There was a calculation error in my recent posting. Thanks to Tom, K7ITM
for spotting it and letting me know. It doesn't affect the conclusions.

Roy Lewallen wrote:
. . .
Now connect another current source across the biased diode, but make it
an AC source and connect it through a capacitor. Make the AC source
current 0.1 mA RMS, which is much less than the bias current of 10 mA.
What AC voltage should we expect? Well, it's not 0.1 mA * 70 ohms,
because 70 ohms is the DC resistance V/I, and the AC current is
encountering a resistance of dV/dI, where dV/dI is the ratio of the
*change* in voltage to the *change* in current. This is the dynamic, or
AC resistance, and it equals the slope of the V/I curve of the diode.
The AC and DC resistance are the same for a simple resistor, but not the
nonlinear diode. The dynamic resistance actually changes for each
instantaneous value of the AC waveform, but as long as the AC current
waveform is much less than the DC bias current, the change won't be
much. It turns out that the AC resistance for a diode is about 0.026/Idc
where Idc is the bias current. So in our case, the AC resistance is
about 0.026/0.01, or about 2.6 ohms. Therefore we'd see an AC voltage
across the diode of 0.1 mA * 2.6 = 0.26 volt.

The AC voltage across the diode is 0.26 mV, not 0.26 V.

The AC voltage is in phase with the AC current through the capacitor, so
the AC current source is supplying power to the diode, an amount equal
to Iac * Vac = 0.1 mA * 0.26 V = 0.026 mW. . .

The power supplied by the AC current source to the diode is 0.1 mA *
0.26 mV = 0.026 uW = 26 nW, not 0.026 mW. This is, of course, also Iac^2
* Rac = (0.1 mA)^2 * 2.6.

Roy Lewallen, W7EL