In our microelectronic circuits class we have been taught about the following simplified model for the diode,
\$(v_D < V_k \implies i_D = 0) \land (v_D = V_k \implies i_D > 0)\$
where \$i_D\$ is the current on the diode, \$v_D\$ is the voltage across it and \$V_k\$ is its knee voltage (neglecting the breakdown effect). We usually use this model for quick pen and paper calculations, which our teacher sometimes doesn't do, however.
I was reviewing one (basic) circuit the other day, which is one of those for which the teacher didn't do all the calculations during the lecture. The LTSpice schematic of the circuit is the following:
The voltage on the capacitor is initially 0, and the input voltage is \$v_{in}(t) = 3sin(2\pi 50t)\cdot1(t)\$ (of period \$T = 20\$ \$ms\$). By applying KCL, KVL, etc. I quickly got the equations
\begin{equation}\tag{1}
v_{in}(t) = v_D + v_{out}\end{equation}
\begin{equation}\tag{2}i_D = Cv'_{out}\end{equation}
where \$v_{out}\$ is the output voltage, and \$v_{out}'\$ is its derivative.
When the diode is forward biased, I get
\$
\begin{equation}
\tag{1F}
v_{out} = V_{in}(t) - V_k
\end{equation}
\$\begin{equation}
\tag{2F}
i_D = Cv'_{out} > 0
\end{equation}
When the diode is reverse biased, I get
\begin{equation}
\tag{1R}
v_{out} > v_{in}(t) - V_k
\end{equation}
\begin{equation}
\tag{2R}
i_D = Cv'_{out} = 0
\end{equation}
The result of the LTSpice simulation is the following:
During the lecture, our teacher, who didn't do the calculation explicitly, asserted that, from time \$T/4\$ onward, the output voltage is constant because the capacitor cannot discharge, which is only approximately true: the diode appears to conduct a small (but of greater order than the saturation current of the diode) amount of current in certain instants. When looking at the simulation for the first time, I thought that this discrepancy was because we were using a simplified model.
However, by the direct calculation using the simplified model above, I found out that, for time instants \$t_h = T/4 + hT\$, where \$h > 0\$, the diode is not forward biased or reverse biased: if it was reverse biased, \$v_{out}(t_h)\$ would be equal to \$v_{out}({t_h}^-)\$, as its derivative would be constant. However, (2R) would not hold, as \$v_{out}({t_h}^-) = v_{in}(t_h) - V_k\$.
On the other hand, if (1F) were to be true, \$v_{out}\$ would have a maximum in \$t_h\$, thus making (2F) false.
Since the only discontinuity in the diode model happens when it switches on / off, I assumed this means that in \$t_h\$ it is switching on and then off (because after this instant the diode turns out to be reverse biased again). The simulation also confirms this.
The question is: in cases such as this one, when using the simplified model above, I find some instants for which the diode appears not to be in any state, can I assume that it's in the middle of switching on/off? Or am I relying too much on a simplified model?
Thanks.
