# Digital Electronics - Counters - Discussion

### Discussion :: Counters - General Questions (Q.No.1)

1.

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

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

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.