ION IMPLANTATION We saw how dopants were introduced into a wafer by using diffusion (predepositionanddrive-in') This process is limited cannot exceed solid solubility of dopant -difficult to achieve light dopin Ion implantation is preferred because controlled, low or high dose can be introduced(101-1018 cm-2) depth of implant can be controlled Used since 1980, despite substrate damage low throughput, and cost Plummer Ch 8, Campbell Ch 5 3.155J/6.152J.2003
ION IMPLANTATION We saw how dopants were introduced into a wafer by using diffusion (‘predeposition’ and ‘drive-in’). This process is limited: -cannot exceed solid solubility of dopant -difficult to achieve light doping Ion implantation is preferred because: -controlled, low or high dose can be introduced (1011 - 1018 cm-2) -depth of implant can be controlled. Used since 1980, despite substrate damage; low throughput, and cost. Plummer Ch. 8, Campbell Ch. 5 3.155J/6.152J, 2003 1
RemInder Analytical Solutlons to Dlifuslon Equatlons Solution for a limitless source of dopant(constant surface concentration Bound cond d dc C(0,t)=Surf Const C urf Bound cond dt dz. C(∞,t)=0 C(z,t=Curfc t>0 Initial cond: C(Z, 0)=0 where erfc(x)=1-erf() and t,=predep time Dose Q-(2/ Curt vDt Dose in sample increases as ti/2 3.155J/6.152J.2003
Reminder: Analytical Solutions to Diffusion Equations Solution for a limitless source of dopant (constant surface concentration): Bound cond: ∂C = D ∂2C C (0, t ) = Csurf C (∞, t ) = 0 t Bound cond: Const Csurf ∂t ∂z 2 È z ˘ C z( , t) = Csurf erfcÍ2 Dt ˙˚ , t > 0 Î z Initial cond: C (z, 0) = 0 where erfc(x) = 1-erf(x) and tp = predep time Dose Q = (2/√p) Csurf √Dtp Dose in sample increases as t1/2 3.155J/6.152J, 2003 2
Diffusion of a thin surface layer into a solid When a thin surface layer diffuses into a solid, what is c(z, t)? Q= initial amount of dopant 'dose) assumed to be a delta-function ∫c(a t dz=Q=const. (#/area) C|t=0 Bound cond dC(0) Bound cond 0 Solutlon Is a Gausslan C(a,)=0 C(x)= √mDr4Dr 0 Initial cond: C(z0)=0(z diffusion length a=2√Dr Dose in sample constant in time 3.155J/6.152J.2003
Diffusion of a thin surface layer into a solid When a thin surface layer diffuses into a solid, what is C(z,t)? Q = initial amount of dopant (‘dose’), • Ú C z, t)dz = Q = const. (# /area) assumed to be a delta-function ( -• t1 t2 t = 0 C Bound cond: dC(0, t) Bound cond: = 0 Solution is a Ga ussia n. dz C ( •,t ) = 0 2 Q È z ˘ C z( , t ) = exp - pDt ÎÍ 4 Dt˚˙ z 0 Initial cond: C z( ,0 ) = 0 ( z 0 ) diffusion length a = 2 Dt Dose in sample constant in time 3.155J/6.152J, 2003 3
Example: N wafer originally has a unifo 5×10N/nwee dopant level, e.g. donor p-type gau e n- type background Predep plus drive-in 3X10 introduces a second dopal an acceptor At a certain depth, a p-n junction is formed N a third pre-dep of donor can then be done to make an npn transistor N Problem 0 *eb *be can only make profiles consisting of st Emitter Base Collector Gausslans centered at the 01,01.4 xm一 substrate surface Bipolar transistor formation by successive diffusions 3.155J/6.152J.2003
Example: wafer originally has a uniform dopant level, e.g. donor. Predep plus drive-in introduces a second dopant, an acceptor. At a certain depth, a p-n junction is formed. A third pre-dep of donor can then be done to make an npn transistor. Problem: can only make profiles consisting of superposed Gaussians centered at the substrate surface. 3.155J/6.152J, 2003 4
Shce the madmum amount of a dopant that can dissolve hn the S ls glven by the sold solublity, you may be muted h the amount of dopant that can be Incorporated. 3.155J/6.152J.2003
Since the maximum amount of a dopant that can dissolve in the Si is given by the solid solubility, you may be limited in the amount of dopant that can be incorporated. 3.155J/6.152J, 2003 5
