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    <p>Great question! Most asynchronous communication also specifies a <em>stop</em> bit, which is the complement of the start bit, ensuring each new symbol begins with a stop-to-start transition.</p> <p>Example: let's transmit the characters <code>ABC</code>, which are ASCII 65, 66, and 67:</p> <pre><code>A = 65 = 0x41 = 0100 0001 B = 66 = 0x42 = 0100 0010 C = 67 = 0x43 = 0100 0011 </code></pre> <p>Let's also assume (arbitrarily) that the start bit is <code>0</code> and the stop bit is <code>1</code>, and the data is transmitted from MSB to LSB. The transmitter will be in the stop (<code>1</code>) state when no data is transmitted. So the receiver might see this:</p> <pre><code>Data: ....1111 0010000011 111 0010000101 0010000111 11111.... (quiet) ^ A $ ^ B $ ^ C $ (quiet) </code></pre> <p>With apologies for the ASCII graphics, the data consists of a series of stop (<code>1</code>) bits while the channel is idle. When the transmitter is ready to send a character, it sends a start (<code>0</code>) bit (marked with <code>^</code>), followed by the character code, and ending with a stop (<code>1</code>) bit (marked with <code>$</code>). It continues to send stop bits until the next character is transmitted, beginning with another start bit.</p> <p>The reason we use start <em>bits</em> instead of <em>bytes</em> is efficiency. The scheme above requires 10 bits (1<sub>start</sub> + 8<sub>data</sub> + 1<sub>stop</sub>) to transmit 8 bits of data, resulting in an <em>overhead</em> of (10 - 8) / 8 = 1/4 = 25%. If we used start and stop <em>bytes</em>, we'd need to transmit 3 bytes for each byte of data, which would be an overhead of (3 - 1)/1 = 2 = 200%. If the start, data, and stop bytes were each 8 bits, we'd have to transmit 24 bits instead of 10 for each character, so it would take almost 2 1/2 times as long to send the data!</p>
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    1. USAdam Liss
    1. USAdam Liss
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    1. COIt seems like in your example that there is a "quiet"/idle phase. In a program, would I check for this with a timeout?
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    2. COI have never seen an implementation behave the way you describe, continually sending stop bits while the line is idle, that doesn't make any sense.
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      1. USRobert Gamble
    3. COI'm a hardware guy. If you check the voltage on the receive line of your PC's serial port, it will *always* be either at the 0 or 1 level (or rapidly transitioning between them). The point of the start bit is to force a transition from "stop" -- which is the same as "idle."
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