The time delay obtained through an 8-bit serial register with 400 MHz clock is:
2023
The time delay obtained through an 8-bit serial register with 400 MHz clock is:
- A.
20 ns
- B.
2.5 µs
- C.
20 µs
- D.
2.5 µs
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Correct answer: A
The clock period is calculated as the inverse of the frequency: T = 1/f = 1 / 400 MHz = 2.5 ns. For an 8-bit serial register, the signal must pass through all 8 bits, requiring 8 clock cycles. Therefore, the total time delay is 8 × 2.5 ns = 20 ns.