John R. Strohm wrote:
"Clifton T. Sharp Jr." wrote...
(http://uweb.superlink.net/bhtongue/7diodeCv/7diodeCv.html).

In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

Think about the V-I curve on an ideal diode. Observe that, if the diode is
reverse-biased, no conduction takes place. In principle, you can crank the
reverse voltage up as high as you like, and STILL no conduction takes place.
(In practice, you run into avalanche and zener breakdown.) If the diode is
forward-biased, the diode conducts like crazy. There is a CORNER in the
curve, i.e. a change in the slope of the V/I curve, at NOMINALLY V=0V.
(Actually, V=0.7V for silicon, V=0.3V for germanium, V=1.7V for red LEDs,
about 2.2V for yellow, about 2.4V for green, and so on and so forth...)

I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

....while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

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