One method of introducing dopant impurities into silicon (in order to control th

Question

One method of introducing dopant impurities into silicon (in order to control the electrical and
electromechanical properties) is to deposit a thin layer on the surface and allow the impurity to
di use into the Si at a high temperature. Suppose we deposit a thin layer of boron containing
2x10^15 atoms/cm2 on the surface of a pure Si wafer, and then anneal the wafer for 3 hours at 1200
C. For boron di using in silicon, the diffusion coefficient can be calculated using the following:
D0 = 14 cm2/s, QD = 3:69 eV.
(a) To what depth will the boron concentration be greater than 10^16 atoms/cm3?
(b) To what depth will the boron concentration be greater than 10^18 atoms/cm3?

Explanation / Answer

A.) The Boron in first Diffused into pure silicon, so lets calculate the diffusion co-efficient as,

diffusion coefficient, D = Do * exp( -Eg / Kb*T )

here given Do = 14 cm2/sec and Eg = 3.69ev

we know that , Kb= 8.617*10-5 and T = 1200.c = 1200+273=1473K

now calculate D = 14 * exp ( -3.69 / 8.617*10-5 * 1473 )

= 14 * exp ( -3.69 / 0.1269)

= 14 * exp ( - 29.078 )

= 14 * 2.352*10-13

= 3.293 * 10 -12 cm2 / sec

now calculate Dt ,

Dt = D * t

= 3.293 * 10 -12   * 3 * 60 * 60

= 3.293 * 10 -12 * 10800

= 3.556 * 10-8   cm2

i.) now we ned to find the junction depth xj for back ground concentration of 2*1015

back ground concentration = Cs * erfc ( xj / 2*sqrt(Dt) ) where Cs is solid solubility of boron

2*1015 = 1016 * erfc ( xj / 2*sqrt( 3.556 * 10-8 ) )

erfc ( xj / 2*sqrt( 3.556 * 10-8 ) ) = 0.2

erfc ( xj / 0.00018857 ) ) = 0.2

takin inverse complementary error function on both sides,

xj / 0.00018857 = erfc-1 ( 0.2)

xj / 0.00018857 = 0.906

xj = 0.906 * 0.00018857

= 0.171 *10-3 cm

=1.71 um

ii.) now we need to find Xj when Cs = 1018

ie proceed same process with Cs = 1018

2*1015 = 1018 * erfc ( xj / 2*sqrt( 3.556 * 10-8 ) )

erfc ( xj / 2*sqrt( 3.556 * 10-8 ) ) = 0.002

erfc ( xj / 0.00018857 ) ) = 0.002

takin inverse complementary error function on both sides,

xj / 0.00018857 = erfc-1 ( 0.002)

xj / 0.00018857 = 2.185

xj = 2.185 * 0.00018857

= 0.412 * 10-3 cm

= 4.12 um

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