Electronic – Dynamic resistance of a p-n junction diode

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

In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current.