analog bicmos design practices and pitfalls phần 1 potx

[...]... ln 10 15 10 16 = 638 mV (1. 5x1 010 )2 b) From Equation 1. 22, we have, for the p-region 0.638 + 10 = 2 qx2 NA 1 1 p + 16 15 2 10 10 10 .638 2 = x2 p 1. 1 qNA xp = 3.5x10−4 cm = 3.5µm From Equation 1. 13, we have xn = xp NA = 0.35µm ND c) From Equation 1. 18, we have Emax = − V 1. 6x10 19 10 15 3.5x10−4 = −5.4x104 1. 04x10 12 cm d) If we assume the depletion region exists entirely within the p-region, the depletion... 10 15 atoms per cm3 and ND = 10 16 atoms per cm3 Calculate the built-in potential, the junction depths in both regions, and the maximum electric field with VR = 10 V Calculate the depletion width assuming a singlesided junction How much error is created using this approximation? Answer a) From Equation 1. 12, we have Ψo = 26 mV ∗ ln 10 15 10 16 = 638 mV (1. 5x1 010 )2 b) From Equation 1. 22, we have, for the... as V2 , we obtain xn qND x2 n V2 = − Edx = (1. 20) 2 0 The voltage across the depletion region is then the sum of V1 and V2 and may be written Ψo + VR = V1 + V2 = q NA x2 + ND x2 p n 2 (1. 21) Factoring and using Equation 1. 13: Ψo + VR = 2 qx2 NA 1 1 p + 2 NA ND (1. 22) Recall Equation 1. 13, NA xp = ND xn If one region is much more heavily doped than the other, the depletion region exists almost entirely... KT (1. 6) Example If a silicon sample is doped with 10 17 acceptors per cm3 , calculate the position of the Fermi level relative to the intrinsic level at room temperature At normal operating temperatures, all acceptors will be ionized and the hole concentration p will equal the acceptor concentration p = NA = 10 17 holes per cm3 From Equation 1. 6: Ei − Ff = KT ln NA ni = 0.0259 ln 10 17 1. 5x1 010 = 0. 41. .. found at x = 0 and has a value Emax = − where qNA xp is the permittivity of silicon =− qND xn (1. 18) We have assumed the depletion region and junction boundaries are sharp and well defined Defining the potential between x = −xp and x = 0 as V1 0 qNA x2 p V1 = − Edx = (1. 19) 2 −xp Similarly, if we define the potential between x = 0 and x = xn as V2 , we obtain xn qND x2 n V2 = − Edx = (1. 20) 2 0 The voltage... + xn , we can approximate xd ≈ xn NA , Equation 1. 22 becomes Since ND Ψo + VR = 2 qx2 ND n 2 1 ND (1. 23) The voltage across the junction exists across xn , and is approximately Ψo + VR ≈ V2 Also, from Equations 1. 23 and 1. 18, the width of the depletion region and the maximum electric field are xd ≈ xn = Emax = 2 (Ψo + VR ) qND 2qND (Ψo + VR ) (1. 24) (1. 25) The width of the depletion region is an important... equal in magnitude and opposite in sign to the charge on the opposite side of the depletion region From Gauss’ Law we have ∇·D =ρ (1. 14) In one dimension, this reduces to dD =ρ dx (1. 15) dE ρ = dx (1. 16) Since D = E, we have Electric field can then be defined E =− dV dx (1. 17) Within the confines of the depletion region, the charge distribution ρ is equal to qNA coul/cm3 in the p-region, and is equal to... attached to silicon atoms The conduction band is occupied by conduction electrons that are free to move about the crystal If all electrons are in their lowest energy states, they are occupying states in the valence band The difference between the conduction band edge and the valence band edge is EG = 1. 12 eV , the band gap When a silicon atom loses an electron, it takes 1. 12 electron volts of energy for the... density for holes is given by Jp = −qDp dp dx (1. 9) where Jp is the current density, amperes/cm2 , Dp is the diffusion constant and p is the hole density, holes/cm2 Einstein’s relation shows the diffusion constant for holes to be proportional to mobility [3, page 38]: Dp = µp VT (1. 10) Dn = µn VT (1. 11) and for electrons Figure 1. 5 The nonuniform distribution of randomly moving positive charges results in... and n is the number of conduction electrons per cubic cm Example A silicon sample is doped with ND = 5x1 017 donors/cm3 What are the majority and minority carrier concentrations? The sample is n-type where electrons are the majority carriers Assuming all donors are ionized, the electron density is equal to the donor concentration, n = 5x1 017 cm−3 The minority (hole) concentration is p= n2 (1. 5x1 010 . Galipeau Contents 1Devices 1. 1Introduction 1. 2SiliconConductivity 1. 2.1DriftCurrent 1. 2.2EnergyBands 1. 2.3SheetResistance 1. 2.4DiffusionCurrent 1. 3PnJunctions 1. 3.1BreakdownVoltage 1. 3.2JunctionCapacitance 1. 3.3TheLawoftheJunction 1. 3.4DiffusionCapacitance 1. 4DiodeCurrent 1. 5BipolarTransistors 1. 5.1CollectorCurrent 1. 5.2BaseCurrent 1. 5.3Ebers-MollModel 1. 5.4Breakdown 1. 6MOSTransistors 1. 6.1SimpleMOSModel 1. 7DMOSTransistors 1. 8ZenerDiodes 1. 9EpiFETs 1. 10ChapterExercises 2DeviceModels 2.1Introduction 2.2BipolarTransistors 2.2.1EarlyEffect 2.2.2HighLevelInjection 2.2.3Gummel-PoonModel 2.3MOSTransistors 2.3.1BipolarSPICEImplementation 2.4SimpleSmallSignalModelsforHand Calculations 2.4.1BipolarSmallSignalModel 2.4.2OutputImpedance 2.4.3SimpleMOSSmallSignalModel 2.5ChapterExercises 3CurrentSources 3.1CurrentMirrorsinBipolarTechnology 3.2CurrentMirrorsinMOSTechnology ChapterExercises 4VoltageReferences 4.1SimpleVoltageReferences 4.2VbeMultiplier 4.3ZenerVoltageReference 4.4TemperatureCharacteristicsofI c andV be 4.5BandgapVoltageReference 5Amplifiers 5.1TheCommon-EmitterAmplifier 5.2TheCommon-BaseAmplifier 5.3Common-CollectorAmplifiers(Emitter Followers) 5.4Two-TransistorAmplifiers 5.5CC-CEandCC-CCAmplifiers 5.6TheDarlingtonConfiguration 5.7TheCE-CBAmplifier,orCascode 5.8Emitter-CoupledPairs 5.9TheMOSCase:TheCommon-Source Amplifier 5 .10 TheCMOSInverter 5 .11 TheCommon-SourceAmplifierwithSourceDegeneration 5 .12 TheMOSCascodeAmplifier 5 .13 TheCommon-Drain(SourceFollower) Amplifier 5 .14 Source-CoupledPairs 5 .15 ChapterExercises 6. Semiconductor Corp. Practices and Pitfalls Analog BiCMOS DESIGN