**4.2.3 Mathematical Model: Dopant Profiling**

After pre-deposition, the wafer is usually placed in a furnace at a higher
temperature, first to grow a sealing silicon dioxide layer and then to achieve
dopant concentration profiling. Let the diffusion coefficient at this higher
temperature, T_{2} be D_{2}. The diffusion of dopant during
this phase of operation is also described by an equation similar to Eq (3-7
) with the modification given below. Subscript A is dropped in all further
developments. That is: C will refer to concentration of A.

(3-15)

At the top surface, x = 0, flux of A will be zero because of the sealing
oxide layer. We express this condition mathematically as the gradient of
C is zero. Recall that flux is directly proportional to the gradient. That
is

(3-15 a)

Far away (more than 1 or 2 microns) from the top surface, we expect concentration
of A not to change during the course of drive-in diffusion. This expressed
as:

(3-15 b)

The initial condition is the profile that existed at the end of pre-deposition
step, and is given by Equation (3-11b) and is given below for a pre-deposition
time of t_{1} and temperature of T_{1}.

(3-15 c)

The solution to Eq (3-15) with the boundary conditions stated above is
given by

(3-16)

where t is drive-in diffusion time. In many applications, dopant initial
concentration, C_{0} is so small in comparison to the dopant concentration
introduced that we can set it to zero. Further, if time is set to drive-in
diffusion time, t_{2}, then the above simplifies to:

(3-17)

The junction depth, x_{j} is the distance at which concentration
of dopant is approximately equal to C_{0}. Setting C to C_{0}
in the above, and rearranging junction depth can be expressed as:

(3-18)

** **

**Example 3** A junction in silicon is made by
doping with Boron using pre-deposition followed by drive-in diffusion.

a) Five minutes at 1100o C are required to deposit the dopant. At what distance from the surface is the concentration of Boron raised to 3 x 10

^{18}atoms/cm^{3}. Assume that silicon is pure.

b) How much boron ( per cm2 of silicon surface) will have been taken up by the silicon during the deposition step?

c) To prevent loss of Boron during drive-in step, surface is masked with silica. Now calculate tie required to achieve a Boron concentration of 3 x 10

^{18}atoms/cm^{3}at a depth of 6 microns. Temperature of operation is: 1150 C

**Part a**

Since silicon is pure, C

_{0}= o. Time of diffusion = t = 5 minutes

Desired concentration = C = 3 x 10

^{18}atoms/cm^{3}

Diffusion Coefficient of B at 1100 C is (from Hand Book) = D = 5.8 x 10

^{-2}µm^{2}/h

Surface concentration of B at 1100 C is (from Hand Book) = C

_{s}= 5.1 x 10^{20}atoms/cm^{3}

The relevant diffusion equation is:

Substitute the knowns in the above.

The unknown, x is obtained by determining the referring to the error function graph. Solution: x ~ 0.27 m.

** **

**Part b**

The amount deposited is calculated using the formula given in Equation (3-13)

**Part c**

Time required for the drive-in diffusion step can be obtained from Equation (3-17)

Here C = 3 x 10

^{18}, x = 6 µm, C_{s}= 5.1 x 10^{20}

Diffusion Coefficient of B at 1100 C is (from Hand Book) = D

_{1}= 5.8 x 10^{-2}µm^{2}/h

Diffusion Coefficient of B at 1150 C is (from Hand Book) = D

_{2}= 1.6 x 10^{-1}µm^{2}/h

time, t

_{1}= (5/60) h

Substituting in the above:

Solving, t_{2} = 203 h (this is not a typical drive-in diffusion
time, which is typically an hour; the depth of 6 microns is much larger
than a typical practical need of about 1 micron. )