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.
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.
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.
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.
Manisha said:
10 years ago
What is full form of MOD?
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).
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.
AKr-@2007 said:
9 years ago
32 means 10000 (5 digits); So 32 mod need 5 FFs.
Nikitha said:
9 years ago
How many flip-flops are required for the modulo-19 counter?
Paramesh said:
9 years ago
How many flip-flaps are required to construct Mod-10 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.
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.
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