1Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Electrical Resistancewhere ρis the electrical resistivityResistanceWtLIVRρ=≡(Unit: ohms)V+_LtWIMaterial with resistivity ρ2Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Resistivity Range of MaterialsAdding parts/billionto parts/thousandof “dopants” to pure Si can changeresistivity by 8 orders of magnitude ! Si with dopantsSiO2, Si3N41 Ω-m = 100 Ω-cm3Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06The Si AtomThe Si CrystalHigh-performance semiconductor devices require defect-free crystals“diamond” structure4Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06-+Top of valence bandBottom of conduction bandelectronholeEnergy gap=1.12 eVCarrier Concentrations of Intrinsic (undoped) Sin (electron conc)= p (hole conc) = ni5Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Donors: P, As, SbAcceptors: B, Al, Ga, InDopants in SiBy substituting a Si atom with a special impurity atom (Column Vor Column III element), a conduction electron or hole is created.6Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Semiconductor with both acceptors and donors has 4 kinds of charge carriersIonized DonorIonizedAcceptorImmobile Chargesthey DO NOTcontribute to current flow with electric field is applied. However, they affect the local electric fieldHoleElectronMobile Charge Carriersthey contribute to current flow with electric field is applied.7Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Charge Neutrality ConditionValid for homogeneously doped semiconductor at thermal equilibriumEven NAis not equal to ND,microscopic volume surroundingany position x has zero net chargeSi atom (neutral)Ionized DonorIonizedAcceptorHoleElectronElectrons and holes created by Si atoms with conc ni8Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Electron and Hole Concentrationsfor homogeneous semiconductor at thermal equilibriumn: electron concentration (cm-3)p : hole concentration (cm-3)ND: donor concentration (cm-3)NA: acceptor concentration (cm-3)1) Charge neutrality condition: ND+ p = NA+ n2) Law of Mass Action : n•p = ni2Note: Carrier concentrations depend on NET dopant concentration (ND- NA) !Assume completely ionized to form ND+and NA-9Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06n-type SemiconductorIf ND>> NA(such that ND– NA≥ 10 ni):ADNNn −≅ADiNNnp−≅2andNote n >> p+=ND /cm3NA /cm3n-type10Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06p-type SemiconductorIf NA>> ND (such that NA – ND≥ 10 ni):Note p >> n+=ND /cm3NA /cm3p-typeDANNp−≅DAiNNnn−≅2and11Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Carrier Drift• When an electric field is applied to a semiconductor, mobile carriers will be accelerated by the electrostatic force. This force superimposes on the random thermal motion of carriers:E.g. Electrons drift in the direction opposite to the E-fieldÆ Current flowsAverage drift velocity = | v | = µECarrier mobility12345electronE12345electronE =012Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Carrier Mobility• Mobile carriers are always in random thermal motion. If no electric field is applied, the average current in any direction is zero. • Mobility is reduced by1) collisions with the vibrating atoms“phonon scattering”2) deflection by ionized impurity atoms “Coulombicscattering”-Si-As+-B--13Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Mobile charge-carrier drift velocity v is proportional to applied E-field:µnµpCarrier Mobility µ| v | = µEMobility depends on (ND+ NA) !(Unit: cm2/V•s)14Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Electrical Conductivity σWhen an electric field is applied, current flows due to drift of mobile electrons and holes:EqnnvqJnnnµ=−=)(electron current density:hole current density:EqppvqJpppµ=+=)(total current density:pnpnpnqpqnEJEqpqnJJJµµσσµµ+≡=+=+=)(conductivity15Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06(Unit: ohm-cm)Electrical Resistivityρpnqpqnµµσρ+=≡11for n-typenqnµρ1≅for p-typepqpµρ1≅Note: This plot does not apply for compensated material!16Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06What are n and p values?What is its electrical resistivity ?Answer:NA= 1016/cm3, ND= 0 (NA>> NDÆ p-type)Æ p ≈ 1016/cm3and n ≈ 104/cm3Example Calculation 1 []cm 4.1)450)(10)(106.1(1111619−Ω=×=≅+=−−ppnqpqpqnµµµρFrom µpvs. ( NA+ ND) plot1016Boron/cm3Si17Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06* The p-type sample is converted to n-type material by adding more donors than acceptors, and is said to be “compensated”.Example Calculation 2: Dopant CompensationAnswer:NA= 1016/cm3, ND= 1017/cm3 (ND>>NAÆ n-type)Æ n ≈ 9x1016/cm3and p ≈ 1.1x103/cm3[]cm 12.0)600)(109)(106.1(1111619−Ω=××=≅+=−−npnqnqpqnµµµρFrom µnvs. ( NA+ ND) plot1016Boron/cm3Si1017Arsenic/cm3+What are n and p values?What is its electrical resistivity ?18Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Summary of Doping Terminologyintrinsic semiconductor: undoped semiconductorextrinsicsemiconductor: doped semiconductordonor: impurity atom that increases the electron concentrationgroup V elements (P, As)in Siacceptor: impurity atom that increases the hole concentrationgroup III elements (B, In) in Sin-type material: semiconductor containing more electrons than holesp-typematerial: semiconductor containing more holes than electronsmajority carrier: the most abundant mobile carrier in a semiconductor minority carrier: the least abundant mobile carrier in a semiconductormobile carriers: Charge carriers that contribute to current flow when electric field is applied.19Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06Sheet Resistance RS• Rsvalue for a given conductive layer (e.g. dopedSi, metals) in IC or MEMS technology is used– for design and layout of resistors– for estimating values of parasitic resistance in a device or circuitWLRWtLRs==ρtRsρ≡Rsis the resistance when W = L(unit of RSin ohms/square)if ρ is independent of depth x V+_LtWIMaterial with resistivity ρV+_LtWIMaterial with resistivity ρV+_LtWIMaterial with resistivity ρ20Professor N Cheung, U.C. BerkeleyLecture 3EE143 S06RS when ρ(x) is function of depth xV+_LtWIρ1, dxρ2, dxρ3, dx….ρn, dxdx)..(dx....dxdxdxR1n21n321Sσ+σ+σ=ρ++ρ+ρ+ρ=[]∫∫+==tpntSdxxpxqxnxqdxxR00)()()()(1)(1µµσFor a continuous σ(x) function: depth x21Professor N Cheung, U.C.
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