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Serial Throughput in UART: Bytes Sent in 15 Seconds at 9600 baud (Odd Parity, 2 Stop Bits)

Serial Throughput in UART: Bytes Sent in 15 Seconds at 9600 baud (Odd Parity, 2 Stop Bits)

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Sep 12, 2026

In an asynchronous UART/serial link, each transmitted character is wrapped in a frame consisting of a start bit, a configurable number of data bits, an optional parity bit, and configurable stop bits. The receiver and transmitter must agree on these framing parameters and the baud rate (bit timing).2

Given:

  • baud rate = 9600 bits/second
  • asynchronous mode
  • odd parity (so a parity bit is present)
  • two stop bits

So the total number of transmitted bits per character (frame) is:

  • 1 start bit
  • 8 data bits (typical UART byte framing; UART frames commonly use 5–9 data bits, with 8 being the standard “byte” width)2
  • 1 parity bit
  • 2 stop bits

Total bits per frame: 1+8+1+2=12 bits/character1 + 8 + 1 + 2 = 12 \text{ bits/character}

A key idea is throughput:

  • bits transmitted in time TT = (baud rate) ×T\times T
  • characters transmitted = (total bits) / (bits per frame)

Footnotes

  1. Asynchronous serial communication - Describes UART framing: start, data bits, optional parity, stop bits; 10 or 11-bit character frames depending on parity/stop. ↩

  2. A Guide to Serial Communications - Shows typical asynchronous serial structure: start bit, 5–8 data bits, optional parity, 1–2 stop bits. ↩ ↩2

  3. Understanding UART - Explains UART frame elements: 5–9 data bits, optional parity, 1–2 stop bits; parity depends on odd/even rules. ↩

Understanding UART (start, stop, parity framing)

Frame-bit budget for this UART configuration

We model one byte as one UART character with the frame structure described above.2

Mermaid overview:

Baud rate UART frame Parity bit Stop bits Throughput

Footnotes

  1. A Guide to Serial Communications - Shows typical asynchronous serial structure: start bit, 5–8 data bits, optional parity, 1–2 stop bits. ↩

  2. Understanding UART - Explains UART frame elements: 5–9 data bits, optional parity, 1–2 stop bits; parity depends on odd/even rules. ↩

Compute bytes sent in 15 seconds at 9600 baud with odd parity and 2 stop bits

  1. 1
    Step 1

    Total bits = 9600 bits/s×15 s=144000 bits9600\,\text{bits/s} \times 15\,\text{s} = 144000\,\text{bits}.

  2. 2
    Step 2

    Bits/frame = start 11 + data 88 + parity 11 + stop 22 = 1212 bits per byte.

  3. 3
    Step 3

    Bytes = total bits / bits per frame = 144000/12=12000 bytes144000 / 12 = 12000\,\text{bytes}.

  4. 4
    Step 4

    The computed result is 1200012000 bytes → option (ii).

Bytes transmitted in 15 seconds for each option (derived from UART framing)

Only (ii) matches the UART framing with 1 start + 8 data + 1 parity + 2 stop bits.

Common confusions in UART throughput problems

Pro Tip

Always compute bits/frame=start+data+parity?+stop\text{bits/frame} = \text{start} + \text{data} + \text{parity?} + \text{stop} first. Throughput then follows by dividing total transmitted bits by bits/frame\text{bits/frame}.

Warning about hidden assumptions

If the UART used 7 data bits instead of 8, bits/frame would be 1+7+1+2=111+7+1+2=11 and the byte throughput assumption would change. In most byte-based UART questions, 8 data bits is implied by the use of “bytes.”2

Footnotes

  1. A Guide to Serial Communications - Shows typical asynchronous serial structure: start bit, 5–8 data bits, optional parity, 1–2 stop bits. ↩

  2. Understanding UART - Explains UART frame elements: 5–9 data bits, optional parity, 1–2 stop bits; parity depends on odd/even rules. ↩

Final Answer

From the calculation:

  • bits in 15 s = 9600times15=1440009600 \\times 15 = 144000
  • bits per frame = 1+8+1+2=121+8+1+2=12
  • bytes in 15 s = 144000/12=12000144000/12 = 12000

✅ Correct option: (ii) 12,000 bytes

Knowledge Check

Question 1 of 4
Q1Single choice

In UART asynchronous framing, with 1 start bit, 8 data bits, parity enabled, and 2 stop bits, how many bits are sent per character?

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