Takashi Ohira
The circuit diagram from last month’s problem is shown in Figure 1, where the three voltages are related as \[{v}_{s}\left({t}\right) + {v}_{D}\left({t}\right) + {v}_{L}\left({t}\right) = {48}, \tag{1} \]
Figure 1. The battery-charging circuit in question.
to meet Kirchhoff’s voltage law. The time-domain analysis starts at ${t} = {0}$. The source voltage ${v}_{s}{(t)}$ linearly increases with t up until it reaches the peak, as shown in Figure 2(a). This is simply formulated as \[{v}_{s}\left({t}\right) = {100}{t}{.} \tag{2} \]
Figure 2. The voltage waveforms of (a) the RF source, (b) the diode, and (c) the inductor.
The very moment ${v}_{s}{(t)}$ crosses 48 V, the diode D turns on. As long as diode D stays on, ${v}_{D}{(t)}$ is kept at zero. Therefore, (1) leads the inductor voltage to \[{v}_{L}\left({t}\right) = {48}{-}{100}{t}\,{\text{ for }}\,{0.48}\,{<}\,{t}\,{<}\,{1}{.} \tag{3} \]
After passing the peak, ${v}_{s}{(t)}$ decreases as \[{v}_{s}\left({t}\right) = {200}{-}{100}{t}{.} \tag{4} \]
Because diode D still remains on, (1) leads to \[{v}_{L}\left({t}\right) = {100}{t}{-}{152}\,{\text{ for }}\,{1}\,{<}\,{t}\,{<}\,{\tau}, \tag{5} \] where ${\tau}$ denotes the turn-off moment.
Looking at the downhill slope of (4), one might have thought that the moment ${v}_{s}{(t)}$ falls across 48 V, it makes diode D turn off. However, that is not true. Because of the inductor’s inherent property, it is hard for the current to change quickly. To find where ${\tau}$ is exactly located, we use another inherent property of lossless inductors: the voltage across L contains no dc component. This is formulated as \[\oint{{v}_{L}\left({t}\right){dt} = {0}}, \tag{6} \] where the circular symbol means an integral for one cycle.
Applying (6) to (3) and (5) gives \[\mathop{\int}\nolimits_{0.48}\nolimits^{1}{\left({{48}{-}{100}{t}}\right){dt} + \mathop{\int}\nolimits_{1}\nolimits^{\tau}{\left({{100}{t}{-}{152}}\right){dt} = {0}{.}}} \tag{7} \]
Note that the integration is performed only when diode D stays on because ${v}_{L}{(t)}$ is kept at zero when diode D stays off. Solving (7), we find ${\tau}\approx{2.255}$, which is located before the peak. As a result, the pyramid of ${v}_{D}{(t)}$ is truncated, as shown in Figure 2(b). Therefore, the correct answer to last month’s quiz is (b). The zero-mean waveform of ${v}_{L}{(t)}$ is also worth viewing from graphics. Focusing on the two triangles indicated by ${S}_{1}$ and ${S}_{2}$ in Figure 2(c), we notice that (7) is translated into the area balance as ${S}_{1} = {S}_{2}$.
Paralleling the waveforms of ${v}_{D}{(t)}$ and ${v}_{L}{(t)}$, a simultaneous discontinuity is observed with vertical broken lines. That is to say, the linearly increasing ${v}_{L}{(t)}$ suddenly transfers to ${v}_{D}{(t)}$ at ${t} = {\tau}$. By contrast, ${v}_{D}{(t)}$ holds continuity back at ${t} = {0.48}$, which is called zero-voltage switching (ZVS); namely, ${v}_{D}{(}{0.48}{)} = {0}$. It is a general feature of ${L} + {D}$ in series that D turns on in ZVS, whereas it turns off in non-ZVS. This law applies not just to the simple battery charger but also to any other diode-oriented RF power electronics, such as rectennas for wireless power transfer.
Digital Object Identifier 10.1109/MMM.2022.3218178