How to Determine EF the Fermi Level in Semiconductors HD

https://www.fiberoptics4sale.com In this video, we will discuss the calculation of Fermi level EF in a nondegenerate, dopants totally ionized semiconductor material. The intrinsic Fermi level Ei is located somewhere near the middle of the band gap. But we need to know the precise positioning of Ei. We also have learned the formulas for carrier concentrations n and p for nondegenerate semiconductors, and the value or positioning of Fermi level EF determines whether a doped semiconductor is nondegenerate or degenerate. Actually the formulas for carrier concentrations n and p provide a one-to-one correspondence between the Fermi level EF and the carrier concentrations n and p. So, by having any one of the three variables – n, p, or EF – we can determine the remaining two variables under equilibrium conditions. On this slide, we will determine the exact positioning of intrinsic Fermi level Ei. In an intrinsic semiconductor material, n equals p, we can also use the nondegenerate formulas for n and p as shown. Where NC is the “effective” density of conduction band states, and NV is the “effective” density of valence band states. Another fact is that EF equals Ei in an intrinsic semiconductor. So by substituting for n and p and setting EF = Ei, we then get the second equation NC times e to the power of Ei minus EC over kT equals NV times e to the power of EV minus Ei over kT. We can then solve the equation for Ei and get the result - Ei equals EC plus EV divided by 2, plus kT divided by 2 times ln(NV over NC). The ratio between NV over NC can be expressed as the relationship of the electron effective mass mn asterisk and the hole effective mass mp asterisk. And then we get the final result of Ei as EC plus Ev over 2, plus 3 quarters times kT times ln(Mp asterisk over mn asterisk). So Ei lies precisely at midgap only if mp asterisk equals mn asterisk or if the temperature T is absolute 0. For silicon at room temperature, mp asterisk over mn asterisk equals 0.69, and 3 quarters times kT times ln(Mp asterisk over mn asterisk) is -0.0073 electron volt, and Ei lies 0.0073 eV below mid-gap. This small deviation from midgap is typically neglected in drawing energy band diagrams, etc. Now let’s discuss the calculation of Fermi level EF in doped semiconductors. The general assumption is that donor- and acceptor-doped semiconductors to be nondegenerate, in equilibrium condition, and maintained at temperatures where the dopants are fully ionized. Under these conditions, we can use the following carrier concentration formula: n equals ni times e to the power of EF minus Ei over kT, p equals ni times e to the power of Ei minus EF over kT. We can then solve (EF - Ei) in terms of n, p and ni. Depending on the particular problem, the appropriate carrier concentration formulas for n and p is then used to determine EF. In typical donor-doped semiconductors, ND is far greater than NA, ND is also far greater than ni, thus we get EF minus Ei equals kT t