3_NanoSCFabrication_chap4 (Лекции Цветкова)

Chap.4 OxidationRequired Reading :Sze – Chapter 3Supplementary reading :Wolf & Tauber – Chapter 7Chandi – Chapter 7Colclaser – Chapter 5Topics :Growth mechanisms and kineticsEffects of oxidationWafer cleaningGrowth techniques and systemsOxide propertiesThickness measurementNational Research Laboratoryfor Nano-SOI ProcessHanyang University1Silicon dioxide• Silicon dioxide is a key material in modern VLSI.It is the simple formation of a thermal oxide that has helped makesilicon the predominant semiconductor material for VLSI. Thermaloxidation of silicon proceeds via a simple chemical reaction :Si(s) + O 2 (g) → SiO 2 (s)A high quality, stable oxide results.Oxide purity depends on gas and substrate purity.• SiO2 is fundamental to VLSI because of a myriad of uses:-Masking (implantation and diffusion)Surface passivation (junctions)Electrical isolation (interlevel dielectric)Device isolation (refilled trenches, SOI)Gate oxide material (MOS)Nano-SOI Process LaboratoryHanyang University2Formation of silicon dioxide••••Thermal oxidationCVDAnodization (electrochemical oxidation)Plasma oxidation (RF and microwave)By far, thermal oxidation is the most common technique.Oxidation reactionsBasically two reactions are used :Dry oxidation : Si (s) + O 2 (g ) → SiO 2 (s)Wet oxidation : Si(s) + 2H 2 O(g ) → SiO 2 (s) + 2H 2These two reactions are typically carried out from 800°C~1200°CNano-SOI Process LaboratoryHanyang University3Oxidation reactionsOxidation of silicon involves consumption of silicon substrate andincorporation into amorphous oxide.Thus, the original silicon surface recedes as thermal oxidation proceeds.How much silicon is consumed?Let’s assume we grow an oxide of thickness = xoxFor an oxide of thickness, xox, we need :Noxxox Si atoms/cm2Nox = # of Si atoms or SiO2 molecules/cm3 in SiO2All of these Si atoms must come from the substrate where we have :NsxsSi atoms available/cm2Ns = # of Si atoms/cm3 in crystalline Sixs = Depth of silicon substrate consumptionNano-SOI Process LaboratoryHanyang University4Oxidation reactions (cont’)Equating the tow Si atom densities we find :N ox x ox = N s x sorxs =N ox x oxNsUsing Nox = 2.2*1022/cm3 (SiO2 molecular density)Ns= 5*1022/cm3We find : x S ≈ 0.44 x oxOr, for every 100nm oxide grown, 44nm of Si is consumed.ExampleA silicon wafer 100㎛ thick is oxidized in O2 to an oxide thicknessof 2.

What is the total thickness of substrate + Oxide?Sol) Not 102μm, Because we consume 0.44 × 2μm of Si.0.44 × 2μm SiO 2 = 0.88μm of SiliconThus the total thickness is 102.24μm.Nano-SOI Process LaboratoryHanyang University5Thermal oxidation model• As oxide grows we need a mechanism for Si atoms toreach O2 or H2O Molecules.• One of two things must take place :1. Si diffuses to oxide surface2. Oxidant diffuses to Si-SiO2 interfaceExperimental evidence has established that:Oxidant diffuses to the silicon-oxide interfaceNano-SOI Process LaboratoryHanyang University6Deal-Grove modelWe have a situation similar to the CVD process, but withthe addition of diffusion across a growing film.The key steps in the thermal oxidation process are:1. Diffusion of oxidant from gas stream to oxide surface(F1)2. Diffusion of oxidant through existing oxide film (F2)3. Reaction at the silicon interface (F3)F1=F2=F3 in steady stateNano-SOI Process LaboratoryHanyang University7Model for thermal oxidation of silicon[Chap.3] Fig.2 Basic model for thermal oxidation of siliconNano-SOI Process LaboratoryHanyang University8Silicon oxidation model• We considered the first flux in our treatment of CVD:F1 = hG (CG-CS)hG = gas-phase mass transport coefficientCG = oxidant concentration in gasCS = oxidant concentration just outside oxideRecall hG = D/y ∝ 1/boundary layer thickness.• Invoking the ideal gas law:PV=nRT or PV=NkTn = # molesN = # atoms or molecules• Concentration of a species is given by:C=N/V=#/cm3Then C=N/V=P/kTPPCG = GCS = SkTkTNano-SOI Process LaboratoryHanyang University9Silicon oxidation model (cont’)Rewrite F1 as :F1 =hG(PG − PS )kTNow use Henry’s law which relates the concentration of a species in asolid to the partial pressure of the species in the surrounding gas (Nomolecular dissociation)i.e.

