In reply to Stephen A. Lawrence's message of Thu, 01 Mar 2007 10:30:06 -0500: Hi, [snip] >> 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.
I'm a bit out of my depth here, but since noise voltage is thermal in origin, wouldn't q*v = k*T by definition? If so, then the current is constant, and equal to 1.718 * I_S. I'm assuming that q is the charge on a free electron, v is the thermal voltage, and k*T is the average thermal energy of these electrons. IOW q*v is just charge x voltage = electron energy, but k*T is supposed to represent the thermal energy of the electron. Since the electron only has one energy at any given time, it seems they must both be different expressions of the same entity, and hence identical. (Note that this is only true when the voltage in question is the thermal voltage, not when it is an externally imposed voltage). [snip] >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. If I am correct hereabove, then the only requirement for charging the cap would be that the leakage current < 1.718 x I_S. (My suspicious nature leads me to suspect however that this may very well be the very definition of diode leakage current.) :( > >> 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. If you are using the power as it is provided, then this ceases to be a problem. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition (capitalism) provides the motivation, Cooperation (communism) provides the means.
