Tricky Circuit

UART: an 8N1 frame

Turn a byte into a timed message

UART means Universal Asynchronous Receiver/Transmitter, hardware that sends and receives bits in sequence without a shared clock wire. A byte is eight bits. Baud is symbols per second; one binary symbol carries one bit here. 8N1 means eight data bits, No parity check bit, one stop bit. LSB means Least Significant Bit, the smallest-weight bit, sent first. ASCII means American Standard Code for Information Interchange, a 7-bit character code: 65 is A, 85 is U; 255 is outside ASCII.

Open interactive experiment

EXPERIMENT BENCH

Change the circuit settings, predict the readings, then check a challenge. These are learning models, with the assumptions explained in each lesson.

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How it works

Static logic frame, not a live transmitter or receiver. µs means microsecond, 0.000001 s. ms means millisecond, 0.001 s. Timing mismatch, voltage interfaces, noise and error checking are omitted.

Idle is 1. A frame is start 0, eight data bits b0 through b7, then stop 1: ten bit periods. Bit time = 1000000/baud in µs; frame time = 10000/baud in ms. Idle time between frames is not included.

Difficulty

  1. Build · Observe

    Power on with byte 65 and 9600 baud. Bit time ≈ 104.1667 µs; frame ≈ 1.041667 ms.

    Need a hint?

    65; 9600 baud → ≈104.1667 µs; ≈1.041667 ms

  2. Explore · Explore

    Choose byte 85 and 4800 baud. Bit time ≈ 208.3333 µs; frame ≈ 2.083333 ms.

    Need a hint?

    85; 4800 baud → ≈208.3333 µs; ≈2.083333 ms

  3. Solve it · Solve

    Choose byte 255 and 1200 baud. Bit time ≈ 833.3333 µs; frame ≈ 8.333333 ms.

    Need a hint?

    255; 1200 baud → ≈833.3333 µs; ≈8.333333 ms

Extra reading and circuit recipes

UART: an 8N1 frame

Devices need to agree on timing and format before a row of highs and lows becomes a message. The familiar letter A is just one possible interpretation of byte 65. UART transports a byte; character encoding determines whether that byte represents a letter.

tbit = 1000000/baud µs; tframe = 10 × tbit = 10000/baud ms

Start marks the beginning so a receiver can align its own timing; stop returns to the idle level. “Asynchronous” means no shared clock wire, not no timing rules. Both ends still need compatible baud and frame settings.

A useful connection

Eight payload bits occupy ten periods in 8N1. At 9600 baud, consecutive frames can carry 960 bytes/s with no extra gaps.

Beyond the model

Changing the byte changes its data pattern, not the duration of the fixed-format frame.

A useful trap to avoid

The binary number is normally written highest-weight bit first, but this frame sends b0 first. UART framing does not make every byte an ASCII character and 8N1 provides no parity bit.

Only baud doubles for the same byte. Frame time becomes…

  1. twice as long
  2. half as long
  3. unchanged
Read more

Ten bit periods remain, but every period becomes half as long.

Send the shape of A

Choose byte 65 at 9600 baud.

Data b0…b7: 1 0 0 0 0 0 1 0; frame ≈ 1.041667 ms.

All data bits high

Choose byte 255 at 1200 baud.

Eight data ones follow the low start; frame ≈ 8.333333 ms.

Try another change

Compare bytes 65 and 85 at 9600 baud, then keep 85 and halve baud. Pattern changes first; duration doubles only when baud changes.

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