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digital question update

Hello,

I am still working on my digital cicuit. I wanted to know what would be the best way to do this. I need to output

001 then 010 then 100, then it repeat itself.. So a counter with no input, and a single clock that will change the output. Some people told me to use flip flop other say asychronous. I do not know hat to do. here is what i did. would it be posible configuration ?

Present state Next state Ta Tb Tc Output Y1 Y2 Y3 Y1 Y2 Y3 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1 0 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 0 1 1 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 1 1 1 0 1 0 0

output: Ta =A+BC' Tb= A'+B+B'C' Tc=B'C+A'C+ABC' Y1=BC+AB'C Y2=A'BC'+AB'C Y3=A'C+ABC'

thanks

ken

Reply to
Ken O
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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.

Reply to
John Popelish

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

ken

Reply to
Ken O

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.

Reply to
hhc314

On 2006-05-30, Ken O

Reply to
Jasen Betts

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

vic

Reply to
vic

here is my truth table

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

0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 1 x x x x x x x x x 1 0 0 x x x x x x x x x 1 0 1 x x x x x x x x x 1 1 0 x x x x x x x x x 1 1 1 x x x x x x x x x

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

Reply to
Ken O

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

Ken

Reply to
Ken O

ya thats it, thanks a lot guys

ken

Reply to
Ken O

Yes. That's how I used to do when I needed less than a 10-counter.

-- Sincerely, |

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Rikard Bosnjakovic | Code chef - will cook for food

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Reply to
Rikard Bosnjakovic

On 2006-05-31, Ken O

Reply to
Jasen Betts

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