半导体物理器件与工艺英文原版Semiconductor Physics, Devices, and Fabrication.Introduction.Semiconductors are materials with electricalconductivity between that of conductors and insulators.This unique property makes them essential for a wide rangeof electronic devices, including transistors, diodes, and solar cells.Semiconductor Physics.The electrical properties of semiconductors are determined by their electronic band structure. In an insulator, the valence band (the band of electrons that are tightly bound to the atoms) is filled, and the conduction band (the band of electrons that are free to move) is empty. In a conductor, the conduction band is partially filled. In a semiconductor, the conduction band is empty and thevalence band is filled, but there is a small energy gap between the two bands.When a semiconductor is exposed to light or heat, electrons can be excited from the valence band to the conduction band. These electrons are then free to move, and the semiconductor becomes more conductive. This phenomenon is known as intrinsic conduction.Semiconductors can also be doped with impurities to increase their conductivity. Donor impurities add electrons to the semiconductor, while acceptor impurities remove electrons. Doped semiconductors are used to create transistors, diodes, and other electronic devices.Semiconductor Devices.Transistors are the basic building blocks of electronic circuits. They can be used to amplify signals, switch currents, and store data. Transistors are made from three layers of semiconductor material: the emitter, the base, and the collector.Diodes are another important type of semiconductor device. They allow current to flow in one direction but not the other. Diodes are used in a variety of applications, including rectifying AC currents and protecting circuits from overvoltage.Solar cells are semiconductor devices that convertlight into electricity. Solar cells are made from photovoltaic materials, which are materials that generate an electrical current when exposed to light. Solar cells are used to power a variety of devices, including calculators, watches, and satellites.Semiconductor Fabrication.Semiconductors are fabricated using a variety of processes, including lithography, etching, and deposition.Lithography is the process of creating patterns in semiconductor materials. Lithography is used to create the features of transistors, diodes, and other electronicdevices.Etching is the process of removing material from semiconductor wafers. Etching is used to create the trenches and vias that connect the different layers of a semiconductor device.Deposition is the process of adding material to semiconductor wafers. Deposition is used to create the metal layers that connect the different parts of a semiconductor device.Semiconductor fabrication is a complex and precise process. The quality of the final product depends on the accuracy of each step in the fabrication process.Conclusion.Semiconductors are essential for a wide range of electronic devices. The physics of semiconductors and the processes used to fabricate semiconductor devices are complex and challenging, but they are also essential forthe development of new and innovative electronic technologies.。
