What is the Fermi energy of a n-type semiconductor?
O A E
O B. E(F) = (EC + Ev)/2
OC. EF = (Ec + Ed)/2
D. EF = (EV + Ea)/2
Answers
Answer:
By multiplying the density of states and probability distribution functions the existence of electrons and holes in the conduction and valence band states of a given semiconductor can be found out.
The probability function distributes over conduction and valence bands, crossing the forbidden band. When the probability to find electrons and holes in the conduction and valence band, respectively, is one at 300 K, then it is half at the mid of the forbidden gap.
But because there is no energy state you cannot find either en electron or hole at the mid of the forbidden gap of the semiconductor.
But let us imaging that a defect state is created at the mid of the forbidden gap and now the chance of finding an electron or hole in this defect state is half.
When temperature raises then probability (of occupation) distribution function shifts.
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.