Electronic – How is dynamic resistance of a diode modeled for large voltage variations

It just looks like magic doesn't it?

If you haven't taken a course in semiconductor device physics, I suppose it does look like magic.

PN junctions are not resistors. Whereas the voltage across a resistor is proportional to the current through

\$v_R =R \cdot i_R \$

the current through a diode is approximately:

\$i_D = I_S(e^{\frac{qv_D}{kT}}-1) \$

In words, the diode current goes up exponentially with the diode voltage.

The derivation of the diode equation isn't trivial but it isn't mysterious either; it's based on well understood device physics.

For zener and avalanche diodes, the physical explanation is a bit easier to grasp. From Wiki:

Under a high reverse-bias voltage, the p-n junction's depletion region expands, leading to a high strength electric field across the junction. A sufficiently strong electric field enables tunneling of electrons from the valence to the conduction band of a semiconductor leading to a large number of free charge carriers. This sudden generation of carriers rapidly increases the reverse current and gives rise to the high slope resistance of the Zener diode.

The Zener effect is distinct from avalanche breakdown which involves minority carrier electrons in the transition region which are accelerated by the electric field to energies sufficient to free electron-hole pairs via collisions with bound electrons. Either the Zener or the avalanche effect may occur independently, or both may occur simultaneously. In general, diode junctions which break down below 5 V are caused by the Zener effect, while junctions which experience breakdown above 5 V are caused by the avalanche effect.

Your misunderstanding is around how the small-signal AC characteristics combine with the large DC characteristics. If V_D is the DC (ie: average) voltage across the diode, you've already assumed that to be 0.6V.

The value r_d represents only the incremental resistance of the diode if you add or subtract a small amount of current (ie: add or subtract a small amount of voltage at the input, like 0.1 cos(wt) ). So you use r_d to calculate the voltage resulting only from that small change in current.

So:

v_in = 0.1 cos(wt), and the resulting small signal effect across the diode is:

$$v_d = \frac{v_{in} r_d}{r_d + 1\mathrm{\ k\Omega}}$$

and V_{diode_total} = V_D + v_d