Counting bad parts

[gblack 0]

I have an application where I need to shut down a machine if 5 of the last 10 parts were bad. I wrote some code that uses a bit shift to set bits in a word for every bad part, but I can not find a simple way to count the bit on in a word. I did come up with a method of using these bits to count into another word and it works fine for 10 parts but would get pretty complicated if I have to expand it to 20 or 30 parts. Does anyone know a simple way to get a percentage of bad vs good parts? The application is using a micrologix 1000 with an input to show a part is in the test area and an input that identifies the part as good.

[TConnolly 13]

OK, I haven't tested this, but RSLogix will let you enter an address like N7:0/35 which is N7:2/3. I'm pretty sure that as long as you don't index past the end of the file you won't get an error. So you should be able to use a FOR loop to count through the entire shift register using an indirect address Let the bit shift register be 3 words long N7:0 to N7:2. We will use a FOR NEXT loop for 0 to 47 (48 loops) Create the following: SOR CLR N7:10 EOR SOR CLR N7:11 EOR SOR LBL 1 XIC N7:0/[N7:10] ADD N7:11 1 N7:11 EOR SOR ADD N7:10 1 N7:10 EOR SOR LES N7:10 47 JMP 1 EOR I'm pretty sure this will work. I'll load it up and check it tomorrow. For a 10 bit shift register, modify the loop termination value from 47 to 9. EDIT: Nevermind. I forgot that the ML1K does not support indirect addressing. It won't work on the ML1K. But it will work on ML1500 and above. Edited 19 Oct 2005 by Alaric

[TechJunki 0]

Take a look at something like this, count the bits with a "Counter" or "ADD" function that are loaded into the bit shift, then subtract them back out after part 10.. in this case at bit B3:0/11. Edited 19 Oct 2005 by TechJunki

[Spedley 0]

I would do something like this ... SOR BST XIC PART_PASSED NXB XIC PART_FAILED BND OSR OSR_BIT BST XIC HISTORY/0 SUB COUNT 1 COUNT NXB OTU HISTRY/0 NXB DIV HISTORY 2 HISTORY NXB XIC PART_FAILED BST ADD COUNT 1 COUNT NXB OTL HISTORY/9 BND BND EOR SOR GEQ COUNT 5 OTE STOP EOR where: HISTORY is a word containg the last 10 results COUNT is a word containg the number of fails in the last 10 parts It uses divide by 2 instead of bitshift (making sure bit 0 is clear otherwise it will round-up) and stores the latest result in bit 9 to reset clear HISTORY and COUNT Edited 19 Oct 2005 by Spedley

[jstolaruk 9]

I would just copy and paste a bunch of rungs inline. Who cares if there is a rung for each bit? CLR n7:0 XIC b3/0 ADD n7:0 1 n7:0 XIC b3/1 ADD n7:0 1 n7:0 XIC b3/2 ADD n7:0 1 n7:0 XIC b3/3 ADD n7:0 1 n7:0 XIC b3/4 ADD n7:0 1 n7:0 XIC b3/5 ADD n7:0 1 n7:0 XIC b3/6 ADD n7:0 1 n7:0 XIC b3/7 ADD n7:0 1 n7:0 XIC b3/8 ADD n7:0 1 n7:0 XIC b3/9 ADD n7:0 1 n7:0 GEQ n7:0 5 OTE bad_cond Very easily read by the guy that has to support it later on down the road on the shop floor. At a glance he knows exactly what the goal is. I won't get any phone calls asking me about this. Edited 19 Oct 2005 by jstolaruk

[Spedley 0]

How would you set the bits in B3:0 ?

[jstolaruk 9]

Oh, you need the bit shifing logic too? For anything less than 16 parts you can use XIC bad_part OTE B3/0 XIC shift_parts OSR temp MUL B3:0 2 B3:0 If more then 15 XIC bad_part OTE B3/0 XIC shift_parts OSR temp OTE many_shifts XIC many_shifts BSL #B3:0 R6:0 B3/0 1 XIC many_shifts BSL #B3:1 R6:1 R6:0.UL 1 ... XIC many_shifts BSL #B3:n R6:n R6:n-1.UL 1 then test CLR n7:0 XIC b3/1 ADD n7:0 1 n7:0 XIC b3/2 ADD n7:0 1 n7:0 XIC b3/3 ADD n7:0 1 n7:0 XIC b3/4 ADD n7:0 1 n7:0 XIC b3/5 ADD n7:0 1 n7:0 XIC b3/6 ADD n7:0 1 n7:0 XIC b3/7 ADD n7:0 1 n7:0 XIC b3/8 ADD n7:0 1 n7:0 XIC b3/9 ADD n7:0 1 n7:0 XIC b3/10 ADD n7:0 1 n7:0 .... XIC b3/n ADD n7:0 1 n7:0 GEQ n7:0 ??? OTE bad_cond Edited 19 Oct 2005 by jstolaruk

[TConnolly 13]

The following program will work in a ML1000 to count the number of bits set in a 16 bit word. It has been tested. Rung 0: SOR BST MOV N7:0 N7:1 NXB CLR N7:5 NXB CLR N7:3 BND EOR Rung 1: SOR LBL 1 BST XIC N7:1/0 BST ADD 1 N7:3 N7:3 NXB OTU N7:1/0 BND NXB DIV N7:1 2 N7:1 NXB ADD 1 N7:5 N7:5 NXB LEQ N7:5 15 JMP 1 BND EOR Rung 2: SOR MOV N7:3 N7:4 EOR You can adjust it for 10 bits by changing the LEQ instruction in rung 3 from LEQ N7:5 15 to LEQ N7:5 9 The contents of N7:0 are copied into N7:1. Next a For loop is executed against N7:1. This is the general format: For I:= 0 to 15 If the least significant bit of N7:1 (N7:1/0) is set, increment the bit count by 1 and clear the bit Divide N7:1 by 2 End For The final bit count result is copied into N7:4.