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Though I want to address a specific aspect which is about normalization I also would like to see short answers/reasons about the purpose of normalization. Maybe this will answer the next:

I got thousands of diodes to analyze which are mainly characterized by their M-V behavior. M is an amplification factor with an arbitrary unit. Simply speaking M is just rising with V as following:

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What I wonder now is: Of high interest is the slope at a specific point (M=150). Now, to compare the diodes among each other the slope @ M=150 is normalized with 1/M. Thus, the slope in total is calculated via 1/M * dM/dV. As M is only 150 here, I wonder about the sense of this. It is just a constant I multiply to the slope of each diode. What does it change? Could someone please shed some light on this and/or me? Thank you!

Ben
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  • This is a physics question, not a mathematics query. – Nij Jan 09 '23 at 08:46
  • Why physics? It is not about the meaning of any variable or so, it could be any arbitrary variables like price vs house size.. – Ben Jan 09 '23 at 08:51
  • Asking why the specific physical ratio is normalised in this way is not a question that applies to arbitrary variables, so it is necessarily related to the actual physical variables involved, and this must be for a physical reason. – Nij Jan 09 '23 at 09:17
  • I doubt that as a diode behavior, like shown above, is always just rising.. hence I consider it as a pure mathematical approach. – Ben Jan 09 '23 at 09:47

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I don't agree with @Nij : it is indeed a mathematical question about the so called "logarithmic derivative" of function $M(V)$ ($M$ as a function of $V$). Indeed $(1/M)(dM/dV)$ can be written

$$\dfrac{M'(V)}{M(V)}=\frac{d}{dV}\log(M(V))$$

I think that the interest of all that is connected to the fact that function $M(V)$ is a "kind of" exponential

$$M(V)=ke^{aV}\tag{1}$$ (for some positive contants $k$ and $a$)

Taking the logarithm of (1) gives a linear behavior:

$$\log(M(V))=aV+\log(k)$$

Fitting a straight line to the obtained measure points gives access to the value of parameters $a$ and $\log(k)$.

Jean Marie
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  • Thanks a lot for your input! Tbh, I guess it's over my head. Could you maybe please boil it more down in which way 1/M comes into play? What is the benefit of it (as it is in my eyes simply a constant)? – Ben Jan 09 '23 at 10:31
  • Is it clearer now with my extended explanations ? In order to answer to your comment, if you have only one sample, $M$ has only one value, yes ; but we work in the case where there can be several values of $M$... But maybe I haven't well understood what you mean... – Jean Marie Jan 09 '23 at 10:34
  • Yes, M can take several values respectively it is able to vary. So, let's consider only one sample (which means a single photodiode), then 1/M changes with M, obviously. But what do I gain by this? I mean the slope changes throughout the entire curve anyway? – Ben Jan 09 '23 at 10:41
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    In fact, as in many case in mathematics, there are formulas where certain terms "uprise" in the calculations. This is the case here with this division by $M$ (through the use of $\log$). Interpretating this division by $M$ as a normalization is maybe not the best way to see it. – Jean Marie Jan 09 '23 at 11:13
  • Ah, well, ok, I guess I got it now or at least the idea. Thank you very much! – Ben Jan 09 '23 at 11:19