Describe the UART frame structure and calculate its efficiency.

A UART frame consists of four parts transmitted in sequence: (1) a start bit (always logic low) that signals the beginning of a character, (2) 5 to 9 data bits (8 is standard), sent LSB first, (3) an optional parity bit for single-bit error detection, and (4) 1 or 2 stop bits (always logic high) that mark the end of the frame. Between frames the line idles high — this is what makes the falling edge of the start bit detectable.

For the most common configuration, 8N1 (8 data bits, no parity, 1 stop bit), each character requires 10 bit periods: 1 start + 8 data + 1 stop. The protocol efficiency is 8/10 = 80%. At 115200 baud, this gives a maximum throughput of 115200 / 10 = 11520 bytes per second, or approximately 11.25 KB/s. Adding parity (8E1 or 8O1) makes it 11 bit periods per character, dropping efficiency to ~73%. Some protocols like DMX512 use two stop bits (8N2), further reducing throughput to 8/12 = 67%.

A common interview trap: candidates confuse "baud rate" with "data rate." Baud rate is the number of symbol transitions per second (bits per second for binary signaling). Data rate is lower because of the framing overhead.