Question 1.15: DIFFUSION OF DOPANTS IN SILICON The diffusion coefficient of...
DIFFUSION OF DOPANTS IN SILICON The diffusion coefficient of P atoms in the Si crystal follows Equation 1.41 with D_o = 10.5 cm² s^{−1} and E_A = 3.69 eV. What is the diffusion coefficient at a temperature of 1100 °C at which dopants such as P are diffused into Si to fabricate various devices? What is the rms distance diffused by P atoms in 5 minutes? Estimate, as an order of magnitude, how many jumps the P atom makes in 1 second if you take the jump distance to be roughly the mean interatomic separation, ∼0.27 nm.
From Equation 1.41,
D=D_{o} \exp \left(-\frac{E_{A}}{k T}\right)=\left(10.5 cm ^{2} s ^{-1}\right) \exp \left[-\frac{(3.69 eV )\left(1.602 \times 10^{-19} J eV ^{-1}\right)}{\left(1.381 \times 10^{-23} J K ^{-1}\right)(1100+273 K )}\right]
=3.0 \times 10^{-13} cm ^{2} s ^{-1}
The rms distance L diffused in a time t = 5 min = 5 × 60 seconds is
L=\sqrt{2 D t}=\left[2\left(3.0 \times 10^{-13} cm ^{2} s ^{-1}\right)(5 \times 60 s )\right]^{1 / 2}=1.3 \times 10^{-5} cm \quad \text { or } \quad 0.13 \mu m
Equation 1.40 was derived for a two-dimensional crystal as in Figure 1.31, and for an impurity diffusion. Nonetheless, we can still use it to estimate how many jumps a P atom makes in 1 second. From Equation 1.40,
f \approx 2 D / a^{2} \approx 2\left(3.0 \times 10^{-17} m ^{2} s ^{-1}\right) /\left(0.27 \times 10^{-9} m \right)^{2}=823
jumps per second. It takes roughly 1 ms to make one jump. It is left as an exercise to show that at room temperature it will take a P atom 10^{46} years to make a jump! (Scientists and engineers know how to use thermally activated processes.)
L^{2}=a^{2} f t=2 D t [1.40]
