Electro-thermal modeling of large-surface OLED

HBM靜電放電測試系統之二次電性過應力事件的形成與抑制作者Tom Meuse (1), Larry Ting (2), Joe Schichl (2), Robert Barrett (1), David Bennett (1), Roger Cline (2), Charvaka Duvvury (2), Michael Hopkins (1), Hans Kunz (2), JohnLeiserson (1), Robert Steinhoff (2)(1) Thermo Electron Corporation (2) Texas Instruments Inc.譯者何正江訊程實業股份有限公司目錄1.背景介紹2. 元件失效的描述和測試系統間的相互作用3. 異常尾波(trailing pulse)的發現4. 異常尾波(trailing pulse)形成的機制5. 測試系統的修改以抑制異常尾波(trailing pulse)6. 對於HBM測試標準改進之建議以及將來之挑戰7.結論HBM靜電放電測試系統之二次電性過應力事件的形成與抑制前言以往在人體帶電模型(Human Body Model，HBM)靜電放電測試系統中，未被偵測到的異常尾波(trailing pulse)，最近已被證實對於較先進技術下所製造出來的半導體產品，會在其閘極氧化層(gate oxide)部位產生預期之外的破壞。

此第二脈衝是由於放電繼電器以及充電電路寄生特性所共同引起的電性過應力(Electrical Over-Stress，EOS)自然現象。

本文件主要在探討此一重要現象以及建立測試系統如何來抑制異常尾波的機制。

1. 背景介紹自從1979年美國軍方首度制定人體帶電模型(Human Body Model，HBM)耐靜電測試標準MIL-STD-883 Method 3015以來許多商品化的HBM測試系統相繼在半導體工業界被採用。

Band structures and band offsets of high K dielectrics on SiJ.Robertson *Engineering Department,Cambridge University,Trumpington Street,Cambridge CB21PZ,UKAbstractVarious high dielectric constant oxides will be used as insulator in ferroelectric memories,dynamic random access memories,and as the gate dielectric material in future complementary metal oxide semiconductor (CMOS)technology.These oxides which have moderately wide bandgaps provide a good test of our understanding of Schottky barrier heights and band offsets at semiconductor interfaces.Metal induced gap states (MIGS)are found to give a good description of these interfaces.The electronic structure and band offsets of these oxides are calculated.It is found that Ta 2O 5and SrTiO 3have small or vanishing conduction band offsets on 2O 3,Y 2O 3,ZrO 2,HfO 2,Al 2O 3and silicates like ZrSiO 4have offsets over 1.4eV for both electrons and holes,making them better gate dielectrics.#2002Elsevier Science B.V .All rights reserved.Keywords:Band structures;Band offsets;Dielectric constant oxides1.IntroductionThe closed shell transition metal (TM)oxides like SrTiO 3have been extensively studied for their ferro-electric properties,phase transitions and soft modes [1].They are now of great technological importance for electronic devices such as dynamic random access memories (DRAMs),ferroelectric non-volatile mem-ories (FeRAMs),and as alternative gate oxides in future complementary metal oxide semiconductor (CMOS)transistors [2±4].This requires them to be considered in terms of their electronic properties,by treating them as wide bandgap semiconductors [5].This paper reviews the band structures of these oxides,and then considers important electronic prop-erties such as their band offsets and Schottky barrier heights (SBHs).It turns out that the oxides have intermediate bandgaps and so they provide a goodtest of our present models of Schottky barriers and band offsets.2.Band structuresThe simplest band structures are those of the cubic ABO 3perovskites such as SrTiO 3or BaTiO 3.The bandgap is direct at G (Fig.1)[6].The valence band consists mainly of the 2p states of the O 2Àions,and the conduction band of the Ti 4 3d (t 2g )states [7].The Sr s and p states lie higher in the conduction band.However,the bonding is 60±70%ionic,and so there is signi®cant mixing of Ti d states in the valence band.The bands of the Pb perovskites differ in that Pb is divalent and it retains its 6s electrons [8].The ®lled Pb s states form an additional valence band at about À7eV as in PbTiO 3(Fig.2)[6,9].There is also some Pb s admixture in the upper valence band.The empty Pb 6p states now lie near the lowest conduction band.When Zr replaces Ti in SrTiO 3(or BaTiO 3),the bandgap increases strongly by 2eV,because itisApplied Surface Science 190(2002)2±10*Tel.: 44-1223-33-2689;fax: 44-1223-33-2662.E-mail address:jr@ (J.Robertson).0169-4332/02/$±see front matter #2002Elsevier Science B.V .All rights reserved.PII:S 0169-4332(01)00832-7controlled by the energy of the Zr d states.In contrast,in PZT,the Pb 6p states form the conduction band minimum,so the gap barely increases from 3.3to 3.7eV [10].It is recognised that the resonant covalence of Ti-d/O-p states is the origin of ferroelectricity in SrTiO 3type perovskites [11].In Pb perovskites,there is additional resonant covalence between Pb s and O p states which increases the ferroelectric polarity.SrBi 2Ta 2O 9is a layered crystal built from perovs-kite blocks separated by Bi 2O 2layers.It turns out that the Bi s and p states form the highest valence band and lowest conduction bands,respectively,while the ferro-electric response originates mainly from the TaO 3perovskite blocks [12].There is therefore an interest-ing separation of the functionality onto the Ta and Bi sub-lattices.Cubic ZrO 2has the ¯uorite structure.It has a simple band structure,as shown in Fig.3.The O p states form the valence band with a maximum at X [13].The conduction band minimum is at G ,and consists of Zr d states.The Zr d x 2Ày 2and d z 2states lie below the d xy states.The Zr s state lies midway between these at G ,but it disperses rapidly upwards.2.1.Models of Schottky barriers and semiconductor heterojunctionsThe band line-up of two semiconductors is deter-mined,like the SBH of a semiconductor on a metal,by charge transfer across the interface and the presence of any dipole layer at the interface.The charge transfer is that between the metal and the interface states of the semiconductor (Fig.4)[14].The charge