Write Instruction

[JDHammond 0]

I am writing from one DO6 to another DO6. Im trying to write whats in v2500,V2501,V2502,V2503,V2504,V2505,V2600,V2601,V2602,V2603,V2604,V2605,V2700,V2 701,V2702,V2703,V2704,V2705 to the other DO6 in the same data files. Im writing it as LD K158 LD K? How many bytes would I need to transfer? LDA ? When I get here it will not let me enter V2500 for my starting address? WX ? Will it acecpt the V2500? Thanks

[b_carlton 61]

The first operand is problematic. The main CPU manual says that the address should be in BCD but the D0DCM manual (which you are apparently using given the '1') says the address should be in hex. How many bytes? The total for the entire span would be 268 bytes which I think is beyond one message capability. You may want to break this up into 3 instructions. Each would transfer 12 bytes. The LDA command take an Octal value. In this case you would write O2500 (that's the letter 'O'). Of course if you are breaking it up then the other two LDA's would be O2600 and O2700. Finally the WX would take V2500 (then V2600, finally V2700). Of course this whole group would be conditioned by SP120. I would recommend using stages to seperate the 3 WX groups so that they transfer sequentially.

[JDHammond 0]

I am already writing to another DO6 with this plc and Im using K0163 where the zero is the base # of the master,1 is the ECOM slot number, and the 63 is the module ID. The module ID on this computer is 58 is why i was going to use the K158. Ok I was trying to enter OV2500. So just use the O2500. The WX would be the V2500. Im updating these registers every 8 hours for data logging. So I could just write every 8 hours for each shift. Ill give it a try! Thanks