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MS 852 Advance Surface Coatings

Final Term Examination (80% Paper)

DC-Sputtering & DC-glow discharge (M.Aamir Hassan)

MS 852 Advance Surface Coatings Final Term Examination (80% Paper)

1. The dc-sputtering is carried out at a pressure of ___________________

(a) 1 - 0.1 Torr

(b) 10 - 100 Torr

(c) 1 - 0.1 mTorr

2. The dc-sputtering is used to deposit films of ________________

(a) Conducting polymers

(b) Insulator materials only (d) any material

(c) Conducting metals/alloys only

9/3/2010 School of Chemical and Materials Engineering (SCME-NUST) Muhammad Khaliq U Rehman (2009-NUST-MS PhD-MS-E-06)

(d) 10 - 100 mTorr

3. Argon gas is used a s sputter gas in DC sputtering because __________________ (a) Argon is an inert gas

(b) No special reason

(c) Highly available

(d) None

4. In DC-sputtering of the ________ target, ion bombardment quickly builds up a positive charge.

(a) Metal

(b) non-metal

(c) both (a) & (b)

(d) insulating

5. If the insulating target is used in DC-sputtering the glow-discharge will sustain false) for a longer time but the fil m will not be deposited (true/ false 6. In DC-glow discharge, a very _________ current flows at first due to the _________ initial charge carriers in the system.

(a) small, enormous

(b) small, collisions between

(d) None

(c) small, few

7. When the voltage is increased in DC glow discharge system, which of the following process is the initiating step to create more ions: (a) Ion collisions with cathode

(b) generation of secondary electrons (c) ionization of gas

(d) None

8. With charge multiplication in DC glow discharge, the current increases rapidly and the voltage __________ (a) Remains constant

(b) increases significantly

(c) decreases sharply

(d) None

9. If when enough of the electrons generated produce sufficient ions to regenerate the same number of initial electrons, the discharge becomes_______________ (a) Self-sustaining

(b) Townsend discharge

(c) Cathode discharge

(d) None

10. When the gas begins to glow now and the voltage drops, accompanied by a sharp rise in current, we call it ________________

(a) Townsend glow

(b) normal glow

(c) abnormal glow

(d) None

11. The operating domain for sputtering is _______________ regime.

(a) Normal glow

(b) cathode glow

(c) Townsend

(d) abnormal glow



Assignment 1

2. Both Au, which is FCC, and W, which is body-centered cubic (BCC) have a density of 19.3 g/cm3. Their respective atomic weights are 197.0 and 183.9 a. What is the lattice parameter of each metal? b. Assuming both contain hard sphere atoms, what is the ratio of their diameters? Solution: Density of Au = 19.3 g/cm3 Density of W = 19.3 g/cm 3 Atomic weight of Au = AAu = 197.0 Atomic weight of W = AW = 183.9 a)

   

Where n = no. of atoms per unit cell A = atomic weight VC = volume of unit cell NA = avagadro’s number ρ = density Rearranging equation 1

     19.36.4197.023100  6.779  10    6.779  10    .    . Å   19.36.2183.023109  3.164  10    3.164 10   .    . Å 2√2 √2  

For FCC Au n = 4, since there are four atoms per FCC unit cell



Rearranging equation 2

For BCC W Where R = radius of atoms D = Diameter of atoms Rearranging equation 3

Ratio of diameters =

3. a. Comment on the thermodynamic stability of a thin-film super lattice composite consisting of alternating Si and Ge 0.4Si0.6, film layers shown in Fig. 1417 given the Ge-Si phase diagram (Fig. 1-12). b. Speculate on whether the composite is a single phase (because it is a single crystal) or consists of two phases (because there are visible film interfaces). Solution:

Since volume of cubic unit cell = V C = a3, where a is the lattice constant

For BCC W n = 2, since there are two atoms per BCC unit cell

b) Assuming both contain hard sphere atoms For FCC Au Where R = radius of atoms D = Diameter of atoms

a) Ge and Si are both completely soluble in each other in liquid as well in solid state. Germanium and Silicon both have Diamond crystal structure. Also Ge 0.4Si0.6 has same Diamond crystal structure. For Ge0.4Si0.6, variance is 1 since the phase diagram is at atmospheric pressure f=n+1-Ψ f = Degree of freedom Ψ = no. of phases = 1 n = no. of components = 2 f=2+1–1=2 For Si which is a single component system in a P-V diagram

MS 852 Advance Surface Coatings Si + O2  SiO2;

ΔGoSiO2

2Mg + O2  2MgO;

ΔGoMgO



 4  √   410 3600 1010  10 3600  5.142  10      .  