transfertendsFig.1.Band structure of BaTiO 3calculated by pseudo-potential method [6].J.Robertson /Applied Surface Science 190(2002)2±103to align the Fermi level E F of the metal to the energy level of the interface states.The SBH for electrons f n between a semiconductor S and a metal M is f n S F M ÀF S F S Àw S(1)Here,F M is the work function of the metal,F S the energy of the semiconductor interface states,w S the semiconductor's electron af®nity (EA)and S the Schottky pinning parameter.S is given by [15]S11 e 2N d =ee 0(2)where e is the electronic charge,e 0the permittivity of free space,N the areal density of the interface states and d their decay length in the semiconductor.The dimensionless pinning factor S describes if the barrieris `pinned'or not.S varies between the limits S 1for unpinned Schottky barriers,and S 0for `Bardeen'barriers pinned by a high density of interface states in which the SBH is f n F S Àw S .There are numerous models of the origins of inter-face states,both intrinsic and extrinsic.In the intrinsic model originating from Bardeen and Heine,a semi-in®nite semiconductor in contact with a metal pos-sesses intrinsic states which are now called metal-induced gap states (MIGS)by Tersoff [14].F S is then the charge neutrality level (CNL)of the interface states,de®ned as the energy above which the states are empty for a neutral surface [16±18].On the other hand,the extrinsic models stress that the metal can react with the semiconductor [19].Brillson correlated the heat of reaction with S .This reaction maycreateFig.2.Band structure of PbTiO 3calculated by pseudo-potential method [6].4J.Robertson /Applied Surface Science 190(2002)2±10interface defects such as vacancies,whose gap states can pin the metal Fermi level,as noted by Spicer [20]and Dow [21].These models were supported by theobservation that pinning occurs even for monolayer coverage of metal,before the MIGS could be estab-lished.It is now believed that,overall,the intrinsic model gives a better description of Schottky barriers,because intrinsic states have a larger pinning dipole,N d ,than surface defects.The pinning parameter S has been in¯uential in our empirical understanding of Schottky barriers.Some years ago,Kurtin et al.[22]noted that S seemed to vary sharply with the ionicity of semi-conductor (Fig.5),from near 0for low ionicity semiconductors like Si and GaAs to 1for higher ionicity solids like SiO 2,SrTiO 3and KTaO 3.S is a dimensionless slope of barrier height to metal work function,S@f n @F M(3)Fig.3.Band structure of ZrO 2calculated by pseudo-potential method[6].Fig.4.Schematic diagram of SBHs.J.Robertson /Applied Surface Science 190(2002)2±105However,Louie [23]and Schluter [24]noted that Kurtin [22]had actually correlated the barrier heights to S H :S H@f n @X(4)which is the slope of barrier height to the Pauling electronegativity of the metal,and not the dimension-less S in (4).The work function and electronegativity vary roughly as [25,26]:F M 2:27X M 0:34(5)Thus,S H 2:27S ,and the Schottky limit should be S H 2:27.The data rarely reach this limit and Schluter [24]observed that S had a better correlation with the dielectric constant of the semiconductor e 0.Empiri-cally,Mo Ènch [14,27]found that S varied with e I as S11 0:1 e I À1 2(6)Certain materials are key tests of Schottky barriermodels.Diamond and xenon [14,28]have zero ioni-city but small e I ,and so their large S values show that S depends on e not on ionicity.This is tested by plotting log 1= S À1 against log e I À1 as in Fig.6.The wide gap oxides provide another key test,because they have intermediate e I values.SrTiO 3and KTaO 3were taken as high ionicity solids in the original Kurtin plot,with S H $1.However,this wasbefore data was actually known.When data [29]became available for SrTiO 3,showing S lying between 0.25and 0.4(Fig.6),it was clear that S is much lower.SrTiO 3falls well on the trend in Fig.3.The reason for this is that the SBHs depend on e I .e I is controlled by the states closest to the bandgap [5].In SrTiO 3,these are the moderately ionic Ti±O states of Ti±O bonds,not the highly ionic Sr±O states which lie well away from the gap and provide a much smaller contribution to e I .This can be seen in the partial density of states (DOS)of SrTiO 3in Fig.6.Thus,SrTiO 3and KTaO 3were misplaced in Fig.5as highly ionic solids.A lesser point is that the moderate value of S of SrTiO 3clearly correlates with e I ,and not with the low frequency dielectric constant e 0,which has a very large value for ferroelectrics and would give S %0from (6).SrTiO 3also serves as an evidence against the defect model,in that the barrier lies some way into the gap,not at the conduction band edge where the O vacancy states lie and would cause pinning.In sum-mary,the MIGS model of Schottky barriers holds for a wide range of solids of various ionicity and dielectric constants [5].The band alignment between two semiconductors is controlled by charge transfer and interface dipoles,just as Schottky barriers [30].For no dipoles,the Schottky limit,the conduction band offset isgivenFig.5.Schottky barrier pinning factor S H in the (incorrect)model of Kurtin etal.Fig. 6.Log±log plot of 1= S À1 vs.e I À1for various semiconductors and insulators to verify the MIGS model of Schottky barrier pinning factor S .6J.Robertson /Applied Surface Science 190(2002)2±10by the difference