Serial loading time of a shift register: With a clock frequency of 50 kHz, how long does it take to serially enter six bits into a shift register (assume 1 bit per clock)?

Introduction / Context:
Shift registers load or transmit one bit per clock edge in simple serial schemes. Timing estimates rely on the reciprocal relationship between frequency and period, then multiplying by the number of bits.

Given Data / Assumptions:

• Clock frequency f_clk = 50 kHz.
• Bits to shift N = 6.
• Assume one bit shifts per clock cycle and no extra overhead.

Concept / Approach:
Compute the clock period T = 1 / f_clk. Total time T_total = N * T. Convert seconds to microseconds for the final answer.

Step-by-Step Solution:
Find T: T = 1 / 50,000 s = 0.00002 s = 20 µs.Multiply by bits: T_total = 6 * 20 µs = 120 µs.Therefore, six serial bits require 120 µs at 50 kHz.

Verification / Alternative check:
Reasonableness: At 100 kHz, 6 bits would take 60 µs; halving the frequency doubles the time, yielding 120 µs at 50 kHz—consistent.

Why Other Options Are Wrong:
12 µs and 3 µs imply much higher clock rates.12 ms and 120 ms are off by factors of 100–1000 (milliseconds instead of microseconds).

Common Pitfalls:
Mixing units (kHz vs. Hz) or forgetting that total serial time scales linearly with the number of bits.

Final Answer:
120 µs