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I have been measuring the reverse bias IV characteristics of a Zener Diode BZX (2.7 V) at three different temperatures, -196.2 celcius (liquid nitrogen temperature), 22.0 celcius (room temperature) and 99.0 celcius. Now I need to determine the breakdown voltages at said temperatures. Below is a plot of my results, which I processed on python using matplotlib.

enter image description here

I understand it is quite easy to roughly determine the breakdown voltage by eye but I was wondering if there was a more sophisticated and accurate way of determining the breakdown voltage. One thought I had was to fit an exponential of the form $y = -e^{-kx}$ but then I'm not sure which parameter of the exponential would correspond to the breakdown voltage. Perhaps I could then (numerically) take the second derivative of the data, set it to equal 0, which would then give me the voltage at which the slope is steepest (and thus corresponds to the breakdown region?) Is there another method I'm missing which could be more accurate?

Also, as an aside, my next step is to calculate the temperature coefficient for the zener diode. Is this simply a matter of calculating the following: $$T_{c} = \frac{V_2 - V_1}{T_2 - T_1}$$ the numerator is the difference between the breakdown voltage measured at temperature 2 and temperature 1, and the denominator is the change in temperature between temperature 2 and temperature 1?

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  • $\begingroup$ In practice, you decide on an operating current, and your "breakdown voltage" is what you measure at that current. $\endgroup$
    – John Doty
    Oct 24, 2022 at 13:20

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Function at extremum point (minimum or maximum) does not change, so you need to evaluate differential :

$$ \frac {d}{dx} y = \frac {d}{dx} \left(-e^{-kx}\right) = k e^{-k x} \tag 1$$

Now expand (1) into Taylor series, take couple of terms and set it to zero :

$$ k - k^2 x = 0 \tag 2$$

Solving (2) for $x$, gives approximate breaking voltage :

$$ V = \frac 1k~~~~~~\text{for}~k \ne 0 \tag 3$$

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  • $\begingroup$ This is very clever. Many thanks! $\endgroup$ Oct 24, 2022 at 14:08
  • $\begingroup$ You are welcome. $\endgroup$ Oct 24, 2022 at 15:13
  • $\begingroup$ Sorry, is there any reason why this method is preferable compared to evaluating the second derivative = 0 $\endgroup$ Oct 24, 2022 at 19:43
  • $\begingroup$ First order derivative test gives you extrema, while second order derivative test can say what kind of extrema is there- maximum or minimum. $\endgroup$ Oct 24, 2022 at 21:12

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