Electronic – Charging an ideal capacitor through an ideal diode, forever

The phenomenon you are seeing is called reverse recovery time. Look that up and you will see it is due to carriers in the junction still being there when the voltage reverses. Until those carriers are "used up", the diode will continue to conduct.

Modeling is all about knowing which characteristics actually matter and ignoring the rest. If you didn't do that, it would be reality instead of a model, but then it would also be too complex to implement.

At first approximation, just assume the diode will conduct in reverse for a fixed amount of time. Diodes meant for applications where this matters will have the maximum reverse recovery time listed in the datasheet. If the purpose of the model is to want to make sure your circuit still works, this is a good model because it represents the worst case conditions.

More accurate models take into account the current immediately before the voltage reversal and look at total charge leaked backwards. There are fancy equations for all that you will have to look them up in semiconductor physics texts if you want this level of detail.

Shockly diode equation is given by: $$\style{}{I=I_S(e^{V_D/nV_T}-1)}\tag1$$

Where,

• \$I_S\$ is the reverse bias saturation current.
• \$V_D\$ is the voltage across the diode
• \$V_T\$ is the thermal voltage
• \$n\$ is the ideality factor, also known as the quality factor or emission coefficient.

I see two material dependent parameters here:

1. The reverse saturation current, \$I_S\$ depends on the material.
2. The parameter \$n\$ depends on fabrication process and semiconductor material.

EDIT
From equation (1), $$V_D = n \cdot V_T \ln\left(\frac{I}{I_S}+1\right) \approx n \cdot V_T\ln\left(\frac{I}{I_S}\right)$$ $$V_D \approx n \cdot V_T \cdot \ln10 \cdot \log_{10}\left(\frac{I}{I_S}\right)$$

Assuming room temperature and \$n=1\$, $$V_D \approx 0.05916 \cdot \log_{10}\left(\frac{I}{I_S}\right)\tag2$$

Typical values of the saturation current at room temperature are:

• \$I_S = 10^{-12}\$ for silicon diodes;
• \$I_S = 10^{-6}\$ for germanium diodes.

For a current of 1.0 mA:

• \$V_D \approx 0.53 V\$ for silicon diodes (9 orders of magnitude)
• \$V_D \approx 0.18 V\$ for germanium diodes (3 orders of magnitude)

For a current of 100 mA:

• \$V_D \approx 0.65 V\$ for silicon diodes (11 orders of magnitude)
• \$V_D \approx 0.30 V\$ for germanium diodes (5 orders of magnitude)

Further increase in current won't cause much increase in \$V_D\$ (0.05916V per decade is the rate of change of \$V_D\$ with respect to \$I\$). Hence in the common cases (current in the range of mA), the voltage drop remains constant around 0.6V for silicon diodes.

And hence values of 0.6 or 0.7 Volts are commonly used as voltage drop for silicon diodes and 0.3 for germanium diodes.

source: wikipedia