Physics 301 25-Nov-2002 29-1ReadingFinish K&K Chapter 13.Electron Distribution in Degenerate SemiconductorsWhen the chemical potential is sufficiently close to one of the band edges that theclassical approximation no longer applies, we have to use the full treatment as we did (tosome level of approximation) for a Fermi gas in lectures 16 and 17. In particular, our oldexpression for the number of particles wasN =∞0f( ) D( ) d ,and µ appearing in f( ) is adjusted to produce the correct number of particles, and D( ) d is the number of states with energies in the range → + d .Foraspin1/2particleinabox we hadD( )=V2π22m¯h2 3/2√ .As we’ve discussed, to get the right density of states we should replace m by m∗e.Inaddition, the energy that appears in the square root in the density of states really camefrom changing variables from momentum to energy. At the bottom of the conduction band,the electrons presumably have very little momentum. That is, the first state to be filledcorresponds to an electron wave which is a standing wave made up of the smallest possiblemomentum components. The energy cis negative and is mostly due the interactions withthe atoms (remember, before the atoms formed a crystal, the atomic state that gave riseto the conduction band was a bound electronic state). The upshot of all this is that weshould replace the energy in the density of states expression by − cand the integralshould start at c.Wehavene=NeV=12π22m∗e¯h2 3/2∞c√ − cd e( − µ)/τ+1.Recall the definition of nc,nc=2m∗eτ2π¯h2 3/2,from which we getnenc=2√π1τ3/2∞c√ − cd e( − µ)/τ+1.Now let x =( − c)/τ and η =(µ − c)/τ, and we can write the result in dimensionlessformnenc=2√π∞0√xdxex − η+1= I(η) ,Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 25-Nov-2002 29-2where I(η) is called the Fermi-Dirac integral. In the classical limit, where η −1, theFermi-Dirac integral reduces to our previous resultnenc= eη= e(µ − c)/τ.If this limit doesn’t apply, then an excellent approximation isη =lognenc+1√8nenc.Figure 13.4 of K&K shows how well this Joyce-Dixon approximation works.Ionization of Donors and AcceptorsWe were working out the electron and hole concentrations in a semiconductor. Wefound that the concentrations depended on the concentration of donor and acceptor ions.Up to now, we’ve assumed that these were “givens.” Of course, the true givens are theconcentrations of the donors and acceptors and an overall equilibrium will be establishedbetween donor and acceptor ion concentrations, the concentration of electrons in the con-duction band, and the concentration of holes in the valence band.When a donor is introduced into a semiconductor, the extra electron may stay boundto the donor, or the donor may ionize and contribute the extra electron to the conductionband. When bound to the donor, the energy of the electron is d. This must be less thanthe lower edge of the conduction band c(otherwise the donor will always be ionized).However, dshould not be a lot less than cor the donor will hardly ever be ionized andthere wasn’t much point in introducing this donor in the first place. Thus donor boundstates are “in the gap,” but close to the conduction band edge. Similarly, acceptor boundstates, with energy aare in the gap but close to the valence band edge at v. Table 13.2of K&K gives some values of ∆ d= c− dand ∆ a= a− v. The values are a few tensof milli-eV.Next we want to calculate the probability that a donor is ionized or neutral. There aretwo bound states corresponding to the two spin states of the electron. There is one ionizedstate and this state has no particles and no energy. Then the grand partition function forthis system isZ =1+e(µ − d)/τ+ e(µ − d)/τ.The probabilities that the donor is ionized or neutral aref(D+)=11+2e(µ − d)/τ,f(D) =11+12e( d− µ)/τ.Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 25-Nov-2002 29-3In the case of an acceptor, there’s one way to ionize it—adding an electron (which meansproducing the hole in the valence band) gives four covalent bonds with the four neighbors.Since all these electrons are equivalent, there’s only one way to produce this state. Ithas one extra electron and it has energy a. When it’s neutral, the added hole (missingelectron) has two spin states, and the system has energy 0.Z = e(µ − a)/τ+1+1f(A−)=e(µ − a)/τe(µ − a)/τ+2=11+2e( a− µ)/τ,f(A) =2e(µ − a)/τ+2=11+12e(µ − a)/τ.To get the concentrations of ions, we multiply by the total concentrations.n+d= ndf(D+)=nd1+2e(µ − d)/τ,n−a= naf(A−)=na1+2e( a− µ)/τ,Recall our expressions for neand nhin the classical limitne= nce−( c− µ)/τ,nh= nve−(µ − v)/τ.We now have expressions for four concentrations in terms of givens and one unknown.The givens are the temperature, the concentrations of donors and acceptors, the quantumconcentrations of electrons and holes in the conduction and valence bands (which dependon temperature and effective masses), the energies at the band edges, and the donorand acceptor ionization energies (whew!). The unknown is the chemical potential. Thechemical potential must be chosen so that the system is electrically neutral. To help withthis, note that the total negative charge concentration must equal the total positive chargeconcentration,n−= ne+ n−a= nh+ n+d= n+.We can make a semi-log plot of all four concentrations versus µ. The electron and holeconcentrations are just straight lines, and the ionized donor and acceptor concentrationsare essentially two intersecting straight lines. Then we can plot n+and n−,andwherethey intersect determines µ and all the concentrations. All this is shown in the figure. Therange of µ shown in the figure is from vto c.Ofcourse,µ must remain a few τ awayfrom either band edge in order that our classical approximation remain valid.The figure is drawn with µ scaled to run between 0 and 1 at the valence and conductionband edges. The curves have been drawn for nd=1017cm−3, na=1015cm−3, nv=Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 25-Nov-2002 29-41019cm−3,andnc=1020cm−3. The donor ionization concentration is flat when µ  dand slopes downward at greater µ. Similarly, the acceptor ion