You have a silicon wafer and diffuse osmium (Os), which has a difusion coefficie

Question

You have a silicon wafer and diffuse osmium (Os), which has a difusion coefficient of 10-189cm2/s at 1000°C, an activation energy for diffusion of 3.6 eV, and a surface solubility Co of 1019/cm3 You carry out the diffusion for 20 minutes at 1000°C. The diffusion length -2(Dt)2 a. What is the diffusion length and total number of Qs/cm2 in the sample at the end of the diffusion? Consider the Os to have a single charge, and assume linear diffusion. What electric field would be required to give a current equal to the flux at a distance one-half the diffusion length at t 20 minutes? (2pts) What temperature would you need to increase the junction depth a factor of 10 at t-20 minutes? b. c.

Explanation / Answer

Given,

Diffusion Coefficient of Osmium (D) = 10-13 cm2/s

Activation Energy for diffusion = 3.6 eV

Surface solubility (Co) = 1019/cm3

Diffusion Time (t)= 20 min = 1200 s

The answer for Question a:

Diffusion Length = 2(Dt)1/2 = 2(10-13 cm2/s * 1200 s) = 2.191 * 10-5 cm

Number of Os/cm2 at the end of diffusion = Diffusion length * Surface Solubility = 2.191 * 10-5 cm * 1019/cm3 = 2.191 * 1014/cm2

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