Sqo

Hi,

I have on a rung SQO #B39:0 0FFFFH B35:0 R6:0 20 0.

1) With each scan, this rung is activated. Is that a true statement?
2) If so, does the next scan of this rung start at position 0 again?
3) If not, then would you need to advance the position pointer in a different part of the logic (ie the last variable of SQO)?

Thanks,
Dai

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The SQO instruction is edge triggered so you need a false to true transition leading up to it to get the instruction to do anything. I have seen people try to cheat edge triggered instructions by putting them on a rung unconditionally and unlatching the .EN bit in a branch after the instruction. I don't know if that will work here. But as a general rule you need something on the rung before the SQO to give it a false to true transition.
The position pointer will incrememnt by one for every false to true transition. The pointer resets to one on the transition when the pointer equals the length. That's one to be a bit careful of. The position resets to 1, not wht is in the position field when you program the instruction.

Keith

Dai_Ly said:

3) If not, then would you need to advance the position pointer in a different part of the logic (ie the last variable of SQO)?

[email protected]

Click to expand...

Yes. What you've got here is a 'one-of-twenty one' data selector. Set up as it is you can selectively (read randomly) bring out any of the file words to the output and you don't need and transition logic in front of it to do this. Using the instruction with conditional logic to trigger it you can only advance the POS value of the R6 register.