Conduction band. B -. B - Acceptor level for B. h +. h + Valence band

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1 Sesson #14: Homework s Problem #1 (a) Determne the amount (n grams) of boron (B) that, substtutonally ncorporated nto 1 kg of germanum (Ge), wll establsh a charge carrer densty of x /cm 3. (b) Draw a schematc energy band dagram for ths materal, and label all crtcal features. (a) The perodc table gves the molar volume of Ge as cm 3 and 1 mole of Ge weghs g, so set up the rato = 1000 g and solve for x to get x cm 3 for the total volume. The addton of boron gves 1 charge carrer/b atom. B concentraton n S must be x B/cm 3 N A of B atoms weghs g x B atoms wegh = g for every 1 cm 3 of Ge, add 5.55 x 10 6 g B (b) for cm 3 of Ge, add x 5.55 x 10 6 = 1.04 x 10 3 g B Conducton band Band gap, E g B -. B - Acceptor level for B h +. h + Valence band x B - ons n the band gap just above the top of the v.b x holes n the v.b. Problem #2 (a) An electron beam strkes a crystal of cadmum sulfde (CdS). Electrons scattered by the crystal move at a velocty of 4.4 x 10 5 m/s. Calculate the energy of the ncdent beam. Express your result n ev. CdS s a semconductor wth a band gap, E g, of 2.45 ev. (b) Cadmum tellurde (CdTe) s also a semconductor. Do you expect the band gap of ths materal to be greater or less than the band gap of CdS? Explan.

2 (a) mv E - = Eemtted γ + E - = Eg + ncdent e scattered e 2 2 ( ) kg m/s = 2.45 ev ev/j 2 = 2.45 ev ev = 3.00 ev (b) E g (CdTe) < E g (CdS) Problem #3 The Cd S bond s stronger than Cd Te bond because although both S and Te are group 16, Te s much larger than S. AlN and GaSb are compounds, sold at room temperature. On the bass of bondng consderatons and data provded n the perodc table, attempt to predct dfferences n the propertes of these solds. Both compounds are of the III V famly, whch hybrdze and form adamantne (damond lke) structures whch places them nto the category of semconductor. AlN ΔEN = The covalent rad of the consttuents are small and, combned wth the large EN, the bonds (polar covalences) are very strong the semconductor s expected to exhbt a large band gap (lkely transparent). GaSb ΔEN = The covalent rad of both consttuents are sgnfcantly larger (than those of AlN), the onc contrbuton to bondng s small the semconductor s expected to exhbt a much smaller band gap than AlN. AlN: E g = 3.8 ev GaSb: E g = 0.8 ev

3 Problem #4 Problem #5 Explan the dfference between extrnsc and ntrnsc semconductors. An extrnsc semconductor s a semconductor whch contans foregn elements capable of contrbutng moble charge carrers, electrons, to the conducton band (n type) or holes to the valence band (p type). An ntrnsc semconductor contans no foregn elements. The number of electron-hole pars n ntrnsc germanum (Ge) s gven by: 15 3/2 -E g /2KT 3 n = T e [cm ] (E = 0.72 ev) (a) What s the densty of pars at T = 20 C? (b) Wll undoped Ge be a good conductor at 200 C? If so, why? (a) Recall: T n thermally actvated processes s the absolute temperature: T o K = ( t o C); Boltzmann s constant = k = 1.38 x J/ o K T = K: (b) 200 o C = K n = e 15-7 = n = / cm n = e n (200K) = n = / cm The number of conductng electrons (n the conducton band) at 200 o C s by about fve orders of magntude less than that of a good conductor. The materal wll not be a good conductor. (There are addtonal factors whch contrbute to the relatvely poor conductvty of Ge at ths temperature.) g

