EECS130 Integrated Circuit DevicesAnnouncementsLast Lecture: Energy Band DiagramCarrier ConcentrationsExample: 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 13EXAMPLE: Dopant CompensationCarrier Concentrations at Extremely High and Low TemperaturesInfrared Detector Based on Freeze-outChapter 2 Summary Energy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef. Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Effective MassSlide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33EECS130 Integrated Circuit DevicesProfessor Ali Javey9/04/2007Semiconductor FundamentalsLecture 3Reading: finish chapter 2 and begin chapter 3Announcements•HW 1 is due next Tuesday, at the beginning of the class. Late HWs will not be accepted.•Midterm 1 is on Oct. 4th, in class. It’s closed book, but you may bring and use one sheet of note.•Check EE130 web site often for lecture notes and announcements. (http://inst.eecs.berkeley.edu/~ee130/fa07/)Last Lecture: Energy Band DiagramEcEv•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 . EdCarrier ConcentrationskTEEcfceNn/)( −−=kTEEvvfeNp/)( −−=At E=Ef , f(E)=1/2From density of states and Fermi function, we obtain:Example: 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 ionized donors /cm3I. (i.e., N-type) If , II.(i.e., P-type) If, adNNn −=nnpi2=iadnNN >>−adNN >>dNn=diNnp2=andidanNN >>−daNNp −=pnni2=daNN >>aNp=aiNnn2=andGeneral Effects of Doping on n and pEXAMPLE: Dopant CompensationWhat are n and p in Si with (a) Nd = 6×1016 cm-3 and Na = 2×1016 cm-3 and (b) additional 6×1016 cm-3 of Na ?(a)(b) Na = 2×1016 + 6×1016 = 8×1016 cm-3 > Nd !+ + + + + + + + + + + +. . . . . .. . . . . . . . . . .--------. . . . . .. . . . . .316cm104−×=−=adNNn3316202cm105.2104/10/−×=×== nnpi3161616cm102106108−×=×−×=−=daNNp3316202cm105102/10/−×=×== pnniNd = 6×1016 cm-3Na = 2×1016 cm-3n = 4×1016 cm-3Nd = 6×1016 cm-3Na = 8×1016 cm-3p = 2×1016 cm-3Carrier Concentrations at Extremely High and Low Temperaturesintrinsic regimen = Ndfreeze-out regimelnn1/Thigh temp.roomtemperaturecryogenictemperatureInfrared Detector Based on Freeze-outTo image the black-body radiation emitted by tumors requires a photodetector that responds to hν’s around 0.1 eV. In doped Si operating in the freeze-out mode, conduction electrons are created when the infrared photons provide the energy to ionized the donor atoms.photonEcEvelectronEdChapter 2 Summary Energy band diagram. Acceptor. Donor. mn , mp . Fermi function. Ef .kTEEcfceNn/)( −−=kTEEvvfeNp/)( −−=adNNn−=daNNp−=2innp =Thermal Motion• Zig-zag motion is due to collisions or scatteringwith imperfections in the crystal. • Net thermal velocity is zero.• Mean time between collisions (mean free time) is τm ~ 0.1psThermal Energy and Thermal Velocityelectron or hole kinetic energy 22123thmvkT ==kg101.926.0K300JK1038.13331123−−−×××××==effthmkTvcm/s103.2m/s103.275×=×=~8.3 X 105 km/hrDriftElectron and Hole Mobilities• Drift is the motion caused by an electric field.Effective 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=-qERemember :F=ma=mV/t = -qEElectron and Hole Mobilitiespmppmqτμ=nmnnmqτμ=•μp is the hole mobility and μn is the electron mobilitymppqvmτ=pmpmqvτ=pvμ=nvμ−=Electron and hole mobilities of selected semiconductorsElectron and Hole MobilitiesSi Ge GaAs InAsμn (cm2/V·s) 1400 3900 8500 30000μp (cm2/V·s) 470 1900 400 500. sVcmV/cmcm/s2⎥⎦⎤⎢⎣⎡⋅=v =μ ; μhas the dimensions of v/Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices?EXAMPLE: Given μp = 470 cm2/V·s, what is the hole drift velocity at = 103 V/cm? What is τmp
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