One integrated circuit design calls for the diffusion of arsenic into silicon wafers; the background concentration of As in Si is 2.5 x 1020 atoms/m. The predeposition heat treatment is to be conducted at 1000°C for 45 minutes, with a constant surface concentration of 8x 10^26 As atoms/m³. At a drive in temperature of 1100°C, determine the diffusion time required for a junction depth of 1.21m. For this system, values of Q and Do are 4.10 eV and 2.29x10^3m/s, respectively.
Answers
The diffusion time that is needed for the junction depth of 1.21 m is given as 13875 seconds
How to solve for the diffusion timebackground concentration of As in Si is 2.5 x 10²⁰ atoms/m
tp = 1000°c
tp = 45 minutes
= 45 x 60 seconds
surface concentration = 8x 10^26
Td = 1100°c
Junction depth = 1.2
= 1.2 x 10⁻⁶m
qd = 4.10ev
Do = 2.29 x 10⁻³
We have to use the equation
[tex]Dp = Doexp(-\frac{qd}{kTP} )\\\\= (2.23 * 10^-^1^3 m^2/s) exp [-\frac{4.10}{8.62*10^-^5 *(1000+273)} ][/tex]
= 1.36 x 10⁻¹³
From the relation that we have here
[tex]Qo = 2cs\sqrt{\frac{Dpte}{pi} } \\\\= 2 * 8* 10^2^6*\sqrt{\frac{1.36*10^-^1^3(45*60)}{pi} }[/tex]
Qo = 1.73 x 10¹³
Dd at 1100
[tex]Dd = 2.29*10^-^3exp(\frac{-4.10}{8.62x10^-^5)(1100 + 273)}[/tex]
Dd = 2.06 x 10⁻¹⁸m²/s
From the relation we have
[tex]exp[\frac{(1.2 * 10^-^6)^2}{4*(2.06x10^-^1^8)+d} ]=\frac{1.73*10^1^9}{2.5x10^2^0*\sqrt{pi*2.06*10^-^1^8x+d} }[/tex]
[tex]ln\frac{2.72*10^7}{\sqrt{+d} } =\frac{1.75*10^5}{+d}[/tex]
Using heat and trial, we would have the value of +d = 13875 seconds
The diffusion time is 13875 seconds
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