lon Implantation Beam of energetic dopant ions is fired into surface of wafer Energies are 5-200 keV This leads to implantation(burial) of the ions into the substrate What happens at the substrate Tons can: bounce off absorb sputter atoms(10ev -10 kev) implant into surface(5 kev -200 kev) and do tremendous damage 3.155J/6.152J.2003
Ion Implantation Beam of energetic dopant ions is fired into surface of wafer. Energies are 5 - 200 keV. This leads to implantation (burial) of the ions into the substrate. What happens at the substrate? Ions can: bounce off absorb sputter atoms (10 eV - 10 keV) implant into surface (5 keV - 200 keV)… and do tremendous damage 3.155J/6.152J, 2003
lon Implantaton Eq山mnt Ions generated in a source(from feed gas, e.g. BF3, ASH3, PH3.or heated solid source, then ionized in arc chamber by electrons from hot filament select desired species by g/m, using a magnet, accelerated by an e-field and focused using electrostatic lenses and impact substrate a bend removes neutrals )in raster pattern Resolving Aperature Focus Analyzing Magnet 0-200ke Neutral Trap X &Y scan Plates Wafer Neutral Beam 0-30keV Ion S Faraday Cup 3.155J/6.152J,2003
Ion Implantation Equipment Ions generated in a source (from feed gas, e.g. BF3, AsH3, PH3 ... or heated solid source, then ionized in arc chamber by electrons from hot filament) select desired species by q/m, using a magnet, accelerated by an E-field and focused using electrostatic lenses and impact substrate (a bend removes neutrals) in raster pattern. 3.155J/6.152J, 2003 7
What happens to lons Inside the matera? lons lose energy, dE/dx, interacting elastically with nuclei and inelastically with electrons E)+SE d x S(E)is Stopping power(eVcm lon range in target de r= dx= N S,(E)+SE) What can we say about nuclear and electronIc stoppIng 3.155J/6.152J.2003
What happens to ions inside the material? Ions lose energy, dE/dx, interacting elastically with nuclei and inelastically with electrons dE = -N S [ n (E ) + S (E)] dx e S (E) is Stopping power (eVcm2) i R 0 dE Ion range in target: R = Ú dx = 1 E Ú N 0 0 S n e (E) + S (E ) What can we say about nuclear and electronic stopping… 3.155J/6.152J, 2003 8
Nuclear stopping power: Coulomb scattering(assumed elastic Incident ion interacts with nucleus of stationary ion b= impact parameter 1.0 0.8 dE/E 04 2 0.0 2468 Energy lost by incoming ion(microscopic) Mo/M △E=E1{1- cos esin+ cos p sin e The angles depend on masses and on b 4M,M, △E=E Max energy loss is when b=0, 0=0 (M,+M 3.155J/6.152J.2003
Nuclear stopping power: Coulomb scattering (assumed elastic) Incident ion interacts with b q f b E1, M1 M2 Energy lost by incoming ion (microscopic) DE = E1 1 - sin2 f cosq sinf + cosf sinq Ï Ì Ó ¸ ˝ ˛ 0 2 4 6 8 10 M2/M1 1.0 0.8 dE/E 0.4 0.2 0.0 nucleus of stationary ion = impact parameter The angles depend on masses and on b. Max. energy loss is when b = 0, f = 0: 4 M1M2 DE = E1 (M1 + M2 )2 3.155J/6.152J, 2003 9
Nuclear stopping power: Coulomb scattering(assumed elastic At 100 kev an ion of 15 amu has velocit ≈106m/s! E. M This is 1000 times faster than speed of sound in solids So ion is far past nucleus before nucleus can displace in response to Coulomb force So nuclear scattering is not strong at high ion velocity There are also only significant inelastic collisions when ion slows down that transfer energy 3.155J/6.152J.2003
q f b E1, M1 M2 q f b E1, M1 M2 F µ Q1Q2 r 2 Nuclear stopping power: Coulomb scattering (assumed elastic) q f b E1, M1 M2 So nuclear scattering At 100 keV an ion of 15 amu has velocity , vion ≈ 106 m/s! This is 1000 times faster than speed of sound in solids… So ion is far past nucleus before nucleus can displace in response to Coulomb force is not strong at high ion velocity; There are also only significant inelastic collisions when ion slows down. that transfer energy… 3.155J/6.152J, 2003 10
