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: enter image description here
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:

enter image description here

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?

  • \$\begingroup\$ The diode has an exponential change in current, as the forward voltage increases. Assume 1mA at 0.7 volts across the diode (others will suggest 0.6 volts; you can consult a diode datasheet, eg. Fairchild's). As the forward voltage drops 0.018 volts, the current drops 2:1. For0.036 volts less, the current drop 4:1, for 0.054 volts, the current drops 8:1, for 0.72 volts, the current drops 16:1. On a per-decade basis, 0.058 volts causes either 10:1 decrease or 10:1 increase. Thus 0.600 + 0.058 will produce 10 milliamp forward current; welcome to diodes. \$\endgroup\$ Commented May 5, 2019 at 21:54

1 Answer 1


Yes you are relying too much on the simple model.

Since Ipk is 1mA but then drops to 0 after 5ms, the cap with no load R, is able to charge up to about 2.55 V from 3V . This means the diode voltage may have started near 0.6V from the 1N4148 typ. datasheet but reduced to 3-2.55= 0.45 V after the 1st sine peak.

This corresponds to a forward current of approx 50uA which is approx, what you simulated.

Then on the 2nd cycle another 40uA peak is conducted and the voltage rises another xx mV and on the 3rd peak it reaches ~2.6V or a diode Vf =0.4V

For this test the knee voltage Vk appears to reduce with current but we know it also rises with current. So the thing to remember is what is the Vk & tolerance @ some specified forward current. I use 0.6V@ 1mA @ 25’C

Get used to reading datasheets for the VI curve and keep in mind, the tolerance @25’C is due to a series bulk resistance due to process controls and Vk (@ x mA) is fixed, but can shift negatively with rising temp (it’s also a thermometer)

For low currents it has a log VI straight line from Shockley’s Equation then a linear straight line scale from bulk series resistance near rated current.

  • \$\begingroup\$ Ok, I get it's a lot more complicated than that. Thank you! \$\endgroup\$
    – Lucen
    Commented May 14, 2019 at 9:43

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