Dynamic resistance of diode

I have been given a circuit and asked to find the voltage across the diode (both the DC and AC parts).

I believe the right approach is the ignore the ac source initially and find the I_DQ. I know there is 0.6 V across the diode so I found I_DQ to be 14.4 V/1 kΩ = 0.0144 A by analyzing the resistor.

I know there is an equation for dynamic resistance: $$r_d = \frac{nV_T}{I_DQ}$$ I am given n = 1 and V_T = 26 mV so I get r_d = 1.86 Ω. I interpret this as: in this setup, because the ac signal is small and the diode will always be forward biased, we can effectively treat the diode as a resistor of value r_d.

But if I look at the voltage across the diode at the Q point using the voltage divider: $$V_D = \frac{15r_d}{r_d + 1\mathrm{\ k\Omega}} = 0.028\mathrm{\ V}$$ But it should equal 0.6 V. Have I done something wrong in my calculations? Or am I interpreting r_d incorrectly?

I know I haven't finished the given question yet, but I thought I would make this check and it would show me if I am on the right track or if I have already made a mistake.

Answer

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