C0=HPs, where H=Henry’s constantAlso we can define C* as the equilibrium concentration of oxidant in thebulk of the oxide :C*=HPGhGFlux of oxidant through boundary layerIf we define h =then :HkTF1=h(C*-C0)We assume that the diffusion of oxidant in the solid obeys Fick’s law:F2 = − DNano-SOI Process Laboratory∂C D(C 0 − C i )=∂xd0Hanyang University10Silicon oxidation model (cont’)Assuming first order kinetics for the reaction at the interface:F3 = ksCiAgain we have F1=F2=F3D(C o − C i )h (C * − C o ) =doD(C o − C i ) = k s C idoSolving the simultaneous equations for Ci and Co gives:Ci =C*k s k sd1++hDk sdoC * (1 +)DCo =k s k sd1++hDNano-SOI Process LaboratoryHanyang University11Silicon oxidation model (cont’)Consider two limiting cases :1.

D is very small; diffusion through oxide slowCi -> 0Co -> C*Diffusion Controlled2. D is very largeC*Ci = Co ->ks1+hReaction ControlledNano-SOI Process LaboratoryHanyang University12Growth rate of oxide• We must determine the quantity of oxidant incorporatedinto a unit volume of SiO2=N1.• We know for SiO2 we have 2.2X1022 SiO2 molecules/cm3• Two cases must be considered:O2 OxidationH2O Oxidation- Case 1 : O2 Oxidation :1 Oxidant molecule/SiO2 molecule.Thus N1=2.2X1022 oxidant molecules/cm3 SiO2.- Case 2: H2O Oxidation2 Oxidant molecules/SiO2 molecule.Thus N1=4.4X1022 oxidant molecules/cm3 SiO2.Nano-SOI Process LaboratoryHanyang University13Growth rate of oxide (cont’)• The oxide growth rate is:#oxidant molecules2∂d oF3cm•scm===∂tsN1 #oxidant moleculescm 3Recall that :k sC *F3 = k s C i =kk d1+ s + s ohDThen we have a differential equation :∂d o=N1∂tNano-SOI Process Laboratoryk sC *k s k sdo+1+hDHanyang University14Growth rate of oxide (cont’)We can solve the differential equation with some initial condition.Assume we have initial oxide thickness, di, at t=0:do(0)=diThe solution under these boundary conditions gives:d o 2 + Ad o = B( t + τ )A = 2D(1 / k s + 1 / h )B = 2DC * / N1d i 2 + Ad iτ=Bτ accounts for time shift due to presence of initial oxide.Solving for do gives:A ⎧⎪⎡t +τ ⎤d o = ⎨⎢1 + 22 ⎪⎩⎣ A / 4B ⎥⎦Nano-SOI Process Laboratory1/ 2⎫⎪− 1⎬⎪⎭Hanyang University15Limiting casesLimiting case #1For long oxidation times : t+τ>>A2/4B and t>> τThus:AtA4Btdo ≈ •≈•2A 2 / 4B 2A2- do2=Bt : Parabolic growth- B is the parabolic rate constant -> Diffusion controlledLimiting case #2For short oxidation times (t+τ)<<A2/4B taking the first term of thebinomial expansion of the square root term from the equation for do, andsimplifying gives :Bd o ≈ (t + τ )AB/A = Linear rate constant -> Reaction rate controlledNano-SOI Process LaboratoryHanyang University16Pressure dependence of oxidation rateFrom B =2DC *we see B ∝ C *N1We also know C*=HPG; Thus we find B∝C*The parabolic rate constant increases linearly with oxidation pressure.Because the linear rate constant = B/A, and A has no significant pressuredependence:The linear rate constant increases linearly with oxidation pressureHigher pressure = Faster oxidationNano-SOI Process LaboratoryHanyang University17Temperature dependence of rate constantsNano-SOI Process LaboratoryHanyang University18Validity of the Deal-Grove model• Experimental fit is very good except for:Thin oxide growth(<30nm) in dry O2• The extrapolated initial oxide thickness for O2 oxidationis ~25nm.• We know that clean silicon has a native oxide of only1~2nm• Initial oxide growth rate is faster than predicted by DealGrove modelA different mechanism must existNano-SOI Process LaboratoryHanyang University19Orientation dependence of oxidation rate• Oxidation rate depends on substrate crystal orientation.Why should we expect this?• We know that F1=F2=F3=ksCiBut, perhaps we should write :F = ks*CiCsitesWhere Csites = concentration of available oxidation sites.Csites depends on the areal density of Si atoms (bonds).Thus, we expect the following orientation dependence based on thesurface silicon bond density :(111) > (110) > (100)FasterSlowerOxidationOxidationNano-SOI Process LaboratoryHanyang University20Wet vs.

