0.4 Tutorial on chapter 1-crystal and crystal growth.  (Page 2/2)

Therefore density of Ge=4.4495×10 ²² atoms/cc×1.206×10 ^-22 gm/atom=5.366gm/cc.

Therefore density of GaAs=4.42776×10 ²² atoms/cc×1.2012×10 ^-22 gm/atom=5.318gm/cc gm/cc.

1. A Silicon Ingot should have a P-Type doping of 10 ¹⁶ Boron Atoms/cc. In a Czocharlaski Crystal Growth set-up what amount of Boron element by weight should be added to a Si melt of 60Kg. Given:Atomic Weight of Boron =10.8, Segregation Coefficient= 0.8.

[Answer; About 6 milli gm]

Method:

Let Solid Phase concentration of Boron = C _S ;

Let Liquid Phase concentration of Boron = C _L ;

By Definition, segregation coefficient = C _S / C _L = 0.8.

Therefore C _L = (10 ¹⁶ /cc)/0.8= 1.25×10 ¹⁶ atoms/cc.

Volume of the Melt = V ₀ = (Weight of the Si-Melt)/ρ _Si = (60×10 ³ gm)/(2.33gm/cc).

Therefore V ₀ = 25.75×10 ³ cc.

No. of Boron atoms to be added= V ₀ × C _L = 25.75×10 ³ cc ×1.25×10 ¹⁶ atoms/cc=N ₀ .

Therefore N ₀ = 32.2×10 ¹⁹ Boron atoms to be added to the whole melt.

Number of gm-moles of Boron required = N ₀ /N _Avogadro

= 32.2×10 ¹⁹ Boron Atoms/6.02×10 ²³ Boron Atoms/gm-mole = 5.349×10 ^-4 gm-mole.

Weight of Boron required=

gm-mole×Atomic Weight =5.349×10 ^-4 gm-mole×10.8gm/gm-mole= 5.78 milligm.

1. A layer of 10μm thick epitaxial film is to be deposited on a Si Substrate by CVD method. For how long should the substrate stay in the CVD oven at 1270°C if the gas mixture of gases is in the following concentrations in atoms/cc: H ₂ is 2.94×10 ¹⁹ and SiCl ₄ is 6×10 ¹⁷ .

Given:

The above figure gives Growth Rate(microns/min) vs the Mole Fraction(y) of SiCl ₄ . At higher concentrations of HCl i.e. at mole fraction>0.275 in-situ etching occurs.

From the Figure the following Table is generated:

In the above table the growth rate is the net result of two competing reactions:

Forward reaction where SiCl ₄ is decomposing at 1270°C and aiding the Si epitaxial growth and Backward reaction where HCl vapour is etching the Si surface in-situ.

From low mole fraction to 0.275, forward reaction dominates. At 0.275 the two competing reactions are balanced. Above mole fraction 0.275 backward reaction prevails resulting into etching of the Si surface of the substrate.

In our problem the Mole Fraction is :

From the Graph, corresponding to y= 0.02, the growth rate is 1.5microns per minute.

Therefore 10 microns will require 6.7 minutes of exposure in CVD Furnace.

1. In a Molecular Beam Epitaxy System, find the time required to form a monolayer of O ₂ at pressure 1 Torr, 10 ^-6 Torr and 10 ^-10 Torr. Given Oxygen Molecular Weight = M=32 and Oxygen Molecular Diameter = 3.45A°. Assume that the mono-layer is a close pack structure.

[Answer: 2.43×10 ^-6 sec, 2.43 sec, 6 hours]

Method:

Surface Concentration of the closely packed monolayer of Oxygen Molecules= N _S .

Let the time required for monolayer of Oxygen formation be t ₀ where t ₀ =N _S /Φ

Where Φ=Impingement Rate of Oxygen molecules on Substrate Surface in MBE Bell Jar.

Here M=32 and T =300K and P is the corresponding pressure in Torr at which the time for monolayer growth is to be calculated.

This problem clarifies why MBE requires a super Vacuum to fabricate stable and reproducible devices. This is precisely why using Compound Semiconductors is prohibitively costly and we are looking for alternatives to Compound Semiconductors.