Diode Equation needs a Knee Transplant

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

Reply to
jalbers
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Adjust N, which is a constant that depends upon the manufacturing process and semiconductor material. A value somewhere in the neighborhood of 0.5 will probably do the trick.

Reply to
Greg Neill

9,

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
Reply to
George Herold

/

-19,

n

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

Reply to
extremesoundandlight

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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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Reply to
Tim Wescott

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.

Reply to
Michael Robinson

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