Omron's MOVD (read with caution)

CAUTION: This will be a long post.

I get a little confuse with MOVD (MOVE Digit) instruction of omron PLC.

According to an explanation I googled, a MOVD instruction has 3 elements: Source (S), Control word (C), Destination (D). The Source (S) is the value to be copied, control word (C) is the manipulator of the "movement", destination (D) is the destination of the copied value starting from the digit specified in C and filling in to the left.

Example:

MOVD DM0 #0132 DM10, DM0 = 5678 (D3=5, D2=6, D1=7, D0=8)

S = DM0
D = DM10
C = 0132 (D3=0, D2=1, D1=3, D0=2)
D0 = 2 (Specifies the 1st digit copied from source, the value must be
from 0 - 3)
Hence, the D2 of DM0, which 6, will be the first to be transferred
to the location specified by D2 of C.
D1 = 3 (Indicates the number of digits to be copied, 0=1 digit, 1=2
digits, 2=3 digits, 3=4 digits)
Since D1 of C = 3, hence we will move all 4 digits of Source to
Destination.
D2 = 1 (Indicates the starting digit that the data will be moved into
the Destination)
The D2 of DM0, which is 6, will be moved to D1 of the
Destination. The rest of the digit will be copied in the same pattern and
wrapping back around to D0.
D3 = 0 (The MSB must always be zero)

Hence result of the "movement" will be:

MOVD DM0 #0132 DM10 (DM0 = 5678, DM10 = 8567)

Continue to next post.........

If I have:

MOVD DM0 #0000 101, DM0 = 3 (D3=0, D2=0, D1=0, D0=3)

Hence,

101 = 3
102 = 0
103 = 0
104 = 0

However, 101 is a bit. So how it is going to function having a value of 3?

MOVD is only moving digits within a single word.
You need to look at the word as B.C.D so 4 digits per word

hexdec

theripley said:

If I have:

MOVD DM0 #0000 101, DM0 = 3 (D3=0, D2=0, D1=0, D0=3)

Hence,

101 = 3
102 = 0
103 = 0
104 = 0

However, 101 is a bit. So how it is going to function having a value of 3?

Click to expand...

101 is a CIO address 101.00 - 101.15 would be a bit address

I see.

Hence,

D0 = 101.00-101.03
D1 = 101.04-101.07
D2 = 101.08-101.11
D3 = 101.12-101.15

Meaning if DM0 = 3 with C = 0000, then

101.00 - 1
101.01 - 0
101.02 - 1
101.03 - 0

theripley said:

I see.

Hence,

D0 = 101.00-101.03
D1 = 101.04-101.07
D2 = 101.08-101.11
D3 = 101.12-101.15

Meaning if DM0 = 3 with C = 0000, then

101.00 - 1
101.01 - 0
101.02 - 1
101.03 - 0

Click to expand...

Almost
3 in binary is 0011
So :
101.00 - 1
101.01 - 1
101.02 - 0
101.03 - 0

chelton said:

Almost
3 in binary is 0011
So :
101.00 - 1
101.01 - 1
101.02 - 0
101.03 - 0

Click to expand...

Thank you for the correction, I get my binary wrong.