Digital Electronics - Counters - Discussion

1. 

How many flip-flops are required to make a MOD-32 binary counter?

[A]. 3
[B]. 45
[C]. 5
[D]. 6

Answer: Option C

Explanation:

No answer description available for this question.

Ashwin said: (Dec 17, 2010)  
Mod-32 means divide by 32 counter.

2^n=32. => n=5.

i.e. 5 flip flops.

Nikita said: (Jun 4, 2011)  
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.

T.Milhaan said: (Aug 5, 2011)  
2^n>N>2^n-1
n-number of flip flops
N-number of states ie in mod32 N=32

Abdul Wahab Khokhar said: (Oct 7, 2011)  
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.

Ankita said: (Jul 15, 2012)  
2^5=32.

Hence, it required 5 flip flops.

Gopi said: (Sep 20, 2012)  
Mod-32 means divide by 32 counter.

2^n=32. => n=5.

i.e 5 flip flops.

Sampath said: (Sep 30, 2012)  
How can you people say that and represent no of flip flops ? CAN anybody explain it?

Neha said: (Oct 12, 2012)  
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: (Sep 19, 2013)  
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.

Neelima said: (Oct 30, 2013)  
N <= 2^n.
Where N is modulus of counter.

n is no.of flip flops.

So when n=5 condition is satisfied.

Swathi said: (Nov 8, 2013)  
2^n = N.

Where n is no.of flip-flops.

N is no.of states.

Mallikarjuna said: (Nov 16, 2013)  
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: (Aug 27, 2015)  
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.

Manisha said: (Sep 10, 2015)  
What is full form of MOD?

Rahul Sharma said: (Jan 14, 2016)  
MOD 32 means that counter can justify 32 states.

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

Niraj Sharma said: (Apr 29, 2016)  
32 mod means till 5 digit number.

So 5 flip flop is required.

Akr-@2007 said: (Jul 7, 2016)  
32 means 10000 (5 digits); So 32 mod need 5 FFs.

Nikitha said: (Sep 14, 2016)  
How many flip-flops are required for the modulo-19 counter?

Paramesh said: (Sep 21, 2016)  
How many flip-flaps are required to construct Mod-10 counter?

Usha Kiran said: (Nov 18, 2016)  
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.

Daniel said: (Apr 3, 2017)  
What is the least number of jk flip flops it would take to create a mod 500 counter?

N.Kiranmai said: (Apr 18, 2018)  
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.

Piyush Kumar said: (Jul 2, 2019)  
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.

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