Binary counter

I am trying to make a 4 bit counter with jk flip flops with the sequence of 0, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5. what will the state table and the k maps look like

by me45lol
November 16, 2021

If you look at the sequence, except for the first few terms, it is counting by 2 ( 1, 3, 5, 7, 9, 11, 13, 15 repeat). So, basically, it is a counter 2 by 2.

If you look at the outputs { Bit0, Bit1, Bit2, Bit2 }, those make the repeated sequence (Bit0 is the least significant bit).

Note that initially Bit0 is at 0, but the OR gate latches to a logical High after the first clock front. So, the generated numbers will all be odd, except the initial zero.

The same technique is used to skip the initial 1 and 3: An OR gate latches to High as soon as the first 5 is reached. AND gates are added to silence the initial Bit1 and Bit0, so, if we read { Bit0_prime, Bit1_prime, Bit2, Bit3} we get the exact required sequence (Ok, the initial zero stays on for a "long" time, but I assumed that we can live with that hick-up.)

That is the kind of thing that we could more easily do, and customize, through software though. But the question was about doing it using flip-flop. Well, at least, that is what I understood. Sorry for the "table" and Karnaugh map, my wrong, but I failed to see their usefulness here.

by vanderghast
November 16, 2021

1 Answer

Answer by unaflores

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0 votes
by unaflores
November 20, 2021

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