September 21, 2005, Wednesday
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1 , Wednesday Doping and diffusion I Faster MOSFET requires shorter channel P + Poly Al Al Motivation Requires shallower source, drain Al P + Poly Al source drain Shorter channel length; yes, but same source & drain depth means drain field dominates gate field => drain-induced barrier lowering DIBL Shallower source, drain depth demands better control in doping & diffusion. CHANNEL ASPECT RATIO =>ρ s 1 How are shallow doped layers made? 1) Predeposition: controlled number of dopant species at surface 6s: film or gas phase of dopant at surface c(x) Surface concentration is limited by equilibrium solubility Now: Ion implant (non-equilibrium), heat substrate to diffuse dopant but ions damage target requires anneal, x changes doping, c( ) Soon: return to film or gas phase of dopant at surface ) Drive-in process: heat substrate after predeposition, diffusion determines junction depth, sharpness c(x) Need sharper diffusion profiles: c x depth 1
2 Figure removed for copyright reasons. Please see: Figures in Ghandi, S. VLSI Fabrication Principles: Silicon and Gallium Arsenide. nd ed. New York, NY: Wiley-Interscience, ISBN:
3 , Wednesday c) Mass (or heat) flow J, d) due to concentration gradient J ( area t)= D c Fick I c(, t ) J = + Inflow out dc() J in = J () =+J ()J+ ( ) J out = J+ ( ) dt area t dc = J ( ) J+ ( ) dj dt d dc() dt or if D is constant = d d D C Fick II dc(, t) = D C,t ( ) dt These Eqs => time evolution of some initial conditions & boundary conditions t > Time dep. Schrödinger Eq. +ih t Diffusing species must be soluble In host = h m + V 5 Atomistic picture of diffusion in solids See web site for movies: En Most important is vacancy diffusion. i f E a Initial and final states have same energy Also possible is direct exchange ( = broken bond) Higher energy barrier or break more bonds => lower value of D = D exp( E kt ) 6 3
4 , Wednesday steps Atomistic picture of vacancy diffusion for diffusion: 1) create vacancy ) achieve energy E a n v = N v = exp!.6 % '! N kt & v =! exp E & % a ( kt ' Vacancy diffusion ν a E VD D ~ a x v = a! exp E v + E a & % kt ' (! cm & s % D = D exp! E VD kt % Contains ν Debye frequency! 3 k B T h = 9 11 s s 1 7 Solubility limits 1. Sample phase diagram Liquid From phase diagram: 1. T/T m Liq + Sol Solubility dopant in Si Solid T/T m B As P Si at% dopant 1 at% 1 at% Impurity content (cm -3 ) Impurity content (cm -3 ) 8 4
5 , Wednesday Solubility limits Figure removed for copyright reasons. Please see: Figure -4 in Campbell, S. The Science and Engineering of Microelectronic Fabrication. 1st ed. New York, NY: Oxford University Press, ISBN: Analytic Solution to Diffusion Equations, Fick II:!C!t = D! C! There are many different solutions to this or any DE; the correct solution is the one that satisfies the BC. Steady state, dc/dt = Implies either a) D = C() c() may be curved or b) d c/d = c(x) linear In oxidation we assumed steady state : C() = a + b O diffusion through SiO, where flux,j =!D C is same everywhere. =!Db, Not necessarily so in diffusion where non-linear c() can exist and be froen in at D = 1 5
6 , Wednesday For solutions, boundary condition, consider classical experiment: Diffusion couple: thindopant layer on rod face, press identical pieces together, heat. Then study diffusion profile in sections. symmetry ( ) BC ( ) d dc,t C, t d = Q = const. ( /area) = t = IC C, ( )= C(,t)= ( ) BC 11 Analytic Solution to Diffusion Equations J = D c, c t = D c I. Drive in of fixed amount, Q, of dopant; solution is Gaussian c,t ( )= const exp 4Dt Predeposition is delta function, (). C,t ( )= dc(,t) = d Q Dt Width of Gaussian = t = Dose, Q, amount of dopant in sample, is constant. 1 exp 4Dt a = Dt c, ( )= ( ) t > c(,t)= ( ) = diffusion length a (a is large relative to width of predeposition) C, t d = Q = const. ( /area) Units /vol = Q/(length ) 1/ So Q has units /area 1 6
7 , Wednesday Solutions for other I.C./B.C. can be obtained by superposition: II. Limitless source of dopant (e.g. growth in presence of vapor) C, t IC C, BC ( ) = ( ) = C Const C t ( ) = C C!,t (, t ) BC C imp (, t) =! u= Dt exp d % erf u [ ] ( ) = erf & ( ' Dt ) C t =! = 4Dt ( ) 4Dt erf 13 Other I.C./B.C. (cont.): erf 1 ( ) erfc( u) =1! erf ( u) c ( ) = C surf erfc C,t! Dt %&, t > a = diffusion length = Dt (D 1-15 ) (t = 1 3 ) a. µm Dose! Q = C(,t)d = Dt C = a! C C(, t) = C C(,) = Const c t c(, t) = c limited by solid solubility Dose, Q, amount of dopant in sample, increases as t 1/. 14 7
