Hi .
I have performed a Capacitance-voltage experiment for my p-n diode sample. I plotted the (1/C)^2 - V and found that the Cut-in voltage is very high (more than 10V). It should come like ~0.8V. Can anyone tell me the reason of such high cut-in voltage from the (1/C)^2 - V.
From what I understand : C is dependent on the
1.Energy band gap of the material(s) used.
2. The dimension that form capacitance (say depletion width !)
3. Fermi levels after doping
4. Worst case - voltage applied (switching speed )
To the best of my knowledge, the cut-in voltage is given by the formula :-
V= VT ln( NA X ND / ni2)where V is the cut-in voltage ;
VT is the volt equivalent of temperature; given by T / 11600; T being temperature in degrees kelvin;
NA is the acceptor impurity concentration (P- side)
ND is the donor impurity concentration (N- side)
ni is the intrinsic concentration
I feel that you better substiute the above values pertaining to your diode, and check it.
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