Serial loading time vs. clock rate At a 200 kHz clock frequency, how long does it take to serially load eight bits into a shift register?

Introduction / Context:Serial interfaces and shift registers are clocked devices, so throughput and latency are directly tied to the clock frequency. Converting between frequency, period, and total transfer time is a routine skill in digital systems.

Given Data / Assumptions:

• Clock frequency f = 200 kHz.
• Word length = 8 bits (8 clock pulses to load).
• One bit is entered per clock pulse.

Concept / Approach:The time per clock is the period T = 1 / f. The total time to clock N bits serially is N * T. With f in kilohertz, convert to seconds or microseconds carefully to avoid powers-of-ten mistakes.

Step-by-Step Solution:Compute period: T = 1 / 200,000 s = 5 μs.Total time for 8 bits: 8 * 5 μs = 40 μs.

Verification / Alternative check:Use frequency scaling: at 100 kHz, 10 μs/bit; doubling to 200 kHz halves the per-bit time to 5 μs, and 8 bits then take 40 μs. The result is consistent.

Why Other Options Are Wrong:4 μs / 0.4 μs: off by factors of 10 or 100.400 μs / 40 ms: orders of magnitude too slow for 200 kHz operation.

Common Pitfalls:Confusing kHz with kbits/s or forgetting that 1 kHz = 1000 Hz (not 1024) when doing rough timing estimates.

Final Answer:40 μs