8 Figure removed for copyright reasons. Please see: Figure 7-8 in Plummer, J., M. Deal, and P. Griffin. Silicon VLSI Technology: Fundamentals, Practice, and Modeling. Upper Saddle River, NJ: Prentice Hall,. ISBN:
9 , Wednesday Internal E fields alter Fick s Law Heavily doped layer can generate its own field due to displacement of mobile carriers from ionied dopants: A A p-type Si h h h + A h h A A C A C C Mobile hole A (dopant concentration + + ions) Net charge - + E E enhances diffusion of A - to right, (also down concentration gradient). J mass =!D C + CµE! D! C + CqE! ' & ) % kt ( A - drift diffusion µ = Dq kt (units) Einstein relation from Brownian motion 17 Neutral and charged impurities, dopants If impurity is Gp. IV (e.g. Ge): uncharged, no e or h Si But if impurity = B, P As it will be charged: So vacancies can be charged As e - For small dopant concentration, different diffusion processes are independent, but generally: (these D o are NOT same as single activation energy values) Ea kt D e = D! D + D 1 n + D % n & ( +... D + % p & ( + D + % p & ( +... electrons n i n i ' p i ' p i ' Intrinsic Diffusivities and Activation Energies of Substitutional Diffusers in Silicon* D i o D i + D i - D i D o E o D o E o D o E o D o E o P As Sb B Al Ga * D in cm /s; E in ev. holes Power series representation; higher orders in n describe dopant-dopant interactions [ Higher activation energy for charged vacancy diffusion; prefactor is greater] 18 Figure by MIT OCW. 9
10 , Wednesday What is n? D off = D + D! n % ' & + n % D! ' D + p % ' & & + p % D+ ' +... & n i n i n is local free electron concentration in host. n > n i always n! N D + N D % ' + n & i So clearly, D eff = D + D - (n/n i ) + can be >> D = D exp(-e VD /kt) (provided D 1- etc not too small) For intrinsic semiconductor or N D << n i, n = p = n i n i n i See example Plummer, p. 41, As n i = 7.14 x 1 18 at 1 C D eff As = D + D - (n/n i ) +.67 x x 1-15 (n/n i ) N D = 1 18 : D eff As = 1.43 x 1-15 N D = 1 : D eff As = 16.6 x 1-15 D eff = D + D - (1) + Caution: these numbers not from table on prior slide (Single-activation-energy value: D = 1.5 x 1-15 ) 19 Internal E field C h h h C A h + h Mobile hole concentration Is related to concentration dependent diffusion n D off = D + D! % ' + D! n % ' +... & & n i n i Net charge - + E E enhances diffusion of A - to right, (also down concentration gradient). Higher dopant concentration means greater charge separation, Δq, thus larger internal electric fields. 1
11 , Wednesday Exercise Calculate diffusivity of P in Si at 1 C for a) c P < n i b) c P = cm -3 c) compare diffusion length b) with uncharged estimate a) c P < n i = 1 19 from Fig 1.16 Figure removed for copyright reasons. Please see: Figure 1.16 in Plummer et al.,. 1 Exercise Calculate diffusivity of P in Si at 1 C for a) c P < n i b) c P = cm -3 c) compare diffusion length b) with uncharged estimate a) c P < n i = 1 19 D P = 3.85exp! 3.66 % Diffusion of P in Si kt & ' =1.3(1!14 (cm s!1 ) D! P = 4.4exp! 4 D kt% = 6.63&1!16 (cm s!1 ) E a D - E - a (cm /s) (ev) (cm /s) (ev) D = D! n P + D % P ' =1.37 (1!14 (cm s!1 ) & n i 1 8 n i 6 4 n! N D + N D % ' + n & i Plummer, Eq N D /n i Linear approximation Single-valued D Better than single-activation-energy : D P = 4.7exp! 3.68 % kt & ' =1.3(1!14 (cm s!1 ) 11
12 , Wednesday Exercise Calculate diffusivity of P in Si at 1 C for a) c P < n i b) c P = cm -3 c) compare diffusion length b) with uncharged estimate b) c P = N D = n! N D + N D % ' + n & i =! ! = 4.4! 1 19 cm 3 Plummer, Eq D = D! n % P + D P & ' = 1.3 ( 4.4 1! ( 1!16 % 1 & ' = 1.57 ( 1!14 (cm s!1 ) vs 1.37 n i c) a = Dt, a = 1.3! 1 14! 36 =.137 µm 1 hr a P = 1.57! 1 14! 36 =.151 µm (This is a measure of the depth of doping.) 3 Consequence: Diffusion is enhanced at high dopant concentrations, giving sharper diffusion profile D eff (T) D ev D eff 3.99 ev Graph removed for copyright reasons. 1 3 /T D 3.44 ev 4 1
