Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

Topic 11-3: Fermi Levels of Intrinsic Semiconductors with Effective Mass in Temperature

Summary: In this video we aim to get an expression for carrier concentration in an intrinsic semiconductor. To do this we begin by looking at the density of states and Fermi Dirac distribution of electrons and holes. Using this we solve for the concentration of each and develop an expression for the Fermi level. We then explore what happens to the Fermi level at different temperatures. Finally, we develop the np product rule and bring it together with a real world example.

• Recall from the free electron model ! !! ! ! o Electron density of states in a parabolic dispersion ! ! = ( ) !! ! !!! ħ! !!!! o Conductivity related to free carrier concentration (n) ! = !! • Goal: develop an expression for the carrier concentration in an intrinsic semiconductor • Need to be mindful of holes o One concentration for electrons, n o Once concentration for holes, p • Only electrons in the conduction band and only holes in the valence band are included in these concentrations • If we have a low concentration of conduction band electrons we can approximate the conduction band edge as a parabola

• Density of states near band edge is similar to the free electron model except the energy is offset by the energy of the conduction band and we have a correction for the band mass

! !! ∗ ! ! ! ! = ( ! ) !(! − ! ) ! [1] !!! ħ! ! Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

• Can simplify the Fermi Dirac distribution when µ is within the gap ! ! ! = ≅ !(!!!)/!!! [2] !(!!!)/!!!!! • Plugging this all into Mathematica we get ! ! ∗! ! ! ! ! (!!!!)/!!! ! = !!(!) !!(!) !" → ! = 2( ) ! ! [3] !! !!ħ! • Now for holes: ! ! = ! ! (!) ! (!) !" [4] !! ! ! o Density of states now has an energy offset of the valence band and a correction for this band mass

! !! ∗ ! ! ! ! = ( ! ) !(! − !) ! [5] !!! ħ! ! o Fermi Dirac distribution will be different because we are counting empty states, this means that it will be one minus the Fermi Dirac distribution for the conduction band

(!!!)/!!! !! ! = 1 − !! ! = ! [6] ! ∗! ! ! ! = 2( ! ! ) !!(!!!!)/!!! [7] !!ħ! • We know that n=p for an intrinsic semiconductor • Setting n=p ! ∗ ! !!/!!! ! ! !!/!!! ! = ( ∗) ! [8] !!

. where !! = !! − !! o Take the natural log of both sides ∗ ! ! !! ! = !! + !!! ln( ∗) [9] ! ! !! . with the energy scale zeroed to the energy of the valence band • What happens at 0 kelvin ! o ! = ! (second term disappears) ! ! o Fermi level is halfway in the gap o Invoking that the Fermi level and chemical potential, µ, are effectively the same term • Case 1: What happens at temperatures greater than 0 kelvin ∗ ∗ o Assume band edges have the same curvature: !! = !! Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

o !! !! + ! = !! −! • Where is µ o Try putting it in the center of the gap again

• Equal areas on the far right side of the above picture so n=p

• Case 2: !! !! + ! ≫ !! −!

• Try Fermi level at the center of the gap again

• In this case n and p are not equal which is bad • Need to shift Fermi Dirac function until the areas are equal Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

• n=p in the above diagram so the asymmetry of the band masses shifts the Fermi energy off the center of the band gap • In equilibrium at some temperature we expect some value for n and p for a given material o n and p reflect a balance of electron excitation and recombination to and from the conduction band • Let A be the excitation rate o Independent of n and p for an intrinsic semiconductor o Lots of empty states in the conduction band and lots of electrons in the valence band makes it easy to make a transition • Let B be the recombination rate o Depends on n and p because the excited electrons must find holes to recombine with

!" = !ℎ!"#$ !" ! !" ! !"#$%&'# !" !"#$ = ! − !"# = !" [10] !" !" !" !" • In equilibrium = = 0 !" !" ! • This means !" = ! • Using our previously found expressions for n and p we get ! ! ! ! ∙ ! = 4( ! )!(! ∗! ∗) !!!!!/!!! [11] !!ħ! ! ! ! ∗ ∗ !!!/!!! o The last part of the above expression, (!! !! ) !! , is the band structure • Example: Silicon !" ! * * o ! ∙ ! = 2!10 !"##$%#&/!" at 300 kelvin form knowledge of mh , me , and Eg o ! = ! = !" = 4.5!10! !"##$%#&/!"! o Not a lot of free carriers given that 1 cm3 of silicon holds about 0.1 moles or 6x1022 silicon atoms Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

o This gives one excited electron per 1013 atoms o Makes sense as you are asking an electron to jump a 1.1 eV gap at 300 kelvin

Colorado School of Mines Solid State Physics in a Nutshell solidstate.mines.edu

Questions to Ponder 1. Would the np product rule continue to hold if the semiconductor was placed in sunlight or if we doped the semiconductor?

2. Say I create a band structure with 2 minima in the valence band and one maxima in the conduction band. Assuming all three band edges have the same effective mass where is the equilibrium location of the Fermi level for temperatures greater than 0 K?

3. What is the temperature dependence of the Fermi level when the two band masses are the same?