Download Differentiate between a) Chemical vapor deposition (CVD) and

Problem 1
 Differentiate between
a) Chemical vapor deposition (CVD) and Electrodeposition
 CVD
 Chemical process in gas
 Precursor gases react or decompose and
precipitate on the substrate to form the coating
 Sometimes precursor gases can react with the
substrate as well to form a coating
 Mostly metal compounds, Si compounds and
carbon nanotubes are deposited
Chemical vapor deposition
 Electrodeposition
 Chemical process in solution
 The anode and cathode are connected to an external
supply of direct current, and immersed in a solution
called an electrolyte.
 Metals or conductive polymers are deposited on the
cathode plate.
Electrodeposition
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Problem 1
 Differentiate between
b) Wet etching and dry etching
 Wet etching
 The etch reactants come from a liquid source
 The etching is purely a chemical reaction
 Dry etching
 The etch reactants are in a gas or vapor phase and mostly
ionized
 The etching is due to both physical and chemical processes
 The ions being high energy particles are able to physically
knock off the material without having to react with it
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Problem 1
 Differentiate between
c) Electrophoresis and Dielectrophoresis (DEP)
 Electrophoresis
 A dispersed charged particle in a fluid moves and
migrates under the action of a uniform electric field
towards the electrode of opposite charge.
Electrophoresis
 Dielectrophoresis (DEP)
 If the electric field is not uniform, the interaction between
the induced dipole and the non-uniform electric field
generates a force.
 Due to presence of the electric field gradient, these forces
are not equal in magnitude and generate net movement
of the charged particle.
Dielectrophoresis
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Problem 2
 Although in theory it is possible to obtain features in the size range of a few angstroms using ebeam lithography, why is it not possible to obtain them in practice?
 Proximity Effect of E-Beam Lithography
(A)
 Electron scattering in the resist and the substrate
limit the theoretical resolution
 Back scattered and secondary electrons also
expose the resist
 This effect results in beam spreading and
reduces the resolution
 100 Å e-beam become 0.2 µm line
(B)
A) Scattering of the e-beam in the resist
and on the substrate resulting in forward
scattered, back scattered and
secondary electrons. B) Spreading of
forward- and back scattered electrons
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Problem 3
 Is it possible to apply the technique of Dip-Pen Lithography in a vacuum environment? Why?
 No, it is not possible
 In Dip-Pen lithography, an AFM tip is “inked” by dipping the tip in a solution containing a small
concentration of the molecules of interest
 The water meniscus that naturally forms between the tip and the surface enables the diffusion
and transport of the molecules to the surface.
 At low pressure (or vacuum) the water would immediately evaporate.
Principle of dip-pen nanolithography using a
AFM tip wetted with the molecules of interest
The phase diagram of water
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Problem 4
 The fabrication steps for elliptical shapes are given below. The wafer is electrodeposited with iron
from a solution containing iron sulfate by applying a current density J of 60 mA∙cm -2, calculate the
time necessary to obtain elliptical shapes (a = 3 mm, b = 2 mm) of 50 μm thickness.
with
𝑚: Mass of the deposited solid metal
𝑄: Total electric charge used in the
deposition process (𝑄 = I∙t = J∙A∙t)
𝑀: Molar mass of the substance
being deposited = 55.845 g∙mol-1
𝑧: Number of electrons per metal ion
𝐹: Faraday constant = 96487 C∙mol-1
Area of ellipse is given by A = 𝜋𝑎𝑏
Density of iron (𝐹𝑒) is 7.874 g∙cm-3
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Problem 4
 Given the Faraday’s law of electrodeposition:
𝑚=
𝑄∙𝑀
𝐹∙𝑧
(Electrochemical reaction: 𝐹𝑒2+ + 2𝑒− → 𝐹𝑒)
 With the total deposited mass:
𝑚 = 𝜌𝐹𝑒 ∙ 𝐴 ∙ ℎ
(where 𝜌𝐹𝑒 is to the density, 𝐴 the area
of the ellipse and ℎ is the thickness)
𝑄 =𝐽∙𝐴∙𝑡
(where 𝐽 is to the current density, 𝐴 the
area of the ellipse and 𝑡 is the time)
 And with the total Charge:
 Combining these three formula gives:
𝐽∙𝐴∙𝑡∙𝑀
𝜌𝐹𝑒 ∙ 𝐴 ∙ ℎ =
𝐹∙𝑧
𝜌𝐹𝑒 ∙ ℎ ∙ 𝐹 ∙ 𝑧
→𝑡=
𝐽∙𝑀
with 𝜌𝐹𝑒 = 7.874 g∙cm-3, ℎ = 50 μm, 𝐹 =
96487 C∙mol-1, 𝑧 = 2, 𝐽 = 60 mA∙cm-2
and 𝑀 = 55.845 g∙mol-1
→ 𝑡 = 37.8 𝑚𝑖𝑛
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