Rcochran@ee,
I realized that I did not answer your question, I just gave you another way to do the same thing.
Your way will work but with a couple of changes.
1st change:
add a XIO (normally closed) contact after each counter dn bit from the respective control word. R6:0.EN R6:1.EN R6:2.EN
This is required because the counter needs to see a transition from off to on in order to count. If it is enabled all the time it will only count the first scan.
The R6:X.EN will open the logic to the counter after the BSR is enabled.
2nd change:
A Bit Shift shifts data into the register, not out of it. So in order to shift the bits to the right to align the nibbles to the least significant nibble of each word you need the length in the BSR to be 16.
Each time the BSR is executed it will insert what ever bit is at b10:49/0 into the most left bit (16) in the word and then move the bits one bit to the right.
One thing about doing it this way, it will require 24 scans to complete the BSR that is shifting 12 bits.
On the first scan (once I:2/15 in enabled) counter C5:2.CU in enabled.
With C5:2.CU enabled the BSR instruction will execute. It's done bit R6:2.DN will enable.
On the second scan with R6:2.DN enabled C5:2.CU will disable.
With C5:2.CU disabled, the BSR will not be enabled and R6:2.DN is disabled.
On the third scan with R6:2.DN disabled, we start the whole sequence of counting, bit shifting and then resetting.
Keep this in mind when using this logic that it will take 24 scans to complete. I don't know what else you are doing in the logic as you did not show all rungs. Since you are triggering from an input just make sure the data doesn't change while you are doing the bit shifts.
My previous example will complete in 1 scan.
...LaRoy
I am wanting to take a word and say chop up the hex value ie. word b9:0 = DCA9 
