→ gone20080919:112果然閒 05/15 02:23
推 chihow0520:謝謝! 05/15 13:31
推 milord:推 05/19 11:06
推 myemail:第一題可以用Fick's second law model嗎? 05/27 15:18
※ 引述《chihow0520 (異鄉人)》之銘言：
: 最近上到diffusion..
: 這個章節不是很瞭解
: 老師出了一個題目不知道該怎麼解
: 請各位大大 賜教
: 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
: 希望各位大大能幫忙小弟
: 解這一題....
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※ 編輯: NBL123 來自: 140.112.244.20 (05/15 01:05)