______________________________________________________________________________________Chapter 66.115105⨯==d o N n cm 3-()4152102105.4105105.1⨯=⨯⨯==d i o N n p cm 3- (a) Minority carrier hole lifetime is a constant.70102-⨯==p pt ττs117401025.2102105.4⨯=⨯⨯==-p opo p R τcm 3-s 1- (b)7144010210105.4-⨯+⨯=+='p o po pp R τδ 20105⨯=cm 3-s 1-_______________________________________ 6.216102⨯==a o N p cm 3-()4162621062.1102108.1-⨯=⨯⨯==o i o p n n cm 3-(a) 21714010105105=⨯⨯=='-n n R τδcm 3-s 1- (b)n ontoptop n n p R τττ===()()()741601051062.1102--⨯⋅⨯⨯=⋅=n o o pt n p ττ131017.6⨯=s_______________________________________ 6.3 (a) Recombination rates are equalpOo nO o pn ττ=1610==d o N n cm 3-()41621021025.210105.1⨯=⨯==o i o n n p cm 3-Then641610201025.210-⨯⨯=nO τ which yields61089.8+⨯=nO τs (b) Generation rate = recombination rate Then96410125.110201025.2⨯=⨯⨯=-G cm 3-s 1- (c)910125.1⨯==G R cm 3-s 1-_______________________________________ 6.4 (a) ()()1083410630010310625.6--⨯⨯⨯===λνhch E or191015.3-⨯=E J; energy of one photon Now1 W = 1 J/s 181017.3⨯⇒photons/s Volume = (1)(0.1) = 0.1 cm 3 Then1.01017.318⨯=g191017.3⨯= e-h pairs/cm 3-s (b)()()61910101017.3-⨯⨯===τδδg p n or141017.3⨯==p n δδcm 3-_______________________________________ 6.5We havepp p p g F t p τ-+∙-∇=∂∂+ andp eD p e J p p p ∇-E =μ The hole particle current density is()p D p e J F p ppp ∇-E =+=+μ Now______________________________________________________________________________________ ()p D p F p p p ∇∙∇-E ∙∇=∙∇+μ We can write()E ∙∇+∇∙E =E ∙∇p p p andp p 2∇=∇∙∇ so()p D p p F p p p 2∇-E ∙∇+∇∙E =∙∇+μThen()E ∙∇+∇∙E -=∂∂p p tpp μpp p p g p D τ-+∇+2 We can then write ()E ∙∇+∇∙E -∇p p p D p p μ2tpp g p p ∂∂=-+τ _______________________________________6.6 From Equation (6.18), pp p pg F t p τ-+∙-∇=∂∂+ For steady-state,0=∂∂tpThenp p p R g F -+∙-∇=+0For a one-dimensional case,192010210⨯-=-=+p p p R g dxdF or19108⨯=+dxdF p cm 3-s 1- _______________________________________ 6.7From Equation (6.18),1910200⨯-+-=+dxdF p or19102⨯-=+dxdF p cm 3-s 1-_______________________________________ 6.8We have the continuity equations (1) ()()[]E ∙∇+∇∙E -∇p p p D p p δμδ2()tp p g p p ∂∂=-+δτ and (2) ()()[]E ∙∇+∇∙E +∇n n n D n n δμδ2()t n n g n n ∂∂=-+δτBy charge neutrality,()()p n n p n δδδδδ∇=∇⇒≡= and()()p n δδ22∇=∇ and ()()t p t n ∂∂=∂∂δδ A lso g g g p n ≡=, R np np ≡=ττ Then we have (1) ()()[]E ∙∇+∇∙E -∇p n n D p p δμδ2()tn R g ∂∂=-+δ and (2) ()()[]E ∙∇+∇∙E +∇n n n D n n δμδ2()tn R g ∂∂=-+δ Multiply Equation (1) by n n μ and Equation(2) by p p μ, and add the two equations. We find()()n pD nD n p p n δμμ2∇+()()n n p p n δμμ∇∙E -+ ()()R g p n p n -++μμ______________________________________________________________________________________ ()()tn pn p n ∂∂+=δμμ Divide by ()p n p n μμ+, then()n p n pD nD p n n p p n δμμμμ2∇⎪⎪⎭⎫⎝⎛+++()()n p n n p p n p n δμμμμ∇∙E ⎥⎥⎦⎤⎢⎢⎣⎡+-()()tn R g ∂∂=-+δ Define()pD n D p n D D p n pD nD D p n p n p n n p p n ++=++='μμμμ and ()pn n p p n p n μμμμμ+-='Then we have()()()R g n n D -+∇∙E '+∇'δμδ2()tn∂∂=δ Q.E.D. _______________________________________6.9 p-type material;minority carriers are electrons(a)n μμ=' From Figure 5.3, 1300≅n μcm 2/V-s (b)()()13000259.0=⋅⎪⎭⎫⎝⎛=='n n e kT D D μ 67.33=cm 2/s (c)7010-==n nt ττs15107⨯==a o N p cm 3-()152102107105.1⨯⨯==a i o N n n41021.3⨯=cm 3-pto nt o pn ττ=pt τ15741071010214.3⨯=⨯- so 41018.2⨯=pt τs_______________________________________6.10 For Ge: 13104.2⨯=i n cm 3- 2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+=()21321313104.221042104⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1310124.5⨯=cm 3-()1313213210124.110124.5104.2⨯=⨯⨯==o i o n n p cm 3- (a) We have:3900=n μcm 2/V-s, 101=n D cm 2/s1900=p μcm 2/V-s, 2.49=p D cm 2/s For very, very low injection, ()p D n D p n D D D p n p n ++='()()()()()()()1313131310124.12.4910124.510110124.110124.52.49101⨯+⨯⨯+⨯= 2.54=cm 2/s and()pn n p p n p n