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:
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!
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 $\log(k)$.