Through elimination of O2 the reaction 2Mg + SiO2  2MgO + Si;

ΔGo

Where ΔGo = ΔGoMgO - ΔGoSiO2 Since the ΔGO-T curve shown in Ellingham Diagram for MgO is more negative or lower than that for SiO 2, the reaction is thermodynamically favored as written. Therefore Mg films tend to reduce SiO2 films, leaving free Si behind. Thus during thermal evaporation of Mg to deposit Mg thin films using Quartz crucible is not a wise choice. 9. A solar cell is fabricated by diffusing phosphorous (N dopant) from a constant surface source of 10 20 atoms/cm3 into a P-type Si wafer containing 10 l6 B atoms/cm3. The diffusivity of phosphorous is 10 -12 cm2/sec, and the diffusion time is 1 hour. How far from the surface is the junction depth-i.e., where C N = CP? Solution: Concentration of Boron atoms in P-type silicon = C P = 10l6 atoms/cm3 Since at junction CP =CN = Concentration of Phosphorous at junction depth x = 10 l6 atoms/cm3 Concentration of phosphors at source surface = C i= 1020 atoms/cm3 Diffusion coefficient = 10 -12 cm2/sec Time = t = 1 hour = 3600 seconds Junction depth = x =? Since

Rearranging


11. Measurements on the electrical resistivity of Au films reveal a three order of magnitude reduction in the equilibrium vacancy concentration as the temperature drops from 600 to 300 oC. a. What is the vacancy formation energy? b. What fraction of sites will be vacant at 1080 oC? Solution:

       fvacccrancyeaticoncon enerentgryaton E vaccancy Where

k = Boltzmann constant = 1.380 × 10-23 J/k T = temperature in Kelvin Part (a): Given that T1 = 600 oC = 873 K T2 = 300 oC = 573 K

3   3  3          3  3

Since there is a three order of magnitude reduction in vacancy concentration at 573 K



3  113   13 1   1.380  101 5733 1.380  101  873   1.265101.09868.3010 0986   4.31.410   2.5276 10   0.158 

Since 1 eV = 1.602 × 10−19 J

Part (b):

T = 1080 OC = 1353 K

     1.32.80 5281010 1353  0.258

f is a unit less quantity since it is the fraction of vacancies present in the lattice.



Assignment 2

2. A 1-m3 cubical-shaped vacuum chamber contains O 2 molecules at a pressure of 10-4 atm at 300 K. a. How many molecules are there in the chamber? b. What is the ratio of maximum potential energy to average kinetic energy of these molecules? c. What fraction of gas molecules has a kinetic energy in the x direction exceeding RT? What fraction exceeds 2RT? Solution: Given that: Pressure of O2 molecules = 10 -4 atm = 10.1325 Pa Temperature = T = 300K Molecular weight of Oxygen = 32 g/mol = 0.032 Kg/mol Volume of vacuum chamber = V = 1 m 3 a) Since we know that

         6.023  2. eculPa eKsmol 108.314mol4m4610 mol10.3100325Pa1 m   3     38.3140.032300  233831.25 / .   12   . 0.00406  12 0.032   233831.25   15.19 Joules   12  0.00406 52 8.314   300 25.32 

Where R = gas constant = 8.314 m3 Pa K-1 mol-1 n = no. of moles = N/N A

N = no. of molecules NA = Avogadro’s number = 6.023 × 1023 molecules/mol

b) Average kinetic energy is calculated as follows Mean square velocity is given as

No. of moles of O2 = n = 0.00406 mol

Total internal energy is calculated as

As degree of freedom for a diatomic gas = f = 5

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