Hi,

I am just beginning to use the Modbus protocol and I need some final informations. I just want to be sure I have well understood what I have just read in the docs.

I am using this configuration: RTU and no parity and want to send a request to one slave.

What I have understood is that I have to send 1 start bit and 2 stop bits every time I want to send 8 bits of data (I mean slave, function...).

But what I didn't find in the doc is what is the start/stop/idle bit value. I think it would be logic that the start/stop bits are different from the idle one. Is that true? If it is, is the idle value set to 1 or 0?

Thank you for your help.

 

Don't worry about the start/stop/idle bits - your hardware will handle them correctly. Just configure the port to use two stop bits (the same way as you configure the baud rate) and it'll do it for you.

Jiri
--
Jiri Baum <[email protected]> http://www.csse.monash.edu.au/~jirib
MAT LinuxPLC project --- http://mat.sf.net --- Machine Automation Tools

 

Hi

The start, stop bits are rs232 standard. You can learn about this at:
http://www.camiresearch.com/Data_Com_Basics/data_com_tutorial.html

Regards
Andrzej
http://www.modbus.pl

 

A note about "No Parity" per the modbus spec. It is actually MARK parity, so it looks like 2 stop bits. But you can just use MARK parity and transmit 1 stop bit. Most UARTS do this.

 

As others have mentioned you are actually sending a 1 Mark parity bit and 1 stop bits. AB/DF1 in 8,N,1 also does this.

However, direct to your question: in normal async an idle line is a "1" bit - RS-485 generally needs bias to force an idle line into this state when all transmitters are off/in tri-state.

The Start bit is defined by the fall from "1" to "0" (ie: is always "0" and this starts your async baud-rate machine to clock out the bits for this one character. The Stop bits will be "1" , but you'll only see a rise to it if the last data bit is "0" (or a "0" in parity field).

- LynnL, www.digi.com