Discussion
A UART driver receiving at 115 200 baud, 8N1, must accept a byte every
Answer: D
Correct answer: about 87 \( \mu \)s
8N1 sends ten bits per byte - a start bit, eight data bits and a stop bit - so the byte rate is \[ \frac{115200}{10} = 11{,}520\ \mathrm{bytes\ per\ second}, \] one every 86.8 microseconds. Dividing the baud rate by eight instead of ten gives a figure that is ten per cent too optimistic, which matters when sizing the interrupt budget.
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