digital question update

It looks to me that you are simply shifting a pattern across the three bits, so a shift register will probably be the simplest way to move the pattern. Then you just have to come up with the logic that looks at the present state of the shift register (including all the states you never expect to see, once the pattern has shifted a couple times) and produce the new bit value to shift into the end of the register on the next clock.

I think i should do just a 0 to 5 counter and at the output a small translation circuit.

ken

Ken, I thought that I understood what you are trying to do, as as John posted, a simple shift register would be the simplest solution. Still, on your post with more information, it becomes far more complicated...but I am having a devil of a time trying to understand your example data. I suppose this is because you are showing a sequence of 12 bits, not three, and rather than a simple 3-bit shift, some sort of a binary counter seems to be at play. I believe this is where you lost me.

What had envisioned from your original post was that:

State 1 = 001 State 2 = 010 State 3 = 100 State 4 = 001 Etc.

Quite obviously, this result is easily achieve with a simple recirculating 3-bit shift register.

I don't understand where, in light of your original statement, values of 011 are consistent with your originally stated need, possibly because I don't understand how to read your state table and/or what you are trying to do.

Harry C.

On 2006-05-30, Ken O

You can simply use a CD4017 with bit 3 wired to the reset input.

vic

here is my truth table

Present state Next state Output A B C A B C Da Db Dc Y3 Y2 Y1

here is the result from a 3 shift adder:

Y3=B Y2=C Y1=B'C'

Da = Db=C+B Dc=C'

I am using three D flip flop, but what i find weird is that there is no output from the first flip flop

Ken

What you are saying this chip gives the output at pin 1, then after a clock outputs at pin2 , then pin 3 etc... ?

Ken

ya thats it, thanks a lot guys

ken

On 2006-05-31, Ken O