Physics 301 2-Dec-2005 27-1Electron Distribution in Degenerate SemiconductorsWhen the chemical potential is sufficiently close to one of the band edges t hat 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 =Z∞0f(ǫ) D(ǫ) dǫ ,and µ appearing in f(ǫ) is adjusted to produce the correct number of particles, and D(ǫ) dǫis the number of states with energies in the ra nge ǫ → ǫ + dǫ. For a spin 1/2 particle in abox we hadD(ǫ) =V2π22m¯h23/2√ǫ .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 filledcorrespo nds to an electron wave which is a standing wave ma de up of the smallest possiblemomentum components. The energy ǫcis negative and i s mostly due the interactions withthe atoms (remember, before the at o ms 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 ǫ − ǫcand the integralshould start at ǫc. We havene=NeV=12π22m∗e¯h23/2Z∞ǫc√ǫ − ǫcdǫe(ǫ − µ)/τ+ 1.Recall the definition of nc,nc= 2m∗eτ2π¯h23/2,from which we g etnenc=2√π1τ3/2Z∞ǫc√ǫ − ǫcdǫe(ǫ − µ)/τ+ 1.Now let x = (ǫ − ǫc)/τ and η = (µ − ǫc)/τ, and we can write the result in dimensionlessformnenc=2√πZ∞0√x dxex − η+ 1= I(η) ,where I(η) is called the Fermi-Dirac integral. In the classical l imit, where η ≪ −1, theFermi-Dirac integral reduces to our previous resultnenc= eη= e(µ − ǫc)/τ.Copyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 2-Dec-2005 27-2If t his limit doesn’t apply, then an excellent approximation isη = lognenc+1√8nenc.Figure 13.4 of K&K shows how well this Joyce-Dixon a pproximation works.Ionization of Donors and AcceptorsWe were working out the electron and hol e concentrations in a semiconductor. Wefound that the concentrations depended on the concentration of donor and a cceptor 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 t he 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 ǫd. This must be less thanthe lower edge of the conduction band ǫc(otherwise the donor will always be ionized).However, ǫdshould not be a lot less than ǫcor 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 ǫaare in the gap but close to the valence band edge at ǫv. Table 13. 2of K&K gives some values of ∆ǫd= ǫc− ǫdand ∆ǫa= ǫa− ǫv. The values are a few tensof milli-eV.Next we want t o calculate the probabili ty that a donor is ionized or neutral. There aretwo bound states corresponding to the two spin states of the electron. There is o ne ionizedstate and this state has no particles and no energy. Then the grand partitio n function forthis system isZ = 1 + e(µ − ǫd)/τ+ e(µ − ǫd)/τ.The probabilities that the donor is ionized or neutral aref(D+) =11 + 2e(µ − ǫd)/τ,f(D) =11 +12e(ǫd− µ)/τ.In the case of an acceptor, there’s one way to i onize it— adding an electron (which meansproducing the hole in the va lence band) gives four covalent bonds with the four neighbors.Since all these electrons are equivalent, there’s only one way to produce this state. ItCopyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 2-Dec-2005 27-3has one extra electron and it has energy ǫa. When it’s neutral, t he added hole (missingelectron) has two spin states, and the system has energy 0.Z = e(µ − ǫa)/τ+ 1 + 1f(A−) =e(µ − ǫa)/τe(µ − ǫa)/τ+ 2=11 + 2e(ǫa− µ)/τ,f(A) =2e(µ − ǫa)/τ+ 2=11 +12e(µ − ǫa)/τ.To get the concentrations of ions, we multiply by the tot al concentrations.n+d= ndf(D+) =nd1 + 2e(µ − ǫd)/τ,n−a= naf(A−) =na1 + 2e(ǫa− µ)/τ,Recall our expressions for neand nhin the classical limitne= nce−(ǫc− µ)/τ,nh= nve−(µ − ǫv)/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 a nd 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 tota l negative charge concentration must equal the total positi ve 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 plo t n+and n−, and wherethey intersect determines µ and all the concentrations. All this is shown in the figure. Therange of µ shown in the figure is from ǫvto ǫc. Of course, µ must remain a few τ awayfrom either band edge in order that our cl assical 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=1019cm−3, and nc= 1020cm−3. The donor ionization concentration is flat when µ ≪ ǫdand slopes downward at greater µ. Similarly, the acceptor i on