Topic: Digital Logic
The time delay obtained through an 8 bit serial register with 400 MHz clock is:
- A. 20 ns
- B. 2.5 $\mu$s
- C. 20 $\mu$s
- D. 2.5 ns
Answer: A. 20 ns
Explanation: The clock frequency is $400$ MHz, so the clock period $T = 1 / (400 \times 10^{6})$ seconds $= 2.5$ ns. An 8-bit serial register requires 8 clock cycles for a bit to pass through. Thus, total delay $= 8 \times 2.5$ ns $= 20$ ns.