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After reading the appendix on serial communication, I’m a bit confused about the baud rate and how it relates to the bytes per second transfer speed in serial communication. If I understand it correctly in serial communication the baud rate is the same as the bits per second because the signal can only represent one bit (0V for 0, and 5V for 1) But in the book it says:

“With a baud rate set to 9600 you can then send 9600 / 10 / 8 = 120 bytes per second.”

We can send 9600 bits per second but for each byte of data we have to send an extra two bits (start and stop) so for each byte of data we have to send 10 bits. So we get 9600 / 10 = 960 bytes per second. So shouldn’t the transfer rate be 960 bytes per second and not 120? Why is 960 divided by 8?

Kalle:

Thank you very much for pointing this out!

You are right and the correct result is 960 bytes per second. At least if every bit gets transferred in exactly one transmission step.

I have fixed it.

Best, Maik

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