in their electron af®nities,the `elec-tron af®nity rule'.A similar idea was that for no charge transfer,the band line-ups are derived by placing each semiconductor's band on an absolute energy scale such as those of the free atom energy levels [31].Tersoff [16]showed that the band offset between two semiconductors a and b is controlled by interface dipoles as in the Schottky barrier,and so the conduc-tion band offset is given by f n w a ÀF CNL ;a À w b ÀF CNL ;bS F CNL ;a ÀF CNL ;b(7)The offsets are now described by aligning the CNLs of each semiconductor,modi®ed by the S factor.For simple semiconductors like Si,e I is large,and so S is small and the third term was negligible in the original formulation,but it is retained here for wide gap oxides.For strong pinning,the alignment is just given by the alignment of the two CNLs.The CNL energy below the vacuum level is a measure of the mean electronegativity of the semiconductor,in the same way that the work function of a metal is propor-tional to the metal's electronegativity.Thus,Eq.(7)says that the band alignment is the difference in electronegativity screened by the S factor.A wide ranging quantitative comparison found that the CNL models gives a good description of the band offsets [30].The CNL is the branch point of the semiconductor interface states.It is the integral of the Green's func-tion of the band structure,taken over the Brillouin zone [17],G E ZBZ N E H d H EE ÀE H0(8)Cardona and Christensen later provided a quicker method using a sum over special points of the Bril-louin zone [5,32].G E X i 1E ÀE i (9)2.2.Application to oxidesThe band alignments for the various wide gapoxides in contact with metal or silicon are found by calculating their CNLs and S parameters.The S factors are found from (6)using the experimental values of e Iand are shown in Table 1.The CNLs were found by calculating the oxide band structures by the tight-binding method [5,6,8,33].The tight-binding para-meters are found by ®tting to existing band structures [9,10,34],photoemission spectra and optical data [2,35±37].The CNLs for the various oxides are given in Table 1,together with the experimental values of their bandgaps and electron af®nities [2,38].SrTiO 3is an important oxide for future DRAM capacitor dielectrics.SrTiO 3is also the most studied system and the best test of our calculations.Fig.7compares the predicted SBHs of SrTiO 3on various metals with the experimental values [30,39±43].The experimental data are quite scattered but are quite consistent with S !1and our calculated value of 0.28.This shows that SrTiO 3is a key oxide in the tests of Schottky barrier models.The calculated barrier height for SrTiO 3on Pt is 0.9eV ,which is close to the 0.8eV found by photoemission by Copel et al.[43].However we cannot account for the much larger S value found by Shimizu et al.[42].BaTiO 3has similar band offsets to SrTiO 3.PbTi x Zr 1Àx O 3or PZT is an important ferroelectric for non-volatile memories,optical memories and other applications.The predicted barrier height for Pt onTable 1Calculated values for various oxides of their CNL and conduction band (CB)offset with Si aGap (eV)EA (eV)CNL (eV)e I S CB offset (eV)SiO 290.9 2.250.86 3.5b Si 3N 4 5.3 2.1 4.10.51 2.4b Ta 2O 5 4.4 3.3 3.3 4.840.40.3BaTiO 3 3.3 3.9 2.6 6.10.28À0.1BaZrO 3 5.3 2.6 3.740.530.8TiO 2 3.05 3.9 2.27.80.180.05ZrO 2 5.8 2.5 3.6 4.80.41 1.4HfO 26 2.5 3.740.53 1.5Al 2O 38.81c 5.5 3.40.63 2.8Y 2O 362c 2.4 4.40.46 2.3La 2O 36c 2c 2.440.53 2.3ZrSiO 46.5 2.4 3.6 3.80.56 1.5SrBi 2Ta 2O 94.13.53.35.30.4aExperimental values [36,37]of the bandgap,EA [2,38],dielectric constant e I [37]are also given.In Eqs.(2)and (5),F S is the energy of the CNL below the vacuum level,in this table,it is its energy above the valence band.bExperimental values.cEstimated values.J.Robertson /Applied Surface Science 190(2002)2±107PZT (Pb 0.55Zr 0.45O 3)is 1.45eV ,which is close to the 1.5eV measured by Dey et al.[44].The electron barrier of Pt on PZT is larger than that on BST because its CNL lies lower in the gap.This is because of the different band structure of PZT,in which the Pb 6s and 6p states form the band edges and this tends to lower the CNL.The larger value of the hole barrier than the electron barrier means that PZT thin ®lms can have predominantly electron injection,even though bulk PZT tends to be p-type.SrBi 2Ta 2O 9(SBT)is an important ferroelectric for non-volatile memories [2,45].It does not suffer from the loss of switchable polarisation (fatigue)when used with Pt electrodes,which is a problem for PZT.Note that more recent optical data ®nd that the bandgap of SBT is 4.1eV [2].The Schottky barrier of Pt is predicted to be 1.2eV ,which is essentially the same as that found by photoemission [46].There is an important need for high dielectric constant oxides to act as gate oxides instead of silicon dioxide [3,4].This is because the SiO 2layer is now so thin (2nm),that it no longer acts as a good insulator because of direct tunnelling across it.The solution is to replace SiO 2with a thicker layer of a medium k oxide,with the same equivalent capacitance or `equivalence oxide thickness't ox .The oxides must also satisfy certain other conditions,including chemi-cal stability in contact with Si [47].This rules out Ti and Ta which both react with Si to form SiO 2.The other key requirement is that they act as barriers toboth electrons and holes [5,32].This requires that both their valence and conduction band offsets be over 1eV .There is presently considerable effort to identify the most effective oxide,from a choice of ZrO 2,HfO 2,La 2O 3,Y 2O 3,Al 2O 3and the silicates ZrSiO 4and HfSiO 4.The calculated CB band offsets with Si are given in Table 1and summarised in Fig.8.They are compared in Table 2with recent experimental values [48±53],which is seen to be in good agreement.The important feature of Ta 2O 5and SrTiO 3is that both of them have CB offsets on Si under 1eV ,in fact 0in the case of SrTiO 3.This prediction was recently con®rmed by photoemission data of Chambers et al.