Topic: Analog Circuits
An ideal comparator is fed with a sine wave of 4Vpp (peak-to-peak) with zero DC component as shown in Figure. The reference voltage of the comparator is 1 V. What is the duty cycle of the output $V_{out}$?
- A. 33.33%
- B. 66.66%
- C. 25%
- D. 50%
Answer: A. 33.33%
Explanation: The input is a sine wave with 0 DC offset and 4V peak-to-peak, meaning its peak amplitude is $V_m = 2V$. Thus, $V_{in}(t) = 2\sin(\theta)$.
The output of the non-inverting comparator is high when $V_{in} > V_{ref}$, i.e., $2\sin(\theta) > 1 \Rightarrow \sin(\theta) > 0.5$.
The angles where $\sin(\theta) = 0.5$ are $\theta_1 = 30^\circ$ and $\theta_2 = 150^\circ$.
The signal is above the threshold for $150^\circ - 30^\circ = 120^\circ$ within a full $360^\circ$ cycle.
Duty Cycle = $\frac{120^\circ}{360^\circ} \times 100\% = 33.33\%$.