Takashi Ohira
The circuit shown in Figure 1 is often called a half-wave rectifier because the diode is supposed to stay on for 50% of one cycle. However, that view is untrue since the on-duty ratio depends on the circuit parameters. Assuming the RF voltage source waveform as \[{v}_{s}{(}{t}{)} = {\left[{V}_{P} \quad {V}_{Q}\right]}\,{\left[\begin{array}{c}{\sin}\,{\omega}{t}\\{\cos}\,{\omega}{t} \end{array}\right]} \]
Figure 1. The single-series diode rectifier from the last puzzle, now solving for the specific vs(t) that makes a 50% duty of the diode. Recall that the diode turns on at t = 0.
a certain relationship exists among ${V}_{P}$, ${V}_{Q}$, and the on-duty ratio. To exactly expect a 50% duty cycle, which of the following should be equal to ${V}_{P}$?
(a) ${V}_{Q}$ (b) ${\frac{1}{2}}\,{V}_{Q}$ (c) ${\frac{\pi}{2}}\,{V}_{Q}$ (d) ${\frac{\pi}{4}}\,{V}_{Q}$
Digital Object Identifier 10.1109/MMM.2023.3242520