RS232 Baud Rate Question - How do I calculate the transmit time for 8 bits of data?

[idealpcsolutions]

Hi All,

Ref. RS-232

My question is at 19200 baud, 8 data bits, 1 stop bit, and no parity - how do I calculate the transmission time for a character (8 bits) accounting for the overhead of the start and stop bits?

is 19200 baud simply 1/19200 of a second, and therefore time for a single character = (1/19200)x(8+1+0) seconds or is the stop bit a longer duration?

Thank You

 

[barry]

Yes, the baud rate is the number of bits-per-second. Stop bits can generally be selected to be 1, 1.5, or 2 bits long. It's a bit confusing because the selection of number of stop bits only affects the transmission. If you expect to receive 1 stop bit, but get 2, there's no harm done, but it will take 1 bit time longer to receive each character.

 

[idealpcsolutions]

Ah just thought though - one start bit too?

So is it fair to say that at 19200 baud, 8 data bit, no parity, 1 stop bit equates to 19200/10 characters per second?

 

[horace1]

Yes, the baud rate is the number of bits-per-second. Stop bits can generally be selected to be 1, 1.5, or 2 bits long. It's a bit confusing because the selection of number of stop bits only affects the transmission. If you expect to receive 1 stop bit, but get 2, there's no harm done, but it will take 1 bit time longer to receive each character.

the problem is when data is returned and the receiver expects two stop bits and the received character only contains one you will get corrupt data if it is immediatly followed by the start bit of the next character

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Ah just thought though - one start bit too?

So is it fair to say that at 19200 baud, 8 data bit, no parity, 1 stop bit equates to 19200/10 characters per second?

yes, there is only one start bit but can be 7 or 8 (sometimes 9) data bits, parity bit if required and one or two stop bits