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Re: Ohm's Law



Ludwik makes a comment that needs correction:

Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU> 11/08/99 09:48AM >>>
Consider a diode whose I=f(V) looks like y=x^2 between V=0 and
V=6 volts. You apply 4 V and the current of 2 mA is flowing. You
say R=2000 ohms. This is a dc circuit. Add an ac source in series.
Assume its inner resistance is negligible and the dop amplitude
is 1 volt. The alternating current is superimposed on the dc. Divide
your 1 volt by the amplitude of the current and you have the dynamic
resistance. It is smaller than 2000 ohms. That is how I see it. The ac
and the dc encounter different resistances.
<snip>
Ludwik Kowalski


First of all, the functional relation between current and voltage drop across a diode is an exponential, not a power. An exponential increases much fast that a quadratic. Second, the actual f(V) function is not simply an exponential, since there is turn-on point, below which the diode hardly conducts at all. This turn-on is around 0.2 V for Ge and 0.5 V for Si devices. LEDs have a wide range of turn-on voltages, depending on the color and therefore the chemical composition and band-gap. The function that closely approximates the I-V relation of a diode is I = Irs(exp(Vd/Vt - 1), where Irs is called the reverse saturation current (or current for reverse bias, a few nA for Si, micro-A for Ge), Vd is the drop across the diode (obviously delta-Vd), and Vt is approximately nkT/q (n approx. 2 for Si, 1 for Ge, so Vt = 50 mV for Si, 25 mV for Ge).

In the example Ludwik cites, if you had a 4 volt drop across the diode, the current would not be 2 mA, but something like (2nA)(exp(4/.05)) = 1 E26 A !!

Sorry, Ludwik. Your diode blew up!

The difference between any power curve and the diode I-V curve is that the power takes off from zero and goes up right away, whereas the diode curve is zero until it reaches a turn-on point. After that it goes up very steeply, though not with infinite slope. The dynamic resistance (see comment under thread "delta-V = IR") is the inverse of the slope of the curve at a given point. This decreases the farther up the curve you go, as the curve becomes steeper.

Rondo Jeffery
Weber State University
Ogden, UT 84408-2508
rjeffery@weber.edu