Duty cycle of the MSB in a 4-bit BCD (0–9) counter: Over one full 0 to 9 counting cycle, what is the duty cycle of the most significant bit (the 8’s place) of the BCD output?

Introduction / Context:
Duty cycle of a counter output describes the fraction of time a particular bit remains HIGH during a complete counting period. For BCD counters that count 0–9, the bit patterns differ from straight binary because six states (10–15) do not occur. This affects the duty cycle of each bit.

Given Data / Assumptions:

• We have a 4-bit BCD output representing digits 0..9.
• MSB is the 8’s bit (weight 8).
• One complete cycle includes ten counts (0 to 9).

Concept / Approach:
Determine for which BCD digits the 8’s bit is HIGH. In BCD, the 8’s bit is 1 for values 8 (1000) and 9 (1001) only. That is two counts out of the ten total in the cycle.

Step-by-Step Solution:

Verification / Alternative check:
Plot the MSB across the sequence or simulate a BCD counter. You will see the MSB LOW for counts 0..7, then HIGH for 8 and 9, then back LOW at 0, confirming a 20% HIGH fraction.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing BCD with straight binary; assuming each bit has roughly 50% duty without checking the actual sequence length and valid codes.

Final Answer:
20%