μμμμμ+-='()()()()()()()1313131310124.1190010124.5390010124.510124.119003900⨯+⨯⨯-⨯=1340-=cm 2/V-s (b)For holes, 60102-⨯==p pt ττs For electrons,______________________________________________________________________________________p nt p nττ=6131310210124.110124.5-⨯⨯=⨯ntτ61012.9-⨯=⇒nt τs_______________________________________ 6.11p e n e p n μμσ+= W ith excess carriersn n n o δ+= and p p p o δ+=For an n-type semiconductor, we can write p p n δδδ≡= Then()()p p e p n e o p o n δμδμσ+++= or()()p e p e n e p n o p o n δμμμμσ+++= so()()p e p n δμμσ+=∆In steady-state, pO g p τδ'= So that()()pO p n g e τμμσ'+=∆_______________________________________ 6.12 (a) 1610==a o N p cm 3-()41621021025.210105.1⨯=⨯==o i o p n n cm 3- ()()p p e n n e o p o n δμδμσ+++= ()n e p e p n o p δμμμ++≅ Now ()0/01n t n e g p n ττδδ--'==()()()0/7201105108n t e τ---⨯⨯=()0/141104n t eτ--⨯=cm 3-Then σ()()()161910380106.1-⨯= ()()380900106.119+⨯+-()()0/141104n t e τ--⨯⨯ ()0/10819.0608.0n t e τσ--+= (Ω-cm)1- (b) (i)()608.00=σ(Ω-cm)1-(ii)()690.0=∞σ(Ω-cm)1-_______________________________________ 6.13 (a) For 6100-≤≤t s, ()0/01p t p e g p n ττδδ--'==()()()0/8211105104p t e τ---⨯⨯=()()0/141102p t e τ--⨯= cm 3-At 610-=t s,()()()86105/10146110210--⨯---⨯=e p δ14102⨯=cm 3-Then for 610-≥t s,()()6/1014102p t ep τδ---⨯=cm 3-(b)15105⨯=o n cm 3-()p e n e p n o n δμμμσ++=For 6100-≤≤t s,()()()15191057500106.1⨯⨯=-σ ()()3107500106.119+⨯+-()()0/141102p t eτ--⨯⨯()0/1250.00.6p t eτ--+=(Ω-cm)1-For 610-≥t s,()6/10250.00.6p t eτσ---+=(Ω-cm)1-_______________________________________ 6.14R V I =; ALR σ=______________________________________________________________________________________V L A I ⋅=⇒σFor 1515102108⨯+⨯=+=a d I N N N1610=cm 3- Then, 1300≅n μcm 2/V-s400≅p μcm 2/V-s()p e n e p n o n δμμμσ++≅where 0/0p t p e g p ττδ-'=()()0/720105108p t e τ--⨯⨯= 0/14104p t e τ-⨯=cm 3- ()()()1515191021081300106.1⨯-⨯⨯=-σ()()4001300106.119+⨯+-()0/14104p t e τ-⨯⨯ 0/109.0248.1p t e τσ-+= []()()05.01010109.0248.15/0--+=p t e I τ 0/431018.210496.2p t e τ---⨯+⨯= Aor 0/218.0496.2p t e I τ-+=mA_______________________________________6.151516106102⨯-⨯=-=d a o N N p 16104.1⨯=cm 3-(a)0n g p n τδδ'==021********n τ⨯=⨯ 70105.2-⨯=⇒n τs (b)()0/01n t n eg p n ττδδ--'==()0/141105n t e τ--⨯=()0/71401105.2105n t n e n R ττδ---⨯⨯==' ()not e τ/211102--⨯=cm 3-s 1- (c)(i)()()0/1414110510541n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛()801019.73333.1ln -⨯==n t τs(ii) ()()0/1414110510521n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛ ()701073.12ln -⨯==n t τs (iii)()()0/1414110510543n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛ ()701047.34ln -⨯==n t τs(iv)()()()0/1414110510595.0n t e τ--⨯=⨯()701049.720ln -⨯==n t τs_______________________________________6.161515102108⨯-⨯=-=a d o N N n 15106⨯=cm 3-()415262104.5106108.1-⨯=⨯⨯==o i o n n p cm 3- (a) 0440104.5104p p o o p R ττ-⨯=⨯⇒= so 801035.1-⨯=p τs (b) ()()82101035.1102-⨯⨯='=p g p τδ 13107.2⨯=cm 3- (c) 801035.1-⨯==p ττs _______________________________________6.17 (a)(i)For 71050-⨯≤≤t s()()0/01p t p e g t p ττδ--'=()()()0/7201105105p t e τ---⨯⨯=()0/141105.2p t eτ--⨯= cm3-At 7105-⨯=t s,()1/1141105.2--⨯=e p δ141058.1⨯=cm 3-For 7105-⨯≥t s()()p Ot et p τδ/1051471058.1-⨯--⨯=cm 3-______________________________________________________________________________________ (ii) ()1471058.1105⨯=⨯-p δcm 3- (b) (i) For 61020-⨯≤≤t s()()0/141105.2p t e t p τδ--⨯= cm3-At 6102-⨯=t s,()()()76105/102141105.2--⨯⨯--⨯=e p δ1410454.2⨯=cm 3-For 6102-⨯≥t s,()()p O t et p τδ/10214610454.2-⨯--⨯=cm 3-(ii)()14610454.2102⨯=⨯-p δcm 