13 , Wednesday Effect of oxidation of Si on diffusion Plummer Fig B and P observed to diffuse faster when Si surface is oxidied, Sb slower. Why? So far we have concentrated on diffusion by vacancy mechanism Figure removed for copyright reasons. Please see: Figure 7-36 in Plummer et al.,. Different behavior of B and Sb under oxidation suggests a different mechanism may dominate in these two dopants 5 Effect of surface oxidation on diffusion in Si Si + Oxygen SiO x + Si interstitial Interstitial oxygen => lattice expansion, stacking faults form along (111) planes Because B and P diffuse mainly by an interstitial process; their diffusion is enhanced by oxidation. Si But Sb is large and diffuses only by vacancies. Si interstitials created by oxidation, Si recombine and reduce concentration of vacancies Si suppressing diffusion of Sb atoms. Si Si Sb Si Si Si Si 6 13
14 , Wednesday Diffusion process is different for different species. Figure removed for copyright reasons. Please see: Figure 7-15 in Plummer et al.,. 7 Review: Doping and diffusion, small dose dc(, t) dt = d! d D dc d Plus I.C. and B.C.s C δ fn If predeposition is small dose, followed by a higher T, longer t ( larger a ~ Dt ) drive-in process. C (,) C (,t) inc Dt 1) I.C. C (, ) = > ) B.C. C (, t ) =, dc(,t)/d = 3) Fixed dose Q = a! c Q C(,t) =!Dt exp ' % * ) ( a&, + C(, t) = Q decreases like t -1/!Dt 8 14
15 , Wednesday Review: Doping and diffusion, large dose Diffusion preceded by pre-deposition to deliver a large amount of impurity. If pre-dep is inexhaustible or equivalently, if a ~ Dt is small, then C 1) I.C. C (, ) = > ) B.C. C (, t ) = Might be SiO + dopant inc Dt 3) B.C. C (, t ) = C! C(,t) = C erfc a a = Dt Dose! Q = C(,t)d = Dt C = a! C C limited by solid solubility 9 Junctions between different doped regions Diffuse B at high concentration, into n-type Si, (uniformly doped, N D, with P). Want to know depth of p-n junction ( N A = N D ) Consider limitless dopant source (i.e. erfc) boron phosphorus Small Dt C Boron Log C C B! (, t) = C erfc Dt %& N D (n-type Si) junction! N D = C erfc jct a %&! jct = a erfc -1 N D C % & a = Dt 3 15
16 Figure removed for copyright reasons. Please see: Figure 4.14 in Ghandi, 1994.
17 , Wednesday Internal E fields alter Fick s Law D is not a constant, but depends on c(x) Figure removed for copyright reasons. Please see: Figure 7-6 in Plummer et al.,. 33 Exercise n-type Si, N D = 1 16 cm -3 is doped with boron from a const source with C (boron) = 1 18 cm -3 C (boron) 1 18 /cm 3 ( ) = C erfc a C,t!, N D = 1 16 cm -3 Question: If exposed to c at 1 C for 1 hr, what is junction depth?! jct = a erfc -1 N D C % & = Dt erfc -1 1 ' [ ] Let erfc -1 [1 - ] = x, erfc[x] =.1 = 1-erf[x], erf [x] =.99 From appendix, x = 1.8 = /[(Dt) 1/ ] ( ) =.37exp! 3.46 D boron % ' 173 K kt & cm /s a = Dt = 3.31!1 6 cm =.33µm jct =.33!1.8 =.6µm 34 17
18 , Wednesday jct =.33!1.8 =.6µm jct D (boron) From jct (t) and jct () =, you can calculate junction depth at different time: Question: Now constant source is removed and this dose,c(, 1hr), is driven in farther for 1 hr at 11 C. Now where is junction? 1373 K Q = 7.57!1 15 cm s 1 hr t! C(,1hr) d = ac = cm - Dt = 5.!1 6 cm jct = Dt erfc -1 [ 1! ] jct = a ln N D Q % ' =1.8 (1 )5 cm =.18µm!Dt & 35 R =! L A (), Measuring diffusion profiles Resistance resistivity conductivity! = RA L (m)! = 1 = nqµ I L Define sheet resistance A t! t R s = RA Lt = R W L ( sq ) But n, µ are functions of position due to doping t! = q n( ) µ ( n)d t! = 1 = t q nµd an average measurement of n + _ Spreading resistance probe: (Developed at Bell Labs in 4s) φ 36 18
19 , Wednesday 4-point probe Measuring diffusion profiles I Also square array (Van der Pauw method) 5 micron spacing Equipotentials V I V These average n if done from surface. These are most useful if done on beveled wafer: sample Polish off depth profile 37 Hall effect: electrical transport in magnetic field. B I v _ e + + _ - _ + e - e -e- + + I B + + h + v h + F = qv B J = nq v V I F q = E = J H nq B E H = R H ( J B) I V Hall coefficient => charge sign and concentration R H is slope of V vs B data R H = 1 nq Again an average measurement V H Large hole Concentration p B Small electron concentration n 38 19
20 , Wednesday Capacitance Use MOS structure, gate & substrate are electrodes C =!A d Useful for lightly doped regions Depletion width SIMS (Secondary ion mass spectroscopy) Ion source 1-5 k ev Sputters surface Secondary ions to mass spectrometer Depth profile RBS (Rutherford back scattering) He + Backscatter energy depends on depth and impurities 1 MeV Ions penetrate Backscatter intensity (mass impurities ) 39
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