Parallel transfer timing — 8-bit word at 2.25 MHz The hexadecimal number 4B (8-bit word) is sent in <em>parallel</em>. If the system clock is 2.25 MHz, how long does one parallel word transfer take?

Introduction / Context:Parallel buses transfer an entire word per clock. Understanding the relationship between clock frequency and transfer time is essential for bus bandwidth calculations and timing budgeting.

Given Data / Assumptions:

• Data width is 8 bits (hex 4B), but width does not affect time per transfer in parallel—one word per clock.
• Clock frequency f_clk = 2.25 MHz.
• Assume one clock per transfer (no wait states).

Concept / Approach:

In parallel transfer, the time to send one word typically equals one clock period. Period T = 1 / f_clk. Convert units carefully from megahertz to seconds per cycle.

Step-by-Step Solution:

Verification / Alternative check:

Quick scale check: A 1 MHz clock has T = 1 µs. At 2.25 MHz, period should be ~0.44 µs, matching 444 ns.

Why Other Options Are Wrong:

• 444 µs / 3.55 ms: off by factors of 10^3–10^6 due to unit mistakes.
• 3.55 µs: corresponds to ~281 kHz, not 2.25 MHz.
• 2.25 ns: that would be a 444 MHz clock, not 2.25 MHz.

Common Pitfalls:

• Confusing serial timing (bits per second) with parallel word timing.
• Including protocol overhead; the problem states a direct per-clock transfer.

Final Answer:

444 ns