Serial shift timing: At a 1.5 MHz serial clock, how long does it take to shift out the 8-bit hexadecimal value C3₁₆?

Introduction / Context:
Serial interfaces (shift registers, SPI, simple synchronous links) clock out one bit per clock period. The total time to transmit a fixed-width value equals the number of bits multiplied by the clock period.

Given Data / Assumptions:

• Serial clock frequency f_clk = 1.5 MHz.
• Data size: one byte (C3₁₆ → 8 bits).
• Assume one bit is transferred per clock (no start/stop/parity overhead).

Concept / Approach:
Time per bit T_bit = 1 / f_clk. Total time T_total = N_bits * T_bit. With 8 bits and 1.5×10^6 Hz clock, compute precisely and convert to microseconds.

Step-by-Step Solution:
Compute T_bit: 1 / 1.5×10^6 s ≈ 0.6667 µs per bit.Number of bits N_bits = 8 (for one byte).Total time T_total = 8 * 0.6667 µs ≈ 5.333 µs.Therefore, the closest option is 5.33 µs.

Verification / Alternative check:
Convert to seconds: 8 / (1.5×10^6) = 5.333… × 10^-6 s = 5.33 µs. Matches the chosen option.

Why Other Options Are Wrong:
4.69 µs: Corresponds to a faster clock (~1.705 MHz) or fewer bits.10.7 µs: Would imply 16 bits at 1.5 MHz or 8 bits at ~0.75 MHz.Millisecond options (4.69 ms, 5.33 ms): Off by a factor of 1000 (wrong unit).

Common Pitfalls:
Forgetting protocol overhead (start/stop) when present; mixing up kHz/MHz; misinterpreting hex digits as decimal count of bits instead of 4 bits per hex digit.

Final Answer:
5.33 µs