[48].This means that SrTiO 3or BST cannot be a good gate oxide.The calculated CB offset for Ta 2O 5is only 0.36eV for Ta 2O 5on Si.This is consistent with recent photoemission data of Miyazaki and Hirose [49].Data for Ta 2O 5gate FETS also showed only a small elec-tron barrier [50].The CB offsets for BST and Ta 2O 5and BST are small or negligible because the bandgap is quite small and the band offsets are so asymmetric.To increasetheparison of calculated and observed SBHs of SrTiO 3on variousmetals.Fig.8.Predicted band offsets of various oxides on Si.Table 2Comparison of calculated and experimental values [48±53]of conduction band offsets on SiCalculatedExperiment References Ta 2O 50.350Miyazaki SrTiO 3À0.1<0.1Chambers ZrO 2 1.4 1.4Miyazaki 2.0Houssa Al 2O 32.82.8Ludeke8J.Robertson /Applied Surface Science 190(2002)2±10CB offset,we must either increase the bandgap or lower the CNL.The gap can be increased by raising the TM d levels,by using4d or5d metals instead of3d metals or using group IIIB metals instead of group IV. We should use zirconates,not titanates.The gap of BaZrO3is2eV wider than BaTiO3.Its offset is0.8eV.A better strategy is to lower the CNL.The CNL is lowered if the metal valence is lowered from4to3. Indeed,in Y2O3and La2O3,the CNL is much lower in the bandgap.Y2O3and La2O3are the oxides with largest CB offsets for reasonable dielectric constants. ZrO2has a bandgap of5.8eV,which is slightly wider than BaZrO3,and it also has a lower metal/ oxygen stoichiometry.This gives a larger CB offset for ZrO2(1.4eV)than BaZrO3,and indeed one which is just high enough.HfO2is similar.The calculated CB offset of1.4eV for ZrO2compares with an experi-mental value of1.4eV from photoemission[51]and a value of2eV by internal photoemission[52].This CB offset is large enough for devices.Zirconium silicate ZrSiO4and hafnium silicate HfSiO4are glassy oxides with bandgaps of $6.5eV.ZrSiO4consists of chains of alternate edge-sharing ZrO4and SiO2tetrahedra,with addi-tional Zr±O bonds between the chains,leading to an overall six-fold Zr coordination.We estimate the bandgap of ZrSiO4to be6.5eV.The calculated CB offsets are1.5eV,slightly more than ZrO2.Al2O3has a bandgap of8eV close to SiO2but with a higher k($9).Its calculated CB offset is2.8eV, which compares exactly with that measured by Ludeke et al.[53].Overall,the agreement between the calculated and subsequent experimental values for CB offsets in Table2is surprisingly good.References[1]M.E.Lines,X.Glass,Ferroelectrics,Oxford UniversityPress,Oxford,1990.[2]J.F.Scott,Ferroelectrics Rev.1(1998)1.[3]G.D.Wilk,R.M.Wallace,J.M.Anthony,J.Appl.Phys.89(2001)5243.[4]A.I.Kingon,J.P.Maria,S.K.Streiffer,Nature406(2000)1032.[5]J.Robertson,J.Vac.Sci.Technol.B18(2000)1785.[6]P.W.Peacock,J.Robertson,Unpublished work.[7]L.F.Mattheis,Phys.Rev.B6(1972)4718.[8]J.Robertson,W.L.Warren,B.A.Tuttle,D.Dimos,D.M.Smyth,Appl.Phys.Lett.63(1993)1519.[9]R.D.King-Smith,D.Vanderbilt,Phys.Rev.B49(1994)5828.[10]J.Robertson,W.L.Warren,B.A.Tuttle,J.Appl.Phys.77(1995)3975.[11]R.E.Cohen,Nature358(1992)136.[12]J.Robertson,C.W.Chen,W.L.Warren,C.D.Gutleben,Appl.Phys.Lett.69(1996)1704.[13]R.H.French,S.J.Glass,F.S.Ohuchi,Y.N.Xu,W.Y.Ching,Phys.Rev.B49(1994)5133.[14]W.MoÈnch,Phys.Rev.Lett.58(1987)1260.[15]W.MoÈnch,Surf.Sci.300(1994)928.[16]A.W.Cowley,S.M.Sze,J.Appl.Phys.36(1965)3212.[17]C.Tejedor,F.Flores,E.Louis,J.Phys.C10(1977)2163.[18]J.Tersoff,Phys.Rev.Lett.52(1984)465.[19]J.Tersoff,Phys.Rev.B30(1984)4874;J.Tersoff,Phys.Rev.B32(1985)6989.[20]L.J.Brillson,Surf.Sci.300(1994)909.[21]W.E.Spicer,T.Kendelewicz,N.Newman,K.K.Chin,I.Lindau,Surf.Sci.168(1986)240.[22]R.E.Allen,O.F.Sankey,J.D.Dow,Surf.Sci.168(1986)376.[23]S.Kurtin,T.C.McGill,C.A.Mead,Phys.Rev.Lett.30(1969)1433.[24]S.G.Louie,J.R.Chelikowsky,M.L.Cohen,Phys.Rev.B15(1977)2154.[25]M.Schluter,Phys.Rev.B17(1978)5044;M.Schluter,Thin Solid Films93(1982)3.[26]W.Gordy,W.J.O.Thomas,Phys.Rev.24(1956)439.[27]H.B.Michaelson,J.Appl.Phys.48(1977)4729.[28]W.MoÈnch,Phys.Rev.Lett.58(1986)1260.[29]W.MoÈnch,Europhys.Lett.27(1994)479.[30]R.C.Neville,C.A.Mead,J.Appl.Phys.43(1972)4657.[31]W.A.Harrison,J.Vac.Sci.Technol.14(1977)1016.[32]M.Cardona,N.E.Christensen,Phys.Rev.B35(1987)6182.[33]E.T.Yu,J.O.McCaldin,T.C.McGill,Solid State Phys.46(1992)1.[34]J.Robertson,C.W.Chen,Appl.Phys.Lett.74(1999)1168.[35]G.M.Rignanese,X.Gonze,A.Pasquarello,Phys.Rev.B63(2001)104305.[36]R.H.French,J.Am.Ceram.Soc.73(1990)477.[37]E.D.Palik,Handbook of Optical Properties of Solids,V ol.1±3,Academic Press,New York,1985.[38]W.Schmickler,J.W.Schultze,in:J.M.O'Bockris(Ed.),Modern Aspects of Electrochemistry,V ol.17,Plenum Press, London,1986.[39]G.W.Dietz,W.Antpohler,M.Klee,R.Waser,J.Appl.Phys.78(1995)6113.[40]H.Hasegawa,T.Nishino,J.Appl.Phys.69(1991)1501.