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Q1. How many flip-flops are required to construct mod 4 counter? 

Ans - 2  right ? Alway it should be 2 or it may not be 2.

Q2 . If We want to design a synchronous counter that counts the sequence 0−1−0−2−0−3 and then repeats.

        A) What is mod of this counter ?  Ans -6 right ?

        B) The minimum number of J-K flip-flops required to implement this counter is ?

Now if we follow Q1 then ans for Q2. B ) should be log26 = 3. But the correct ans should be 4.

So, Please confirm ans for Q1.  ?

As we know that Modulus counters, or simply MOD counters, are defined based on the number of states that the counter will sequence through before returning back to its original value.

Some problems related to this -

https://gateoverflow.in/12684/how-many-flip-flops-are-required-to-design-modulo-272-counter

https://gateoverflow.in/82111/is-bcd-or-mod-10-counter-are-same

https://gateoverflow.in/39670/gate-2016-1-8

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Answer to 1 - 2  flip flops will be sufficient. but it's not always the case. A ring counter has 4 flipflops still mod 4 counter. It's because some states are not involved in counting. A mod counter is a counter which can actually count something from it's counting states. a counter of 2 bit which has all states like . 0 goes to 0, 1 goes to 1 , 2 goes to 2, and 3 goes to 3. is useless and can't count anything.

Answer to 2 - special type of question and should look at it and understand it from the best answer there.

I think I answered all the questions.

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ajaysoni1924 asked May 22, 2018
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