Prove that the fermi level lies just in the middle of forbidden energy gap
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Answer:
In general ,we say that Ef corresponds to that level which has probability 1/2 being occupied. So for semiconductors Ef must located between Ev &Ec. It is assumed that width of Vb and Cb are very small as compared to Eg. Let each band consist of Z number of possible states per unit volume. At T=0 ,all the states in Ev are filled while all the states in the Cb are empty. At T>0 ,density of electron in Cb is nc = z / {exp{[Ec-Ef] /KT} +1} and density of electrons in the Vb is given as
nv = z / {exp{[Ev-Ef] /KT} +1}. Practically all the electrons in Cb are from Vb. So nc +nv = z. So adding the equations we have ( z / {exp{[Ec-Ef] /KT} +1} ) + ( z / {exp{[Ev-Ef] /KT} +1}) = z . Solving this equations we get Ef =( Ec+Ev )/2 so it is midway between the Ev and Ec. Actually this is valid approximation. If we consider detailed calculation of nc & nv then effective values of elecrton mass (Me) , hole mass (Mh) and T affect on calculation giving result
Ef = { ( Ec+Ev)/2} + (3/4) KT log (Mh / Me) under the condition ne = nh for intrinsic semiconductor.
If we assume Me = Mh the second term vanishes giving same result Ef =( Ec+Ev )/2 .
Ref- Chapter 12 , The Electron Distribution in insulators and semiconductors. from Solid State Physics by A. J. Dekker. MACMILLAN INDIA LIMITED.
From - Sanjay Gadekar , PESJMJ , University of pune.
Answer:
Hence, the probability of occupation of energy levels in conduction band and valence band are equal. Therefore, the Fermi level for the intrinsic semiconductor lies in the middle of forbidden band. Fermi level in the middle of forbidden band indicates equal concentration of free electrons and holes.
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