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