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Hi, How do you calculate how fast bytes will arrive for a particular bause rate? Someone told me somewhere that a baud rate of 115200 means 1 byte will arrive every 51.2us. I cant find any relationship between the 2 numbers though. I read a wiki page about baud rates but i still dont have a clue
This may help
http://www.tech-faq.com/difference-b...aud-rate.shtml
Unless you are dealing with a modem, bit rate and baud rate are pretty much the same.
115200 is BITS per second. With one start bit and one stop bit you have 10 bits per data byte. 115200 / 10 = 11520 BYTES per second. Each byte takes 86.8uS. The 51.2uS figure is about the period of a BIT at 19200 bps.
I still dont quite understand how baud and bits are related. That one sounds confusing. Im not dealing with a modem. Its just a normal RS232 connection betweeen my PC and a PIC chip (with RS485 in the middle).
dhouston, if the baud and bps are the same then that looks quite easy. Its usefull to know that each byte takes 10 bits too instead of 8
This might clarify (or not) the difference between baud and bps.In your simple application baud = bps so the calculation is straightforward.
Using 8N1, there are ten bits transmitted per byte. Using other settings there may be more or fewer bits.
I think it just clicked. I had already read that page but i missed 1 important thing. I was confused because it seems that 1 change of state can represent more than 1 bit of data. I just noticed on that wiki page that it says "a symbol may have more than two states".
As far as i knew there were only 2 states in serial. Thats what im using anyway so yeah, as you say "In your simple application baud = bps". I am also using 8N1.
Thanx for your help. I can make more reliable calculations now