I haven't followed this thread very thoroughly, so this might not be
directly relevant. But it should be of interest to anyone trying to
detect small signals with a diode.

There are several reasons why diodes do poorly with small AC signals.

The first is, of course, the forward drop. However, this can in theory
be reduced to an arbitrarily low value by reducing the current to a low
enough value (by, for example, making the load impedance high enough).

The second is that the ratio of reverse to forward current increases as
the signal gets smaller and smaller, reaching one at the limit. This can
be observed by looking at the I-V curve of a diode. At the origin, the
curve is a straight line -- the diode behaves just like a resistor.

The third reason is the diode capacitance. This shunts the diode,
effectively lowering the reverse impedance. It also lowers the forward
impedance, but when the forward Z is lower than the reverse Z, the net
effect is to further degrade the forward/reverse impedance ratio.

You can make all the DC measurements you want, but they only tell half
the story. When you apply AC, you charge the load capacitor during half
the cycle according to the diode's forward impedance, and charge is
removed from it during the other half according to the diode's reverse
impedance. As the forward/reverse impedance ratio degrades due to the
two effects mentioned above, the net charge you get in the load
capacitance decreases, hence the voltage it's charged to decreases. This
ends up looking like a larger diode forward drop.

I spent a lot of time thinking about this some years ago when designing
a QRP wattmeter, and some of the conclusions I came to appear in the
resulting article, "A Simple and Accurate QRP Directional Wattmeter",
published in QST, February 1990. See the analysis on p. 20, "Ac v Dc:
Why the Difference?"

Roy Lewallen, W7EL