Robin van Spaandonk wrote:
In reply to Stephen A. Lawrence's message of Wed, 28 Feb 2007 23:49:54 -0500:
Hi,
[snip]
In any case there's also thermal noise in the diode, as I believe I also
pointed out (though I didn't phrase it that way), and that is surely
where you should be hunting for the flaw in the design.
[snip]
If you look at the voltage current curve of a diode for both positive and
negative currents, then it is clearly asymmetric. That's why we use them.
In short they convert AC into DC, and it doesn't matter whether the AC is a
perfect sine wave or a random mess. Even the latter will result in at least some
DC component. I agree with you that a diode should produce the same sort of
thermal AC voltage as a resistor, however it should also rectify it's own
voltage.
What you say is true, but there is an issue, which is that real diodes
are not "perfect diodes": they have a nonzero forward voltage drop.
What's more, they're not even "ideal diodes": the forward voltage drop
is not constant. As you suggest, let's look at the current curve for a
hypothetical "real diode", taken from Senturia and Wedlock, "Electronic
Cicuits and Applications", p184 in my copy. It's given as
i = I_S(e^(qv/kT) - 1)
where I_S, q, and k are constants.
The Taylor series in v for values near 0 is
i ~ v * I_S * (q/kT) + v^2 * I_S * (q/kT)^2/2 + order(v^3)
At very low voltages, such as noise produces, we can ignore the square
term and all higher order terms, and the diode "looks" linear, just like
a resistor. Furthermore, if we look at the square term, we see that
it's proportional to the inverse square of the temperature -- the diode
gets more "linear-looking" at 0 volts as the temperature (and, hence,
noise magnitude) increase.
I don't pretend to be able to analyze this situation in detail, but it
appears to me, from the above formula, that for any realistic level of
noise-induced charge on a cap hooked up across the diode, the charge is
going to leak away through the diode long before the next probabilistic
noise crest of sufficient magnitude to charge it up any farther comes along.
IOW a diode connected across a capacitor should eventually charge the capacitor,
if it's thermal voltage is current independent.
But the more charge you get, the longer you have to wait for another
noise "pulse" which exceeds that voltage, and the longer you have for
the charge you already had accumulated to leak away through the diode.
Regards,
Robin van Spaandonk
http://users.bigpond.net.au/rvanspaa/
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