3- _______________________________________ 6.18 (a) For 61020-⨯≤≤t s ()0/0n t n e g t n ττδ-'=()()0/72110510n t e τ--⨯= 0/14105n t e τ-⨯=cm 3-At 6102-⨯=t s,()()76105/102141105--⨯⨯-⨯=e n δ121016.9⨯= cm 3-For 6102-⨯≥t s ()()0/121411016.9105n t e n τδ--⨯-⨯=121016.9⨯+()12/141016.9110908.40⨯+-⨯=-n t e τcm 3-(b) (i)()141050⨯=n δcm 3-(ii)()1261016.9102⨯=⨯-n δcm 3- (iii)()14105⨯=∞n δcm 3-_______________________________________ 6.19p-type; minority carriers - electrons()()12000259.0=⎪⎭⎫⎝⎛=n n e kT D μ 08.31=cm 2/s()()[]2/1601008.31-==n n n D L τ310575.5-⨯=cm(a) ()()n L x e x p x n /14102-⨯==δδcm 3-(b)()[]n L x n nn e dx deD dx n d eD J /14102-⨯==δ ()n L x nne L eD /14102-⨯-=()()()()n L x e /3141910575.510208.31106.1---⨯⨯⨯-=n L x n e J /1784.0--=A/cm 2Holes diffuse at same rate as minority carrier electrons, son L x p e J /1784.0-+=A/cm 2_______________________________________ 6.20 (a)p-type; 1410=pO p cm 3-and()61421021025.210105.1⨯=⨯==pO i pOp n n cm 3-(b) Excess minority carrier concentrationpO p n n n -=δ At 0=x , 0=p n so that()61025.200⨯-=-=pO n n δcm 3-(c) For the one-dimensional case,()022=-nOnndx n d D τδδ or()0222=-nLndx n d δδ where nO n n D L τ=2The general solution is of the form⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-=n n L x B L x A n exp exp δ For ∞→x , n δ remains finite, so0=B .Then the solution is⎪⎪⎭⎫⎝⎛--=n pO L x n n exp δ _____________________________________________________________________________________________________________________________ 6.21()n L x e x n /14105-⨯=δcm 3-where ()()[]2/1601025-==n n n D L τ 3105-⨯=cm()()[]n L x n n n e dxdeD dx n d eD J /14105-⨯==δ()n L x nne L eD /14105-⨯-=()()()()nL x e /3141910*********.1---⨯⨯⨯-=n L x n e J /4.0--=A/cm 2(a) For 0=x ,()141050⨯=n δcm 3- ()4.00-=n J A/cm 2 ()4.00+=p J A/cm 2 (b) For 3105-⨯==n L x cm,()()141141084.1105⨯=⨯=-e L n n δcm 3-()147.04.01-=-=-e L J n n A/cm 2()147.04.01+=+=-e L J n p A/cm 2(c)For 31015-⨯=x cm n L 3=()()133141049.21053⨯=⨯=-e L n n δcm 3-()020.04.033-=-=-e L J n n A/cm 2()020.04.033+=+=-e L J n p A/cm 2_______________________________________6.22n-type, so we have()()02=-E -pO o p ppdx p d dx p d D τδδμδ Assume the solution is of the form()sx A p exp =δ Then()()sx As dxp d exp =δ, ()()sx As dx p d exp 222=δ Substituting into the differential equation()()sx As sx As D o p p exp exp 2E -μ ()0exp =-pOsx A τor012=-E -pOo p p s s D τμDividing by p D , we have 0122=-E -ppop L s D s μ The solution for s is⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛E ±E =22421p o p p o p pL D D s μμ which can be rewritten as⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛E ±E =12212p o p p pop p p D L D L L s μμ Define po p p D L 2E ≡μβThen ⎥⎦⎤⎢⎣⎡+±=211ββpL s In order that p δ0=as +∞→x , use the minus sign for 0>x and the plus sign for0<x . Then the solution is ()x s A p -=exp δ for 0>x ()x s A p +=exp δ for 0<x where⎥⎦⎤⎢⎣⎡+±=±211ββp L s _____________________________________________________________________________________________________________________________ 6.23 Plot_______________________________________6.24 (a) From Equation (6.55)()()022=-E +nO on n ndx n d dx n d D τδδμδ or()()0222=-E +no n n L ndx n d D dx n d δδμδ We have that⎪⎭⎫⎝⎛=ekTD n n μ so we can define ()L e kT D o o nn '≡E =E 1μ Then we can write()()01222=-⋅'+nL ndx n d L dx n d δδδ The solution is of the form()()x n n αδδ-=exp 0 where 0>α Then()()n dxn d δαδ-= and ()()n dxn d δαδ222= Substituting into the differential equation, wefind()()[]0122=--'+nL nn L n δδαδαor0122=-'-nL L ααwhich yields⎥⎥⎦⎤⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛'+'=12212L