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 1 of 3.
Piyush Kumar said:
6 years ago
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.
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.
(1)
RISHI NIGAM said:
3 years ago
Yes, 2^n = 32 from where we get the answer is 5 is a generalized way of getting no of flip-flops and remember id mod is in between means not multiple of 2.
So we just higher 2 multiple for example 14 then we take 4 flipflops as we take count as 16.
So we just higher 2 multiple for example 14 then we take 4 flipflops as we take count as 16.
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.
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.
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.
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).
Nikita said:
1 decade ago
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.
=> 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.
N.kiranmai said:
7 years ago
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.
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.
Abdul wahab khokhar said:
1 decade ago
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.
2^5=32
That is required only 5 flip flop.
If there is mod-30 then also it requires 5 flip flop.
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