Four Mutually exclusive bits.

gbradley · May 26, 2011

How can I turn off the other three bits when I select one of these four bits?
I’ve done this before, and never did like the results.
If Bit_A is on, Bit_A wins.
I can’t turn on any other bit unless I turn off Bit_A first

If Bit_B is on, Bit_B wins, over Bit_C, or Bit_D.
I can’t turn on Bit_C, or Bit_D unless I turn off Bit_B first, but…
If I turn on Bit_A, Bit_B turns Off and Bit_A turns on.

So on and so on until Bit_D, that one is last, so if it is on, turning on any other bit will shut it off.

Rube · May 26, 2011

What Processor are you using?

harryting · May 26, 2011

I didn't put a lot of thought into it but I think you need a buffer register or bit. Along the line of..

B3:6 - New
B3:5 - Now

then compare B3:5 and B3:6, if different, wipe out B3:5 and copy B3:6 into it.

Rube · May 26, 2011

Here's what I did in a CLogix:

it works for "a" and "b" but I'm still working on "c"....

TConnolly · May 26, 2011

I'm not sure if I completely understand your question. Are you saying that you need a bit priority, or complaining that its unsatisfactory because there is a bit priority and you have to undo a bit to change?

Assuming (you know what that means) it is the latter:
Why not use some one shots. That way there is no priority unless two bits simultaneously go true on the same scan.



    A                  B    C    D

---] [---[ONS]--------(U)--(U)--(U)--





    B                  A    C    D

---] [---[ONS]--------(U)--(U)--(U)--



    C                  A    B    D

---] [---[ONS]--------(U)--(U)--(U)--



    D                  A    B    C

---] [---[ONS]--------(U)--(U)--(U)--



note: OTUs are Controllogix style for simplicity

Also, don't forget the DCD (decode) instruction as another option.

gbradley · May 26, 2011

Rube said:
What Processor are you using?

Micrologix (1400)

TConnolly · May 26, 2011

Ooops, I meant to say "Don't forget the DCD instruction as another option" (I went back and edited). I have used it in the past when I wanted to guarantee exclusive bits for a state engine.

Rube · May 26, 2011

Looks like I uploaded the wrong PDF on my last reply...

gbradley · May 26, 2011

Alaric said:
I'm not sure if I completely understand your question.

I knew I wasn't explaining it well.

Remember the old time radio with the preset stations.
You could select any station that you wanted.
You click off station1, and then go to station6, then back to station 2. Any order is possible. That's what I'm trying to do.

harryting · May 26, 2011

Alaric said:
..

That way there is no priority unless two bits simultaneously go true on the same scan.

..

That's the problem as I see it and I don't see any way to do this without using a buffer (unless it's binary, that's any why for OP a/b works but not a/b/c/..)

A buffer is need to tell the PLC what's new vs what was before. It's what I call the "evil twin" problem: one day the evil twin of your friend shows up, both claiming to be your friend. How can you tell which one is telling the truth

Peter Nachtwey · May 26, 2011

Bit hacks

search for bit hacks or bit twiddling. I will let you know when you have the right answer. It should be simple, cleanand doable in one rung.

harryting · May 26, 2011

Well, on 2nd thought, Alaric's solution works if using a oneshot along with move the numeric value into the word.

Code:

--[A]----[ONS]-----[MOV 1]
--[b]----[ONS]-----[MOV 2]
--[C]----[ONS]-----[MOV 4]
--[D]----[ONS]-----[MOV 8]

Peter, from I can see on wikipedia, bit hax use a buffer as well

sthompson · May 26, 2011

If I understand what you are trying to do correctly. Would something like this work for you?

Stu...

TConnolly · May 26, 2011

In a radio button set you want the last button pressed.

Hacks can isolate a single bit, for example I & !(I - 1) isolates the low order bit. But what if the low order bit was not the last to be set? What if the high order bit was not the last one to be pressed? How to tell which was last?

Detecting the rising edge of the bit state will tell you which one was last pressed.

ndzied1 · May 26, 2011

Here's what I do.

Then:

B3:0/0 -> Station 1
B3:0/1 -> Station 2
B3:0/2 -> Station 3
B3:0/3 -> Station 4

Also, you cannot use the rest of the word B3:0 for anything else but usually there is enough room for you to burn the extra bits.

Four Mutually exclusive bits.

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