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, T2 be D2. 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.



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 t1 and temperature of T1.

                                (3-15 c)

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



where t is drive-in diffusion time. In many applications, dopant initial concentration, C0 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, t2, then the above simplifies to:



The junction depth, xj is the distance at which concentration of dopant is approximately equal to C0. Setting C to C0 in the above, and rearranging junction depth can be expressed as:






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 1018 atoms/cm3. 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 1018 atoms/cm3 at a depth of 6 microns. Temperature of operation is: 1150 C

Part a

Since silicon is pure, C0 = o. Time of diffusion = t = 5 minutes

Desired concentration = C = 3 x 1018 atoms/cm3

Diffusion Coefficient of B at 1100 C is (from Hand Book) = D = 5.8 x 10-2 µm2/h

Surface concentration of B at 1100 C is (from Hand Book) = Cs = 5.1 x 1020 atoms/cm3

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 1018, x = 6 µm, Cs = 5.1 x 1020

Diffusion Coefficient of B at 1100 C is (from Hand Book) = D1 = 5.8 x 10-2 µm2/h

Diffusion Coefficient of B at 1150 C is (from Hand Book) = D2 = 1.6 x 10-1 µm2/h

time, t1 = (5/60) h

Substituting in the above:

Solving, t2 = 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. )