[41]K.Abe,S.Komatsu,Jpn.J.Appl.Phys.31(1992)2985.[42]T.Shimizu,N.Gotoh,N.Shinozaki,H.Okushi,App.Surf.Sci.117(1997)400;()T.Shimizu,N.Gotoh,N.Shinozaki,H.Okushi,Mat.Res.Soc.Symp.Proc.(2000).[43]M.Copel,P.R.Duncombe,D.A.Neumayer,T.M.Shaw,R.M.Tromp,Appl.Phys.Lett.70(1997)3227.[44]S.K.Dey,J.J.Lee,P.Alluri,Jpn.J.Appl.Phys.34(1995)3134.[45]C.A.Paz de Araujo,J.D.Cuchiaro,L.D.McMillan,M.C.Scott,J.F.Scott,Nature374(1995)627.[46]C.D.Gutleben,Appl.Phys.Lett.71(1997)3444.[47]H.J.Hubbard,D.G.Schlom,J.Mater.Res.11(1996)2757.J.Robertson/Applied Surface Science190(2002)2±109[48]S.A.Chambers,Y.Liang,Z.Yu,R.Dropad,J.Ramdani,K.Eisenbeiser,Appl.Phys.Lett.77(2000)1662.[49]S.Miyazaki,Appl.Surface Science(2002)``these proceed-ings''.[50]S.Miyazaki,M.Narasaki,M.Ogasawara,M.Hirose,Microelec.Eng.59(2001)373.[51]A.Chatterjee,et al.,IEDM Tech Digest,1998,p.777.[52]M.Houssa,M.Tuominen,M.Nailli,V.Afansev, A.Stesmans,J.Appl.Phys.87(2000)8615.[53]R.Ludeke,M.T.Cuberes,E.Cartier,Appl.Phys.Lett.76(2000)2886;D.J.Maria,J.Appl.Phys.45(1974)5454.10J.Robertson/Applied Surface Science190(2002)2±10。

LCM 模組介紹一、 各部零件介紹二、 製程方法三、 LCM組成方式四、 LCD介紹五、 背光介紹六、 壓縮比七、 手機背光各零件組成:八、 ㈴詞解釋㆒、各部零件介紹: LCM 零件大致可分為: 1. 鐵框黑框, 鍍鋅低框, 鍍鋅高框 2. LCD TN 全透 半透 反射 6 點鐘 12 點鐘 9點鐘 3 點鐘 STN 全透 半透 反射 6 點鐘 12 點鐘 9點鐘 3 點鐘 FSTN全透 半透 反射 6 點鐘 12 點鐘9點鐘3 點鐘3. 導電橡皮為作LCD 與PCB 導通之介面導電橡皮的PITCH寬不可大於LCDITO 或PCB ㈮手指的PITCH 寬,否則會造成短路4. PCB(含軟板)單層板,雙層板,多層板(㈥層板) 5. 背光 CCFL,LED,EL 6. FFC(軟排) 7. CONNECTOR 8. ICLCD 導電橡皮 PCB㆓、依製程可分為:1. COB – Chip on Board Construction Of Wire BondingPC Board & Layout2. COF – Chip on Film3. COG – Chip on GlassOuter Lead Bonding By ACF4. TAB(Tape Automated Bonding)㆔、LCM 組成方式: (1) 字、圖形產品(2) TAB 產品(3) COF 產品鐵框T/PPCB Light-GuideLCD導電橡皮背光㆕、LCD介紹: 依種類:(1) TN(2) STN(3) FSTN(4) TFT依透光度:(1) 全透(2) 半透(3) 反射依點鐘:(1) 6 點鐘(2) 12點鐘(3) 3點鐘(4) 9點鐘㈤、背光介紹:EL BacklightFlat surface light source offers simple and even illumination over large area.Features:Max.1.3mm thickness ( Max. 1.5mm for lead portion )Wide driving condition, 60-1,000Hz at 150V AC Max. With inverter, step-up voltage from1.5V battery is available.Emitted colors are blue-green, yellow-green and white.Operating characteristics of PC2002-A SERIES is 110V, 400Hz, 8mA, ( Ta=20oC, 60% RH ) Temperature Range:Operating 0oC~ +50oCStorage -20oC~ +60oCCCFL Backlight( Cold Cathode Fluorescent Lamp )Bright white color of light source offers clear and even illumination over large viewing area. Features:High BrightnessLong life time.Low Power consumptionWhite color emittedLED BacklightLong life, low power consumption and requires a simple power supply. Available colors are red, green and orange, available in array type illumination or edge illumination.FeaturesLow driving voltage ( DC ) and does not require an inverter.Long life of 100,000 hours ( average )No noise occurrence.Various colors available in red, green and orange etc. ( multi-color by alternative switch isalso available )㈥、壓縮比:字、圖形產品因㈲導電橡皮為作導電介面,然而因為橡皮㈲伸縮性,故壓縮的比率將影響顯示畫面㆒般而言,未鎖角前:L(LCD厚度)+Z(導電橡皮高)+B(PCB厚度) > H (鐵框內徑高)鎖角後:L + Z1 + B = HZ1: 為導電橡皮壓縮後的高壓縮比計算的方式為: Z-Z1/Z最好的壓縮比約為 10%(1)壓縮比愈高,表示組裝愈緊密,但容易造成LCD局部泛色或單點紅.(2)壓縮比愈低,表示組裝不緊密,容易造成顯示異常或斷線㈦、手機背光各零件組成:㆒般而言,背光廠光生產背光的程序為:進行DICE 的分光分色→貼片(SMT)→組裝→壓合(1) 手機背光通常為定電壓,因為DICE 的內電阻各不同,若使用定電流時:依 V = I * R ,V 將不同,隨著內電阻不同,電壓也不同,因此定電流將會使背光DICE 燒毀,背光模組呈現燈死或燈暗的現象(2) 若DICE 分光分色不完全,將使的背光模組點燈時,發現顏色不㆒,亮度不㆒ (3) 導光板若放置位置不正確,將使模組點亮呈現部份燈較暗,部份燈較亮的現象 (4) 若擴散片與錂鏡片放置位置不正確, 將使模組點亮呈現光源散射的現象(5) 若壓合時間不足時,將使的背光各層無法緊密貼付,點燈時呈現光源散射及B/M 銘板與錂鏡片被撕起的現象塑座貼片上擴散片上錂鏡片㈧、㈴詞解釋:COB (Chip On Board )An IC chip is mounted onto a printed circuit board by wire bonding.SMT (Surface Mount Technology)Mounting on the surface of PCBTAB (Tape Automated Bonding)Tape Automatic Bonding technology for LCD module construction With LCD and TCP LSICOG (chip On Glass)A bumped chip directly mounted on LCD panelCOF (Chip On Flex/Film)A bumped chip mounted on a flexible printed circuit filmILB (Inner Lead Bonding)IC Bonding to Flex or TapeOLB (Outer Lead Bonding)Laminating flex or Tape with LCD panelTN (Twisted Nematic)A nematic liquid crystal with a twist angle about 90 degrees, and the type of display thatSTN (Super Twisted Nematic )A nematic liquid crystal with a twist of roughly 180 degrees to 270 degrees, and the type Of display that uses itFSTN (Formulated STN )An optically compensated film is added to STN, and is used as a monochrome displayColor STNSTN display with R/G/B color filter to perform different colorTDF (Transflective Display Film)To replace the bottom polarizer to improve the readability and styleRDF (Reflective Display Film)To replace the bottom polarizer to improve the readability and styleEL (Electro Luminescence)Light generated by an electric field. An EL layer is formed on a high molecular weightLED (Light Emitting Diode )A diode which emits lightCCFL (Cold Cathode Fluorescent Lamp)High brightness suited with graphic LCM displayQFP (Quad Flat Package)A package formed with leads on surfaces oriented in four directionsTQFP (Thin Quad Flat Package)A low profile package formed with leads on surfaces oriented in four directionTCP (Tape carrier Package)A flexible board printed with a circuit pattern, with IC chips moutned on it.Tf (Fall Time)Response time of falling edge .Tr (Rise Time)Response time of rising edge .Vop (Operating Voltage)Liquid crystal driving voltage.Vth (Threhold Voltage)A threshold voltage.ACF (Anisotropic Circuit )Adhesive film for laminating COF & TAB moduleFPC (Flex Printed Circuit )A flexible PCB.FFC (Flex Flat cable)A flexible Cable.TFT (Thin Film Transistor)Active type LCD display for Notebook and other Color application.OLED (Organic Electro Luminescent )Organic EL to provide dot-matrix display function wide viewing angleand high.LTPS (Low Temperature Poly Siiicon)New type TFT display with low power consumption & high contrast.。

