A p-n junction is obtained at a depth of 3 x 10-3 cm below the surface by diffusion of antimony (Sb) into p-type germanium. What is the donor concentration in the bulk germanium if the diffusion is carried out for three hours at 790 C? Express your answer in number of donor atoms per cubic centimeter. Assume the surface concentration of antimony is held constant at 8 x 1018 cm-3 and that the diffusivity at 790 C is 4.8 x 10-11 cm2/s.
unanswered
To find the donor concentration in the bulk germanium, we can use Fick's Second Law of Diffusion:
D = (1/2)dC/dx
Where:
D is the diffusion coefficient
dC/dx is the concentration gradient
In this case, we have the diffusivity (D) and we need to find dC/dx.
To find dC/dx, we can use the equation:
dC/dx = (Cs - C0)/x
Where:
Cs is the surface concentration of antimony (8 x 10^18 cm^-3)
C0 is the concentration of antimony in the bulk
x is the distance from the surface (depth)
Given that the diffusion is carried out at a depth of 3 x 10^(-3) cm and the surface concentration is constant, we can substitute the values into the equation:
dC/dx = (8 x 10^18 cm^-3 - C0)/(3 x 10^-3 cm)
Next, we can substitute the value of the diffusivity (D = 4.8 x 10^-11 cm^2/s) into Fick's Second Law to solve for dC/dx:
4.8 x 10^-11 cm^2/s = (1/2)dC/dx
Now, we can solve for dC/dx:
dC/dx = 2 * 4.8 x 10^-11 cm^2/s
Finally, we can substitute this value back into the previous equation to find the donor concentration in the bulk germanium:
2 * 4.8 x 10^-11 cm^2/s = (8 x 10^18 cm^-3 - C0)/(3 x 10^-3 cm)
Solving for C0, the concentration of antimony in the bulk:
C0 = 8 x 10^18 cm^-3 - (2 * 4.8 x 10^-11 cm^2/s * 3 x 10^-3 cm)
C0 = 8 x 10^18 cm^-3 - 2.88 x 10^5 cm^-1
C0 = 7.99888 x 10^18 cm^-3
Therefore, the donor concentration in the bulk germanium is approximately 7.99888 x 10^18 atoms/cm^3.