Dry oxidation• Wet oxidation is much faster than dry oxidation.However, H2O diffuses more slowly than O2 in SiO2.• The increased growth rate of wet oxides is due to the greater solidsolubility of H2O in SiO2 compared to O2.i.e. C*(H2O)≈1000C*(O2)• We might expect that dry oxidation would tend to exhibit parabolicgrowth at most temperatures.Nano-SOI Process LaboratoryHanyang University21Oxide thickness vs.

time for thermal oxidation(a) Dry oxidation(b) Wet oxidationExperimental results of silicon dioxide thickness as a function ofreaction time and temperature for two substrate orientation.Nano-SOI Process LaboratoryHanyang University22Problems with the model• Too simplistic• Initial rapid growth regime not understood• Some debate over nature of oxidation species molecularO2 or atomic O2• Debate over the role (if any) of ionic oxidant species andspace charge effectsWork is still needed to fully understand the thermaloxidation processNano-SOI Process LaboratoryHanyang University23Effects of impurities on oxidation rate• H2OWet oxidation is much faster than dry oxidation.As little as 25 ppm H2O contamination can increase “dry” oxidethickness significantly.• Substrate dopantsRedistribution of dopants at growth interface affects oxidation rate.Typically, oxidation rate increases with increasing substrate dopantlevel.• HClAddition of 1%~5% HCl to oxidizing ambient has the followingeffects:– Increases oxidation rate partially via 4HCl+O2−>2H2O+2Cl2– Improves oxide properties : Stacking fault suppressionGettering of metallicsLifetime enhancementImproved breakdown strengthIonic Na+ passivation• TCE : Similar to HClNano-SOI Process LaboratoryHanyang University24Oxidation of impurity doped siliconBoronHigher bulk impurity concentration→ Weakening SiO 2 bonding→ Oxide growth ↑Nano-SOI Process LaboratoryPhosphorusFor oxidation of phosphorus-doped silicon: a concentration dependence is observedonly at the lower temperatures(where the surface reaction becomesimportant)Hanyang University25Thin oxide growthFor current and future demands of VLSI, we need gateoxides < 10nm thick.Reproducible Growth requires new techniques:•••••Low pressure oxidation – lower growth rateDiluted oxidizing ambient – lower PGRapid thermal oxidation – short time controlLow temperature oxidation – lower growth rateLow temperature, high pressure combinationsNano-SOI Process LaboratoryHanyang University26Dopant redistribution during oxidationDefine a segregation coefficient, m, as follows:m=equilibrium impurity conc.