L L L L n nn α We may note that if 0=E o , then ∞→'Land nL 1=α(b)nO n n D L τ= where ⎪⎭⎫ ⎝⎛=e kT D n n μso ()()1.310259.01200==n D cm 2/s and ()()47104.391051.31--⨯=⨯=n L cmorμ4.39=n L m For 12=E o V/cm, then ()4106.21120259.0-⨯==E ='oe kT L cmand21075.5⨯=αcm 1-(c) Force on the electrons due to the electric field is in the negative x-direction. Therefore, the effective diffusion of the electrons is reduced and the concentration drops off faster with the applied electric field. _______________________________________ 6.25p-type so the minority carriers are electrons and()()()t n n g n n D nO n n ∂∂=-'+∇∙E +∇δτδδμδ2 Uniform illumination means that()()02=∇=∇n n δδ. For ∞=nO τ, we areleft with()g dtn d '=δ which gives 1C t g n +'=δFor 0≤t , 001=⇒=C n δ Thent G n o'=δ for T t ≤≤0 For T t >, 0='g so that()0=dtn d δ AndT G n o'=δ (no recombination) _______________________________________ 6.26______________________________________________________________________________________ n-type, so minority carriers are holes and ()()()t p p g p p D pO p p ∂∂=-'+∇∙E -∇δτδδμδ2 We have ∞=pO τ , 0=E , and()0=∂∂tp δ(steady-state). Then we have ()022='+g dxp d D p δ or ()pD g dx p d '-=22δFor L x L +<<-, oG g '='= constant. Then()1C x D G dx p d p o +'-=δand 2122C x C x D G p po++'-=δ For L x L 3<<, 0='g so we have()022=dx p d δ so that()3C dxp d =δ and 43C x C p +=δFor L x L -<<-3, 0='g so that()022=dx p d δ so that()5C dxp d =δ and 65C x C p +=δThe boundary conditions are: (1) 0=p δ at L x 3+= (2) 0=p δat L x 3-=(3) p δ continuous at L x = (4) p δ continuous at L x -=(5)()dx p d δ continuous at L x = (6) ()dxp d δ continuous at L x -= Applying the boundary conditions, we find()2252x L D G p po -'=δ forL x L +<<-()x L D L G p po-'=3δ for L x L 3<< ()x L D L G p po+'=3δ for L x L -<<-3 _______________________________________6.27 204.080===E L V V/cm ()()60010322025.0-⨯=E =t d p μ 6.390=cm 2/V-s ()()2216t t D pp ∆E =μ()()[]()()62621032161035.9206.390--⨯⨯=42.10=p D cm 2/sWe find02668.06.39042.10==p p D μV This value is very close to 0.0259 for 300=T K._______________________________________ 6.28 (a)Assume that()()⎪⎪⎭⎫ ⎝⎛-=-Dt x Dt t x f 4exp 4,22/1π is the solution to the differential equationt fx f D ∂∂=⎪⎪⎭⎫ ⎝⎛∂∂22 To prove: we can write()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=∂∂-Dt x Dt x Dt x f 4exp 42422/1π and()⎪⎪⎭⎫ ⎝⎛-⎢⎣⎡⎪⎭⎫⎝⎛-=∂∂-Dt x Dt x Dt x f 4exp 424222/122π______________________________________________________________________________________ ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-+Dt x Dt 4exp 422 A lso()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-=∂∂-Dt x t D x Dt t f 4exp 1442222/1π()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-+--Dt x t D 4exp 21422/32/1π Substituting the expressions for 22x f∂∂ and t f ∂∂into the differential equation, we find0 = 0.Q.E.D.(b)Considerdx Dt x ⎰+∞∞-⎪⎪⎭⎫ ⎝⎛-4exp 2Let 2x u =, then dx x du ⋅=2 or udu x du dx 22==Let Dta 41= Nowdx Dt x dx Dt x ⎰⎰∞+∞∞-⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛-0224exp 24exp()()du au u du au u -=-=⎰⎰∞∞exp 1exp 2120t D a ππ4==Thent D t D dx Dt x t D πππ444exp 412=⎪⎪⎭⎫ ⎝⎛-⎰+∞∞-1= _______________________________________6.29Plot_______________________________________6.30 (a) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1016105.1104ln 0259.0383225.0=eV(b) ()()7210105102-⨯⨯='==p g p n τδδ 1510=cm 3- ⎪⎪⎭⎫ ⎝⎛+=-i o Fi Fn n n n kT E E δln ()⎪⎪⎭⎫⎝⎛⨯+⨯=101516105.110104ln 0259.0 383865.0=eV ⎪⎪⎭⎫ ⎝⎛+=-i o Fp Fi n p p kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯≅1015105.110ln 0259.0 28768.0=eV (c) 383225.0383865.0-=-F Fn E E 000640.0=eV or 640.0=meV _______________________________________ 6.31 (a) p-type⎪⎪⎭⎫ ⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1015105.1105ln 0259.0 or3294.0=-F Fi E E eV (b)______________________________________________________________________________________14105⨯==p n δδcm 3-and ()4152102105.4105105.1⨯=⨯⨯==o i o p n n cm 3-Then⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=10144105.1105105.4ln 0259.0 or2697.0=-Fi Fn E E eV and⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=101415105.1105105ln 0259.0 or3318.0=-Fp Fi E E eV_______________________________________ 6.32 (a) For n-type,()()Fi F Fi Fn F Fn E E E E E E ---=-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n n kT n n n kT ln ln δ ⎪⎪⎭⎫⎝⎛+=o o n n n kT δln So ()⎪⎪⎭⎫⎝⎛⨯+⨯=1515105105ln 0259.000102.0n δ⎪⎭⎫⎝⎛⨯=+⨯0259.000102.0exp 1051051515n δWhich yields 14102⨯≅n δcm 3-(b) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯⨯+⨯=101415105.1102105ln 0259.0 33038.0=eV(c) ⎪⎪⎭⎫⎝⎛≅-i Fp Fi n p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1102ln 0259.02460.0=eV_______________________________________ 6.33(a) ⎪⎪⎭⎫⎝⎛≅-i Fi Fn n n kT E E δln or ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fn i exp δ ()⎥⎦⎤⎢⎣⎡⨯=0259.0270.0exp 105.110141005.5⨯=cm 3-(b)⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=101415105.11005.5106ln 0259.0 33618.0=eV (c) (i) ()()F Fi Fp Fi Fp F E E E E E E ---=-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n p kT n p p kT ln ln δ ⎪⎪⎭⎫⎝⎛+=o o p p p kT δln (ii) Fp F E E -()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=1514151061005.5106ln 0259.0 310093.2-⨯=eV or 093.2=meV_____________________________________________________________________________________________________________________________ 6.34 (a) (i) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E ln ()()()⎪⎪⎭⎫ ⎝⎛⨯=616108.11002.1ln 0259.0 58166.0=eV (ii) ⎪⎪⎭⎫ ⎝⎛≅-i Fp Fi n p kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=616108.11002.0ln 0259.047982.0=eV (b) (i) ()⎪⎪⎭⎫⎝⎛⨯⨯=-616108.1101.1ln 0259.0Fi Fn E E58361.0=eV (ii)()⎪⎪⎭⎫⎝⎛⨯⨯=-616108.1101.0ln 0259.0Fp Fi E E52151.0=eV _______________________________________ 6.35Quasi-Fermi level for minority carrier electrons:⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()4162621024.310108.1-⨯=⨯==o i o p n n cm 3-We have()⎪⎭⎫⎝⎛=501014x n δThen()⎥⎦⎤⎢⎣⎡⨯+⨯=--6144108.150101024.3ln x kT E E Fi Fn We findx (μm)(Fi Fn E E -) (eV)0 12 1020 50-0.581 +0.361 +0.379 +0.420 +0.438 +0.462Quasi-Fermi level for holes: we have ⎪⎪⎭⎫ ⎝⎛+=-i o Fp Fi n p p kT E E δln We have 1610=o p cm 3- and p n δδ=. We findx (μm)(Fp Fi E E -) (eV)0 50+0.58115 +0.58140 _______________________________________ 6.36 (a) We can write⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln and⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln so that()()Fp F F Fi Fp Fi E E E E E E -=---⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n p kT n p p kT ln ln δ or()kT p p p kT E E o o Fp F 01.0ln =⎪⎪⎭⎫⎝⎛+=-δ Then______________________________________________________________________________________()010.101.0exp ==+oop p p δ or⇒=010.0op pδlow injection, so that 12105⨯=p δcm 3-(b)⎪⎪⎭⎫ ⎝⎛≅-i Fi Fn n p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯=1012105.1105ln 0259.0 or1505.0=-Fi Fn E E eV_______________________________________6.37Plot_______________________________________6.38(a) ⎪⎪⎭⎫⎝⎛≅-i Fp Fi n p kT E E δln ()⎪⎭⎫ ⎝⎛⨯=10105.1ln 0259.0p δ 1110=p δcm 3-,04914.0=-Fp Fi E E eV1210 