Baudrate - Can someone check my math please..

pacman

Ok,

I'm writing a modbus (slave) implementation and I need to figure out the 3.5 character spacing at the end of an RTU frame.

Now, I _think_ I've got the right maths, but can someone please check my thinking before I madly code...

9600 baud = 1 bit in 1/9600 of a second.

A character (7,E,2) is 11 bits long.

Thus a 3.5 character 'frame end' space is 3.5 * 11 * 1/9600 of a second.

Which would be just a shade over 4mS.

Am I correct?

kwinn

7,E,2 would be 7 data bits, one parity bit (even) and 2 stop bits. With one start bit that comes to 11 bits. At 9600 bps each character would take 1.14583333...etc milliseconds.

pacman

Thanks Kwinn. means my 'maffs' is right.

Modbus RTU uses a 3.5 character space of no transmission to signify the end of a message/start of the next message.

So if I get no data for 3.5 * 1.14583333...etc milliseconds, then the message is over and I should actually process it.

Thanks again for the sanity check.....

kwinn

Yes, and you can use something like "char := uarts.rxtime (0, 4)" to see if the time between characters is 3.5 character times (4.01042mS at 9600 baud) if you are using "pcFullDuplexSerial4FC" (the 4 port version of "FullDuplexSerial") for communications.

FullDuplexSerial has a similar function with a slight difference in the parameters.

The call returns the next character or -1 if it timed out.

Oops....uarts.rxtime function accepts only integer mS values for timeout. I've got to stop posting answers so late. Make too many silly mistakes.