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Ordinary diodes obey the Shockley diode equation, but I can't find anything similar for avalanche and Zener diodes.

It's easy to find their VI curves, crude piecewise fits and lots of descriptions of how it works. I could also extract the numerical data from the data sheet graphs.

I'd like an analytic expression for the curve shape, something that models the physics reasonably accurately.

Any help?

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    \$\begingroup\$ You might need to study semiconductor physics to learn the differences in Zener effect and Avalanche effect which conduct with negative incremental resistance due to very high electric fields in kV/mm or V/um , so are quite different in structure and voltages tends to be higher for avalanche mode. Are you familiar with bandgap and Group IV doping characteristics and Chip-integrated van der Waals forces? wikiwand.com/en/Avalanche_transistor \$\endgroup\$ Commented May 2, 2022 at 20:35
  • \$\begingroup\$ My only Rule of Thumb is that the incremental resistance is usually less than the inverse max power capacity of the size of the chip. ESR=k/Pmax for k <=1 typ but k increases with log voltage . What range of power levels are you interested in? \$\endgroup\$ Commented May 2, 2022 at 20:44
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    \$\begingroup\$ Maybe this paper might have a decent model for Zener: ieeexplore.ieee.org/document/1051530 \$\endgroup\$
    – Ste Kulov
    Commented May 3, 2022 at 6:27

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I don't know about exact physics, but it's often fitted by an exponential e.g. as in SPICE breakdown voltage equation.

I suspect you aren't going to find any more pleasing analytical expression, given the exponentials likely to show up; note a series diode and resistor already has no analytic solution but can be solved by numerical approximation (as SPICE does) or the Lambert W function.

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    \$\begingroup\$ Lambert W can form a closed solution. \$\endgroup\$
    – jonk
    Commented May 2, 2022 at 21:09
  • \$\begingroup\$ Ah, also Lambert W is analytic on z > -1/e so that works, it's just not analytic below there, which we probably don't need for these solutions anyway. \$\endgroup\$ Commented May 3, 2022 at 1:11
  • \$\begingroup\$ Yes I've forgotten where, but I've in the past mentioned here that one gets real values for \$z\ge -\frac1{e}\$. For example, in that case the result is just -1. This and a lot more is discussed on Wolfram's site. And for the arguments one is likely to get with positive, real-valued component values, it applies well enough for most uses. \$\endgroup\$
    – jonk
    Commented May 3, 2022 at 2:02
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There is a way to do this, but it might be the long way around. If you have the curves (graphs from the manufacturer), there is software that you can use to capture data points (x,y) from the graph itself. In the olden days, I used software that was free, but you had to click on each point you wanted and have it stored to a data set. Nowadays, there is software that will get the data points for you (Quintessa, for example).

After you have all of the data points on your curve, you can use an equation fitter (easily found online) that will give you a function that fits your curve. The last time I did this, Excel had a function that did this. It made up very nice equations for me that worked very well, with sometimes up to 4th power variables.

Avalanche and Zener diodes are discontinuous in their outputs, and I don't know if other programs will see this, but you can always use a Heaviside function to break it into parts.

So, you have two equations associated with two Heaviside functions that most decent math programs can easily deal with. That is, for V < a, you get one curve and for V >= a, you get a second. This can all be expressed in a single equation, if it's written properly.

There is a harder, but more precise, way to do this, but it will still involve two Heaviside functions. That, of course, is to write the equations according to the device physics and it's characteristics and doping. I think that most device manufacturers keep this information confidential.

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