英文原版教材班“材料科学基础”考试试题试卷一Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. Glossary (2 points for each)1) crystal structure:2) basis (or motif):3) packing fractor:4) slip system:5) critical size:6) homogeneous nucleation:7) coherent precipitate:8) precipitation hardening:9) diffusion coefficient:10) uphill diffusion:2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 13. Determine the crystal structure for the following: (a) a metal with a0 =4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points)4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (4 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (4 points)(3)Plot the isothermal section at 150o C. (7 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(1)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(2)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibi ts a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题答案Solution s of the course of “fundament of materials science”1. Glossary (2 points for each)1) The arrangement of the atoms in a material into a repeatable lattice.2) A group of atoms associated with a lattice.3) The fraction of space in a unit cell occupied by atoms.4) The combination of the slip plane and the slip direction.5) The minimum size that must be formed by atoms clustering together in the liquid before thesolid particle is stable and begins to grow.6) Formation of a critically sized solid from the liquid by the clustering together of a largenumber of atoms at a high undercooling (without an external interface).7) A precipitate whose crystal structure and atomic arrangement have a continuousrelationship with matrix from which precipitate is formed.8) A strengthening mechanism that relies on a sequence of solid state phase transformationsin a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) A temperature-dependent coefficient related to the rate at which atom, ion, or otherspecies diffusion. The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) A diffusion process in which species move from regions of lower concentration to that ofhigher concentration.2. Solution: A(-364), B(-340), C(346).3. Solution: (a)fcc; (b) bcc.4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (5 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (5 points)(3)Plot the isothermal section at 150o C. (5 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(3)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(4)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(3)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(4)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题试卷二Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another. (5 points)2. Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy. (10 points)3.Above 882℃, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? (10 points)4. The density of BCC iron is 7.882 g/cm3and the lattice parameter is 0.2866 nm whenhydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average to contain one hydrogen atom. (15 points)5. A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980℃, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h. (15 points)6. The following data were collected from a standard 0.505-in.-diameter test specimen of acopper alloy (initial length (t o) = 2.0 in.):Load Gage Length Stress Strain(lb) (in.) (psi) (in/in.)0 2.00000 0 0.03,000 2.00167 15,000 0.0008356,000 2.00333 30,000 0.0016657,500 2.00417 37,500 0.0020859,000 2.0090 45,000 0.004510,500 2.040 52,500 0.0212,000 2.26 60,000 0.1312,400 2.50 (max load) 62,000 0.2511,400 3.02 (fracture) 57,000 0.51After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensile test. Thefollowing measurements are made.Change in Force (N) Gage length (cm) Diameter (cm)16,240 0.6642 1.202819,066 1.4754 1.088419,273 2.4663 0.9848Determine the strain hardening coefficient for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.(15 points)8. Based on Hume-Rothery’s conditions, which of the following systems would be expected todisplay unlimited solid solubility? Explain. (15 points)(a) Au-Ag (b) Al-Cu (c) Al-Au (d)U-W(e) Mo-Ta (f) Nb-W (g) Mg-Zn (h) Mg-Cd英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2.T (o C)L i–180.7N a– 97.8K – 63.2R b– 38.9As the atomic number increases, the melting temperature decreases, (10 points)3. We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared.V BCC = (0.332 nm)3 = 0.03659 nm3V HCP= (0.2978 nm)2(0.4735 nm)cos30 = 0.03637 nm3△V=x 100 =×100= -0.6%Therefore titanium contracts 0.6% during cooling. (10 points)4. (a) 7.882 g/cm3 =x = 0.0081 H atoms/cellThe total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus:(10 points)(b) Since there is 0.0081 H/cell, then the number of cells containing H atoms is:cells = 1/0.0081 = 123.5 or 1 H in 123.5 cells (5 points)5. D = 0.23 exp[-32,900/(1.987)(1253)] = 42 × 10-8 cm2/sC x= 0.87% CC x= 0.43% CC x= 0.18% C(15 points)6. σ=FI (π/4)(0.505)2 = F/0.2ε = (l-2)/2(a) 0.2% offset yield strength = 45,000 psi(b)tensile strength = 62,000 psi(c) E = (30,000 - 0) / (0.001665 - 0) = 18 x 106 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 57,000 psi(g)true stress at fracture = 11,400 lb / (TC/4)(0.374)2= 103,770 psi (h) From the graph, yielding begins at about 37,500 psi. Thus:(15 points)7.Force(lb) Gage length(in.) Diameter(in.) True stress(psi) True strain(psi)16,240 3.6642 12.028 143 0.20019,066 4.4754 10.884 205 0.40019,273 5.4663 9.848 249 0.600σt=Kεt2or ln143=ln K + n ln0.2ln 249 = ln K + nln 0.6n=0.51A strain hardening coefficient of 0.51 is typical of FCC metals.