikarjuna said:   1 decade ago
We can count up to 31 in mod 32 so 5 bits are enough to count, because if all 5 bits are 1's its equals to 31. i.e 11111 = 31.

Swathi said:   1 decade ago
2^n = N.

Where n is no.of flip-flops.

N is no.of states.

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.

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.

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.

SAMPATH said:   1 decade ago
How can you people say that and represent no of flip flops ? CAN anybody explain it?


Post your comments here:

Your comments will be displayed after verification.