Dear Harsha,
I must disagree with Hala, you can not obtain the type (electrons/negative or holes/positive) of mobile carriers from Mott-Schottky plots, only their concentration and possibly (only in "good" cases though) the difference in electrochemical potential- the so called "built-in potential". You must do either thermopower measurement and/or Time of Flight measurement.
it is true though that in Schottky-Ohmic sample and depending on the type of the mobile charge carriers, the depletion region either increases or decreases for a given applied voltage polarity.
I hope that this helps
with bets regards
Petr
Dear Harsha,
I am not sure if one can determine the type of conductivity from Mott-Schottky equation. Certainly one can envisage the carrier density, but not the carrier type. There is no magnetic field applied in EIS. In that case, it is difficult to determine the current direction and as a result, the carrier type
Dear Harsha Bantawal,
Generally,If you already carried out a mott-Schottky (M-S) analysis, you can confirm the n-type and p-type conductivity based on the sign of the M-S plot slope, the positive slope of the M-S plot confirms the n-type conductivity of your film. While the negative slope of the M-S plot confirms p-type conductivity.
I hope that this helps
Dear Harsha and Hadi,
no, you can not obtain the type of majority carrier from simple Mott-Schottky plot. To do that , to follow Hadi's suggestion, some other criteria must be satisfied. These are :
1. The system has to be strictly Ohmic-Schottky system, one contact ohmic, the other Schottky. This gives depletioin region (high resistance) at the Schottky contact at zero external applied dc voltage.
2. Knowing which external wire goes from you voltage supply to which contact, you can now apply positive voltage (charge) to your Schottky contact.
3. Following now Hadi, If your majority carrier are electrons, they move towards this contact and the depletion region decreases -> the resistance of your system decreases and the measured capacitance (~depletion with) increases. The hole do the opposite.
With best regards
Petr
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