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How does UARt determine the end of transmission of a series of bits?

The single stop bit could be apart of the transfer data?

How does UARt know the previous 1's were not the end of transmission?

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    \$\begingroup\$ Don't fill every answer with "thank you" notes, there's a button for upvoting and one for accepting. \$\endgroup\$
    – pipe
    Commented Jan 6, 2021 at 15:53

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The UART actually does know it. When you start using the UART, you must configure all parameters, like at what bit rate the bits are sent, and the number of data, parity (if any) and stop bits it is expected to receive, and so the stop bit always is the last bit of an asynchronous start-stop frame.

So the start bit starts a frame, which includes the all the bits and stop bit is always last.

If the receiver is expecting a high stop bit after 8 data bits, but if the stop bit is not high, then most UARTS will signal a framing error, so the user can know that there was an error receiving this byte, and it can then be ignored or reacted upon, like trying to resync after an error.

Which actually is how sending a "break" signal works. The transmitter sends logic 0 for very long time, so that the receiver sees it as start bit, and the data (and parity) bits will be 0 too, and then stop bit is also 0, so receiving a full 0x00 byte with framing error is a sign of receiving a break condition.

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When you configure your UART you need to specify not only the baudrate but also the number of bits. Your example has configured it to 7 data bits and one parity bit, thus the receiver knows that the first 8 bits can not be stop bits.

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A UART receiver has to be "told" beforehand how many data bits there are and whether parity is used or not and, if used, whether it's even parity or odd parity or forced parity. It also has to be "told" beforehand the data rate. Sometimes, it's even necessary to inform a UART that there might be two stop bits before the next byte is received.

So, on this basis, it will know when the end of the transmission is.

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