Takashi Ohira
Figure 1 shows the circuit diagram for last month’s enigma. The rectifier is excited by a purely sinusoidal voltage source vs(t). However, the diode nonlinearity makes the current is(t) contain the dc component Io plus multiple harmonics [denoted by ${\cdots}$ in (2)] in addition to the fundamental wave as follows: \begin{align*}{v}_{s}{\left({t}\right)} & = {\left[\begin{array}{cc}{V}_{P}&{V}_{Q}\end{array}\right]}{\left[\begin{array}{c}{\sin}\,{\omega}{t} \\ {\cos}\,{\omega}{t}\end{array}\right]} \tag{1} \\ {i}_{s}{\left({t}\right)} & = {I}_{o} + {\left[\begin{array}{cc}{I}_{P}&{I}_{Q}\end{array}\right]}{\left[\begin{array}{c}{\sin}\,{\omega}{t} \\ {\cos}\,{\omega}{t}\end{array}\right]} + {\cdots}{.} \tag{2} \end{align*}
Figure 1. The rectifier circuit diagram.
Focusing on the fundamental wave term in (2), the trigonometric Fourier theorem tells us \[{\left[\begin{array}{c}{I}_{P} \\ {I}_{Q} \end{array}\right]} = \frac{2}{T} \mathop{\int}\nolimits_{0}\nolimits^{T / 2}{i}_{s}{\left({t}\right)}{\left[\begin{array}{c}{\sin}\,{\omega}{t} \\ {\cos}\,{\omega}{t}\end{array} \right]}{dt}{.} \tag{3} \]
Since we assume half-wave rectification (50% duty cycle), this integral is not for the entire period T but confined to the half interval 0 < t < T/2 only when the diode stays on.
The previous puzzles [1], [2] remind us that \[{i}_{s}{\left({t}\right)} = \frac{\left[{V}_{P} \quad {V}_{Q}\right]}{{\omega}{L}}{\left[\begin{array}{c}{1} - {\cos}\,{\omega}{t} \\ {\sin}\,{\omega}{t}{-}{\omega}{t}\end{array}\right]} \tag{4} \] \[{V}_{P} = \frac{\pi}{2}{V}_{Q} \tag{5} \] \[{V}_{Q} = {V}_{o} = {R}_{o}{I}_{o}{.} \tag{6} \]
Adopting (4)–(6) into (3), the half-wave integral yields \begin{align*}{\left[\begin{array}{c}{I}_{P} \\ {I}_{Q} \end{array}\right]} & = \frac{1}{2{R}_{o}}{\left[\begin{array}{cc}{4}&{{-}{\pi}}\\{{-}{\pi}}&{4}\end{array}\right]}{\left[\begin{array}{c}{V}_{P}\\{V}_{Q}\end{array}\right]} \\ & = \frac{1}{4}{\left[\begin{array}{c}{2}{\pi} \\ {8}{-}{\pi}^{2}\end{array}\right]}{I}_{o}{.} \tag{7} \end{align*}
The first row of (7) finds IP = (π/2)Io. Therefore, the correct answer to last month’s quiz is (c).
Equations (5), (6), and (7) merge into a single matrix form: \[{\left[\begin{array}{cc}{V_P} & {I_P} \\ {V_Q} & {I_Q} \end{array} \right]} = \frac{1}{4} {\left[\begin{array}{cc}{2}{\pi}{R_o} & {2}{\pi} \\ {4}{R_o} & {8} - {\pi}^{2} \end{array} \right]}{I_o}. \tag{8} \]
As seen in (8), all of the RF input components have been solved back from the dc output current. Thus, we are ready to reveal the rectifier’s practically important properties, namely, input impedance and power conversion efficiency. They will be coming up step by step in forthcoming enigmas.
[1] T. Ohira, “Rectifier output voltage,” IEEE Microw. Mag., vol. 24, no. 3, p. 99, Mar. 2023, doi: 10.1109/MMM.2022.3226632.
[2] T. Ohira, “Half-wave rectification,” IEEE Microw. Mag., vol. 24, no. 5, p. 149, May 2023, doi: 10.1109/MMM.2023.3242520.
Digital Object Identifier 10.1109/MMM.2023.3294882