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 2 of 3.
Niraj sharma said:
9 years ago
32 mod means till 5 digit number.
So 5 flip flop is required.
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).
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.
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.
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.
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.
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?
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