(15 points)8.The Au–Ag, Mo–Ta, and Mg–Cd systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. [The Au –Ag and Mo –T a d o have isomorphous phase diagrams. In addition, the Mg–Cd alloys all solidify like isomorphous alloys; however a number of solid state phase transformations complicate the diagram.] (15 points)英文原版教材班“材料科学基础”考试试题试卷三Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. (5 points)2. Plot the melting temperatures of elements in the 4A to 8-10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energy, (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. (10 points)3. Beryllium has a hexagonal crystal structure, with a o= 0.22858 nm and c o= 0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell.(10 points)4. Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor. (15 points)5. Iron containing 0.05% C is heated to 912oC in an atmosphere that produces 1.20% C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the difference. (15 points)6. The following data were collected from a 0.4-in. diameter test specimen of poly vinyl chloride(l0 = 2.0 in):Load(lb) Gage Length(in.) Stress(psi) Strain(in/in.)0 2.00000 0 0.0300 2.00746 2,387 0.00373600 2.01496 4,773 0.00748900 2.02374 7,160 0.011871200 2.032 9,547 0.0161500 2.046 11,933 0.0231660 2.070 (max load) 13,206 0.0351600 2.094 12,729 0.0471420 2.12 (fracture) 11,297 0.06After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A titanium alloy contains a very fine dispersion of tiny Er203 particles. What will be the effectof these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Explain. (15 points)8. Suppose 1 at% of the following elements is added to copper (forming a separate alloy witheach element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Is any of the alloying elements expected to have unlimited solid solubility in copper?(a) Au (b) Mn (c) Sr (d) Si (e) Co (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2. Ti –1668 Zr – 1852 Hf – 2227V –1900 Nb –2468 Ta – 2996Cr –1875 Mo–2610 W–3410Mn–1244 Tc –2200 Re–3180Fe –1538 Ru –2310 Os–2700Co –1495 Rh –1963 Ir –2447Ni –1453 Pd –1552 Pt –1769For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable. (10 points)3.V= (0.22858 nm)2(0.35842 nm)cos 30 = 0.01622 nm3 = 16.22 × 10-24 cm3(a)From the density equation:1.848 g/cm3 =x = 2 atoms/cell(b)The packing factor (PF) is:PF == 0.77 (10 points)4. There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:(a)(b)(15 points)5. t= (24 h)(3600 s/h) = 86,400 sD BCC = 0.011 exp[-20,900/(1.9871185)] = 1.54 × 10-6 cm2/sD FCC = 0.23 exp[-32,900/(1.987)(1185)] = 1.97×10-7 cm2/sBCC: = erf[0.0685] = 0.077C x= 1.11% CFCC: = erf[0.192] = 0.2139C x = 0.95% CFaster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point “x”. (15 points)6. σ=F /(π/4)(0.4)2 = F/0.1257ε = (l-2)/2(a)0.2% offset yield strength = 11,600 psi(b) tensile strength = 12,729 psi(c) E= (7160 - 0) / (0.01187 - 0) = 603,000 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 11,297 psi(g)true stress at fracture = 1420 lb / (TC/4)(0.393)2= 11,706 psi (h) From the figure, yielding begins near 9550 psi. Thus:(15 points)7. These particles, by helping pin die grain boundaries, will increase the grain growth temperature and decrease the grain size. (15 points)8.The Cu-Sr alloy would be expected to be strongest (largest size difference). The Cu-Au alloy satisfies Hume-Rothery ’s conditions and might be expected to display complete solid solubility—in fact it freezes like an isomorphous series of alloys, but a number of solid state transformations occur at lower temperatures.(15 points)英文原版教材班“材料科学基础”考试试题试卷四Examination problems of the course of “fundament of materials science”姓名：班级：记分：1.Aluminum has a density of2.7 g/cm3. Suppose you would like to produce a compositematerial based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain. (5 points)2. (a) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atomsof aluminum are contained in this sample of foil?(b) Using the densities and atomic weights given in Appendix A, calculate and compare thenumber of atoms per cubic centimeter in (i) lead and (ii) lithium. (10 points)3. The density of potassium, which has the BCC structure and one atom per lattice point, is0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the latticeparameter; and (b) the atomic radius of potassium. (10 points)4. The density of a sample of HCP beryllium is 1.844 g/cm3 and the lattice parameters are a0=0.22858 nm and c0= 0.35842 nm. Calculate (a) the fraction of the lattice points that containvacancies and (b) the total number of vacancies in a cubic centimeter. (15 points)5. A ceramic part made of MgO is sintered successfully at 1700℃in 90 minutes. To minimizethermal stresses during the process, we plan to reduce the temperature to 1500℃. