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diode I,V curve

At 11:22 AM 11/8/99 -0700, Rondo Jeffery wrote:

First of all, the functional relation between current and voltage drop
across a diode is an exponential, not a power.

True.

An exponential increases much fast that a quadratic.

True.

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.

I don't think so. Typical diodes are exponential to a very good
approximation. The so-called turn-on voltage is just the point where the
exponential gets to be big enough to be interesting.

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,

True.

a few nA for Si, micro-A for Ge),

Don't you think that might depend on things like the area of the diode
junction, rather than being some function purely of the
material? Hint: think about the I,V curve for a few diodes in parallel.