4 Problem #6 If no electron-hole pars were produced n germanum (Ge) untl the temperature reached the value correspondng to the energy gap, at what temperature would Ge become conductve? (E th = 3/2 kt) E = th 3KT 2 ; -19 E g = J Problem # o T = = 5565K = C The temperature would have to be 5.3 x 10 3 o C, about 4400 o C above the meltng pont of Ge. (a) How do you expect the conductvty to vary n an ntrnsc semconductor wth ncreasng temperature? Explan your answer. (b) How do you expect the conductvty to vary n a metallc conductor wth ncreasng temperature? (a) From our lmted knowledge of the conducton behavor, we must assume that n semconductors the conductvty wll ncrease wth T snce the number of electrons n the conducton band ncreases. (b) As a frst approxmaton, we can say that the number of free electrons s ndependent of T; that, however, the atoms n the lattce wll, wth ncreasng temperature, be subject to ncreased oscllatons about ther (relatvely) fxed poston. These oscllatons provde greater opportunty for scatterng of conductng electrons and thus wll reduce the moblty of the electrons. We can expect the electrcal conductvty of metals to decrease wth ncreasng temperature. Problem #8 The energy gap (E g ) of ZnSe s 2.3 ev. (a) Is ths materal transparent to vsble radaton? Substantate your answer. (b) How could you ncrease the electrcal conductvty of ths materal? Gve the reasons for the effectveness of your suggested approach. (a) The optcal propertes of ZnSe can be explaned when comparng the energy band of the vsble spectrum wth the energy band dagram of ZnSe. Absorpton takes place va photoexctaton for all radaton wth E 2.3 ev.

5 From the energy dstrbuton of the vsble spectrum we recognze that the blue green porton has photon energes n excess of the band gap (E g = 2.3eV) and thus wll be absorbed. The yellow red poton, on the other hand, has photon energes less than the band gap t wll be transmtted. ZnSe, therefore, s expected to exhbt a yellowsh red color. (b) In prncple there are two ways to ncrease the electrcal conductvty of ZnSe: a. A temperature rse. Any rse n temperature wll ncrease the number of thermally actvated charge carrers n the conducton band (electrons) and n the valence band (holes) and, thus, the electronc conductvty. Asde: The electrcal conductvty of solds, demonstrated by the flow of electronc charge carrers under an appled electrc feld (E), can be formulated through Ohm s law, J = σe, whch states that the current densty (J = number of charges transported through a unt area n a unt tme) s proportonal (σ = conductvty) to the appled electrc feld. Accordngly: J = N e v d where N = number of charge carrers/unt volume, e = electronc charge and v d = average drft velocty of charge carrers n an appled electrc feld. We thus obtan: σ = (N e v d) / E

6 and f we defne σ = N e μ (v d / E) = μ, the charge carrer moblty, we have: In ntrnsc semconductors we have both electrons and holes contrbutng to conducton: σ = Neeμ e + Nneμ h = Ne(μ e + μ h) snce N e = N h. Takng the number of thermally generated charge carrers, gven by the relatonshp -E /2KT 3/2 g N = AT e we obtan the temperature dependence of the conductvty as: 3/2 -E g /2KT σ = AT e e(μ + μ ) e h To assess the temperature dependence of electrcal conductvty we must take nto consderaton that, because of ncreased vbraton of the atoms about ther lattce postons, the charge carrer moblty wll decrease (ncreased scatterng of charge carrers) wth ncreasng temperature. Ths effect explans why the electronc conductvty n metals, where N s constant, wll decrease wth ncreasng temperature. In semconductors, where N ncreases wth temperature, the accompanyng moblty effect s not apparent at low temperatures (conductvty ncreases), but becomes pronounced at hgh temperatures (conductvty decreases). b. Introducton of shallow mpurty (or defect) states close to the conducton or valence band. Ths s accomplshed by the ncorporaton of approprate dopant elements nto the crystal matrx. If these mpurtes are shallow (~0.01 ev from the conducton or valence band), they wll be totally onzed at room temperature and each wll contrbute an electron (donor dopant: K, Na) or holes (acceptor dopant: G, Br), thus ncreasng the electrcal conductvty wthout the necessty of a temperature rse. (Be aware that certan defects n the crystal lattce may also ncrease the electronc conductvty.)

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