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? (15 points)6. (a) A thermosetting polymer containing glass beads is required to deflect 0.5 mm when aforce of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.(b) The flexural modulus of alumina is 45 x 106 psi and its flexural strength is 46,000 psi. Abar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart.Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. (15 points)7. Based on the following observations, construct a phase diagram. Element A melts at 850°Cand element B melts at 1200°C. Element B has a maximum solubility of 5% in element A, and element A has a maximum solubility of 15% in element B. The number of degrees of freedom from the phase rule is zero when the temperature is 725°C and there is 35% B present. At room temperature 1% B is soluble in A and 7% A is soluble in B. (15 points)8.Suppose that age hardening is possible in the Al-Mg system (see Figure 10-11). (a)Recommend an artificial age-hardening heat treatment for each of the following alloys, and(b) compare the amount of the precipitate that forms from your treatment of each alloy. (i)Al-4% Mg (ii) Al-6% Mg (iii) Al-12% Mg (c) Testing of the alloys after the heat treatment reveals that little strengthening occurs as a result of the heat treatment. Which of the requirements for age hardening is likely not satisfied? (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1. In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm 3, we wouldneed to select a material having a density considerably less than 1.5 g/cm 3. While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix —processes such as casting or powder metallurgy would destroy the polyethylene .Therefore polyethylene would NOT be a likely possibility.One approach, however, might be to introduce hollow glass beads .Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. (5 points)2. (a) In a one square inch sample:number ==6.69 × 1021 atoms(b) (i) In lead:= 3.3 × 1022 atoms/cm 3(ii) In lithium:= 4.63 × 1022 atoms/cm 3 (10 points)3. (a) Using Equation 3-5:0.855 g/cm 3 =a o 3 = 1.5189 × 10-22 cm 3 or a o = 5.3355 × 10-8 cm(b) From the relationship between atomic radius and lattice parameter:r == 2.3103 × 10-8cm (10 points)4. V u.c.= (0.22858 nm)2(0.35842 nm)cos30 = 0.01622 nm 3= 1.622 x 10~23 cm 3 (a) From the density equation:x = 1.9984fraction =29984.12 = 0.0008(b) number == 0.986 x 1020 vacancies/cm 3 (15 points)5. Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.D 1700= 0.000043 exp[-82,100/(1.987)(1973)] = 3.455 × 10-14 cm 2/sD1500= 0.000043 exp[-82,100/(1.987)(1773)] = 3.255 × 10-15 cm2/st1500 = D1700 t1700/D1500== 955 min = 15.9 h (15 points)6. (a) Solution:The minimum distance L between the supports can be calculated from the flexural modulus.L3 = 4w/z3δ(flexural modulus)/3FL3 = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPA)(1000 MPa/GPa) / 500 NL3 = 69,000 mm3 or L = 41 mmThe stress acting on the bar when a deflection of 0.5 mm is obtained isσ= WL/2wh2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)2 = 61.5 MPaThe applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.(b) Solution:The force required to break the bar isF = 2w/z2(flexural strength)/3LF= (2)(1 in)(0.3 in)2(46,000 psi / (3)(7 in.) = 394 lbThe deflection just prior to fracture is8 = FZ3/4wh3(flexural modulus)8 = (394 lb)(7 in)3/(4)(l in)(0.3 in)3(45 x 106 psi) = 0.0278 in. (15 points)7.(15 points)8. (a) The heat treatments for each alloy might be:Al-4% Mg Al-6% Mg Al-12% MgT Eutectic451°C 451°C 451°CT Solvs210°C 280°C 390°CSolutionTreat at: 210-451°C 280-451°C 390-451°CQuench Quench QuenchAge at: <210°C <280°C <390°C(b) Answers will vary depending on aging temperature selected. If all threeare aged at 200°C, as an example, the tie line goes from about 3.8 to 35% Mg:A1-4% Mg: %β = (4− 3.82)/(35 − 3.82) X 100 = 0.6%Al-6% Mg: %β = (6 − 3.82)/(35 − 3.82) X 100 = 7.1%Al-12% Mg: %β = (12 −3.82)/(35− 3.82) X 100 = 26.8%(c) Most likely, a coherent precipitate is not formed; simple dispersionstrengthening, rather than age hardening, occurs. (15 points)英文原版教材班“材料科学基础”考试试题试卷五Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to design an aircraft that can be flown by human power nonstop for adistance of 30 km. What types of material properties would you recommend? What materials might be appropriate? (5 points)2. Boron has a much lower coefficient of thermal expansion than aluminum, even though bothare in the 3B column of the periodic table. Explain, based on binding energy, atomic size, and the energy well, why this difference is expected. (10 points)3. Determine the ASTM grain size number if 20 grains/square inch are observed at amagnification of 400. (10 points)4. We currently can successfully perform a carburizing heat treatment at 1200o C in 1 h. In aneffort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950℃. What time will be required to give us a similar carburizing treatment? (15 points)5.The data below were obtained from a series of Charpy impact tests performed on foursteels, each having a different manganese content. Plot the data and determine (a) the transition temperature (defined by the mean of the absorbed energies in theductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0°C?Test temperature°C Impact snergy (J)0.30% Mn 0.39% Mn 1.01% Mn 1.55% Mn-100 2 5 5 15-75 2 5 7 25-50 2 12 20 45-25 10 25 40 700 30 55 75 11025 60 100 110 13550 105 125 130 14075 130 135 135 140100 130 135 135 140(15 points)。