Takashi Ohira
The circuit diagram is shown in Figure 1. For time-domain analysis, recall the three equations \[{v}_{s}{{\left({t}\right)}} = {\left[{{V}_{P}{V}_{Q}}\right]}{\left[{\begin{array}{c}{\sin{\omega}{t}}\\{\cos{\omega}{t}}\end{array}}\right]} \tag{1} \] \[{v}_{s}{\left({t}\right)} + {v}_{L}{\left({t}\right)} + {v}_{D}{\left({t}\right)} = {V}_{o} \tag{2} \] \[{V}_{Q} = {V}_{o} \tag{3} \]
Figure 1. The rectifier circuit diagram along with its internal current and voltages for time-domain analysis.
from the previous puzzle [1], where ${\omega} = {2}{\pi}{f} = {2}{\pi} / {T}$.
Immediately after the diode turns on, vD(t) becomes zero, and vL(t) increases as \begin{align*}{v}_{L}{\left({t}\right)} & = {V}_{o}{-}{v}_{s}{\left({t}\right)} \\ & = \left[{{V}_{P}{V}_{Q}}\right]{\left[{\begin{array}{c}{{-}\sin{\omega}{t}}\\{{1}{-}\cos{\omega}{t}}\end{array}}\right]}{.} \tag{4} \end{align*}
The voltage accumulates the inductor current as \begin{align*}{i}_{s}{\left({t}\right)} & = {-}\frac{1}{L}\mathop{\int}\nolimits_{0}\nolimits^{t}{{v}_{L}{\left({t}\right)}{dt}} \\ & = \frac{\left[{{V}_{P}{V}_{Q}}\right]}{{\omega}{L}}{\left[{\begin{array}{c}{{1}{-}\cos{\omega}{t}}\\{\sin{\omega}{t}{-}{\omega}{t}}\end{array}}\right]} \tag{5} \end{align*} which continues flowing until the diode turns off. Because the on-duty cycle is specified to be 50%, the diode is supposed to turn off at t = T/2. Imposing the turn off condition \[{i}_{s}{\left({\frac{1}{2}T}\right)} = {0} \tag{6} \] on (5), we find \[{V}_{P} = \frac{\pi}{2}{V}_{Q}{. } \tag{7} \]
Therefore, the correct answer to last month’s quiz is (c). Looking at (7), one may wonder how we can put this voltage relation into practice. This will be done by adjusting the load resistance Ro next month.
[1] T. Ohira, “Rectifier output voltage,” IEEE Microw. Mag., vol. 24, no. 3, p. 99, Mar. 2023, doi: 10.1109/MMM.2022.3226632.
Digital Object Identifier 10.1109/MMM.2023.3256861