※ 引述《chihow0520 (异乡人)》之铭言： : 最近上到diffusion.. : 这个章节不是很了解 : 老师出了一个题目不知道该怎麽解 : 请各位大大 赐教 : You are diffusing phosphorus,P(an n-type dopan) into : undoped silicon wafer in a 1100 degree C furnace using : a continuous supply of phosphine gas PH3 that establishes : a surface P concentration of 5.00*10^18 atoms*cm^-3 : 1)is this a case of steady-state or non-steady state diffusion? why? Non-steady state, because in a normal doping problem, the thickness of silicon wafer is infintely large relative to the doping thickness. The steady state will never be reached. : 2)after the wafer has been in the furnace for 80mins, : phosphorus has had a chance to diffuse into the wafer. qualitatively plot : the concentration of phosphorus as a function of depth from the surface. : what is the name of the mathematical function that has the same shape as : this concentration profile? The profile is described by an error function. The mathematical descrption of the concentration profile is C(x,t)=Co*erfc(x/2√(Dt)), while Co is the surface concentration. (This profile can be solved by using Fick's 2nd law and Laplace transform) : 3)at same time as in the graph you drew for part2( after 80mins), there : will be a point withn the wafer where the P concentration is exactly : 0.60 of surface P concentration. Calculate this depth, in nm. : The pre-exponential coefficient Dο, for diffusion of P in Si : is 10.5cm^2S^-1 and the activation energy of this diffusion is 3.69eV : (per atom) D1100 = Doexp(-Q/kT)=10.5exp(-3.69/(8.625*10^-5)/1373)=3.08*10^-13 cm^2/s Solve C(x, 4800s)=Co*erfc(x/2√(3.08*10^-5*4800))=0.6Co x=286.2 nm : 希望各位大大能帮忙小弟 : 解这一题.... --

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