EECS130 Integrated Circuit DevicesLast Lecture: Energy Band DiagramTemperature Effect on Band GapMeasuring the Band Gap Energy by Light AbsorptionSemiconductors, Insulators, and ConductorsDonor and Acceptor Levels in the Band ModelDopants and Free CarriersEffective MassDensity of StatesThermal EquilibriumThermal Equilibrium An Analogy for Thermal EquilibriumSlide Number 12Slide Number 13Effect of T on f(E)QuestionSlide Number 16Slide Number 17Slide Number 18Intrinsic SemiconductorSlide Number 20Slide Number 21Example: The Fermi Level and Carrier ConcentrationsThe np Product and the Intrinsic Carrier ConcentrationEXAMPLE: Carrier ConcentrationsEXAMPLE: Complete ionization of the dopant atomsDoped Si and ChargeBond Model of Electrons and Holes (Intrinsic Si)Dopants in SiliconGeneral Effects of Doping on n and pSlide Number 30EXAMPLE: Dopant CompensationCarrier Concentrations at Extremely High and Low TemperaturesInfrared Detector Based on Freeze-outChapter Summary Energy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef.EECS130 Integrated Circuit DevicesProfessor Ali Javey8/30/2007Semiconductor FundamentalsLecture 2Read: Chapters 1 and 2Last Lecture: Energy Band DiagramConduction bandEcEvEgBand gapValence band •Energy band diagram shows the bottom edge of conduction band, Ec , and top edge of valence band, Ev .•Ec and Ev are separated by the band gap energy, Eg .Temperature Effect on Band GapDecreasing atomic separationEnergypsisolated atomslattice spacingvalence bandconduction bandHow does the band gap change with temperature?Measuring the Band Gap Energy by Light Absorptionphotonsphoton energy: hv > EgEcEvEgelectronholeBandgap energies of selected semiconductors• Eg can be determined from the minimum energy (hν) of photons that are absorbed by the semiconductor.Material PbTe Ge Si GaAs GaP DiamondEg(eV) 0.31 0.67 1.12 1.42 2.25 6.0Semiconductors, Insulators, and Conductors•Totally filled bands and totally empty bands do not allow •Metal conduction band is half-filled.EcEvEg=1.1 eVEcEg= 9 eVemptySi (Semiconductor)SiO2(Insulator)ConductorEcfilledTop ofconduction bandEvcurrent flow. (Just as there is no motion of liquid in a totally filled or totally empty bottle.).•Semiconductors have lower Eg's than insulators and can be doped.Donor and Acceptor Levels in the Band ModelConduction BandEcEvValence BandDonor LevelAcceptor LevelEdEaDonor ionization energyAcceptor ionization energyIonization energy of selected donors and acceptors in siliconAcceptorsDopant Sb P As B Al InIonization energy, Ec–Ed or Ea–Ev (meV)3944544557160DonorsHydrogen: Eionm0 q413.6 eV==8ε02h2Dopants and Free CarriersDopant ionizationenergy ~50meV (very low). Donorsn-typeAcceptorsp-typeEffective MassIn an electric field, , an electron or a hole accelerates.Electron and hole effective massesSi Ge GaAs GaPmn/m00.26 0.12 0.068 0.82mp/m00.39 0.30 0.50 0.60electronsholesRemember :F=ma=-qEDensity of StatesEgcgvEcEvg(E)EcEvΔΕ⎟⎠⎞⎜⎝⎛⋅⋅ΔΔ≡3cmeV1 volumein states ofnumber )(EEEgc() 2)(32**hEEmmEgcnncπ−≡() 2)(32**hEEmmEgvppvπ−≡Thermal EquilibriumThermal Equilibrium An Analogy for Thermal Equilibrium•There is a certain probability for the electrons in theconduction band to occupy high-energy states under the agitation of thermal energy (vibrating atoms, etc.)DishVibrating TableSand particlesAt E=EF , f(E)=1/2Assume the two extremes:High Energy (Large E): E-Ef >> kT, f(E) 0Low Energy (Small E): E-Ef << kT, f(E) 1Effect of T on f(E)T=0KQuestion• If f(E) is the probability of a state being occupied by an electron, what is the probability of a state being occupied by a hole?n = electron density : number of unbound electrons / cm3p = hole density : number of holes / cm3Nc is called the effective density of states (of the conduction band) .Nv is called the effective density of states of the valence band.Intrinsic Semiconductor• Extremely pure semiconductor sample containing an insignificant amount of impurity atoms.n = p = niMaterial Ge Si GaAsEg (eV) 0.67 1.12 1.42ni (1/cm3) 2 x 10131 x 10102 x 106Ef lies in the middle of the band gapintrinsicn-typeRemember: the closer Efmoves up to Ec, the larger n is; the closer Efmoves down to Ev, the larger p is.For Si, Nc= 2.8×1019cm-3and Nv= 1.04×1019cm-3.EcEvEfEcEvEfExample: The Fermi Level and Carrier ConcentrationskTEEcfceNn/)( −−=()()eV 614.010/108.2ln026.0ln1719=×==− nNkTEEcfcEcEfEv0.146 eVWhere is Ef for n =1017 cm-3? Solution:The np Product and the Intrinsic Carrier Concentration• In an intrinsic (undoped) semiconductor, n = p = ni .kTEvcigeNNn2/−=2innp =kTEEcfceNn/)( −−=kTEEvvfeNp/)( −−=andMultiplykTEvckTEEvcgvceNNeNNnp//)(−−−==Question: What is the hole concentration in an N-type semiconductor with 1015 cm-3 of donors?Solution: n = 1015 cm-3. After increasing T by 60°C, n remains the same at 1015 cm-3 while p increases by about a factor of 2300 because .Question: What is n if p = 1017cm-3 in a P-type silicon wafer?Solution:EXAMPLE: Carrier Concentrations3-5315-3202cm10cm10cm10=≈=−nnpikTEigen/2−∝3-3317-3202cm10cm10cm10=≈=−pnniEXAMPLE: Complete ionization of the dopant atomsNd = 1017 cm-3 and Ec -Ed =45 meV. What fraction of the donors are not ionized?Solution: First assume that all the donors are ionized.EcEfEv146 meVEd45meVProbability of non-ionization ≈02.01111meV26/)meV)45146((/)(=+=+−−eekTEEfdTherefore, it is reasonable to assume complete ionization, i.e., n = Nd .meV146cm10317−=⇒==−cfdEENnDoped Si and Charge• What is the net charge of your Si when it is electron and hole doped?Bond Model of Electrons and Holes (Intrinsic Si)•Silicon crystal in a two-dimensionalrepresentation.Si Si SiSi Si SiSi Si SiSi Si SiSi Si SiSi Si SiSi Si SiSi Si SiSi Si Si•When an electron breaks loose and becomes a conduction electron, a hole is also created.Dopants in SiliconSi Si SiSi SiSi Si SiSi Si SiSi SiSi Si SiAs B•As (Arsenic), a Group V element, introduces conduction electrons and creates N-type silicon,•B (Boron), a Group III element, introduces holes and creates P-type silicon, and is called an acceptor.•Donors and acceptors are known as dopants.and is called a donor.N-type SiP-type SiGeneral Effects of Doping on n and pCharge neutrality:daNpNn−−++_= 0daNpNn−−+= 0Assuming total ionization of acceptors and donors:aN_: number of ionized acceptors /cm3dN+: number of ionized donors /cm3aN: number of ionized acceptors /cm3dN+: number of
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