10877.0 1310 16841.0 1410 0.22805 1510 0.28768 (b) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()⎪⎪⎭⎫⎝⎛⨯+⨯=1016105.1102ln 0259.0n δ 1110=n δcm 3-, 365273.0=-Fi Fn E E eV12100.365274 1310 0.365286 1410 0.36540215100.366536_______________________________________ 6.39 (a)()()()p p C n n C n np N C C R p n i t p n '++'+-=2()()()p p n n n np nO pO i '++'+-=ττ2Let i n p n ='='. For 0==p n nO pO i i nO i pO i n n n n R ττττ+-=+-=2 (b) We had defined the net generation rate as()R R g g R g o o '+-'+=-where o o R g =since these are the thermal equilibrium generation and recombination rates.If 0='g , then R R g '-=- andnOpO i n R ττ+-=' so that nOpO in R g ττ++=-Thus a negative recombination rate implies anet positive generation rate._______________________________________6.40 We have that()()()p p C n n C n np N C C R p n i t p n '++'+-=2()()()i nO i pO i n p n n n np +++-=ττ2If n n n o δ+= and n p p o δ+=, then()()()()i o nO i o pO i o o n n p n n n n n p n n R +++++-++=δτδτδδ2______________________________________________________________________________________ ()()()()i o nO i o pO i o o o o n n p n n n n n p n n p n +++++-+++=δτδτδδ22 If i n n <<δ, we can neglect ()2n δ: also 2i o o n p n =Then()()()i o nO i o pO o o n p n n p n n R ++++=ττδ(a) For n-type; O o p n >>, i o n n >> Then 7101+==pOn R τδs 1-(b) For intrinsic, i o o n p n == Then ()()i nO i pO i n n n n R 222ττδ+=or ⇒⨯+=+=--771051011nO pO n R ττδ 61067.1+⨯=nR δs 1- (c) For p-type; o o n p >>, i o n p >>Then6710210511+-⨯=⨯==nO n R τδs 1- _______________________________________ 6.41 (a) From Equation (6.56) ()022=-'+pO p p g dx p d D τδδSolution is of the form⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-+'=p p pO L x B L x A g p exp exp τδAt +∞=x , pO g p τδ'= so that 0=B , Then⎪⎪⎭⎫⎝⎛-+'=p pO L x A g p exp τδ We have()()00===x x p p s dx p d D δδ We can write()p x L A dx p d -==0δ and ()A g p pO x +'==τδ0Then ()A g s L AD pO pp+'=-τSolving for A , we find s L D g s A pp pO +'-=τThe excess concentration is then()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛-⋅+-'=p p p pO L x s L D s g p exp 1τδ where ()()37101010--===pO p p D L τcm Now()()7211010-=p δ ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⋅+-⨯-p L x s s exp 101013 or ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⋅+-=p L x s s p exp 10110414δ (i) For 0=s , 1410=p δcm 3- (ii) For 2000=s cm/s,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=p L x p exp 167.011014δ (iii) For ∞=s ,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=p L x p exp 11014δ______________________________________________________________________________________ (b) (i) For 0=s , ()14100=p δcm 3-(ii) For 2000=s cm/s,()1410833.00⨯=p δcm 3- (iii) For ∞=s , ()00=p δ _______________________________________6.42 ()()710525-⨯==nO n n D L τ4104.35-⨯=cm(a) At 0=x ,()()1572110105102=⨯⨯='-nO g τcm 3- or()15100='=nO g n τδcm 3-For 0>x()()0022222=-⇒=-nnO n L ndx n d n dx n d D δδτδδ The solution is of the form ⎪⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛-=n n L x B L x A n exp exp δ At 0=x , ()B A n n +==0δδ At W x =,⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-==n n L W B L W A n exp exp 0δ Solving these two equations, we find()()()n n L W L W n A 2exp 12exp 0+-+-=δand ()()n L W n B 2exp 10+-=δSubstituting into the general solution, we find()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛+=n n L W L W n n exp exp 0δδ()()⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡-+⨯n n L x W L x W exp exp which can be written