Question

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentration at the surface of Si specimen at the level of 3*10^26 boron atoms/m^3. The process was carried out at 1100 degrees C. The diffusion coefficient of B in Si at 1100 degrees C is 4*10^-13 cm^2/s. Calculate the concentration of B at the depth of 10^-4 cm from the surface after 130 minutes of diffusion at 1100 degrees C

Answer 1

From Fick's law:

`$(D_(AB))/(\Delta z ) * (C_(A1)-C_(A2))=N_a$`

Mass balance: Exits = Accumulation

`-N_A A = (dm)/(dt)`

`-N_A A = (dVp)/(dt)`

`-N_A A = (dV)/(dt)p`

`-N_A A = (dhA)/(dt)`

`-N_A A = (dh)/(dt) * Ap`

From the last step, area cancels out and thus leaves :

`-N_A = (dh)/(dt) * p`

So now we can substitute the
`$N_A$` by the Fick's law

`$(D_(AB))/(\Delta z ) * (C_(A1)-C_(A2))=(dh)/(dt) p$`

Substituting the values we get

`$=(-4 * 10^(-13))/(0.0001) * (3 * 10^(26) - C_(A2)) = (dh)/(dt) * 2.46$`

`$=-4 * 10^(-9) *( 3 * 10^(26) - C_(A2)) = 0.0001 * 2.46$`

`$= -7800 * 4 * 10^(-9) * (3 * 10^(26)-C_(A2))=0.000246$`
`$=-(3 * 10^(26)-C_(A2)) = 7.8846 * 100 * (1)/(69.62) * 6.022 * 10^(23)$`

`$C_(A2) = 6.85 * 10^(28) \ \text{ boron atoms} /m^3$`