Diode Equation needs a Knee Transplant

[[email protected]]

I would like to know how to modify the diode equation: I = Io (Exp(eV/
NkT)-1) to work for a germanium diode. Using Io=10E-12, e=1.602E-19,
T=295, N=1, and k=1.380E-23 has the correct knee 0.6 for a silicon
diode. What needs to be changed (or is there a new equation) to move
the knee back to around 0.3 to model a germanium diode?

Any help would be greatly appreciated. Thanks

 

[George Herold]

I would like to know how to modify the diode equation: I = Io (Exp(eV/
NkT)-1) to work for a germanium diode.  Using Io=10E-12, e=1.602E-19,
T=295, N=1, and k=1.380E-23 has the correct knee 0.6 for a silicon
diode.  What needs to be changed (or is there a new equation) to move
the knee back to around 0.3 to model a germanium diode?

Any help would be greatly appreciated.  Thanks

There is no band gap voltage in the "low voltage" diode equation. As
the applied voltage exceeds the band gap voltage about 0.6 for Si and
0.3 for Ge a different aproximate equation is used.
Sorry but I don't know off hand what that "high voltage" equation is.
George

 

[[email protected]]

There is no band gap voltage in the "low voltage" diode equation.  As
the applied voltage exceeds the band gap voltage about 0.6 for Si and
0.3 for Ge a different aproximate equation is used.
Sorry but I don't know off hand what that "high voltage" equation is.
George

if the equation is not at equilibrium, a potential barrier wont
develope across the P-N junction so its not that important.

 

[Michael Robinson]

Here's one more for all the contradictory advise:

Increase Io. Dramatically.

Perhaps take some measurements of the diode voltage at several different
current values (i.e. 1uA if you can get that low, 10uA, 100uA, 1mA --
don't exceed the diode ratings). Then do a curve fit on semilog paper --
figure that you are going to see Io e^(eV/NkT) only, the "- 1" part will
drop right out.

Yes, go the empirical route. You can use ordinary graph paper if you want,
by graphing log I versus V, or by graphing I versus exp(V). The -1 drops
out to insignificance, as Tim says. You will find that the data points lie
on a very straight line, until self-heating kicks in, whereupon the graph
will start to curve away from the straight line established at low currents
(constant temperature) where you see the purely exponential side of the
relationship. For small diodes I agree with Tim's suggested current range.

You will have to take more than three measurements if you want to see just
where the curve diverges from the ideal, as a result of self-heating
effects.
Try it. I was pretty impressed at how precisely nature mirrors the math.