The practical pin diode is in fact p+ n n+  diode. So, by reverse biasing its space charge layer  between the p+ layer and the lightly doped n-layer will be widen . By  increasing the the reverse voltage, the space charge layer continue widening till the it covers the whole width of the n-layer after which any further increase of the voltage will not affect the the space charge region scr width. In the first phase the diode depletion layer capacitance decreases inversely with the scr width after which it saturates to a constant value..

By plotting 1/c^2 versus the reverse voltage one can get the doping concentration NB of the n- lyer of the diode from the slope of its straight line part:, the slope can be expressed by

d (1/C^2)/ dV = 2/ q epsilons NB ,according to S.M. Sze, physics of semiconductor devices. Where q is the electronic charge, epsilon is the dielectric constant. C is the measured capacitance per unit area. The width of the n-layer will correspond to the fixed value capacitance with the voltage.So, one can estimate also the width of the  n-layer using C= epsilons/ WB provided WB is much large than the width of the space charge region at zero bias.

For more information you can consult SM.Sze.

You can check  out the value through the measurement of the breakdown voltage it is a unique function of  the doping and width of the n-layer.

I received a message from Manoj that there is no straight line part in his 1/C^2 versus V plot. Please Manoj would you  display your curve and tells us any information about the diode construction.

Any how, for any doping profile, there exist a relation between the slope S=d (1/C^2)/ dV  at any distance X from the junction such that

S= 2/ q epsilons N(X) where N(x) is the doping at the distance X,

And X can be calculated from X= epsilons / C(V).In this way you can get the doping profile N(X).

For more r information please see S. M. Sze.

wish you success