as()⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=n n L W L x W n n sinh sinh 0δδ where ()15100=n δcm 3- and μ4.35=n L m (b) If ∞=nO τ, we have ()022=dx n d δ so the solution is of the form D Cx n +=δ Applying the boundary conditions, we find()⎥⎦⎤⎢⎣⎡-=W x n n 10δδ _______________________________________6.43For ∞=pO τ, we have()022=dxp d δ So the solution is of the formB Ax p +=δ At W x = ()()W x W x p p s dx p d D ===-δδor()B AW s A D p +=-which yields()sW D sAB p +-=At 0=x , the flux of excess holes is()A D dxp d D p x p-=-==01910δso that1819101010-=-=A cm 4-and()⎪⎭⎫⎝⎛+=+=W s sW s B 101010101818 The solution is now______________________________________________________________________________________⎪⎭⎫ ⎝⎛+-=s x W p 101018δ(a)For ∞=s ,()x p -⨯=-418102010δ cm 3- Then()dxp d eD J p p δ-= ()()()181********.1-⨯-=- or6.1=p J A/cm 2 (b)For 3102⨯=s cm/s,()x p -⨯=-418107010δ cm 3- A lso6.1=p J A/cm 2_______________________________________ 6.44For 0<<-x W()022='+on G dxn d D δ so that()1C x D G dx n d no +'-=δ and2122C x C x D G n no++'-=δ For W x <<0,()022=dx n d δ so that43C x C n +=δThe boundary conditions are (1) 0=s at W x -= so that()0=-=Wx dx n d δ (2) ∞=s at W x += so that ()0=W n δ(3) n δ continuous at 0=x(4) ()dxn d δ continuous at 0=x Applying the boundary conditions, we findn o D W G C C '-==31 and noD W G C C 242'+==Then for 0<<-x W()22222W W x x D G n no +--'=δand for W x +<<0()x W D WG n no -'=δ_______________________________________ 6.45Plot_______________________________________ 6.48 (a) GaAs:Ω=⨯==-66101022I V R ()ALR σ∆= and ()p e p n δμμσ+=∆()()13821010510510⨯=⨯='=-p g p τδcm 3-For 1610=d N cm 3-, from Figure 5.3, 7000≅n μcm 2/V-s,310≅p μcm 2/V-s()()()13191053107000106.1⨯+⨯=∆-σ 05848.0=(Ω-cm)1- Let μ20=W mThen ()()441041020--⨯⨯==W d A 81080-⨯=cm 2So ()()86108005848.010-⨯==L RWhich yields 21068.4-⨯=L cm (b) Silicon:Ω=610R , 13105⨯=p δcm 3- For 1610=d N cm 3-, from Figure 5.3,______________________________________________________________________________________ 1300≅n μcm 2/V-s,410≅p μcm 2/V-s()()()13191054101300106.1⨯+⨯=∆-σ01368.0=(Ω-cm)1- Let μ20=W mThen ()()441041020--⨯⨯==W d A 81080-⨯=cm 2So ()()86108001368.010-⨯==LRWhich yields 21009.1-⨯=L cm_______________________________________。
半导体物理与器件思政课程案例分享英文版Semiconductor Physics and Device Ideological and Political Course Case SharingIn the field of electronics, semiconductors play a pivotal role, serving as the backbone of modern electronics industry. Understanding the intricacies of semiconductor physics and its applications in devices is crucial for students pursuing careers in this domain. Incorporating ideological and political education (IDE) into the curriculum of semiconductor physics and devices not only enhances the students' understanding of the subject matter but also prepares them to be responsible and ethical professionals.One effective way to integrate IDE into the teaching of semiconductor physics and devices is through case studies. These cases can range from historical perspectives, technological advancements, to societal impacts and ethicalconsiderations. By analyzing real-world scenarios, students can gain a deeper understanding of the practical applications of semiconductor technology and the associated responsibilities.For instance, a case study on the development of the transistor, a semiconductor device, can trace its historical journey from its invention in the 1940s to its present-day applications in various electronic devices. This case highlights the significance of innovation and research in推进科技发展和改善人类生活。
